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+{"text":"\\section{\\def\\@secnumfont{\\mdseries}\\@startsection{section}{1}%\n  \\z@{.7\\linespacing\\@plus\\linespacing}{.5\\linespacing}%\n  {\\normalfont\\scshape\\centering}}\n\\def\\subsection{\\def\\@secnumfont{\\bfseries}\\@startsection{subsection}{2}%\n  {\\parindent}{.5\\linespacing\\@plus.7\\linespacing}{-.5em}%\n  {\\normalfont\\bfseries}}\n\\makeatother\n\\def\\subl#1{\\subsection{}\\label{#1}}\n \\newcommand{\\Hom}{\\operatorname{Hom}}\n  \\newcommand{\\mode}{\\operatorname{mod}}\n\\newcommand{\\End}{\\operatorname{End}} \\newcommand{\\wh}[1]{\\widehat{#1}} \\newcommand{\\Ext}{\\operatorname{Ext}} \\newcommand{\\ch}{\\text{ch}} \\newcommand{\\ev}{\\operatorname{ev}}\n\\newcommand{\\Ob}{\\operatorname{Ob}} \\newcommand{\\soc}{\\operatorname{soc}} \\newcommand{\\rad}{\\operatorname{rad}} \\newcommand{\\head}{\\operatorname{head}}\n\\def\\operatorname{Im}{\\operatorname{Im}}\n\\def\\operatorname{gr}{\\operatorname{gr}}\n\\def\\operatorname{mult}{\\operatorname{mult}}\n\\def\\operatorname{Max}{\\operatorname{Max}}\n\\def\\operatorname{Ann}{\\operatorname{Ann}}\n\\def\\operatorname{sym}{\\operatorname{sym}}\n\\def\\operatorname{loc}{\\operatorname{loc}}\n\\def\\operatorname{\\br^\\lambda_A}{\\operatorname{\\br^\\lambda_A}}\n\\def\\underline{\\underline}\n\\def$A_k(\\lie{g})(\\bsigma,r)${$A_k(\\lie{g})(\\bsigma,r)$}\n\\def\\operatorname{res}{\\operatorname{res}}\n\n \\newcommand{\\Cal}{\\cal} \\newcommand{\\Xp}[1]{X^+(#1)} \\newcommand{\\Xm}[1]{X^-(#1)}\n\\newcommand{\\on}{\\operatorname} \\newcommand{\\Z}{{\\bold Z}} \\newcommand{\\J}{{\\cal J}} \\newcommand{\\C}{{\\bold C}} \\newcommand{\\Q}{{\\bold Q}}\n\\renewcommand{\\P}{{\\cal P}}\n\\newcommand{\\N}{{\\Bbb N}} \\newcommand\\boa{\\bold a} \\newcommand\\bob{\\bold b} \\newcommand\\boc{\\bold c} \\newcommand\\bod{\\bold d} \\newcommand\\boe{\\bold e} \\newcommand\\bof{\\bold f} \\newcommand\\bog{\\bold g}\n\\newcommand\\boh{\\bold h} \\newcommand\\boi{\\bold i} \\newcommand\\boj{\\bold j} \\newcommand\\bok{\\bold k} \\newcommand\\bol{\\bold l} \\newcommand\\bom{\\bold m} \\newcommand\\bon{\\bold n} \\newcommand\\boo{\\bold o}\n\\newcommand\\bop{\\bold p} \\newcommand\\boq{\\bold q} \\newcommand\\bor{\\bold r} \\newcommand\\bos{\\bold s} \\newcommand\\boT{\\bold t} \\newcommand\\boF{\\bold F} \\newcommand\\bou{\\bold u} \\newcommand\\bov{\\bold v}\n\\newcommand\\bow{\\bold w} \\newcommand\\boz{\\bold z} \\newcommand\\boy{\\bold y} \\newcommand\\ba{\\bold A} \\newcommand\\bb{\\bold B} \\newcommand\\bc{\\bold C} \\newcommand\\bd{\\bold D} \\newcommand\\be{\\bold E} \\newcommand\\bg{\\bold\nG} \\newcommand\\bh{\\bold H} \\newcommand\\bi{\\bold I} \\newcommand\\bj{\\bold J} \\newcommand\\bk{\\bold K} \\newcommand\\bl{\\bold L} \\newcommand\\bm{\\bold M} \\newcommand\\bn{\\bold N} \\newcommand\\bo{\\bold O} \\newcommand\\bp{\\bold\nP} \\newcommand\\bq{\\bold Q} \\newcommand\\br{\\bold R} \\newcommand\\bs{\\bold S} \\newcommand\\bt{\\bold T} \\newcommand\\bu{\\bold U} \\newcommand\\bv{\\bold V} \\newcommand\\bw{\\bold W} \\newcommand\\bz{\\bold Z} \\newcommand\\bx{\\bold\nx} \\newcommand\\KR{\\bold{KR}} \\newcommand\\rk{\\bold{rk}} \\newcommand\\het{\\text{ht }}\n\n\n\\newcommand\\toa{\\tilde a} \\newcommand\\tob{\\tilde b} \\newcommand\\toc{\\tilde c} \\newcommand\\tod{\\tilde d} \\newcommand\\toe{\\tilde e} \\newcommand\\tof{\\tilde f} \\newcommand\\tog{\\tilde g} \\newcommand\\toh{\\tilde h}\n\\newcommand\\toi{\\tilde i} \\newcommand\\toj{\\tilde j} \\newcommand\\tok{\\tilde k} \\newcommand\\tol{\\tilde l} \\newcommand\\tom{\\tilde m} \\newcommand\\ton{\\tilde n} \\newcommand\\too{\\tilde o} \\newcommand\\toq{\\tilde q}\n\\newcommand\\tor{\\tilde r} \\newcommand\\tos{\\tilde s} \\newcommand\\toT{\\tilde t} \\newcommand\\tou{\\tilde u} \\newcommand\\tov{\\tilde v} \\newcommand\\tow{\\tilde w} \\newcommand\\toz{\\tilde z} \\newcommand\\woi{w_{\\omega_i}}\n\\newcommand\\chara{\\operatorname{Char}}\n\\title{Conjecture $\\mathcal{O}$ holds for some Horospherical Varieties of Picard Rank 1}\n\\author[Bones, Fowler, Schneider, Shifler]{Lela Bones, Garrett Fowler, Lisa Schneider and Ryan M. Shifler}\\address{\\noindent Department of Mathematics, Salisbury University, MD 21801}\n\\email{lbones1@gulls.salisbury.edu, gfowler2@gulls.salisbury.edu, lmschneider@salisbury.edu, rmshifler@salisbury.edu}\n\\begin{abstract}\nProperty $\\mathcal{O}$ for an arbitrary complex, Fano manifold $X$, is a statement about the eigenvalues of the linear operator obtained from the quantum multiplication of the anticanonical class of $X$. Conjecture $\\mathcal{O}$ is a conjecture that Property $\\mathcal{O}$ holds for any Fano variety. Pasquier listed the smooth non-homogeneous horospherical varieties of Picard rank 1 into five classes. Conjecture $\\mathcal{O}$ has already been shown to hold for the odd symplectic Grassmannians which is one of these classes. We will show that Conjecture $\\mathcal{O}$ holds for two more classes and an example in a third class of Pasquier's list. The theory of Perron-Frobenius reduces our proofs to be graph-theoretic in nature.\n\\end{abstract}\n\\maketitle\n\\section{Introduction}\nThe purpose of this paper is to prove that Conjecture $\\mathcal{O}$ holds for some horospherical varieties of Picard rank 1. We recall the precise statement of Conjecture $\\mathcal{O}$ for varieties of Picard rank 1, following \\cite[Section 3]{GGI}. Let $F$ be a Fano variety, let $K:=K_F$ be the canonical line bundle of $F$, let $F_D$ be a fundamental divisor of $F$, and let $c_1(F):=c_1(-K) \\in H^2(F)$ be the anticanonical class. The Fano index of $F$ is $r$, where $r$ is the greatest integer such that $K_F \\cong -rF_D$. The small quantum cohomology ring $(QH^*(F), \\star)$ is a graded algebra over $\\mathbb{Z}[q]$, where $q$ is the quantum parameter. We define the small quantum cohomology in Section~\\ref{sec:Section 2.1}. Consider the specialization $H^{\\bullet}(F):=QH^*(F)|_{q=1}$ at $q=1$. The quantum multiplication by the first Chern class $c_1(F)$ induces an endomorphism $\\hat{c}_1$ of the finite-dimensional vector space $H^{\\bullet}(F)$: \\[y \\in H^{\\bullet}(F) \\mapsto \\hat{c}_1(y):=(c_1(F)\\star y)|_{q=1}. \\]\n\nDenote by $\\delta_0:=\\max \\{|\\delta| : \\delta \\mbox{ is an eigenvalue of } \\hat{c}_1 \\}.$ Then Property $\\mathcal{O}$ states the following:\n\\begin{enumerate}\n\\item The real number $\\delta_0$ is an eigenvalue of $\\hat{c}_1$ of multiplicity one.\n\\item If $\\delta$ is any eigenvalue of $\\hat{c}_1$ with $|\\delta|=\\delta_0$, then $\\delta=\\delta_0 \\gamma$ for some $r$-th root of unity $\\gamma \\in \\mathbb{C}$, where $r$ is the Fano index of $F$.\n\\end{enumerate}\nThe property $\\mathcal{O}$ was conjectured to hold for any Fano, complex manifold $F$ in \\cite{GGI}. If a Fano, complex, manifold has Property $\\mathcal{O}$ then we say that the space satisfies Conjecture $\\mathcal{O}$. Conjecture $\\mathcal{O}$ underlies Gamma Conjectures I and II of Galkin, Golyshev, and Iritani. The Gamma Conjectures refine earlier conjectures by Dubrovin on Frobenius manifolds and mirror symmetry. Conjecture $\\mathcal{O}$ has already been proved for the homogeneous $G\/P$ case in \\cite{CL}, the odd symplectic Grassmannians in \\cite{LMS}, del Pezzo surfaces in \\cite{delpezzo}, and projective complete intersections in \\cite{projinter}. The Perron-Frobenius theory of nonnegative matrices reduces the proofs that Conjecture $\\mathcal{O}$ holds for the homogeneous and the odd symplectic Grassmannian cases to be a graph-theoretic check. This is because Conjecture $\\mathcal{O}$ is largely reminiscent of Perron-Frobenius Theory. In this manuscript we will use the same graph-theoretic approach to prove that Conjecture $\\mathcal{O}$ holds for some smooth horospherical varieties of Picard rank 1.\n\n\nNext we recall the definition of a horospherical variety following \\cite{GPPS}. Let $G$ be a complex reductive group. A $G$-variety is a reduced scheme of finite type over the field of complex numbers $\\mathbb{C}$, equipped with an algebraic action of $G$. Let $B$ be a Borel subgroup of $G$. A $G$-variety $X$ is called spherical if $X$ has a dense $B$-orbit. Let $X$ be a $G$-spherical variety and let $H$ be the stabilizer of a point in the dense $G$-orbit in $X$. The variety $X$ is called {\\it horospherical} if $H$ contains a conjugate of the maximal unipotent subgroup of $G$ contained in the Borel subgroup $B$. \n\n\nSmooth horospherical varieties of Picard rank 1 were classified by Pasquier in \\cite{P}. These varieties are either homogeneous or can be constructed in a uniform way via a triple (Type($G$),$\\omega_Y$,$\\omega_Z$) of representation-theoretic data, where Type($G$) is the semisimple Lie type of the reductive group $G$ and $\\omega_Y, \\omega_Z$ are fundamental weights. See \\cite[Section 1.3]{P} for details. Pasquier classified the possible triples in five classes:\n\\begin{enumerate}\n\\item $(B_n, \\omega_{n-1},\\omega_n)$ with $n \\geq 3$;\n\\item $(B_3, \\omega_1,\\omega_3)$;\n\\item $(C_n,\\omega_m,\\omega_{m-1})$ with $n \\geq 2$ and $m \\in [2,n]$ (the odd symplectic Grassmannians);\n\\item $(F_4, \\omega_2, \\omega_3)$;\n\\item $(G_2,\\omega_1,\\omega_2)$.\n\\end{enumerate}\n\n In Proposition 3.6 of \\cite{Pthesis}, Pasquier showed the triples in the above list are Fano varieties. We are now able to state the main theorem:\n\n\\begin{thm}\\label{thm:main}\nIf $F$ belongs to the classes (1) for $n=3$, (2), (3), and (5) of Pasquier's list, then Conjecture $\\mathcal{O}$ holds for $F$.\n\\end{thm}\n{\\it Acknowledgements:} We would like to thank the anonymous referees for their useful comments and suggestions to improve the presentation of this paper.\n\\section{Preliminaries}\n\n\n\\subsection{Quantum Cohomology}\n\\label{sec:Section 2.1}\nThe small quantum cohomology is defined as follows. Let $(\\alpha_i)_i$ be a basis of $H^*(F)$, the classical cohomology ring, and let $(\\alpha_i^{\\vee})_i$ be the dual basis for the Poincar\\'e pairing. The multiplication is given by \\[ \\alpha_i \\star \\alpha_j= \\sum_{d\\geq 0, k} c_{i,j}^{k,d}q^d\\alpha_k \\] where $c_{i,j}^{k,d}$ are the 3-point, genus 0, Gromov-Witten invariants corresponding to the classes $\\alpha_i, \\alpha_j$, and $\\alpha_k^{\\vee}$. We will make use of the quantum Chevalley formula which is the multiplication of a hyperplane class $h$ with another class $\\alpha_j$. The result \\cite[Theorem 0.0.3]{GPPS} implies that if $F$ belongs to the classes (1) for $n=3$, (2), or (5) of Pasquier's list, then there is an explicit quantum Chevalley formula. The explicit quantum Chevalley formula is the key ingredient used to prove Property $\\mathcal{O}$ holds.\n\n\n\n\\subsection{Sufficient Criterion for Property $\\mathcal{O}$ to hold} We recall the notion of the (oriented) quantum Bruhat graph of a Fano variety $F$. The vertices of this graph are the basis elements $\\alpha_i \\in H^{\\bullet}(F):=QH^*(F)|_{q=1}$. There is an oriented edge $\\alpha_i \\rightarrow \\alpha_j$ if the class $\\alpha_j$ appears with positive coefficient (where we consider $q>0$) in the quantum Chevalley multiplication $h \\star \\alpha_i$ for some hyperplane class $h$. Using the Perron-Frobenius theory of non-negative matrices, Conjecture $\\mathcal{O}$ reduces to a graph-theoretic check of the quantum Bruhat graph. The techniques involving Perron-Frobenius theory used by Li, Mihalcea, and Shifler in \\cite{LMS} and Cheong and Li in \\cite{CL} imply the following lemma:\n\n\\begin{lemlab}\\label{lemma: propO}\nIf the following conditions hold for a Fano variety $F$:\n\\begin{enumerate}\n\\item the matrix representation of $\\hat{c}_1$ is nonnegative,\n\\item the quantum Bruhat graph of $F$ is strongly connected, and \n\\item there exists a cycle of length $r$, the Fano index, in the quantum Bruhat graph of $F$,\n\\end{enumerate}\nthen Property $\\mathcal{O}$ holds for $F$. We say the matrix representation of $\\hat{c}_1$ is nonnegative if all of the entries are nonnegative.\n\\end{lemlab}\n\nWe refer the reader to \\cite[section 4.3]{Minc} for further details on Perron-Frobenius theory.\n\n\\section{Checking Property $\\mathcal{O}$ Holds}\nLet $X$ be a horospherical variety. We will simplify our notation where the basis of $H^{\\bullet}(X)$ is $\\{1,h,\\alpha_i\\}_{i\\in I}$ for some finite index set $I$.  Observe by \\cite{GPPS} that  the anticanonical classes are \\[ c_1(X) = \\left\\{ \\begin{array}{ll}\n            5h & \\text{ when X is case (1) for }n=3 \\\\\n            7h & \\text{ when X is case (2)}\\\\\n            4h & \\text{ when X is case (5)}\n        \\end{array}\n    \\right.\\]\nand the Fano indices are\n \\[ r = \\left\\{ \\begin{array}{ll}\n            5 & \\text{ when X is case (1) for }n=3 \\\\\n            7 & \\text{ when X is case (2)}\\\\\n            4& \\text{ when X is case (5)}\n        \\end{array}.\n    \\right.\\]   \nThe endomorphism $\\hat{c}_1$ acting on the basis elements of $H^{\\bullet}(X)$ is determined by the Chevalley formula in the following way:\n\\begin{eqnarray*}\n            \\hat{c}_1(\\alpha_i)&=&5(h\\star \\alpha_i)|_{q=1}  \\text{ when X is case (1) for }n=3, \\\\\n            \\hat{c}_1(\\alpha_i)&=&7(h\\star \\alpha_i)|_{q=1}   \\text{ when X is case (2)}, \\text{and}\\\\\n            \\hat{c}_1(\\alpha_i)&=&4(h\\star \\alpha_i)|_{q=1}   \\text{ when X is case (5)}.\n\\end{eqnarray*}\n\nEach of the following three subsections will show that Conjecture $\\mathcal{O}$ holds for case (1) for $n=3$, case (2), and case (5) of Pasquier's list, respectively. In each subsection we will reformulate the quantum Chevalley formulas stated in \\cite{GPPS}, present the quantum Bruhat graph, and argue that each condition of Lemma \\ref{lemma: propO} is satisfied. For each case, we have kept the same format of the equations presented by \\cite{GPPS} with our prescribed basis for ease of identification for the reader. For example, line 3 in \\cite[Proposition 4.3]{GPPS} is \\[h*\\sigma'_{u_2}=\\sigma'_{u_3}+\\sigma'_{u'_3} \\mbox{ } and \\mbox{ } h* \\sigma'_{u'_2}=2\\sigma'_{u'_3}+\\tau_{v_0}. \\] In Proposition 1 below we identify this line with \\[\\hat{c}_1(\\alpha_1)=5\\alpha_3+5\\alpha_4 \\mbox{ } and \\mbox{ } \\hat{c}_1(\\alpha_2)=10\\alpha_3+5\\alpha_5. \\]\n\n\\subsection{Case (1) for $n=3$}  We will reformulate the quantum Chevalley formula stated in \\cite{GPPS} using the basis $\\{1,h, \\alpha_1, \\alpha_2,\\cdots, \\alpha_{18}\\}$.\n\n\\begin{proplab}\\label{Gon4.3} The following equalities hold by \\cite[Proposition 4.3]{GPPS}.\n\\begin{enumerate}\n\\item $\\hat{c}_1(1)=5h$\n\\item $\\hat{c}_1(h)=10\\alpha_1+5\\alpha_2$\n\\item $\\hat{c}_1(\\alpha_1)=5\\alpha_3+5\\alpha_4$ and $\\hat{c}_1(\\alpha_2)=10\\alpha_3+5\\alpha_5$\n\\item $\\hat{c}_1(\\alpha_3) = 10\\alpha_6+5\\alpha_7+5\\alpha_8, \\ \\ \\hat{c}_1(\\alpha_4)=5\\alpha_6 + 10\\alpha_7, $ and $\\hat{c}_1(\\alpha_5)=5\\alpha_8$\n\\item $\\hat{c}_1(\\alpha_6)=10\\alpha_9+5\\alpha_{10}+5\\alpha_{11},\\ \\ \\hat{c}_1(\\alpha_7) = 5\\alpha_{10}$ and $\\hat{c}_1(\\alpha_8)=5\\alpha_{11}+5\\cdot1$\n\\item $\\hat{c}_1(\\alpha_9)=5\\alpha_{12}+5\\alpha_{13}, \\ \\ \\hat{c}_1(\\alpha_{10})=10\\alpha_{13}+5\\alpha_{14}\\ \\ \\hat{c}_1(\\alpha_{11})=5\\alpha_{12}+5\\alpha_{14}+5h\n\\item $\\hat{c}_1(\\alpha_{12})=5\\alpha_{15}+5\\alpha_1, \\ \\ \\hat{c}_1(\\alpha_{13})=5\\alpha_{15} +5 \\alpha_{16},$ and $\\hat{c}_1(\\alpha_{14}) = 5\\alpha_{15}+5\\alpha_2\n\\item $\\hat{c}_1(\\alpha_{15})=5\\alpha_{17}+5\\alpha_3$ and $\\hat{c}_1(\\alpha_{16}) =5 \\alpha_{17}+5\\alpha_5\n\\item $\\hat{c}_1(\\alpha_{17})=5 \\alpha_{18}+5\\alpha_6+5\\alpha_8$\n\\item $\\hat{c}_1(\\alpha_{18}) =5 \\alpha_9+5\\alpha_{11}+10\\cdot1$\n\\end{enumerate}\n\\end{proplab}\n\n\\label{ex: case(1)}\nThe following figure is the quantum Bruhat graph of the Fano variety $X$ in case (1) for $n=3$. Colored edges are introduced in this figure to improve readability. The bold edges indicate a cycle of length $r=5$, the Fano index.\n\\begin{figure}[H]\n\\begin{center}\n\\caption{}\n\\label{qcbg:1-3}\n\\scalebox{.75}{\n\\begin{tikzpicture}\n\\tikzset{edge\/.style = {->,> = latex'}}\n    \\node (0) at (0,0) {$1$};\n    \\node (1) at (0,-1) {$h$};\n    \\node (2) at (-1,-2) {$\\alpha_1$};\n    \\node (3) at (1,-2) {$\\alpha_2$};\n    \\node (4) at (-2,-3) {$\\alpha_3$};\n    \\node (5) at (0,-3) {$\\alpha_4$};\n    \\node (6) at (2,-3) {$\\alpha_5$};\n    \\node (7) at (-2,-4) {$\\alpha_6$};\n    \\node (8) at (0,-4) {$\\alpha_7$};\n    \\node (9) at (2,-4) {$\\alpha_8$};\n    \\node (10) at (-2,-5) {$\\alpha_9$};\n    \\node (11) at (0,-5) {$\\alpha_{10}$};\n    \\node (12) at (2,-5) {$\\alpha_{11}$};\n    \\node (13) at (-2,-6) {$\\alpha_{12}$};\n    \\node (14) at (0,-6) {$\\alpha_{13}$};\n    \\node (15) at (2,-6) {$\\alpha_{14}$};\n    \\node (16) at (-1,-7) {$\\alpha_{15}$};\n    \\node (17) at (1,-7) {$\\alpha_{16}$};\n    \\node (18) at (0,-8) {$\\alpha_{17}$};\n    \\node (19) at (0,-9) {$\\alpha_{18}$};\n    \\draw [edge] (0) to (1);\n    \\draw [edge] (1) to (2);\n    \\draw [edge] (1) to (3);\n    \\draw [edge] (2) to (4);\n    \\draw [edge] (2) to (5);\n    \\draw [edge] (3) to (4);\n    \\draw [edge] (3) to (6);\n    \\draw [edge] (4) to (7);\n    \\draw [edge] (4) to (8);\n    \\draw [edge] (4) to (9);\n    \\draw [edge] (5) to (7);\n    \\draw [edge] (5) to (8);\n    \\draw [edge] (6) to (9);\n    \\draw [edge] (7) to (10);\n    \\draw [edge] (7) to (11);\n    \\draw [edge] (7) to (12);\n    \\draw [edge] (8) to (11);\n    \\draw [edge] (9) to (12);\n    \\draw [blue, edge] (9) to [bend right=100] (0);\n    \\draw [edge] (10) to (13);\n    \\draw [edge] (10) to (14);\n    \\draw [edge] (11) to (14);\n    \\draw [edge] (11) to (15);\n    \\draw [edge] (12) to (13);\n    \\draw [edge, ultra thick] (12) to (15);\n    \\draw [red, edge] (12) to [bend right=100] (1);\n    \\draw [edge] (13) to (16);\n    \\draw [green,edge] (13) to [bend left=100] (2);\n    \\draw [edge] (14) to (16);\n    \\draw [edge] (14) to (17);\n    \\draw [edge, ultra thick] (15) to (16);\n    \\draw [pink, edge] (15) to [bend right=100] (3);\n    \\draw [edge, ultra thick] (16) to (18);\n    \\draw [orange, edge] (16) to [bend left=100] (4);\n    \\draw [edge] (17) to (18);\n    \\draw [yellow, edge] (17) to [bend right=100] (6);\n    \\draw [edge, ultra thick] (18) to (19);\n    \\draw [purple, edge] (18) to [bend left =100] (7);\n    \\draw [teal, edge] (18) to [bend right=100] (9);\n    \\draw [lime, edge] (19) to [bend left=100] (10);\n    \\draw [edge, ultra thick] (19) to [bend right=100] (12);\n    \\draw [cyan, edge] (19) to [bend right=100] (0);\n\\end{tikzpicture}\n}\n\\end{center}\n\\end{figure}\n\n\\begin{lemlab}\\label{lemma:case1}\nProperty $\\mathcal{O}$ holds when $X$ is case (1) with $n=3$ of Pasquier's list.\n\\end{lemlab}\n\\begin{proof}\nThe coefficients that appear in the equations in Proposition \\ref{Gon4.3} are the entries of the matrix representation of $\\hat{c}_1$. Therefore, the matrix representation of $\\hat{c}_1$ is nonnegative. The quantum Bruhat graph is strongly connected by Figure \\ref{qcbg:1-3}, and the cycle $\\alpha_{18}\\alpha_{11}\\alpha_{14}\\alpha_{15}\\alpha_{17}\\alpha_{18}$ has length $r=5$.\n\\end{proof}\n\n\\subsection{Case (2)} Again, we reformulate the quantum Chevalley formula from \\cite{GPPS} using the basis $\\{1,h, \\alpha_1, \\alpha_2,\\cdots, \\alpha_{12}\\}$.\n\\begin{proplab}\\label{Gon4.4} The following equalities hold by \\cite[Proposition 4.4]{GPPS}.\n\\begin{enumerate}\n\\item $\\hat{c}_1(1)=7h$\n\\item $\\hat{c}_1(h)=7\\alpha_1$\n\\item $\\hat{c}_1(\\alpha_1)=14\\alpha_2+7\\alpha_3$\n\\item $\\hat{c}_1(\\alpha_2) = 7\\alpha_4+7\\alpha_5$ and $\\hat{c}_1(\\alpha_3)=7\\alpha_5$\n\\item $\\hat{c}_1(\\alpha_4)=7\\alpha_6+7\\alpha_7$ and $\\hat{c}_1(\\alpha_5)=7\\alpha_7$\n\\item $\\hat{c}_1(\\alpha_6)=7\\alpha_8$ and $\\hat{c}_1(\\alpha_7)=7\\alpha_8+7\\alpha_9\n\\item $\\hat{c}_1(\\alpha_8)=7\\alpha_{10}$ and $\\hat{c}_1(\\alpha_9) =7 \\alpha_{10}+7\\cdot1\n\\item $\\hat{c}_1(\\alpha_{10})=7\\alpha_{11}+7h\n\\item $\\hat{c}_1(\\alpha_{11})=7 \\alpha_{12}+7\\alpha_1$\n\\item $\\hat{c}_1(\\alpha_{12}) =7 \\alpha_2$\n\\end{enumerate}\n\\end{proplab}\n\nThe quantum Bruhat graph is \n\\begin{figure}[H]\n\\caption{}\n\\label{qcbg:2}\n\\begin{center}\n\\scalebox{.75}{\n\\begin{tikzpicture}\n\\tikzset{edge\/.style = {->,> = latex'}}\n    \\node (0) at (0,0) {$1$};\n    \\node (1) at (0,-1) {$h$};\n    \\node (2) at (0,-2) {$a_1$};\n    \\node (3) at (1,-3) {$a_3$};\n    \\node (4) at (-1,-3) {$a_2$};\n    \\node (5) at (1,-4) {$a_5$};\n    \\node (6) at (-1,-4) {$a_4$};\n    \\node (7) at (1,-5) {$a_7$};\n    \\node (8) at (-1,-5) {$a_6$};\n    \\node (9) at (2,-6) {$a_9$};\n    \\node (10) at (1,-6) {$a_8$};\n    \\node (11) at (1,-7) {$a_{10}$};\n    \\node (12) at (1,-8) {$a_{11}$};\n    \\node (13) at (1,-9) {$a_{12}$};\n    \\draw [edge] (0) to (1);\n    \\draw [edge] (1) to (2);\n    \\draw [edge] (2) to (3);\n    \\draw [edge] (2) to (4);\n    \\draw [edge] (3) to (5);\n    \\draw [edge] (4) to (5);\n    \\draw [edge, ultra thick] (4) to (6);\n    \\draw [edge] (5) to (7);\n    \\draw [edge] (6) to (7);\n    \\draw [edge, ultra thick] (6) to (8);\n    \\draw [edge] (7) to (9);\n    \\draw [edge] (7) to (10);\n    \\draw [edge, ultra thick] (8) to (10);\n    \\draw [edge] (9) to [bend right] (0);\n    \\draw [edge] (9) to (11);\n    \\draw [edge, ultra thick] (10) to (11);\n    \\draw [edge] (11) to [bend right=100] (1);\n    \\draw [edge, ultra thick] (11) to (12);\n    \\draw [edge] (12) to [bend right=100] (2);\n    \\draw [edge, ultra thick] (12) to (13);\n    \\draw [edge, ultra thick] (13) to [bend left=100] (4);\n\n\\end{tikzpicture}\n}\n\\end{center}\n\\end{figure}\n\n\n\\begin{lemlab} \\label{lemma:case2}\nProperty $\\mathcal{O}$ holds when $X$ is case (2) of Pasquier's list. \n\\end{lemlab}\n\\begin{proof}\nThe coefficients that appear in the equations in Proposition \\ref{Gon4.4} are the entries of the matrix representation of $\\hat{c}_1$. Therefore, the matrix representation of $\\hat{c}_1$ is nonnegative. The quantum Bruhat graph is strongly connected by Figure \\ref{qcbg:2}, and the cycle $\\alpha_{12}\\alpha_2\\alpha_4\\alpha_6\\alpha_8\\alpha_{10}\\alpha_{11}\\alpha_{12}$ has length $r=7$.\n\\end{proof}\n\n\n\\subsection{Case(5)} Again, we reformulate the quantum Chevalley formula from \\cite{GPPS} using the basis $\\{1,h, \\alpha_1, \\alpha_2,\\cdots, \\alpha_{10}\\}$.\n\\begin{proplab}\\label{Gon4.6} The following equalities hold by \\cite[Proposition 4.6]{GPPS}.\n\\begin{enumerate}\n\\item $\\hat{c}_1(1)=4h$\n\\item $\\hat{c}_1(h)=12\\alpha_1+4\\alpha_2$\n\\item $\\hat{c}_1(\\alpha_1)=8\\alpha_3+4\\alpha_4$ and $\\hat{c}_1(\\alpha_2)=4\\alpha_4$\n\\item $\\hat{c}_1(\\alpha_3) = 12\\alpha_5+4\\alpha_6$ and $\\hat{c}_1(\\alpha_4)=4\\alpha_6+4\\cdot1$\n\\item $\\hat{c}_1(\\alpha_5)=4\\alpha_7+4\\alpha_8$ and $\\hat{c}_1(\\alpha_6)=8\\alpha_7+4h$\n\\item $\\hat{c}_1(\\alpha_7)=4\\alpha_9+4\\alpha_1$ and $\\hat{c}_1(\\alpha_8)=4\\alpha_9+4\\alpha_2\n\\item $\\hat{c}_1(\\alpha_9)=4\\alpha_{10}+4\\alpha_3+4\\alpha_4\n\\item $\\hat{c}_1(\\alpha_{10})=4\\alpha_5+4\\alpha_6+8\\cdot 1\n\\end{enumerate}\n\\end{proplab}\n\n\n\nThe associated quantum Bruhat graph is \n\\begin{figure}[H]\n\\caption{}\n\\label{qcbg:5}\n\\begin{center}\n\\scalebox{.75}{\n\\begin{tikzpicture}\n\\tikzset{edge\/.style = {->,> = latex'}}\n    \\node (0) at (0,0) {$1$};\n    \\node (1) at (0,-1) {$h$};\n    \\node (2) at (-1,-2) {$\\alpha_1$};\n    \\node (3) at (1,-2) {$\\alpha_2$};\n    \\node (4) at (-1,-3) {$\\alpha_3$};\n    \\node (5) at (1,-3) {$\\alpha_4$};\n    \\node (6) at (-1,-4) {$\\alpha_5$};\n    \\node (7) at (1,-4) {$\\alpha_6$};\n    \\node (8) at (-1,-5) {$\\alpha_7$};\n    \\node (9) at (1,-5) {$\\alpha_8$};\n    \\node (10) at (-1,-6) {$\\alpha_9$};\n    \\node (11) at (-1,-7) {$\\alpha_{10}$};\n    \\draw [edge] (0) to (1);\n    \\draw [edge] (1) to (2);\n    \\draw [edge] (1) to (3);\n    \\draw [edge] (2) to (4);\n    \\draw [edge] (2) to (5);\n    \\draw [edge] (3) to (5);\n    \\draw [edge] (4) to (6);\n    \\draw [edge] (4) to (7);\n    \\draw [edge] (5) to (7);\n    \\draw [edge] (5) to [bend right=100] (0);\n    \\draw [edge] (6) to (8);\n    \\draw [edge] (6) to (9);\n    \\draw [edge, ultra thick] (7) to (8);\n    \\draw [edge] (7) to [bend right=100] (1);\n    \\draw [edge] (8) to [bend left=100] (2);\n    \\draw [edge, ultra thick] (8) to (10);\n    \\draw [edge] (9) to [bend right=100] (3);\n    \\draw [edge] (9) to (10);\n    \\draw [edge] (10) to [bend right=100] (5);\n    \\draw [edge, ultra thick] (10) to (11);\n    \\draw [edge] (10) to [bend left=100] (4);\n    \\draw [edge, ultra thick] (11) to [bend right=100] (7);\n    \\draw [edge] (11) to [bend left=100] (6);\n    \\draw [edge] (11) to [bend left=100] (0);\n\n\\end{tikzpicture}\n}\n\\end{center}\n\\end{figure}\n\n\\begin{lemlab}\\label{lemma:case5}\nProperty $\\mathcal{O}$ holds when $X$ is case (5) of Pasquier's list. \n\\end{lemlab}\n\\begin{proof}\nThe coefficients that appear in the equations in Proposition \\ref{Gon4.6} are the entries of the matrix representation of $\\hat{c}_1$. Therefore, the matrix representation of $\\hat{c}_1$ is nonnegative. The quantum Bruhat graph is strongly connected by Figure \\ref{qcbg:5}, and the cycle $\\alpha_{10}\\alpha_6\\alpha_7\\alpha_9\\alpha_{10}$ has length $r=4$.\n\\end{proof}\nTheorem \\ref{thm:main} follows from Lemmas \\ref{lemma:case1}, \\ref{lemma:case2}, \\ref{lemma:case5}, and the previously mentioned work done by Li, Mihalcea, Shifler for the odd symplectic Grassmannian case in \\cite{LMS}.\n\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}}
+{"text":"\\section{Introduction}\nThe heavy ion collisions have the advantage of being able to study the hot medium in detail under controlled environments.\nIt has recently been reported that the strongest magnetic fields of the order of $10^{18}-10^{19}$ Gauss are produced in non-central heavy-ion collisions by the electric current from the positively charged spectators that travel at nearly the speed of light ~\\cite{Bzdak2012,Deng2012,Kharzeev2008}.\nIt is expected that such a huge magnetic field may have important consequences on the dynamics of the quark-gluon matter produced in heavy-ion collisions ~\\cite{Gyulassy2005}.\nIn particular, it has been proposed that the interplay of quantum anomalies with ultra-intense magnetic field results in several special transport phenomena\nthat are closely related to chiral anomaly and thus are called anomalous transports ~\\cite{Xu-Guang Huang2016,Kharzeev2016}.\nThe external magnetic fields may induce charge separation in a chirality-imbalanced medium, namely the chiral magnetic effect (CME) ~\\cite{Kharzeev2008,Fukushima2008}\nwhich has been observed at RHIC and the LHC and the measurements signals indeed consistent with the predictions of the CME\n~\\cite{STAR2010,STAR2009,ALICE2016}.\nAlong with CME, the chiral separation effect (CSE) ~\\cite{Metlitski2005,Son2004} represents the generation of the axial current along the external magnetic field in the presence of finite vector charge density. The duality between CME and CSE leads to the interesting collective effect,\ncalled ``chiral magnetic wave\"(CMW) ~\\cite{Kharzeev2011}, which induces a quadrupole deformation of the electric charge distribution\nthat might be responsible for breaking the degeneracy between the elliptic flows of $\\pi^{\\pm}$ ~\\cite{Burnier2011}.\n\nThe relativistic hydrodynamic models have so far nicely described the thermodynamic evolution of the produced matter and the\nexperimentally measured flow harmonics in heavy-ion collisions ~\\cite{Romatschke2007,Luzum2008,Songprl2008,Songprc2008,Schenke2012,\nRoy2012,Niemi2012}. The influence of strong magnetic fields on the hot and dense nuclear matter have been intensively investigated.\nIn principle, such studies can be accomplished by solving the relativistic magnetohydrodynamics (MHD) equations that takes into account\nthe dynamical coupling of the magnetic field to the fluid.\nAlthough the magnetic field generated in heavy-ion collisions rapidly decays in the vacuum,\nthe hot medium created in the heavy-ion collision as a conducting plasma might substantially\ndelay the decay of the magnetic field through the generation of an induction current due to\nLenz's law ~\\cite{Gursoy2014,Zakharov2014,Tuchin2013}.\n\nIn Refs.~\\cite{Roy2015}, one-dimensional magnetic fluid had been investigated by using the longitudinally boost-invariant Bjorken flow~\\cite{Bjorken1983} with a transverse and time-dependent homogeneous magnetic field.\nIn ideal MHD limits, with the infinite electrical conductivity and neglecting other dissipative effects such as viscosity and thermal conduction, it is extraordinary that the evolution of energy density is the same as the case without magnetic fields due to ``frozen-flux theorem\".\nLater, a nonzero magnetization effect is introduced to the Bjorken flow in MHD~\\cite{Shi2016}.\n\nQGP expanding in the beam direction can be described by considering a boost invariant Bjorken flow, that is known to be a good approximation at mid-rapidity. It leads to a flat rapidity distribution of final particle,\nwhich is inconsistent with observations at RHIC, except for a narrow region around mid-rapidity ~\\cite{Bjorken1983}.\nIt has been pointed out that in realistic situations the energy density at mid-rapidity decreases faster than in the Bjorken flow.\nAlthough the Bjorken-estimation for the initial energy density is widely used, the longitudinal expansion dynamics of hydrodynamics seems ~\\cite{Csorgo:2006ax,Csand:2016arx,Jiang2017,Jiang2018} to be able to offer a more realistic estimation for the initial energy density estimation and the final state description. Acceleration effects are important in the estimation of the initial energy density even at mid-rapidity, if the expanding system is finite: even the most central fluid element exert a force on the volume elements closer to the surface, and this work decreases the internal energy of cells even at mid-rapidity.\n\nThis paper is organized as follows. In Sec.\\ref{section2}, the ideal-MHD framework with acceleration effects is presented.\nIn Sec.\\ref{section3}, we present the evolution of the energy density in ideal transverse MHD with longitudinal expansion dynamics.\nWe consider the decay of the energy density within an external homogeneous magnetic field which decays with a power-law in proper time.\nIn subsection.~\\ref{CNC approximation}, an exact analytic solution under the CNC approximation is obtained.\nIn subsection.~\\ref{numerical solution}, we show the results obtained from numerical method for a realistic equation of state (EoS).\nFinally, we discuss and conclude in the last section.\nThroughout this work, $u^{\\mu}=\\gamma\\left(1,\\overrightarrow{\\boldsymbol{v}}\\right)$\nis the four-velocity field that satisfies $u^{\\mu}u_{\\mu}=1$ and\nthe spatial projection operator $\\Delta^{\\mu\\nu}=g^{\\mu\\nu}-u^{\\mu}u^{\\nu}$ is defined with\nthe Minkowski metric $g^{\\mu\\nu}={\\rm diag}\\left(1,-1,-1,-1\\right)$.\nIt is note-worthy that the orthogonality relation $\\Delta^{\\mu\\nu}u_{\\nu}=0$ is satisfied.\nWe adopt the standard convention for the summation over repeated indices.\n\n\\section{ideal MHD with acceleration}\n\\label{section2}\n\nRelativistic magnetohydrodynamics (RMHD for short) concerns the mutual interaction of fluid flow and magnetic fields.\nThe fluids in question must be electrically conducting and non-magnetic.\nThe RMHD evolution equations describe the dynamics of the overall system based on local conservation of this fluid current\n(associated to the net-baryon current or to any other conserved charge)\nand the total (matter and fields) energy-momentum as well as on the additional assumption of local thermal equilibrium.\n\nConsider an non-viscous fluid coupling with  a magnetic field.\nWe assume that the medium is perfectly conducting and the electric field in the co-moving frame vanishes\nto avoid the onset of huge currents in the plasma.\nThe total energy-momentum tensor of ideal fluid is given by ~\\cite{Huang2010,Giacomazzo2006,Giacomazzo2007}\n\\begin{eqnarray}\nT^{\\mu\\nu}=(e+p+B^{2})u^{\\mu}u^{\\nu}-\\left(p+\\frac{B^{2}}{2}\\right)g^{\\mu\\nu}-B^{\\mu}B^{\\nu},\n\\label{1}\n\\end{eqnarray}\nwhere\n\\begin{eqnarray}\nB^{2}=B^{\\mu}B_{\\mu},\\quad B^{\\mu}=\\frac{1}{2}\\epsilon^{\\mu\\nu\\alpha\\beta}u_{\\nu}F_{\\alpha\\beta},\n\\label{2}\n\\end{eqnarray}\nhere $e,~p$ and $F_{\\alpha\\beta}$ are the fluid energy density, pressure and the Faraday tensor.\nHere, $\\epsilon^{\\mu\\nu\\alpha\\beta}$ is the completely antisymmetric four tensor with $\\epsilon^{0123}=-\\epsilon_{0123}=1$.\nThe magnetic field four-vector $B^{\\mu}$ is a space-like vector with modulus $B^{\\mu}B_{\\mu}=-B^{2}$ and is orthogonal to $u^{\\mu}$, i.e.,~ $B^{\\mu}u_{\\mu}=0$, where $B=|\\vec{\\boldsymbol{B}}|$ and $\\vec{\\boldsymbol{B}}$\nis the magnetic field three-vector in the frame moving with four-velocity $u^{\\mu}$.\n\nIn the present paper we consider the special case of a fluid flow with the external magnetic field\n$\\vec{\\boldsymbol{B}}$ directed along the transverse plane.\nThis setup is consistent with the scenario in non-central heavy ion collision at top RHIC energy~\\cite{Deng2012}.\nThe system of ideal RMHD equations can be closed by choosing the rather general EoS\n\\begin{equation}\np={c}_{s}^{2}\\,e=\\frac{1}{\\kappa}\\,e\\,,\n\\label{3}\n\\end{equation}\nwhere $c_s$ stands for the local speed of sound which is assumed to be a constant.\nIn a fully realistic solution, we should use results form the lattice QCD, with the speed of sound being a function of temperature~\\cite{Gupta2004,Qin2015}.\nHowever, in current work we approximate $c_s(T)$ as a temperature independent constant $c_s$.\nWe postpone the analysis of the case of $c_s(T)$ for a later, more detailed investigation.\n\nWe decompose the covariant derivative as\n\\begin{eqnarray}\n\\partial_{\\mu}=u_{\\mu}D+\\nabla_{\\mu},\n\\label{4}\n\\end{eqnarray}\nwhere $D=u^{\\mu}\\partial_{\\mu}$ indicates the time derivative in the local rest frame,\nand $\\nabla^{\\mu}=\\Delta^{\\mu\\nu}\\partial_{\\nu}$ is the spatial gradient in the local rest frame.\nThe energy conservation equation is derived by projecting the conservation law $\\partial_{\\mu}T^{\\mu\\nu}=0$ along the fluid four-velocity $u^{\\mu}$,\n\\begin{eqnarray}\nu_{\\mu}\\partial_{\\nu}T^{\\mu\\nu}\t&=&\tu_{\\mu}u^{\\mu}u^{\\nu}\\partial_{\\nu}(e+p+B^{2})+(e+p+B^{2})u_{\\mu}\\partial_{\\nu}(u^{\\mu}u^{\\nu})-u_{\\mu}\\partial_{\\nu}\\left[\\left(p+\\frac{B^{2}}{2}\\right)g^{\\mu\\nu}\\right]-u_{\\mu}\\partial_{\\nu}(B^{\\mu}B^{\\nu})\\nonumber\\\\\n\t&=&\tu^{\\nu}\\partial_{\\nu}(e+p+B^{2})+(e+p+B^{2})\\partial_{\\nu}u^{\\nu}+0-u^{\\nu}\\partial_{\\nu}\\left(p+\\frac{B^{2}}{2}\\right)-\\partial_{\\nu}(u_{\\mu}B^{\\mu}B^{\\nu})+B^{\\mu}B^{\\nu}\\partial_{\\nu}u_{\\mu}\n\t\\nonumber\\\\\n\t&=&\t\tD(e+p+B^{2})+(e+p+B^{2})\\theta-D\\left(p+\\frac{B^{2}}{2}\\right)\n\t\\nonumber\\\\\n\t&=&\t\tD\\left(e+\\frac{B^{2}}{2}\\right)+(e+p+B^{2})\\theta\n\t\\nonumber\\\\\n\t&=&\t\t0,\n\\label{5}\n\\end{eqnarray}\nwhere $\\theta\\equiv\\partial_{\\mu}u^{\\mu}=\\nabla_{\\mu}u^{\\mu}$ is the expansion factor\nand we have used relation $u^{\\mu}B_{\\mu}=0$ and $B^{\\nu}\\partial_{\\nu}u_{\\mu}=0$,\nsince $u_{\\mu}=\\left(u_{0},0,0,u_{z}\\right)$ and $B_{\\mu}=\\left(0,B_{x},B_{y},0\\right)$ in our setup.\nThus one obtains the energy-conservation equation as follows,\n\\begin{eqnarray}\nD\\left(e+\\frac{B^{2}}{2}\\right)+(e+p+B^{2})\\theta=0 \\,.\n\\label{6}\n\\end{eqnarray}\n\nThe relativistic version of the MHD Euler equation is retrieved by projecting the energy-momentum conservation equation onto the direction orthogonal to $u^{\\mu}$,\n\\begin{eqnarray}\n\\triangle_{\\mu\\nu}\\partial_{\\alpha}T^{\\alpha\\nu}\t&=&\t\t(e+p+B^{2})\\triangle_{\\mu\\nu}\\partial_{\\alpha}(u^{\\alpha}u^{\\nu})+0-\\triangle_{\\mu\\nu}\\partial^{\\nu}\\left(p+\\frac{B^{2}}{2}\\right)-\\triangle_{\\mu\\nu}\\partial_{\\alpha}(B^{\\alpha}B^{\\nu})\\nonumber\\\\\n\t&=&\t\t(e+p+B^{2})(g_{\\mu\\nu}-u_{\\mu}u_{\\nu})\\partial_{\\alpha}(u^{\\alpha}u^{\\nu})-\\triangle_{\\mu\\nu}\\partial^{\\nu}\\left(p+\\frac{B^{2}}{2}\\right)-(g_{\\mu\\nu}-u_{\\mu}u_{\\nu})\\partial_{\\alpha}(B^{\\alpha}B^{\\nu})\\nonumber\\\\\n\t&=&\t\t(e+p+B^{2})\\left[u^{\\alpha}\\partial_{\\alpha}u_{\\mu}+u_{\\mu}\\partial_{\\alpha}u^{\\alpha}-u_{\\mu}u_{\\nu}u^{\\alpha}\\partial^{\\alpha}u^{\\nu}-u_{\\mu}u_{\\nu}u^{\\nu}\\partial_{\\alpha}u^{\\alpha}\\right]-\\triangle_{\\mu\\nu}\\partial^{\\nu}\\left(p+\\frac{B^{2}}{2}\\right)\\nonumber\\\\\n&&-B^{\\alpha}\\partial_{\\alpha}B_{\\mu}-B_{\\mu}\\partial_{\\alpha}B^{\\alpha}+u_{\\mu}u_{\\nu}B^{\\alpha}\\partial_{\\alpha}B^{\\nu}+u_{\\mu}u_{\\nu}B^{\\nu}\\partial_{\\alpha}B^{\\alpha}\n\\nonumber\\\\\n\t&=&\t\t(e+p+B^{2})Du_{\\mu}-\\triangle_{\\mu\\nu}\\partial^{\\nu}\\left(p+\\frac{B^{2}}{2}\\right)-B^{\\alpha}\\partial_{\\alpha}B_{\\mu}-B_{\\mu}\\partial_{\\alpha}B^{\\alpha}-u_{\\mu}B^{\\alpha}B^{\\nu}\\partial_{\\alpha}u_{\\nu}\\nonumber\\\\\n&=&\t\t0.\n\\label{7}\n\\end{eqnarray}\n\nThe last three terms vanish and leads to the Euler equation as follow,\n\\begin{equation}\n(e+p+B^{2})Du_{\\mu}-\\nabla_{\\mu}\\left(p+\\frac{B^{2}}{2}\\right)=0\\,.\n\\label{8}\n\\end{equation}\n\nWe use the well-known Rindler coordinates $\\tau=\\sqrt{t^{2}-r^{2}}$ and $\\eta_{s}=\\frac{1}{2}{\\rm log}\\left(\\left(t+r\\right)\/\\left(t-r\\right)\\right)$ as independent variables inside the forward lightcone and parametrize the fluid velocity as $v={\\rm tanh}\\Omega$, and the fluid rapidity $\\Omega$ depends only on $\\eta_s$ here. (For simplicity, we will use $\\Omega$ to denotes $\\Omega(\\eta_s)$, and $\\Omega^{\\prime}$ denotes $d\\Omega\/d\\eta_{s}$). One obtains\n\\begin{eqnarray}\nD&=&u^{\\mu}\\partial_{\\mu}=u^{0}\\partial_{t}+u^{z}\\partial_{z}\\nonumber\\\\\n&=&{\\rm cosh}\\Omega\\left({\\rm cosh}\\eta_{s}\\frac{\\partial}{\\partial\\tau}-\\frac{{\\rm sinh}\\eta_{s}}{\\tau}\\frac{\\partial}{\\partial\\eta_{s}}\\right)+{\\rm sinh}\\Omega\\left(-{\\rm sinh}\\eta_{s}\\frac{\\partial}{\\partial\\tau}+\\frac{{\\rm cosh}\\eta_{s}}{\\tau}\\frac{\\partial}{\\partial\\eta_{s}}\\right)\\nonumber\\\\\n&=&{\\rm cosh}(\\Omega-\\eta_{s})\\frac{\\partial}{\\partial\\tau}+\\frac{1}{\\tau}{\\rm sinh}(\\Omega-\\eta_{s})\\frac{\\partial}{\\partial\\eta_{s}}\n\\label{9}\n \\end{eqnarray}\nand\n\\begin{eqnarray}\n\\theta&=&\\partial_{\\mu}u^{\\mu}=\\partial_{t}u^{0}+\\partial_{z}u^{z}\\nonumber\\\\\n&=&\\left({\\rm cosh}\\eta_{s}\\frac{\\partial}{\\partial\\tau}-\\frac{{\\rm sinh}\\eta_{s}}{\\tau}\\frac{\\partial}{\\partial\\eta_{s}}\\right){\\rm cosh}\\Omega+\\left(-{\\rm sinh}\\eta_{s}\\frac{\\partial}{\\partial\\tau}+\\frac{{\\rm cosh}\\eta_{s}}{\\tau}\\frac{\\partial}{\\partial\\eta_{s}}\\right){\\rm sinh}\\Omega\\nonumber\\\\\n&=&{\\rm cosh}\\eta_{s}{\\rm sinh}\\Omega\\frac{\\partial\\Omega}{\\partial\\tau}-\\frac{{\\rm sinh}\\eta_{s}}{\\tau}{\\rm sinh}\\Omega\\frac{\\partial\\Omega}{\\partial\\eta_{s}}-{\\rm sinh}\\eta_{s}{\\rm cosh}\\Omega\\frac{\\partial\\Omega}{\\partial\\tau}+{\\rm \\frac{{\\rm cosh}\\eta_{s}}{\\tau}{\\rm cosh}\\Omega}\\frac{\\partial\\Omega}{\\partial\\eta_{s}}\\nonumber\\\\\n&=&{\\rm sinh}(\\Omega-\\eta_{s})\\frac{\\partial\\Omega}{\\partial\\tau}+\\frac{1}{\\tau}{\\rm cosh}(\\Omega-\\eta_{s})\\frac{\\partial\\Omega}{\\partial\\eta_{s}}.\n\\label{10}\n \\end{eqnarray}\n\nThe peak value of the magnetic field $\\vec{\\boldsymbol{B}}$ is well determined by using event-by-event simulations with in the Monte-Carlo Glauber model~\\cite{Bloczynski2013}.\nNevertheless, the lifetime of magnetic field is still an open question so far. We assume in this paper the homogeneous magnetic field obeys a power-law decay in proper time~\\cite{Roy2015},\n\\begin{equation}\n\\overrightarrow{B}(\\tau)=\\overrightarrow{B}_{0}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{a}.\n\\label{11}\n\\end{equation}\nHere $a=1$ correspond to the ideal-MHD case~\\cite{Davidson2017}, where $a>1$ corresponds to the case with the magnetic field decaying steeper than the ideal-MHD case,\nand $a<1$ corresponds to a decay slower than in the ideal-MHD limit.\n$\\tau_0$ is the initial proper time of the fluid expansion and $B_{0}\\equiv B(\\tau_{0})$ is the initial magnetic field strength.\n\nAbove assumptions allow one to rewrite the conservation equations in Rindler coordinate as follows\n\\begin{eqnarray}\n&&\\tau\\frac{\\partial\\widetilde{e}}{\\partial\\tau}+{\\rm tanh}(\\Omega-\\eta_{s})\\frac{\\partial\\widetilde{e}}{\\partial\\eta_{s}}+\n\\left(\\widetilde{e}(1+c_{s}^{2})+\\sigma_{0}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}\\right)\\Omega'=\\sigma_{0}a\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a},~\n\\label{12}\\\\\n&&\\frac{\\partial\\widetilde{e}}{\\partial\\eta_{s}}={\\rm tanh}(\\Omega-\\eta_{s})\\left[\\frac{\\sigma_{0}}{c_{s}^{2}}\n\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}(a-\\Omega')-\\frac{1+c_{s}^{2}}{c_{s}^{2}}\\widetilde{e}\\Omega'-\\tau\\frac{\\partial\\widetilde{e}}{\\partial\\tau}\\right],\n\\label{13}\n\\end{eqnarray}\nwith the dimensionless quantities $\\widetilde{e}\\equiv e\/e_{0},~\\sigma_{0}\\equiv B_{0}^{2}\/e_{0}$.\n\nThe combination of energy conservation equation Eq.~(\\ref{12}) and the Euler equation Eq.~(\\ref{13}) generates a partial differential equation,\n\n\\begin{eqnarray}\n\\tau\\frac{\\partial\\widetilde{e}}{\\partial\\tau}=\\left(\\frac{{\\rm sinh}^{2}(\\Omega-\\eta_{s})}{c_{s}^{2}}-{\\rm cosh}^{2}(\\Omega-\\eta_{s})\\right)\\left[\\left(\\sigma_{0}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}+\\widetilde{e}(1+c_{s}^{2})\\right)\\Omega'-\\sigma_{0}a\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}\\right].\n\\label{14}\n\\end{eqnarray}\n\\section{energy-density evolution}\n\\label{section3}\nThe exact solution with CNC approximation (Sec.\\ref{CNC approximation}) and numerical solution (Sec.\\ref{numerical solution}) of energy density evolution in MHD are presented in this section step by step.\n\n\\subsection{Exact solution of MHD with CNC approximation}\n\\label{CNC approximation}\nFor a perfect fluid with longitudinal accelerating expansion, one finds $\\Omega\\neq\\eta_{s}$.\nThe exact solution for such longitudinal accelerating hydrodynamics is the well-known CNC solution with $\\Omega=\\lambda\\eta_{s}$ and $\\kappa=1$, $c_{s}^{2}=\\frac{1}{\\kappa}=1,~\\Omega'=\\lambda,~\\Omega''=0$. From Eq.(\\ref{14}), one gets\n\n\\begin{eqnarray}\n\\tau\\frac{\\partial\\widetilde{e}}{\\partial\\tau}=\n\\sigma_{0}a\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}-\\left(\\sigma_{0}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}+2\\widetilde{e}\\right)\\lambda.\n\\label{15}\n\\end{eqnarray}\nThe solution $\\widetilde{e}(\\tau,\\eta_{s})$ is\n\\begin{eqnarray}\n\\widetilde{e}(\\tau,\\eta_{s})=-\\frac{1}{2}\\sigma_{0}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}+\\tau^{-2\\lambda}C(\\eta_{s}),\n\\label{16}\n\\end{eqnarray}\nwhere $C(\\eta_{s})$ is an undetermined function related to the $\\eta_s$ part of the energy density $\\widetilde{e}(\\tau,\\eta_{s})$.\n\nPutting Eq.(\\ref{16}) into the Euler equation Eq.(\\ref{13}), one gets\n\\begin{eqnarray}\nC(\\eta_{s})=C.\n\\label{17}\n\\end{eqnarray}\nThen substituteing Eq.(\\ref{17}) to Eq.(\\ref{16}) and using the initial condition $\\widetilde{e}_{0}(\\tau_0,0)=1$, one obtains\n\\begin{eqnarray}\nC=\\frac{1}{2}(2+\\sigma_{0})\\tau_{0}^{2\\lambda}.\n\\label{18}\n\\end{eqnarray}\nFinally, inputting Eq.(\\ref{18}) and Eq.(\\ref{17}) into Eq.(\\ref{16}), an analytical solution of the fluid energy density with CNC approximation can be written as follow\n\\begin{eqnarray}\n\\widetilde{e}(\\tau,\\eta_{s})=\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2\\lambda}+\\frac{\\sigma_{0}}{2}\\left[\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2\\lambda}-\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}\\right].\n\\label{19}\n\\end{eqnarray}\n\n\\begin{table}[htb]\n\\begin{tabular}{|c|c|c|c|}\n\\hline\n$$ & $a\\to\\lambda$ & $a\\gg\\lambda$ & $a\\ll\\lambda$\n\\\\\n\\hline\n$\\widetilde{e}(\\tau,\\eta_{s})$ & $\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2\\lambda}$ & $\\frac{2+\\sigma_{0}}{2}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2\\lambda}$ & $-\\frac{\\sigma_{0}}{2}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}$\n\\\\\n\\hline\n\\end{tabular}\n\\caption{The fluid energy density in the three kinds of limit conditions}\n\\end{table}\nOnce again, it is possible to see that in the limit of vanishing magnetization $\\sigma_{0}=0$ and $\\lambda=1$, Eq.(\\ref{19}) coincides with the solution for Bjorken flow.\nIf $\\sigma_{0}=0$ and $\\lambda \\neq 1$, the solution coincides with CNC solution.\nFurthermore, for $\\sigma_{0}\\neq 0$ and $\\lambda=1$, one obtains the same solution as the Bjorken-Victor type flow~\\cite{Roy2015}.\n\nOne can obtain the extreme value of the energy density from the following steps\n\\begin{eqnarray}\n\\frac{\\partial\\widetilde{e}(\\tau,\\eta_{s})}{\\partial\\tau}&&=\\frac{a\\sigma_{0}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}-\\lambda(2+\\sigma_{0})\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2\\lambda}}{\\tau}=0\n\\label{20}\\\\\n&&\\Rightarrow\\tau=\\tau_{0}\\left(\\frac{a\\sigma_{0}}{\\lambda\\left(2+\\sigma_{0}\\right)}\\right)^{\\frac{1}{2(a-\\lambda)}}.\n\\label{21}\n\\end{eqnarray}\nUnfortunately, these CNC solutions have a shortcoming, namely the acceleration parameter $\\lambda$ becomes\na free fit parameter only for the superhard EoS of $\\kappa=1,~e=p$.\nIn this case, the speed of sound is equal to the speed of light $c$, so the investigation was thought to be rather academic.\n\n\\subsection{Numerical solution for MHD}\n\\label{numerical solution}\nTo get a realistic solution of the energy density, we consider the case in which $\\Omega\\equiv\\lambda\\eta_{s}\\equiv(1+\\lambda^{*})\\eta_{s}$ with $\\lambda^{*}$ being a very small constant acceleration parameter ($0<\\lambda^{*}\\ll1$) and $\\Omega'=1+\\lambda^{*},~\\Omega''=0$.\n\nThus, the energy equation and Euler equation can be expressed as\n\\begin{eqnarray}\n&&\\tau\\frac{\\partial\\widetilde{e}}{\\partial\\tau}+{\\rm tanh}(\\lambda^{*}\\eta_{s})\\frac{\\partial\\widetilde{e}}{\\partial\\eta_{s}}+\\left(\\widetilde{e}(1+c_{s}^{2})+\\sigma_{0}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}\\right)(1+\\lambda^{*})=\\sigma_{0}a\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a},\\label{22}\\\\\n&&\\frac{\\partial\\widetilde{e}}{\\partial\\eta_{s}}={\\rm tanh}(\\lambda^{*}\\eta_{s})\\left[\\frac{\\sigma_{0}}{c_{s}^{2}}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}(a-1-\\lambda^{*})-\\frac{1+c_{s}^{2}}{c_{s}^{2}}\\widetilde{e}(1+\\lambda^{*})-\\tau\\frac{\\partial\\widetilde{e}}{\\partial\\tau}\\right].\n\\label{23}\n\\end{eqnarray}\n\nThe combination of energy equation Eq.(\\ref{22}) and Euler equation Eq. (\\ref{23}) can be rewritten as follows\n\\begin{eqnarray}\n\\tau\\frac{\\partial\\widetilde{e}}{\\partial\\tau}&=&\\left(\\kappa{\\rm sinh}^{2}(\\lambda^{*}\\eta_{s})-{\\rm cosh}^{2}(\\lambda^{*}\\eta_{s})\\right)\\left[\\left(\\sigma_{0}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}+\\widetilde{e}(1+\\frac{1}{\\kappa})\\right)(1+\\lambda^{*})-\\sigma_{0}a\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}\\right],\n\\label{24}\\\\\n\\frac{\\partial\\widetilde{e}}{\\partial\\eta_{s}}&=&\\frac{1}{2}{\\rm sinh}(2\\lambda^{*}\\eta_{s})\\left[\\kappa\\sigma_{0}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}(a-1-\\lambda^{*})-(1+\\kappa)\\widetilde{e}(1+\\lambda^{*})-\\sigma_{0}a\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}\\right.\\nonumber\\\\\n&&\\left.+\\left(\\widetilde{e}(1+\\frac{1}{\\kappa})+\\sigma_{0}\\left(\\frac{\\tau_{0}}{\\tau}\\right)^{2a}\\right)(1+\\lambda^{*})\\right].\n\\label{25}\n\\end{eqnarray}\nThe main idea of solving the above partial differential equations is to treat this two PDEs as two ordinary differential equations with a given initial condition $\\widetilde{e}(\\tau_{0},0)=1$. Then one can get the relation between $\\widetilde{e}$ and $\\tau$ from Eq. (\\ref{24}). The sets of data obtained above can be taken as the initial conditions for solving Eq.(\\ref{25}). For this purpose, one obtains the full profile of energy density right away and obtains the evolution of temperature by using relation $e\\propto T^{\\kappa+1}$.\n\nFig. \\ref{fig1} reports numerical solution of the fluid energy density and the temperature of accelerating fluid in one-dimensional relativistic magnetohydrodynamics with parameters $a=2,~\\sigma_{0}=1.0,~\\kappa=7,~\\lambda^{*}=0.03$. The profile of $\\widetilde{e}(\\tau,~\\eta_{s})$ is a (1+1) dimensional scaling solution, and it contains not only acceleration but also the magnetic field dependent terms now with the $\\eta_s$ dependence of the Gaussian form.\nNote that (1) if $\\lambda^{*}=0$ and $\\sigma_{0}=0$ one obtains the Bjorken solutions, (2) if $\\lambda^{*}=0$ and $\\sigma_0\\neq0$ one obtains the same solution as the Bjorken-Victor type flow, (3) if $\\lambda^{*}\\neq 0$ and $\\sigma_{0}=0$ one obtains the case of the well-known CNC solution.\n\\begin{figure}[htb]\n\\begin{minipage}[htb]{8.5cm}\n\\centerline{\\epsfig{figure=fig1.pdf,width=8cm}}\n\\end{minipage}\n\\hfill\n\\begin{minipage}[htb]{8.5cm}\n\\centerline{\\epsfig{figure=fig2.pdf,width=8cm}}\n\\end{minipage}\n\\caption{Left panel indicates the fluid energy density $e\/e_{0}$, while the right panel shows the temperature $T\/T_{0} $(right panel) profile, with parameters $a=2,\\sigma_{0}=1.0,\\kappa=7,\\lambda^{*}=0.03$.}\n\\label{fig1}\n\\end{figure}\n\nFor the sake of comparison with Bjorken-Victor type flow,\nwe take the space-time rapidity $\\eta_s=0$ in Fig.\\ref{fig2} and Fig.\\ref{fig3}. In Fig.\\ref{fig2}, we compare the evolution of fluid energy density $\\widetilde{e}$ for following different conditions: (\\textbf{a})~different longitudinal acceleration parameter $\\lambda^*$, (\\textbf{b})~different magnetic field decay parameter $a$, and (\\textbf{c})~different EoS parameter $\\kappa$. In Fig.\\ref{fig2} case (\\textbf{a}), different lines represent to different values of the longitudinal acceleration parameter $\\lambda^{*}$, ranging from $\\lambda^{*}=0$ (Bjorken-Victor type flow without longitudinal acceleration effect; black solid line) up to cases with $\\lambda^*=0.03$ (red dashed line), $\\lambda^*=0.06$ (blue dotted line) and $\\lambda^*=0.1$ (magenta dot-dashed line). In Fig.\\ref{fig2} case (\\textbf{b}), we show the evolution of the normalized energy density $\\widetilde{e}$ for $a=2$ (black solid line), $a=1$ (ideal-MHD limit; red dashed line), and $a=2\/3$ (blue dotted line). In Fig.\\ref{fig2} case (\\textbf{c}), different lines means different values of EoS $\\kappa=1$ (CNC approximation; black solid line), $\\kappa=3$ (red dashed line), $\\kappa=7$ (blue dotted), and $\\kappa=10$ (magenta dot-dashed line). As the graph illustrates, the longitudinal acceleration effect of the fluid increase the decay of the fluid energy density; $\\widetilde{e}$ decays faster for $a=2\/3$ than the ideal-MHD limit $a=1$ case, whereas for $a=2$ it initially decays more slowly and then decays asymptotically at the same rate as for the ideal-MHD $a=1$ case; the evolution of the fluid energy density $\\widetilde{e}$ decays more quickly with decreasing $\\kappa$.\n\n\\begin{figure}[htb]\n\\begin{minipage}[htb]{5cm}\n\\centerline{\\epsfig{figure=fig3.pdf,width=7.5cm}}\n\\end{minipage}\n\\hfill\n\\begin{minipage}[htb]{5cm}\n\\centerline{\\epsfig{figure=fig4.pdf,width=7.5cm}}\n\\end{minipage}\n\\hfill\n\\centering\n\\begin{minipage}[htb]{5cm}\n\\centerline{\\epsfig{figure=fig5.pdf,width=7.5cm}}\n\\end{minipage}\n\\caption{Evolution of fluid energy density $e\/e_0$ as a function of proper time $\\tau$ and we choose initial condition as $\\widetilde{e}_{0}\\equiv\\widetilde{e}(\\tau_{0})=1$. (a)Different lines refer to different levels of longitudinal acceleration parameter: $\\lambda^{*}=0$ (black solid line), $\\lambda^{*}=0.03$ (red dashed line), $\\lambda^{*}=0.06$ (blue dotted line), $\\lambda^{*}=0.1$ (magenta dot-dashed line). Clearly, it is gradually speed up the decay rate of fluid energy density with increasing acceleration parameter $\\lambda^*$. (b)\nDifferent lines refer to different levels of magnetic field decay parameter: $a=2$ (black solid line), $a=1$ (red dashed line), and $a=2\/3$ (blue dotted line). Clearly, the fluid energy density decreases more rapidly for $a=2\/3$ than in the case $a=1$, not to mention $a=2$. (c)Different lines refer to the evolution for $\\kappa=1$ (black solid line), $\\kappa=3$ (red dashed line), $\\kappa=7$ (blue dotted line), and $\\kappa=10$ (magenta dot-dashed line).\n}\n\\label{fig2}\n\\end{figure}\n\nIn Fig.\\ref{fig3}, we consider the evolution of the fluid energy density $\\widetilde{e}$ (upper panel) and the total energy density $e\/e_{0}+\\sigma_{0}(B\/B_{0})^{2}\/2$ (lower panel) in the different cases and when the parameters are set to $a=2\/3$ (left panel) and $a=2$ (right panel). Left panel report the evolution of fluid energy density and the total energy density $e\/e_{0}+\\sigma_{0}(B\/B_{0})^{2}\/2$ for $a=2\/3$ and where different lines refer to different levels of the initial magnetization: $\\sigma_0=0$ (black solid line), $\\sigma_0=0.5$ (red dashed line), $\\sigma_0=1.0$ (blue dotted line), and $\\sigma_0=2$ (magneto dot-dashed line). It is clear that larger values of initial magnetization $\\sigma_0$ will lead to a faster decrease in $\\widetilde{e}$ and $e\/e_{0}+\\sigma_{0}(B\/B_{0})^{2}\/2$. Right panel shows the evolution of fluid energy density $\\widetilde{e}$ and the total energy density $e\/e_{0}+\\sigma_{0}(B\/B_{0})^{2}\/2$ in the case $a=2$. In this case, different lines refer to different levels of the initial magnetization, $\\sigma_0=0.01$ (black solid), $\\sigma_0=1.0$ (red dashed), and $\\sigma_0=10$ (blue dotted). As shown in the Fig.~\\ref{fig3} (upper-right panel), it produces even a temporary increase in the fluid energy density evolution. This interesting phenomena, which can be associated with the resistive \"heating up\" of the fluid, and it depends on the values of the initial magnetization $\\sigma_0$ and the magnetic field decay parameter $a$. This increase in the fluid energy density evolution will be larger for larger magnetic field decay parameter $a$ due to the fact that the Lorentz force allows energy to transfer back and forth between the magnetic field and the fluid. The total energy density of this system decays quickly with increasing $\\sigma_0$ for $a<1$. Increasing $\\sigma_0$ only adds energy density to the system, but does not alter the temporal evolution of the total energy density for the case with $a>1$.\n\n\\begin{figure}[htb]\n\\begin{minipage}[htb]{8.8cm}\n\\centerline{\\epsfig{figure=fig6.pdf,width=8.5cm}}\n\\end{minipage}\n\\hfill\n\\begin{minipage}[htb]{8.8cm}\n\\centerline{\\epsfig{figure=fig7.pdf,width=8.5cm}}\n\\end{minipage}\n\\hfill\n\\begin{minipage}[htb]{8.8cm}\n\\centerline{\\epsfig{figure=fig8.pdf,width=8.5cm}}\n\\end{minipage}\n\\hfill\n\\begin{minipage}[htb]{8.8cm}\n\\centerline{\\epsfig{figure=fig9.pdf,width=8.5cm}}\n\\end{minipage}\n\\caption{The evolution of the fluid energy density $e\/e_0$ (upper panel) and the total energy density $e\/e_{0}+\\sigma_{0}(B\/B_{0})^{2}\/2$ (lower panel) in the different cases and when the parameters are set to $a=2\/3$ (left panel) and $a=2$ (right panel). (Left panel) Different lines refer to different levels of the initial magnetization: $\\sigma=0$ (black solid line), $\\sigma=0.5$ (red dashed line), $\\sigma=1.0$ (blue dotted line), and $\\sigma=2.0$ (magenta dot-dashed line). (Right panel) Different lines refer to different levels of the initial magnetization, ranging from $\\sigma_0=0.01$ (black solid line), $\\sigma_0=1.0$ (red dashed line), and $\\sigma_0=10.0$ (blue dotted line).}\n\\label{fig3}\n\\end{figure}\n\n\\section{discussion and conclusions}\n\\label{section4}\n\nWe have investigated the evolution of the energy density of the QGP generated by the non-central heavy ion collisions by one-dimensional MHD flow in the limit of infinite electrical conductivity with longitudinal acceleration parameter $\\lambda^*$ and got an exact solution under the CNC approximation. Compared with Bjorken-Victor type flow, the longitudinal acceleration effect accelerates the decay of the energy density. For larger $\\kappa$ of EoS, the energy density decays more slowly, thus the temperature dependent EoS should be calculated from lattice QCD simulations.\n\nBased on the definition of the acceleration coordinate (Rindler coordinate, Kottler-M\\o ller coordinates, and Radar coordinates), the \"acceleration parameter\" $\\lambda$ has following physics meaning: (1)$\\lambda<0$, for heavy ion collisions, it means that the fireball system's element flowing into the fireball's core and the system's thermodynamics quantities density is increasing with the time, or in other words, the fireball system does not swell but contracts, which means after enough long time, there will creating a black holw; (2)$\\lambda=0$ correspond to the rest fireball system; (3)$0<\\lambda<1$, the fireball system's expansion speed is decelerating. The energy density deposit to large $\\eta_s$; (4)$\\lambda=1$, the fireball system's expansion speed is average; (5)$\\lambda>1$, the fireball system's expansion is fast and many energy density deposit to the mid-rapidity $\\eta_s$, which is consistent with the experimental data. Thus, we only focus on the case that longitudinal acceleration parameter $\\lambda^*$ is greater than $0$ in the previous discussion.\n\nFor the case that the magnetic field evolution follows a power-law decay in proper time with exponent $a$, we find\nthe magnetic field decays more quickly than in the ideal-MHD case for $a>1$, while the magnetic field with $a<1$ correspond to a decay that is slower than in the ideal-MHD limit.\nIn heavy-ion collisions the remnants of colliding nuclei can give an additional contribution to the magnetic field to slow down its decay.\nThus, considering the case $a<1$ is reasonable in this paper.\nIt is clearly that larger values of initial magnetization $\\sigma_0$ leads to faster decreasing in $\\widetilde{e}$ for $a<1$. But it also results in a temporary increase in the fluid energy density for $a>1$. As we know, the magnetic field energy can be converted to fluid energy via Lorzent force, thus the evolution of the fluid energy density becomes more complex. For $a\\rightarrow0$, the magnetic field is constant in proper time and does not evolve with the fluid.\nThus, the fluid energy density must decay very rapidly to keep this constant magnetic field.\nFor $a\\rightarrow\\infty$, the magnetic field decays fast and the energy is transferred to the fluid-element according to the energy-conservation law.\nThus, one can expect a peak of the energy density near the initial time, which is associated with a \"reheating\" of the fluid with longitudinal acceleration effect.\n\nHowever, the recent estimates both from lattice QCD simulations ~\\cite{Alessandro2013,Amato2013,Greif2014} and fitting of experimental data point~\\cite{Yi Yin2014} toward high, but finite value for the electrical conductivity of the QGP. For a quantitative comparison with experimental data, the effects of the electrical resistivity has to be taken into account.\n\nAs a next step, we try to include the dissipative effects (shear and bulk viscosity and a finite electric conductivity), the rescatterings in the hadronic phase, the decays of hadronic resonance into stable hadrons and anomalous currents. Note it would be necessary to modify the Cooper-Frye formula by taking into account the presence of an electromagnetic field.\n\n\\begin{acknowledgements}\nWe specially thank Dirk H. Rischke for the useful suggestion about the MHD theory at the ATHIC2018. This work is in part supported by the Ministry of Science and Technology of China (MSTC) under the \"973\" Project No. 2015CB856904(4), by NSFC Grant Nos. 11735007, 11890711\nThis work was supported by the Sino-Hungarian bilateral cooperation program, under the Grand No.Te'T 12CN-1-2012-0016, by the financial supported from NNSF of China under grant No.11435004. Z-F. Jiang would like to thank T.~Cs\\\"org\\H{o}, M.~Csan\\'ad, L{\\'e}vai P{\\'e}ter and Gergely G{\\'a}bor Barnafoldi for kind hospitality during his stay at Winger RCP, Budapest, Hungary.\n\n\n\\end{acknowledgements}\n\n\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}}
+{"text":"\\section{Introduction}\nAt present, time invariance violation has only been observed\nindirectly,\nin the CP-violating decay\nof neutral K-mesons \\cite{CPv}. Parity and\ntime invariance violating nuclear and\natomic multipole moments, such\nas the magnetic monopole, electric dipole, magnetic quadrupole,\nand electric octupole moments, are interesting\nbecause, if discovered,\nthey would provide further proof of time invariance violation.\nEven the limits on these moments provide important tests of\ndifferent models of CP-violation.\nParity and time invariance violating nuclear moments induced by\nT- and P-odd nuclear forces have been discussed, e.g., in\nRefs. \\cite{Fein77,Cov83,Hax83,SFK84,FKS86,Khripl}.\n\nIn this paper we consider the electric octupole\nmoments of nuclei. We calculate the EOMs of odd $Z$ nuclei that\nare induced by a T- and P-odd interaction\nbetween the unpaired proton\nand the nuclear core. Note that nuclei with unpaired\nneutrons can\nhave an EOM of comparable magnitude due to the polarization of the\nnuclear core by the T-, P-odd field of the external neutron. A\nsimilar mechanism for the electric dipole and Schiff moments was\nconsidered in Ref. \\cite{FKS86}.\n\nWe give values of the collective electric\noctupole moments of nuclei having a static octupole \ndeformation, using the calculations done\nin Refs. \\cite{AFS96,AFS2}. We also present a\ncalculation of the atomic electric dipole moment (EDM) \nthat would be induced by a nuclear EOM. Finally, we discuss\na possible enhancement mechanism for\nmagnetic quadrupole moments (MQMs) in nuclei with\noctupole deformation.\n\nIn the appendix we present simple estimates of\nthe relative sizes of the contributions\nof various nuclear moments to the atomic EDM. The magnitude of\nthe nuclear EOM is comparable to that of the Schiff moment.\nHowever, the contribution of the EOM to the atomic EDM is\nsmaller than the contribution of the Schiff moment (and the\nmagnetic quadrupole moment) since the EOM interacts with\nhigher angular momentum electron states, whose wave functions\nare suppressed near the nucleus. \n\n\\section{The Hamiltonian of the T-, P-odd nucleon-nucleus\ninteraction and the resulting nucleon wave function}\n\\label{shptnw}\nFor a heavy nucleus, the T- and P-odd interaction between a\nnonrelativistic unpaired nucleon and the nuclear core can be \ndescribed\nby the following effective Hamiltonian (see, e.g.,\nRefs. \\cite{Henley,SFK84,Khripl}):\n\\begin{equation}\nH_{TP} = \\eta \\frac{G}{2 \\sqrt{2} m} \\bbox{\\sigma} \\cdot\n\\bbox{\\nabla} \\rho,\n\\label{ehpta}\n\\end{equation}\nwhere $\\bbox{\\sigma}$ is twice the spin operator for this nucleon,\n$\\rho$ is the density of the nuclear core,\n$G = 1.0 \\times 10^{-5}\/{m_p}^2$ is the Fermi constant,\n$m$ is the mass of the nucleon and $\\eta$ is a dimensionless\nconstant that describes the strength of the interaction.\nIn this paper we\ndeal with odd $Z$ nuclei, and so the nucleon involved is the\nunpaired proton.\n\nLet $U$ be the strong nuclear potential of the core that the \nunpaired proton moves in. The range of\nthe strong nucleon-nucleon interaction\nis small. This means that the potential $U({\\bf r})$ and \nthe nuclear density $\\rho ({\\bf r})$ will be similar in shape.\nIn fact, we assume\nthat they are approximately proportional: $U({\\bf r})\/U({\\bf 0})\n\\approx \\rho({\\bf r})\/\\rho({\\bf 0})$. This allows us to rewrite\nEq.\\ (\\ref{ehpta}) in a form that makes it\neasy to find the perturbed\nwave function due to $H_{TP}$ (we use the method of\nRef. \\cite{SFK84}). We obtain\n\\begin{equation}\nH_{TP} \\approx \\xi \\bbox{\\sigma} \\cdot \\bbox{\\nabla} U,\n\\label{ehptb}\n\\end{equation}\nwhere\n\\begin{equation}\n\\xi = \\eta \\frac{G}{2 \\sqrt{2} m_p} \\frac{\\rho ({\\bf 0})}\n{U ({\\bf 0})} = -2 \\times 10^{-21} \\eta \\mbox{ cm}.\n\\label{ehptc}\n\\end{equation}\nThe total potential that the unpaired proton experiences is\n\\begin{equation}\n\\widetilde{U} = U + H_{TP} \\approx U({\\bf r})\n+ \\xi \\bbox{\\sigma} \\cdot \\bbox{\\nabla} U \\approx\nU({\\bf r} + \\xi \\bbox{\\sigma}).\n\\label{ehptd}\n\\end{equation}\nAs a result, if $\\psi ({\\bf r})$ is the proton's wave function\nwhen only\n$U({\\bf r})$ is present, the perturbed wave function will be\n\\begin{equation}\n\\widetilde{\\psi} ({\\bf r}) \\approx \\psi ({\\bf r} + \\xi\n\\bbox{\\sigma})\n\\approx \\psi({\\bf r}) + \\xi \\bbox{\\sigma} \\cdot \\bbox{\\nabla}\n\\psi({\\bf r}).\n\\label{ehpte}\n\\end{equation}\n\n\\section{The electric octupole moment of a nucleus with\nan unpaired proton}\n\\label{sneom}\nIn this section we calculate the electric octupole moment\n(EOM) of a\nnucleus with an unpaired proton using\nthe perturbed wave function obtained above.\nThe electric octupole moment can be written as\n[see Eq. (\\ref{eneo8caa})]\n\\begin{equation}\nO_{ijk} = \\langle \\widetilde{\\psi} | \\hat{O}_{ijk} |\n\\widetilde{\\psi} \\rangle\n= e \\int \\widetilde{\\psi}^{\\dagger}({\\bf r})\n[r_i r_j r_k\n- {\\textstyle \\frac{1}{5}} r^2 (r_i \\delta_{jk} + r_j \\delta_{ik}\n+ r_k \\delta_{ij})] \\widetilde{\\psi} ({\\bf r}) \\, d^3 r.\n\\label{eaa}\n\\end{equation}\nNote that we use $e > 0$.\nSubstituting $\\widetilde{\\psi} ({\\bf r}) =\n\\psi ({\\bf r}) + \\xi \\sigma_m \n\\frac{\\partial \\psi}{\\partial r_m}$ (from Eq.\\ (\\ref{ehpte}))\ninto the above integral and expanding gives\n\\begin{equation}\nO_{ijk} = 2 e \\xi \\int \\frac{\\partial \\psi^{\\dagger}}{\\partial r_m}\n\\sigma_m [r_i r_j r_k - \n{\\textstyle \\frac{1}{5}} r^2 (r_i \\delta_{jk} + r_j \\delta_{ik}\n+ r_k \\delta_{ij})] \\psi \\, d^3 r.\n\\label{eneom1a}\n\\end{equation}\nNotice that we have discarded the $\\xi^2$ term\nand that the term not\ncontaining $\\xi$ vanishes, as it is an integral of an odd function\nof ${\\bf r}$. Applying integration by parts to the above\nintegral gives \n\\begin{eqnarray}\nO_{ijk} & =  - {\\textstyle \\frac{1}{5}} e \\xi \\langle \\psi\n| & \n5 (\\hat{r}_i \\hat{r}_j \\hat{\\sigma}_k +\n\\hat{r}_i \\hat{r}_k \\hat{\\sigma}_j +\n\\hat{r}_j \\hat{r}_k \\hat{\\sigma}_i)\n-{\\hat{r}}^2 (\\hat{\\sigma}_i \\delta_{jk} +\n\\hat{\\sigma}_j \\delta_{ik}\n+ \\hat{\\sigma}_k \\delta_{ij}) \\nonumber \\\\\n&& - 2 \\hat{r}_m \\hat{\\sigma}_m\n(\\hat{r}_i \\delta_{jk} + \\hat{r}_j \\delta_{ik} + \\hat{r}_k\n\\delta_{ij})\n|\n\\psi \\rangle.\n\\label{eac}\n\\end{eqnarray}\n\nWe can also write the octupole moment tensor in another form.\nSince it is a symmetric, irreducible (traceless) third rank tensor\nand the nuclear angular momentum ${\\bf I}$ is the only \nquantity which defines\na direction in the system, the EOM tensor must be in the form\nof the most general symmetric, irreducible third rank tensor that\ncan be formed from the components of $\\hat{\\bf I}$. That is\n\\begin{equation}\nO_{ijk} = \\langle \\widetilde{\\psi} | \\hat{O}_{ijk} |\n\\widetilde{\\psi} \\rangle ,\n\\label{ead}\n\\end{equation}\nwhere\n\\begin{eqnarray}\n\\hat{O}_{ijk} & = A [ & \\hat{I}_i \\hat{I}_j \\hat{I}_k +\n\\hat{I}_j \\hat{I}_k \\hat{I}_i + \\hat{I}_k \\hat{I}_i \\hat{I}_j\n+ \\hat{I}_k \\hat{I}_j \\hat{I}_i + \\hat{I}_j \\hat{I}_i \\hat{I}_k\n+ \\hat{I}_i \\hat{I}_k \\hat{I}_j \\nonumber \\\\\n&& - {\\textstyle \\frac{6I(I+1)-2}{5}}\n(\\hat{I}_i \\delta_{jk} + \\hat{I}_j \\delta_{ik}\n+ \\hat{I}_k \\delta_{ij} )]\n\\label{eae}\n\\end{eqnarray}\nand $A$ is some constant. The factor of $-[6I(I+1)-2]\/5$ follows\nfrom the requirement\nof tracelessness ($O_{iij}=O_{iji}=O_{jii}=0$).\nThe quantity which is usually referred to as the octupole moment\nis ${\\cal O}$, which is the $O_{zzz}$ component for the nuclear\nstate having\nangular momentum projection $I_z = I$. This is the quantity\nwhich we will calculate. We can write $A$ in terms of ${\\cal O}$\nusing Eqs.\\ (\\ref{ead}) and (\\ref{eae}), with $i=j=k=z$. This gives\n\\begin{eqnarray}\n\\hat{O}_{ijk} & =\n\\frac{5 {\\cal O}}{6I(I-1)(2I-1)}\n[ & \\hat{I}_i \\hat{I}_j \\hat{I}_k +\n\\hat{I}_j \\hat{I}_k \\hat{I}_i + \\hat{I}_k \\hat{I}_i \\hat{I}_j\n+ \\hat{I}_k \\hat{I}_j \\hat{I}_i + \\hat{I}_j \\hat{I}_i \\hat{I}_k\n+ \\hat{I}_i \\hat{I}_k \\hat{I}_j \\nonumber \\\\\n&& - {\\textstyle \\frac{6I(I+1)-2}{5}}\n(\\hat{I}_i \\delta_{jk} + \\hat{I}_j \\delta_{ik}\n+ \\hat{I}_k \\delta_{ij} )].\n\\label{eaf}\n\\end{eqnarray}\nWe now have two expressions for $O_{ijk}$: Eq.\\ (\\ref{eac}) and\nEq.\\ (\\ref{eaf}) (via Eq.\\ (\\ref{ead})). We calculate ${\\cal O}$ by\noperating on both of these equations with $\\hat{I}_i \\hat{I}_j$ on\nthe left and $\\hat{I}_k$ on the right and equating the results.\n\nTo evaluate the results of these operations, the following\ncommutation\nrelations are required: $[\\hat{r}_i,\\hat{\\sigma}_j] = 0$,\n$[\\hat{I}_i,\\hat{r}_j] = i \\varepsilon_{ijk} \\hat{r}_k$, and\n$[\\hat{I}_i,\\hat{\\sigma}_j] = i \\varepsilon_{ijk} \\hat{\\sigma}_k$.\n(Note that $\\hat{I}_i = \\hat{l}_i + \\hat{\\sigma}_i \/ 2$.)\nOther useful relations are\n$[\\hat{I}_i,\\hat{I}_j \\hat{r}_j] = [\\hat{I}_i,\\hat{I}_j\n\\hat{\\sigma}_j]\n= [\\hat{I}_i,\\hat{\\sigma}_j \\hat{r}_j] = 0$ and\n$\\varepsilon_{ijk} A_i A_j = {\\textstyle \\frac{1}{2}}\n\\varepsilon_{ijk} A_i A_j\n- {\\textstyle \\frac{1}{2}} \\varepsilon_{ijk} A_j A_i\n= {\\textstyle \\frac{1}{2}} \\varepsilon_{ijk} [A_i,A_j]$,\nfor an operator $A_i$.\nThe commutation relations\nare used to rearrange the operators so that pairs\nwith the same indices are adjacent.\n\nApplying the operators to\nEq. (\\ref{eac}) gives\n\\begin{eqnarray}\n\\langle \\widetilde{\\psi} | \\hat{I}_i \\hat{I}_j \\hat{O}_{ijk}\n\\hat{I}_k | \\widetilde{\\psi} \\rangle\n& = -{\\textstyle \\frac{1}{5}} e \\xi \\langle \\psi | &\n5 \\hat{I}_i \\hat{I}_j (\\hat{r}_i \\hat{r}_j \\hat{\\sigma}_k +\n\\hat{r}_i \\hat{r}_k \\hat{\\sigma}_j + \\hat{r}_j \\hat{r}_k\n\\hat{\\sigma}_i)\n\\hat{I}_k\n- \\hat{I}_i \\hat{I}_j \\hat{r}^2 (\\hat{\\sigma}_i \\delta_{jk} +\n\\hat{\\sigma}_j \\delta_{ik}\n+ \\hat{\\sigma}_k \\delta_{ij}) \\hat{I}_k \\nonumber \\\\\n&& - 2 \\hat{I}_i \\hat{I}_j \\hat{r}_m \\hat{\\sigma}_m \n(\\hat{r}_i \\delta_{jk} + \\hat{r}_j \\delta_{ik} + \\hat{r}_k\n\\delta_{ij})\n\\hat{I}_k | \\psi \\rangle \\nonumber \\\\\n& = -{\\textstyle \\frac{1}{5}} e \\xi \\langle r^2 \\rangle \\{ &\n\\langle I,I_z,l | (\\hat{\\bbox{\\sigma}} \\cdot {\\hat{\\bf n}})\n[ {\\textstyle \\frac{5}{4}} (\\hat{\\bf I} \\cdot \\hat{\\bbox{\\sigma}})\n- \\hat{I}^2 ] (\\hat{\\bbox{\\sigma}} \\cdot {\\hat{\\bf n}})\n| I,I_z,l \\rangle \\nonumber \\\\\n&&\n+ \\langle I,I_z,l | {\\textstyle \\frac{7}{2}} (\\hat{\\bf I} \\cdot\n\\hat{\\bbox{\\sigma}})\n- 3 \\hat{I}^2 (\\hat{\\bf I} \\cdot \\hat{\\bbox{\\sigma}}) + 1\n- 2 \\hat{I}^2 | I,I_z,l \\rangle \\},\n\\label{eal}\n\\end{eqnarray}\nwhere $\\hat{\\bf n} = \\hat{\\bf r} \/ r$.\nHere the facts that\n$\\hat{\\bf I} \\cdot \\hat{\\bf n} = (\\hat{\\bf l} +\n\\hat{\\bbox{\\sigma}}\/2)\n\\cdot \\hat{\\bf n} = (\\hat{\\bbox{\\sigma}} \\cdot \\hat{\\bf n})\/2$\nand $(\\hat{\\bbox{\\sigma}} \\cdot \\hat{\\bf n})^2 = 1$ \\cite{QMNRT}\nwere used. \nThe radial and angular parts of the\noperators have been separated in Eq.\\ (\\ref{eal});\n$\\langle r^2 \\rangle$\nis the expectation value of $r^2$ and $| I,I_z,l \\rangle$ is the\nangular part of $| \\psi \\rangle$.\nThe second matrix element in Eq.\\ (\\ref{eal}) is\n\\begin{equation}\n\\langle I,I_z,l | {\\textstyle \\frac{7}{2}}\n(\\hat{\\bf I} \\cdot \\hat{\\bbox{\\sigma}})\n- 3 \\hat{I}^2 (\\hat{\\bf I} \\cdot \\hat{\\bbox{\\sigma}}) + 1\n- 2 \\hat{I}^2 | I,I_z,l \\rangle\n= [{\\textstyle \\frac{7}{2}} - 3I(I+1)]\n({\\textstyle \\frac{1}{2}} - \\kappa) + 1 -2I(I+1),\n\\label{eam}\n\\end{equation}\nwhere we have used the fact that\n\\begin{eqnarray}\n(\\hat{\\bf I} \\cdot \\hat{\\bbox{\\sigma}}) | I,I_z,l \\rangle & =\n& [I(I+1) - l(l+1) + 3\/4] | I,I_z,l \\rangle \\nonumber \\\\\n& = & (1\/2 - \\kappa) | I,I_z,l \\rangle,\n\\label{ean}\n\\end{eqnarray}\nwith\n\\begin{equation}\n\\kappa = (I + 1\/2) (-1)^{I+1\/2-l}.\n\\label{eao}\n\\end{equation}\nTo evaluate the first matrix element in Eq.\\ (\\ref{eal}), we use\nthe following identity:\n\\begin{equation}\n(\\hat{\\bbox{\\sigma}} \\cdot \\hat{{\\bf n}}) | I,I_z,l \\rangle\n= - | I,I_z,\\widetilde{l} \\rangle,\n\\label{eap}\n\\end{equation}\nwhere $\\widetilde{l} = 2 I - l$.\nWe then have\n\\begin{eqnarray}\n\\langle I,I_z,l | (\\hat{\\bbox{\\sigma}} \\cdot {\\hat{\\bf n}})\n[ {\\textstyle \\frac{5}{4}} (\\hat{\\bf I} \\cdot \\hat{\\bbox{\\sigma}})\n- \\hat{I}^2 ] (\\hat{\\bbox{\\sigma}} \\cdot {\\hat{\\bf n}})\n| I,I_z,l \\rangle & = &\n\\langle I,I_z,\\widetilde{l} | {\\textstyle \\frac{5}{4}} (\\hat{\\bf I}\n\\cdot \\hat{\\bbox{\\sigma}})\n- \\hat{I}^2 | I,I_z,\\widetilde{l} \\rangle \\nonumber \\\\\n& = & {\\textstyle \\frac{5}{4}} ({\\textstyle \\frac{1}{2}}\n+ \\kappa) - I(I+1).\n\\label{eaq}\n\\end{eqnarray}\n(Note that $(\\hat{\\bf I} \\cdot \\hat{\\bbox{\\sigma}})\n| I,I_z,\\widetilde{l} \\rangle = (1\/2 + \\kappa)\n| I,I_z,\\widetilde{l} \\rangle$.)\nUsing these results in Eq.\\ (\\ref{eal}) gives\n\\begin{equation}\n\\langle \\widetilde{\\psi} | \\hat{I}_i \\hat{I}_j \\hat{O}_{ijk}\n\\hat{I}_k\n| \\widetilde{\\psi} \\rangle = -{\\textstyle \\frac{3}{5}} e \\xi\n\\langle r^2 \\rangle\n(\\kappa - {\\textstyle \\frac{3}{2}})\n(I-{\\textstyle \\frac{1}{2}})(I+{\\textstyle \\frac{3}{2}}).\n\\label{ear}\n\\end{equation}\nNow we find another form of the left hand side of the above\nequation by operating on Eq.\\ (\\ref{eaf}). We have\n\\begin{eqnarray}\n\\langle \\widetilde{\\psi} | \\hat{I}_i\n\\hat{I}_j \\hat{O}_{ijk} \\hat{I}_k\n| \\widetilde{\\psi} \\rangle\n& =\n\\frac{5 {\\cal O}}{6I(I-1)(2I-1)}\n\\langle \\widetilde{\\psi} |\n& \\hat{I}_i \\hat{I}_j (\\hat{I}_i \\hat{I}_j \\hat{I}_k +\n\\hat{I}_j \\hat{I}_k \\hat{I}_i + \\hat{I}_k \\hat{I}_i \\hat{I}_j\n+ \\hat{I}_k \\hat{I}_j \\hat{I}_i + \\hat{I}_j \\hat{I}_i \\hat{I}_k\n+ \\hat{I}_i \\hat{I}_k \\hat{I}_j) \\hat{I}_k \\nonumber \\\\\n&& - {\\textstyle \\frac{6I(I+1)-2}{5}} \\hat{I}_i \\hat{I}_j \n(\\hat{I}_i \\delta_{jk} + \\hat{I}_j \\delta_{ik}\n+ \\hat{I}_k \\delta_{ij} ) \\hat{I}_k | \\widetilde{\\psi} \\rangle.\n\\label{eas}\n\\end{eqnarray}\nThis can be evaluated using the equation\n$[\\hat{I}_i,\\hat{I}_j] = i \\varepsilon_{ijk} \\hat{I}_k$ to\nbring operators with the same indices together\nand $\\varepsilon_{ijk} \\hat{I}_i \\hat{I}_j =\n\\frac{i}{2}\n\\varepsilon_{ijk} \\varepsilon_{ijp} \\hat{I}_p = i \\hat{I}_k$.\nThe result is\n\\begin{equation}\n\\langle \\widetilde{\\psi} |\n\\hat{I}_i \\hat{I}_j \\hat{O}_{ijk} \\hat{I}_k\n| \\widetilde{\\psi} \\rangle = {\\cal O} (I+1)(I+3\/2)(I+2).\n\\label{eat}\n\\end{equation}\nEquating this result with Eq.\\ (\\ref{ear}) gives the following\nresult for the octupole moment:\n\\begin{equation}\n{\\cal O}_{\\rm sing} = \\frac{-3(\\kappa-3\/2)(I-1\/2)}\n{5(I+1)(I+2)} \\langle r^2 \\rangle e \\xi,\n\\label{eau}\n\\end{equation}\n(The subscript refers to the fact that this octupole moment is\ndue to a single particle, as opposed to a collective octupole\nmoment.)\nThe expectation value for $r^2$ can be\napproximated as \\cite{SFK84}\n\\begin{equation}\n\\langle r^2 \\rangle\n\\approx {\\textstyle \\frac{3}{5}} {r_0}^2 A^{2\/3},\n\\label{eaw}\n\\end{equation}\nto an accuracy of about 10\\%,\nwhere $r_0 = 1.1 \\mbox{ fm}$ and $A$ is the mass number of\nthe nucleus. Using the value of $r_0$ and Eqs.\\ (\\ref{ehptc}) and\n(\\ref{eao}) gives\n\\begin{equation}\n{\\cal O_{\\rm sing}} \\approx 8.7 \\times 10^{-9} A^{2\/3}\n\\eta e ({\\rm fm})^3 \\times\n\\left\\{\n\\begin{array}{ll}\n\\frac{-(I-1\/2)}{I+1} & \\mbox{for $I=l+1\/2$} \\\\\n\\frac{(I-1\/2)(I-1)}{(I+1)(I+2)} & \\mbox{for $I=l-1\/2$}\n\\end{array} \\right..\n\\label{eaw2}\n\\end{equation}\nObserve that ${\\cal O}_{\\rm sing}=0$ for $I=1\/2$. This\nis to be expected because\n$\\hat{O}_{ijk}$ is a third rank tensor, and so applying the\ntriangle rule for the addition of angular momenta\nto Eq.\\ ({\\ref{ead})\ngives the result that nuclei having angular momentum less\nthan $3\/2$ cannot have an octupole moment.\nValues of ${\\cal O}_{\\rm sing}$ for various nuclei\nare given in table \\ref{tsingp}, in terms of the parameter $\\eta$.\n  \n\\section{Collective electric octupole moments in nuclei with static\noctupole deformation}\n\\label{sceom}\nIn Refs.\\ \\cite{AFS96,AFS2}, a mechanism was suggested by which\nparity and time invariance\nviolating interactions can produce collective T- and P-odd\nmultipole moments in \neven-odd nuclei having a static octupole deformation\n[i.e. electric octupole moments in their intrinsic\n(or body-fixed) reference frames].\nSuch a deformation has been shown to exist for nuclei in the\nRa--Th and Ba--Sm regions (for a review see, e.g., \\cite{Ahmad93}).\nA similar\nmechanism, for the enhancement of the intrinsic electron EDM and\nother T-, P-odd interactions in polar molecules, was suggested in\nRef. \\cite{SF1978}.\nBelow, we explain the mechanism by which a collective nuclear\nEOM can be produced and provide values of this EOM for various\nnuclei.\n\n\\subsection{Parity and time invariance and octupole moments in\nthe laboratory frame}\nAn electric octupole moment can exist in the nucleus's intrinsic\nframe without parity or time invariance violation. Yet if parity\nand time invariance hold, the expectation value of the octupole\nmoment in the laboratory reference frame will be zero.\n\nConsider\n$|I M K \\rangle$ and $|I M -K \\rangle$, which are two almost\ndegenerate states of the nucleus in the laboratory\nframe. $I$ is the angular momentum of the nucleus,\n$M$ is its projection onto the $z$-axis, and $K$\nis its projection onto\nthe axis of symmetry of the deformed nuclear core\n(the $z^{\\prime}$-axis).\n(${\\bf I}$ is the sum of the unpaired nucleon's angular\nmomentum, ${\\bf j}$ and the nuclear core's orbital angular\nmomentum, ${\\bf R}$.)\nThese states can be written in terms of intrinsic states as\n\\begin{equation}\n|I M \\pm K \\rangle =\n\\sqrt{\\textstyle \\frac{2I+1}{4 \\pi}}\n D^I_{M \\pm K} (\\phi,\\theta,0) \\psi_{\\pm K}\n({\\bf r}^\\prime) \\chi_{\\rm core},\n\\label{ecn21a}\n\\end{equation}\nwhere $D^I_{M \\pm K} (\\phi,\\theta,0)$ is a Wigner $D$-function\n(see, e.g., \\cite{Varshalovich,RelQuant}),\n$\\chi_{\\rm core}$ is the wave function of the nuclear core\nin the intrinsic frame, and\n$\\psi_{\\pm K} ({\\bf r}^\\prime)$ is the wave function of the unpaired\nnucleon in the intrinsic frame, with\na $z^{\\prime}$ angular momentum projection of $\\pm K$. (In the\nintrinsic frame, the nuclear axis plays the role of the usual\n``$z$-axis'' in Quantum Mechanics.) Note that ${\\bf j}$ and\n${\\bf I}$ have the same $z^{\\prime}$ projection.\n\n$|I M K \\rangle$ and $|I M -K \\rangle$ do not\nhave good parity, as $K$\nchanges sign under a parity transformation. However, the following\nstates do, and they form a parity doublet:\n\\begin{equation}\n\\psi^{\\pm} = \\frac{1}{\\sqrt{2}} ( |I M K \\rangle\n\\pm |I M -K \\rangle).\n\\label{eceom1a}\n\\end{equation}\nFor these good parity states \n$\\langle \\psi^{\\pm} | {\\bf I} \\cdot {\\bf n} | \\psi^{\\pm}\n\\rangle = 0$\nbecause $K$ and $-K$ have equal probabilities\nand this means that there is no\naverage orientation of the nuclear axis in the laboratory frame\n($\\langle \\psi^{\\pm} | {\\bf n} | \\psi^{\\pm} \\rangle = {\\bf 0}$).\nThis is a consequence of time invariance and parity\nconservation since\nthe correlation ${\\bf I} \\cdot {\\bf n}$ is T-, P-odd. As a result\nof $\\langle \\psi^{\\pm} | {\\bf n} | \\psi^{\\pm} \\rangle = {\\bf 0}$,\nthe mean value of the octupole moment\n(whose orientation is determined by the direction of the\nnuclear axis)\nis zero in the laboratory frame.  \n\nNow, a T- and P-odd interaction, $H_{TP}$\nwill mix the members of the parity doublet ($\\psi^+$ and $\\psi^-$).\nThe admixed wave function of the predominantly positive parity\nmember of the doublet will be $\\psi = \\psi^+ + \\alpha \\psi^-$ or\n\\begin{equation}\n\\psi = \\frac{1}{\\sqrt{2}} [(1+\\alpha) |I M K \\rangle\n+ (1-\\alpha) |I M -K \\rangle],\n\\label{eceom2c}\n\\end{equation}\nwhere $\\alpha$ is a mixing coefficient:\n\\begin{equation}\n\\alpha = \\frac{\\langle \\psi^- | H_{TP} |\\psi^+ \\rangle}{E_+ - E_-}.\n\\label{enad32}\n\\end{equation}\n$E_+ - E_-$ is the energy splitting between the members of the\nparity doublet.\nThe interaction $H_{TP}$ is given by Eq.\\ (\\ref{ehpta}).\n(A similar expression can be obtained for the predominantly\nnegative parity member of the doublet.)\nThis mixing yields, on average,\nan orientation of the nuclear axis along the direction of the\nangular momentum:\n\\begin{equation}\n\\langle \\psi | {\\bf I} \\cdot {\\bf n} | \\psi \\rangle\n= \\langle \\psi | \\hat{K} | \\psi \\rangle\n= 2 \\alpha K,\n\\label{eceom2d}\n\\end{equation}\nand this means that the octupole moment need no longer vanish in\nthe laboratory frame.\n\n\\subsection{The magnitude of the collective octupole moment}\nThe collective EOM in the laboratory frame\nwas derived in \\cite{AFS96,AFS2},\nwith the following result:\n\\begin{equation}\n{\\cal O}_{\\rm coll} \\approx \\frac{4}{5}\n\\frac{I (I-1) (I - 1\/2)}{(I+1)\n(I+2) (I + 3\/2)}\n\\alpha O_{3,{\\rm intr}}.\n\\label{eceom1}\n\\end{equation}\nOnce again, observe that ${\\cal O}_{\\rm coll} = 0$ for $I=1\/2$.\n$O_{3,{\\rm intr}}$ refers to the octupole moment in the\nintrinsic frame ($ O_{zzz} \\equiv \\frac{2}{5} O_3$)\nand is given by \\cite{BohrMott,LeanderChen}:\n\\begin{equation}\nO_{3,{\\rm intr}} = e Z {R_0}^3 \\frac{3}{2 \\sqrt{7 \\pi}}\n(\\beta_3 + \\frac{2}{3} \\sqrt{\\frac{5}{\\pi}} \\beta_2 \\beta_3 + \n\\frac{15}{11 \\sqrt{\\pi}} \\beta_3 \\beta_4 + \\ldots),\n\\label{eceom2}\n\\end{equation}\nwhere $R_0 = r_0 A^{1\/3}$ ($r_0 = 1.1 \\mbox{ fm}$). \n$\\beta_2$, $\\beta_3$, and $\\beta_4$ are parameters\nthat describe the nuclear\ndeformation; the surface of a deformed nucleus is\n\\begin{equation}\nR = R_0 [1 + \\sum_{l=1}^{\\infty} \\beta_l Y_{l0} (\\theta,\\phi) ].\n\\label{eceom3}\n\\end{equation}\n\nWe will first present an order of magnitude estimate of\n$\\alpha$.\n$\\hat{K} = {\\bf I} \\cdot {\\bf n}$ and $H_{TP}$ are both\nT-, P-odd pseudoscalars. Therefore,\n$\\langle \\psi_{+K} | H_{TP} | \\psi_{+K} \\rangle \\propto K$ and\nso $\\langle \\psi_{-K} | H_{TP} | \\psi_{-K} \\rangle\n= -\\langle \\psi_{+K} | H_{TP} | \\psi_{+K} \\rangle$ (this fact can\nbe easily supported by model calculations). Using this fact\nand Eqs.\\ (\\ref{ecn21a}) and (\\ref{eceom1a}) we get\n$\\langle \\psi^- | H_{TP} | \\psi^+ \\rangle\n= \\langle \\psi_{+K} | H_{TP} | \\psi_{+K} \\rangle$. If\n$\\psi_{+K}$ were a good parity state this matrix element would\nbe zero. However, due to the perturbation caused by the static\noctupole deformation of the nucleus ($V_3$), it is a combination\nof the\nopposite parity spherical orbitals $\\psi_{1,+K}$\nand $\\psi_{2,+K}$ (e.g., $p_{3\/2}$ and $d_{3\/2}$):\n\\begin{equation}\n\\psi_{+K} = \\psi_{1,+K} + \\gamma \\psi_{2,+K},\n\\label{eae11a}\n\\end{equation}\n\\begin{equation}\n\\gamma = \\frac{\\langle \\psi_{2,+K} | V_3 | \\psi_{1,+K}\n\\rangle}{E_1 - E_2},\n\\label{eae11b}\n\\end{equation}\n\\begin{eqnarray}\n\\psi_{1,+K} & = & R_1(r^{\\prime}) \\Omega_{j,l,+K} (\\theta^{\\prime},\n\\phi^{\\prime}), \\nonumber \\\\\n\\psi_{2,+K} & = & R_2(r^{\\prime}) \\Omega_{j,\\widetilde{l},+K}\n(\\theta^{\\prime},\\phi^{\\prime})\n= - R_2(r^{\\prime}) (\\bbox{\\sigma}\n\\cdot {\\bf n}^{\\prime}) \\Omega_{j,l,+K}\n(\\theta^{\\prime},\\phi^{\\prime}),\n\\label{eae11c}\n\\end{eqnarray}\nwhere $\\widetilde{l} = 2j-l$.\n(Of course there will also be an admixture of\nother opposite parity states\nhaving different values of $j$. We neglect these states\nfor simplicity.)\nTherefore, we have\n\\begin{equation}\n\\alpha = \\frac{\\langle \\psi_{+K}|H_{TP}|\\psi_{+K} \\rangle}{E_+-E_-}\n= 2 \\gamma \\frac{\\langle\n\\psi_{1,+K} | H_{TP} | \\psi_{2,+K} \\rangle}{E_+\n- E_-}.\n\\label{ena42}\n\\end{equation}\n\nTo estimate $\\gamma$ we must first derive the form of $V_3$.\nAs in Sec. \\ref{shptnw}, let $U$ be\nthe strong nuclear potential that\nthe unpaired nucleon moves in. $V_3$\nis the difference between the potentials with the octupole\ndeformation present, $U_{\\rm pres}$ and\nabsent, $U_{\\rm abs}$. Once again we have\n$U({\\bf r}^{\\prime}) \\approx U({\\bf 0}) \\rho({\\bf r}^{\\prime}) \/\n\\rho ({\\bf 0})$. We also make the approximation that\n$\\rho({\\bf r}^{\\prime}) \/ \\rho ({\\bf 0}) \\approx\n\\theta (r^{\\prime} - R)$, where $R$ is the nuclear radius and\n$\\theta(x) = 1$ for $x<0$ and $\\theta(x) = 0$ for $x>0$. For an\noctupole deformed nucleus we have $R = R_0 (1 + \\beta_3 Y_{30})$\nand so\n\\begin{equation}\nV_3 = U_{\\rm pres} - U_{\\rm abs}\n\\approx U({\\bf 0}) [\\theta(r^{\\prime}- R_0 - \\beta_3 Y_{30} R_0)\n- \\theta(r^{\\prime} - R_0)]\n\\approx U({\\bf 0}) R_0 \\beta_3 \\delta(r^{\\prime}-R_0) Y_{30},\n\\label{eae21a}\n\\end{equation}\nwhere we have expanded $\\theta$ in a Taylor series, using\n$\\frac{d \\theta}{dx} = -\\delta(x)$.\nUsing Eq.\\ (\\ref{eae11b}) we then\nhave\n\\begin{equation}\n| \\gamma | \\approx \\left| U({\\bf 0})\n\\beta_3 R_1(R_0) R_2(R_0) {R_0}^3\n\\int \\Omega_2^{\\dagger} Y_{30} \\Omega_1 \\, d \\Omega^{\\prime}\n(E_1 - E_2)^{-1} \\right|\n\\sim \\beta_3,\n\\label{eae21b}\n\\end{equation}\nwhere we have used $R_1 (R_0) R_2 (R_0) \\approx 1.4 \/ {R_0}^3$\n\\cite{BohrMott}, $| U({\\bf 0}) | \\approx 50 \\mbox{ MeV}$,\n$|E_1-E_2| \\approx 5 \\mbox{ MeV}$, and\n$\\int \\Omega_2^{\\dagger} Y_{30} \\Omega_1 \\, d \\Omega^{\\prime} \\sim\n0.05$.\n\nFinally, we must estimate\nthe matrix element between the spherical orbitals,\n$\\langle \\psi_{1,+K} | H_{TP} | \\psi_{2,+K} \\rangle$.\nUsing Eq.\\ (\\ref{ehpta}) for\nthe form of $H_{TP}$ and $\\rho(r^{\\prime})\n= \\theta(r^{\\prime}-R_0) \/ ({\\textstyle \\frac{4}{3}} \\pi {r_0}^3)$\nwe get\n\\begin{equation}\nH_{TP} = -\\eta \\frac{3G}{8 \\pi \\sqrt{2} m {r_0}^3} (\\bbox{\\sigma}\n\\cdot {\\bf n}^{\\prime}) \\delta(r^{\\prime}-R_0).\n\\label{eae31c}\n\\end{equation}\nUsing Eq.\\ (\\ref{eae11c}) and $(\\bbox{\\sigma}\n\\cdot {\\bf n}^{\\prime})^2 = 1$\ngives\n\\begin{equation}\n\\langle \\psi_{1,+K} | H_{TP} | \\psi_{2,+K} \\rangle\n= \\eta \\frac{3G}{8 \\pi \\sqrt{2} m {r_0}^3}\nR_1(R_0) R_2(R_0) {R_0}^2\n\\approx \\frac{\\eta}{A^{1\/3}} 1 \\mbox{ eV}.\n\\label{eae31d}\n\\end{equation}\n\nUsing $|E_+ - E_-| \\sim 50 \\mbox{ keV}$\n(see, e.g., \\cite{AFS96,AFS2}),\n$\\beta_3 \\approx 0.1$ (see, e.g., \\cite{Cwiok91}),\nand Eqs.\\ (\\ref{ena42}), (\\ref{eae21b}), and (\\ref{eae31d})\ngives (for $A \\approx 225$)\n$|\\alpha| \\sim 2 \\beta_3 A^{-1\/3} \\eta\n\\, {\\rm eV} \/ |E_+ - E_-|\n\\sim 7 \\times 10^{-7} \\eta$.\nThis provides the following estimate for the collective EOM:\n\\begin{equation}\n| {\\cal O}_{\\rm coll} |\n\\sim 0.05 e {\\beta_3}^2 Z A^{2\/3} {r_0}^3 \\eta\n\\, {\\rm eV} \/ |E_+ - E_-|\n\\sim 4 \\times 10^{-5} \\eta e ({\\rm fm})^3.\n\\label{eeome30}\n\\end{equation}\nWe see that the collective EOM is two orders of magnitude\nlarger than the EOM due to unpaired protons.\n\nWe can do a more accurate calculation of the EOM\nusing Refs. \\cite{AFS96,AFS2},\nwhere the mixing coefficients, $\\alpha$ for various\nnuclei were calculated.\nWe use the values from \\cite{AFS2} that were calculated using\nthe Woods-Saxon potential.\nWe took the values of the $\\beta_i$ parameters from\n\\cite{Cwiok91} and the nuclei's angular momenta were taken\nfrom \\cite{Butler96,LeanderChen}.\nThe results are\nshown in table \\ref{tcoll} for various nuclei.\n\nNote the large value of $^{229}$Pa's collective octupole moment.\nThis is due to its large value of $\\alpha$, which is caused by\nthe small energy splitting between the members of\nits parity doublet \\cite{AFS96,AFS2}.\nThe possible existence of a static octupole deformation\nin $^{229}$Pa was stated in \\cite{Ahmad82}. However, more recent\npapers cast doubt on the existence of such a\ndeformation (see, e.g., \\cite{Grafen91,Levon94}).\nTherefore, the result given for $^{229}$Pa must be\nunderstood as being\nconditional on it having a static octupole deformation.\n\n\\section{The atomic electric dipole moment induced\nby a nuclear electric octupole moment}\n\\label{saedmi}\nIn this section we consider the electric dipole moment of an\natom that is induced by a nuclear electric octupole\nmoment.\nThe electric potential, $\\phi ({\\bf r})$ outside an arbitrary\ncharge distribution can be expanded in\nterms of spherical harmonics as (see, e.g., \\cite{Jackson})\n\\begin{equation}\n\\phi ({\\bf r}) = \\sum_{l=0}^{\\infty} \\sum_{m=-l}^l\n\\frac{4 \\pi}{2l+1} q_{lm} \\frac{Y_{lm} (\\theta,\\phi)}{r^{l+1}}.\n\\label{eedm1}\n\\end{equation}\nHere we neglect that part of the potential that comes\nfrom the screening of the nucleus's Coulomb field by\nthe atomic electrons as it\nis not important in the octupole potential (see the\nappendix).\nThe $q_{lm}$ are spherical electric multipole moments and they\ncan be written in terms of the charge distribution\n$\\rho_c ({\\bf r}^{\\prime})$ as follows \\cite{Jackson}:\n\\begin{equation}\nq_{lm} = \\int Y_{lm}^* (\\theta^\\prime,\\phi^\\prime) \n{r^\\prime}^l \\rho_c ({\\bf r}^{\\prime}) \\, d^3 r^\\prime.\n\\label{eedm2}\n\\end{equation}\nThe octupole term of the above electric potential is\n\\begin{equation}\n\\phi^{(3)} ({\\bf r}) = \\frac{4 \\pi}{7} q_{30} \\frac{1}{r^4}\nY_{30} (\\theta,\\phi),\n\\label{eedm3}\n\\end{equation}\nwhere we have only taken the $m=0$ term, as we will be\ncalculating the matrix element of the perturbation between\nstates with the same angular momentum projections. We can\nwrite this in terms of the octupole moment by using the\nrelation $q_{30} = \\int Y_{30} (\\theta,\\phi) r^3 \\rho_c\n({\\bf r}) \\, d^3 r\n= \\frac{5}{4} \\sqrt{\\frac{7}{\\pi}} O_{zzz}$.\nThe potential energy, $U_{\\rm oct}$ of an electron\n(of charge $-e$) in\n$\\phi^{(3)} ({\\bf r})$ will then be\n\\begin{equation}\nU_{\\rm oct} = -5 \\sqrt{\\frac{\\pi}{7}} e {\\cal O} \\frac{1}{r^4} \nY_{30} (\\theta,\\phi)\n\\label{eedm4}\n\\end{equation}\n(for the nuclear $I_z = I$ state).\n\nUsing perturbation theory, the electric dipole moment of,\nfor example,\nan atom with one electron above closed subshells induced\nby $U_{\\rm oct}$ can be written as\n\\begin{equation}\nd_z = -e \\langle \\widetilde{\\psi} | r_z | \\widetilde{\\psi} \\rangle\n= -2 e \\sum_{| k_2 \\rangle} \\frac{\\langle k_1 | r_z | k_2 \\rangle\n\\langle k_2 | U_{\\rm oct} | k_1 \\rangle}{E_{k_1} - E_{k_2}},\n\\label{eaedm3a}\n\\end{equation}\nwhere $\\widetilde{\\psi}$ denotes the perturbed atomic wave function,\n$| k_1 \\rangle = |n_1,j_1,l_1,m \\rangle$ is the unperturbed\nsingle-electron ground state,\nand $\\{| k_2 \\rangle \\}$ is the set of states with\nwhich $| k_1 \\rangle$ is mixed by the perturbation.\n\nAccording to the triangle rule for the addition of angular\nmomenta, $\\langle k_1 | r_z | k_2 \\rangle$ can only have a nonzero\nvalue if $|j_1 - j_2| \\le 1 \\le j_1 + j_2$. Similarly, for\n$\\langle k_2 | U_{\\rm oct} | k_1 \\rangle$ to be nonzero,\nwe must have\n$|j_1 - j_2| \\le 3 \\le j_1 + j_2$. This implies\nthat the following conditions need to be satisfied\nfor the dipole moment to be nonzero:\n\\begin{equation}\n|j_1 - j_2| \\le 1 \\mbox{ and } j_1 + j_2 \\ge 3.\n\\label{eaedm3b}\n\\end{equation}\nThe lowest pair of values that satisfies this condition\nis $j_1 = 3\/2$ and $j_2 = 3\/2$.\nTherefore, $s$ states cannot contribute\nto the dipole moment induced by the nuclear EOM. Also,\none of $l_1$ and $l_2$ must be even and the other odd, since the\nelectric dipole moment is a parity nonconserving effect.\n\nWe will carry out a relativistic calculation of the matrix\nelement of $U_{\\rm oct}$ between the single-electron states\n$|n_1,j_1,l_1,m \\rangle$ and $|n_2,j_2,l_2,m \\rangle$.\n(Note that although we wrote Eq.\\ (\\ref{eaedm3a}) for an atom with\none electron above closed subshells, this single-electron state\nmatrix element can be used for all atoms to compare various\nsources of atomic EDMs.)\nThe relativistic wave function of an electron\nis (see, e.g., \\cite{RelQuant})\n\\begin{equation}\n\\psi_{njlm} =\n\\left(\n\\begin{array}{c}\nf_{njl} (r) \\Omega_{jlm} (\\theta,\\phi) \\\\\ng_{njl} (r) i (-\\bbox{\\sigma} \\cdot {\\bf n})\n\\Omega_{jlm} (\\theta,\\phi)\n\\end{array}\n\\right),\n\\label{eedm5}\n\\end{equation}\nwhere $\\bbox{\\sigma}$ is twice the spin operator of this electron\nand ${\\bf n} = {\\bf r}\/r$.\nEvaluating the matrix element (using $(\\bbox{\\sigma}\n\\cdot {\\bf n})^2 = 1$) gives\n\\begin{equation}\n\\langle n_1 j_1 l_1 m | U_{\\rm oct} | n_2 j_2 l_2 m \\rangle\n= -5 \\sqrt{\\frac{\\pi}{7}} e {\\cal O}\n\\langle j_1 l_1 m | Y_{30} | j_2 l_2 m \\rangle T,\n\\label{eedm6}\n\\end{equation}\nwhere $T$ is a radial integral:\n\\begin{equation}\nT = \\int_0^{\\infty} \\frac{1}{r^4}\n(f_{n_1 j_1 l_1} f_{n_2 j_2 l_2}\n+ g_{n_1 j_1 l_1} g_{n_2 j_2 l_2}) r^2 \\, dr.\n\\label{eedm7}\n\\end{equation}\nBecause of the factor of $1\/r^4$ in the above integral,\nmost of the contribution to $T$ comes from small values\nof $r$. This allows us to use the following expressions\nfor $f_{njl}$ and $g_{njl}$, for\n$r \\ll a\/Z^{1\/3}$ \\cite{Khripl}:\n\\begin{eqnarray}\nf_{njl} (r) & = & \\frac{c_{njl}}{r}\n[(\\gamma+\\kappa) J_{2\\gamma}(x)\n- \\frac{x}{2} J_{2\\gamma-1}(x)], \\nonumber \\\\\ng_{njl} (r) & = & \\frac{c_{njl}}{r} Z\n\\alpha J_{2\\gamma}(x),\n\\label{eedm8}\n\\end{eqnarray}\nwhere\n\\begin{eqnarray}\nx & = & \\sqrt{\\frac{8 Z r}{a}}, \\nonumber \\\\\n\\gamma & = & \\sqrt{(j+{\\textstyle \\frac{1}{2}})^2\n- Z^2 \\alpha^2}, \\nonumber \\\\\n\\kappa & = & (-1)^{j+1\/2-l} (j+{\\textstyle\n\\frac{1}{2}}), \\nonumber \\\\\nc_{njl} & = & \\frac{\\kappa}{|\\kappa|}\n\\left( \\frac{1}{Z a \\nu^3} \\right)^{1\/2},\n\\label{eedm9}\n\\end{eqnarray}     \nwhere the $J$'s are Bessel functions, $a$ is the\nBohr radius, and $\\nu$ is the effective principal quantum\nnumber ($E_{nl} = -13.6 \\mbox{ eV} \/ \\nu^2$). \nTo avoid confusion, note that $l$ here is the orbital\nangular momentum of the electron, rather than the nucleus, as \nused in Sec. \\ref{sneom}. Also, the $\\kappa$ used\nhere is distinct from\nthe $\\kappa$ defined in Eq.\\ (\\ref{eao}).\nCarrying out the integration in Eq.\\ (\\ref{eedm7}), we obtain\n\\begin{eqnarray}\n\\label{eedm10}\nT & = & \\frac{192 (-1)^{j_2-j_1+1} Z^2}{a^4 \\nu_{1}^{3\/2}\n\\nu_{2}^{3\/2}} \\frac{\\Gamma (-3+\\gamma_1\n+\\gamma_2)}{\\Gamma(4+\\gamma_1+\\gamma_2)\n\\Gamma(4+\\gamma_1-\\gamma_2) \\Gamma(4-\\gamma_1+\\gamma_2)}\n\\\\ \\nonumber\n & \\times & \\{(3+\\gamma_1-\\gamma_2) (3-\\gamma_1+\\gamma_2)\n(3+\\gamma_1+\\gamma_2) (2+\\gamma_1+\\gamma_2) \\\\ \\nonumber\n&&-5 (\\gamma_1+\\kappa_1)(3-\\gamma_1+\\gamma_2)(3+\\gamma_1+\\gamma_2)\n-5 (\\gamma_2+\\kappa_2)(3+\\gamma_1-\\gamma_2)(3+\\gamma_1+\\gamma_2) \\\\\n\\nonumber\n&&+30 [(\\gamma_1+\\kappa_1)(\\gamma_2+\\kappa_2)+(Z \\alpha)^2] \\},\n\\end{eqnarray}\nwhere $\\Gamma$ is the gamma function.\n\nTo illustrate the numerical values involved, we will\nevaluate a value\nof the matrix element for an electron's interaction with the\nsingle particle octupole moment for $^{209}$Bi.\nThis atom has one $6 p_{3\/2}$ electron above closed subshells\n($(6 p_{1\/2})^2 (6 s_{1\/2})^2 \\ldots$). \nWe will consider\nmixing between $d_{5\/2}$ and $p_{3\/2}$ states, each with\nan angular momentum projection of $3\/2$. Using\ntable \\ref{tsingp} and Eqs.\\ (\\ref{eedm6}) and (\\ref{eedm10}) gives\n\\begin{equation}\n\\langle n_1 p_{3\/2},m=3\/2 | U_{\\rm oct} | n_2 d_{5\/2},m=3\/2 \\rangle\n\\approx 1.9 \\times\n10^{-13} (\\nu_1 \\nu_2)^{-3\/2} \\eta {\\rm cm}^{-1}.\n\\label{eaedmme1}\n\\end{equation}\nWe now compare this with the corresponding matrix element for the\ninteraction with a nuclear magnetic quadrupole\nmoment (MQM), which is\nanother possible source of an atomic EDM. The MQM induced by the T-\nand P-odd interaction (\\ref{ehpta}) was calculated\nin Ref. \\cite{SFK84},\nas well as the matrix element of the interaction between\nan electron\nand the MQM field.\nFor $^{209}$Bi the MQM is $\\approx 4.8 \\times 10^{-8} \\eta e\n(\\mbox{fm})^2$. \nWe get the following estimate:\n\\begin{equation}\n\\langle n_1 p_{3\/2},m=3\/2 | U_{\\rm quad} | n_2\nd_{5\/2},m=3\/2 \\rangle\n\\approx 1.3 \\times 10^{-11} (\\nu_1 \\nu_2)^{-3\/2}\n\\eta {\\rm cm}^{-1}.\n\\label{eaedmme2}\n\\end{equation}\nBy comparing the values of these matrix elements it\ncan be seen that the atomic EDM\ninduced by a nuclear EOM will be less than about 1\\% of\nthat induced by a nuclear magnetic quadrupole moment.\nThis same result\nheld for the other atoms and mixing states that we considered.\nAnother mechanism which could produce an atomic EDM is the nuclear\nSchiff moment, which for heavy atoms (which we are\nconsidering) gives a\ncontribution comparable to that of\nthe MQM \\cite{SFK84}. Because\nof this we can conclude that the atomic EDM induced by a\nsingle particle nuclear EOM\nis negligible in comparison with other possible mechanisms.\n(See the appendix for a discussion of the relative contributions\nof the octupole, magnetic quadrupole, and Schiff moments.)\n\nNow we will give example values of the matrix elements for those\nnuclei having a static octupole deformation.\nAs well as possibly having collective EOMs (as shown in Sec.\n\\ref{sceom}), these nuclei can also have collective MQMs that\nare an order of magnitude larger than the single particle MQMs\ndiscussed above \\cite{spinhh}. We will use $^{225}$Ac as an example\nand once again we will consider mixing between $d_{5\/2}$ and\n$p_{3\/2}$ states, each having an angular momentum projection of\n$3\/2$. This nucleus has an MQM of $\\sim 2 \\times 10^{-7} \\eta e\n(\\mbox{fm})^2$ \\cite{spinhh}.  \nWe obtain\n\\begin{equation}\n|\\langle n_1 d_{5\/2},m=3\/2 | U_{\\rm oct} | n_2 p_{3\/2},m=3\/2\n\\rangle|\n\\approx 1.4 \\times 10^{-11} (\\nu_1 \\nu_2)^{-3\/2} \\eta {\\rm cm}^{-1}\n\\label{ecme1}\n\\end{equation}\nand\n\\begin{equation}\n|\\langle n_1 d_{5\/2},m=3\/2 | U_{\\rm quad} | n_2\np_{3\/2},m=3\/2 \\rangle|\n\\sim 6 \\times 10^{-11} (\\nu_1 \\nu_2)^{-3\/2} \\eta {\\rm cm}^{-1}.\n\\label{ecme2}\n\\end{equation}\nThese results show that the contribution of the collective\nnuclear EOM to\nthe atomic EDM is smaller than that of the collective MQM.\nThis also applies to the other isotopes shown in table \\ref{tcoll},\nexcept for $^{229}$Pa. ($^{223}$Rn does have a fairly\nlarge collective\noctupole moment, but since it has a closed electron shell\nthe EOM does not contribute to the\natomic EDM.) For $^{229}$Pa the contribution of the\nEOM may be comparable to that of the MQM, but as stated above, it\nis not certain that this nucleus has a static octupole deformation.\n \n\\section{Possibly enhanced magnetic quadrupole moments in nuclei\nwith an octupole deformation}\n\nFinally, we discuss a mechanism by which single particle\nmagnetic quadrupole moments could be enhanced in even-odd nuclei\nwith an octupole deformation. We will first consider the MQM of\nsuch a nucleus in its intrinsic (body-fixed) frame and then\ntransform the MQM into its laboratory frame.\n\nAs in Sec. \\ref{sceom} the wave function of the external nucleon in\nthe intrinsic frame is $\\psi_{+K} ({\\bf r}^{\\prime})$, defined\nby Eqs. (\\ref{eae11a}), (\\ref{eae11b}), and (\\ref{eae11c}). The\nMQM in the intrinsic frame is then\n\\begin{eqnarray}\nM_{\\rm intr} & = & \\langle \\psi_{+K} | \\hat{M}_{z^{\\prime}\nz^{\\prime}} | \\psi_{+K} \\rangle\n\\label{emqn1a} \\\\\n& = & 2 \\gamma \\langle \\psi_{1,+K} | \\hat{M}_{z^{\\prime}\nz^{\\prime}} | \\psi_{2,+K} \\rangle,\n\\label{emqn1b}\n\\end{eqnarray}\nwhere $\\hat{M}$ is the operator for the MQM (see Ref. \\cite{SFK84}).\nNote that the MQM is defined for the maximum value of the\nprojection of the angular momentum onto the $z^{\\prime}$-axis,\nso $K=j=I$ (for the ground state of the rotational band $I=j$).\nUsing a result from Ref. \\cite{SFK84} we have\n\\begin{equation}\nM_{\\rm intr} = \\gamma \\frac{2I-1}{I+1} \\frac{e}{m}\n(\\mu - q) \\int R_1 R_2 r \\, r^2 dr,\n\\label{emqc}\n\\end{equation}\nwhere $\\mu$ is the magnetic moment of the external nucleon in\nnuclear magnetons and $q = 0$ ($1$) for a neutron (proton).\nIt should be noted that this (intrinsic frame)\nsingle particle MQM differs from\nthe spherical nucleus \nsingle particle EOMs and MQMs (considered in Sec. \\ref{sneom}\nand Ref. \\cite{SFK84}, respectively) in that the former is\ngenerated due to the interaction $V_3$, coming from the nucleus's\noctupole deformed shape [see Eqs. (\\ref{eae11b})\nand (\\ref{eae21a})], while the latter is due to the interaction\n$H_{TP}$ (\\ref{ehpta}). The $H_{TP}$ interaction will\ncome into the current\nsituation when we transform into the lab. frame.\n\nSince we are only working out a rough estimate of the MQM we take\nthe integral in the above equation to be\n\\begin{equation}\n\\int R_1 R_2 r \\, r^2 dr \\sim \\frac{1}{2} r_0 A^{1\/3},\n\\label{eqmd}\n\\end{equation}\nwhere $r_0 = 1.1 \\mbox{ fm}$. For both types of nucleons we\nhave $|\\mu - q| \\approx 1.8$ and so we can write\n\\begin{equation}\n| M_{\\rm intr}| \\sim 0.2 A^{1\/3} \\frac{2I-1}{I+1} |\\gamma| e\n(\\mbox{fm})^2.\n\\label{emqe}\n\\end{equation}\n\nNow we turn to the MQM in the laboratory frame. The wave function\nof the nucleus in the lab. frame is (as in Sec. \\ref{sceom})\n$\\psi = \\psi^+ + \\alpha \\psi^-$, where $\\psi^{\\pm}$ is defined\nby Eqs. (\\ref{eceom1a}) and (\\ref{ecn21a}) and $\\alpha$ is\ngiven by Eq. (\\ref{enad32}) --- this is where the interaction\n$H_{TP}$ comes into the present situation.\nThe MQM in the lab. frame is then\n\\begin{eqnarray}\nM_{\\rm lab} & = & \\langle \\psi | \\hat{M}_{zz} | \\psi\n\\rangle \\nonumber \\\\\n& = & 2 \\alpha \\langle \\psi^+ | \\hat{M}_{zz} | \\psi^-\n\\rangle \\nonumber \\\\\n& = & \\alpha ( \\langle I M K | \\hat{M}_{zz} | I M K \\rangle\n- \\langle I M -K | \\hat{M}_{zz} | I M -K \\rangle) \\nonumber \\\\\n& = & 2 \\alpha \\langle I M K | \\hat{M}_{zz} | I M K \\rangle.\n\\label{emqea2}\n\\end{eqnarray}\n(Once again, the MQM is defined for $M$, the projection of the\nangular momentum onto the $z$-axis, equal to $I$.)\nNote that $\\langle I M -K | \\hat{M}_{zz} | I M -K \\rangle\n= - \\langle I M K | \\hat{M}_{zz} | I M K \\rangle$, as\n$\\psi_{\\pm K} = \\psi_{1, \\pm K} \\pm \\gamma\n\\psi_{2, \\pm K}$ and so\n$\\langle \\psi_{-K} | \\hat{M}_{z^{\\prime} z^{\\prime}}\n| \\psi_{-K} \\rangle = -\\langle \\psi_{+K} |\n\\hat{M}_{z^{\\prime} z^{\\prime}}| \\psi_{+K} \\rangle$. \n\nNow consider $\\langle I M K | \\hat{M}_{zz} | I M K \\rangle$.\nWe have [from Eq. (\\ref{ecn21a})]\n\\begin{equation}\n\\langle I M K  | \\hat{M}_{zz} | I M K \\rangle\n= \\frac{2I+1}{4 \\pi} \\int {D^{I *}_{M K}} (\\phi,\n\\theta,0) \\psi_K^{*} ({\\bf r}^{\\prime}) \\hat{M}_{zz}\n(\\theta,\\phi,0) \\psi_K ({\\bf r}^{\\prime})\nD^I_{M K} (\\phi,\\theta,0) \\, d^3 r^{\\prime} \\, d \\Omega.\n\\label{emqf}\n\\end{equation}\nTransforming $\\hat{M}_{zz}$ from the lab. frame to the intrinsic\n($x^{\\prime},y^{\\prime},z^{\\prime}$) frame gives\n$\\hat{M}_{zz} (\\theta,\\phi,0) = {D^{2 *}_{0 0}} (\\phi,\n\\theta,0) \\hat{M}_{z^{\\prime} z^{\\prime}} (\\theta^{\\prime},\n\\phi^{\\prime},0)$ (see, e.g., \\cite{Varshalovich}). Substituting\nthis into Eq. (\\ref{emqf}) and using Eq. (\\ref{emqn1a}) gives\n(using a formula for the integral of $D$-functions from\n\\cite{Varshalovich})\n\\begin{eqnarray}\n\\langle I M K | \\hat{M}_{zz} | I M K \\rangle & = &\n\\frac{2I+1}{4 \\pi} M_{\\rm intr} \\int {D^{I *}_{M K}} (\\phi,\\theta,\n0) {D^{2 *}_{0 0}} (\\phi,\\theta,0) D^I_{M K} (\\phi,\\theta,0)\n\\, d \\Omega \\nonumber \\\\\n& = & M_{\\rm intr} \\langle I, M ; 2, 0|I, M \\rangle\n\\langle I, K; 2, 0| I, K \\rangle \\nonumber \\\\\n& = & M_{\\rm intr} {\\langle I, I ; 2, 0| I, I \\rangle}^2\n\\nonumber \\\\\n& = & \\frac{I (2I-1)}{(I+1) (2I+3)} M_{\\rm intr}.\n\\label{emqg}\n\\end{eqnarray}\n(Recall that $M=K=I$.) Using Eq. (\\ref{emqea2}) then gives\n\\begin{equation}\nM_{\\rm lab} = 2 \\alpha \\frac{I (2I-1)}{(I+1) (2I+3)}\nM_{\\rm intr}\n\\label{emq5a}\n\\end{equation}\nand so [using Eq. (\\ref{emqe})]\n\\begin{equation}\n|M_{\\rm lab}| \\sim 0.4 A^{1\/3} |\\alpha| |\\gamma|\n\\frac{I (2I-1)^2}{(I+1)^2 (2I+3)} e ({\\rm fm})^2.\n\\label{emq5b}\n\\end{equation}\nNow we have $|\\alpha| \\sim 7 \\times 10^{-7} \\eta$ and\n$|\\gamma| \\sim 0.1$ (see Sec. \\ref{sceom}). For\n$A \\approx 225$ we have\n\\begin{equation}\n|M_{\\rm lab}| \\sim 6 \\times 10^{-8} \\eta e ({\\rm fm})^2.\n\\label{emq5c}\n\\end{equation}\nThis is smaller than the collective MQM due to the spin\nhedgehog mechanism ---\n$\\sim 2 \\times 10^{-7} \\eta e ({\\rm fm})^2$ \\cite{spinhh} and\nso it is not the dominant mechanism. It is, in\nfact, of the same order of magnitude as the single particle\nMQM [e.g., for $^{209}$Bi, approximately\n$4.8 \\times 10^{-8} \\eta e ({\\rm fm})^2$ (see\nSec. \\ref{saedmi})].\nThis means that there is no enhancement.\n\nEnhancement was a possibility here due to the relatively large\nvalue of $\\alpha$ that comes from the smallness of the energy\nsplitting between members of the parity doublet ($E_+ - E_-$).\nHowever, the inclusion of the factors of $|\\gamma| \\sim 0.1$\nand $I (2I-1) \/ [(I+1) (2I+3)] \\sim 0.3$ (this enters on\ntransforming to the lab. frame) ensures that the possible\nenhancement is not realised.\n\nHowever, $^{229}$Pa may be an exception to this, as it has a\nlarge value of $\\alpha$, but it must be remembered that\nit is not certain that this nucleus has a static octupole\ndeformation (see Sec. \\ref{sceom}). If is does have such a\ndeformation then its MQM due to the present mechanism would\nbe $|M_{\\rm lab}| \\sim 3 \\times 10^{-7}\n\\eta e ({\\rm fm})^2$ [using Eq. (\\ref{emq5b}),\nthe value of $\\alpha$ given in Ref. \\cite{AFS2}, and\n$|\\gamma| \\sim 0.1$], which is of the same order of\nmagnitude as the collective MQM due to the spin hedgehog\nmechanism.\n\nNote that the enhancement of MQMs in\ndeformed nuclei with opposite\nparity levels close to each other has also been considered in\nRefs. \\cite{Hax83,SFK84}.\n\n\\begin{acknowledgements}\nOne of us (DWM) is grateful to G.F. Gribakin for\nhelpful discussions.\nThis work was supported by the Australian Research Council\nand by the National Science Foundation through a grant for the\nInstitute for Theoretical Atomic and Molecular Physics at\nHarvard University and the Smithsonian Astrophysical Observatory.\n\\end{acknowledgements}\n\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}}
+{"text":"\\section{Introduction}\nOver the last 10 or 12 years a great deal of outstanding observational\nwork has indicated that the best fit model of our universe is a nearly\nflat Friedmann-Lema\\^{i}tre-Robertson-Walker (FLRW) model with $\\Omega_M \n\\approx 0.27$ and $\\Omega_{\\Lambda} \\approx 0.73 $ (Riess {\\it et al.} 1998;\nPerlmutter {\\it et al.} 1999; Bennett {\\it et al} 2003 (WMAP results);\nPeacock {\\it et al.} 2001; Percival {\\it et al.} 2001; Efstathiou {\\it et al.}\n2002; Spergel {\\it et al.} 2003, and references therein),\nwhere $\\Omega_M$ and $\\Omega_{\\Lambda}$ are the usual density parameters for \nmatter, including nonbaryonic dark matter, and dark energy, modelled here\nas vacuum energy (the cosmological constant $\\Lambda$), respectively. Here\nand throughout this paper $\\Omega_M$ and $\\Omega_{\\Lambda}$ refer to these\nquantities as evaluated at our time now. This\nremarkable concordance is based on WMAP cosmic microwave background\n(CMB) anisotropy measurements, a large number of Supernovae Ia\ndata (see Riess {\\it et al.} 2004), and large scale structure studies, and has\nbeen confirmed by other more recent work. Riess and his collaborators (Riess\n{\\it et al.} 2004), for instance, have recently found a best-fit cosmology \nhaving $\\Omega_M = 0.29$ and $\\Omega_{\\Lambda} = 0.71$ for their sample of\n16 distant ($z > 1$) SN Ia, including 6 with $z > 1.25$, assuming the\nuniverse is exactly flat. Within\nthe errors this is consonant with the ``concordance'' model given above. \\\\\n\nDespite the strength of these results, they will obviously have to undergo\ngradual\nrevision and continual verification, as more precise data from higher redshifts\nare acquired. When $\\Lambda \\neq 0$, there are at present, from a strictly\nmathematical consideration of the Einstein field equations, not yet enough\ncompletely independent observables to constrain all the free parameters\nof the cosmological model (Hellaby, 2006; Stoeger \\& Hellaby, in\npreparation). \\\\  \n\nAssuming that the universe is spherically symmetric on the largest scales\n(FLRW or, more generally, Lema\\^{i}tre-Tolman-Bondi (LTB)), one generally needs\nredshifts, luminosity distances (or angular-diameter distances), and galaxy\nnumber counts, together with a reliable galaxy evolution model, or an \nequivalent set of measurements, to constrain the model fully (see Ellis, {\\it\net al.} 1985). If $\\Lambda \\neq 0$, however, or if there is\nsome other form of dark energy, these data are not enough. We need at least\none other independent parameter -- that is, independent of the observables we\nhave just mentioned and therefore of those which depend upon them. And,\nstrictly speaking, this is what we have not had. Thus, the impressive fittings\nthat have led to the concordance model are still model-dependent in some sense. \\\\\n\nThere is another pair of such independent observables. These would\nimprove and verify our cosmological fitting, when we are able to obtain an\nadequate number of precise luminosity distances -- or angular-diameter \ndistances -- and redshifts for SN Ia, or for other standard candles or standard\nrods , out to $z \\approx 1.8$. These observables are the maximum of the angular-diameter distance (or observer-area distance) $C_{max}$ and the\nredshift $z_{max}$ at which it occurs. It has been realized for many years\n(McCrea 1935, Hoyle 1961, Ellis \\& Tivon 1985) that this distance reaches\na maximum for relatively low redshifts in FLRW universes. For an\nEinstein-deSitter ($ \\Omega =1 $)universe filled with matter, for instance,\nthe observer area distance C has a maximum $C_{max}$ at $z_{max} = 1.25$. This\neffect is due to the global gravitational focusing of light rays\ncaused by the matter in the universe -- in effect the entire universe, filled\nwith homogeneously distributed matter, acts like a gravitational lens. \\\\\n\nKrauss and Schramm (1993) recognized that, for flat FLRW universes, \ndetermination of $z_{max}$ would give us $\\Omega_{\\Lambda}$. They plotted\nand provided a table giving this unique correspondence (see their Table 1),\nand proposed the possibility of using the measurement of compact parsec-scale\nradio jets to observationally exploit it, if the source-evolution problem\ncan be tamed. Since then, there has been little\ndevelopment or discussion of this potentially important connection -- except\nfor Hellaby's (2006) recent closely connected exploration of such\nmeasurements within the more general context of LTB universes (see below). \nCertainly, it is implicit in the Friedmann equation -- most clearly in \nRefsdal, {\\it et al.}'s (1967) numerical results of general cosmological\nmodels, in the brief treatment of cosmic distances by Carroll, {\\it et al.},\n1992 (see pages 510-512, and their Figure 5), and in Peeble's treatment of\nangular diameters in cosmology (Peebles 1993), but not pointed out or\ndiscussed further, until Hellaby's more general treatment. This may be\npartially due to the\ndifficulty of obtaining reliable data at the redshifts where we would expect to\nlocate $C_{max}$ (see below). Now, however, there is the very real prospect of\nobtaining angular diameter distances (indirectly, by measuring luminosity\ndistances of SN Ia) out to $z \\approx 1.8$ using telescopes in space. Thus, it\nis important to point out again and stress this promising connection, which could\neventually be incorporated in the Bayesian-Fisher matrix (see, for example,\nAlbrecht, {\\it et al.}, 2006) fitting of models to data, or be used as an\nindependent consistency check on such fittings. \\\\\n\nRecently, as already mentioned, Hellaby (2006) emphasized the importance\nof such a measurement within\na more general framework. He points out that in any LTB cosmology with\n$\\Lambda = 0$ (which includes all $\\Lambda = 0$ FLRW cosmologies as special\ncases) the measurement of $C_{max}$ is equivalent to a measurement of the\ntotal mass $M_{max}$ within the sphere defined by $C_{max}$. For\n$\\Lambda \\neq 0$ we have for any LTB model, instead, a simple relationship\nbetween the $\\Lambda$, $C_{max}$ and $M_{max}$ (see equation (11) below). So\na measurement of $M_{max}$, or its equivalent, and $C_{max}$ determines\n$\\Lambda$. What becomes apparent is that $C_{max}$ and the redshift $z_{max}$\nat which it occurs constitute independent cosmological observables -- directly\nconstraining $\\Lambda$ and $\\Omega_M$ (see Hellaby's Figure A1 in his Appendix,\nwhich shows how different cosmologcal parameters vary with $z_{max}$.)  \\\\\n\nApplying this directly to flat FLRW models, like those we have good evidence\nrepresent our universe, we quickly see that, since we implicitly have a \nrelation between the total mass-energy density and the matter density, or\nequivalently between the matter density and $\\Omega_{\\Lambda}$ --- i.e.\n$\\Omega_M = 1 - \\Omega_{\\Lambda}$ --- observational determination of \n$z_{max}$ will directly determine $\\Omega_{\\Lambda}$ in a very simple and\nstraightforward way, supporting Krauss and Schramm's results (1993). In\nthis paper we shall integrate and generalize these results, first of all\nverifying Krauss and Schramm's results for flat FLRW universes and writing\ndown that relationship as an algebraic equation in closed form (they\npresented their results numerically), and then generalizing those results\nto non-flat FLRW universes, using the relationship Hellaby (2006) noticed.\nIn this case, $C_{max}$ and $z_{max}$ directly determine both\n$\\Omega_{\\Lambda}$ and $\\Omega_M$, if we know $H_0$ independently.\nIn the course of doing this, we shall, as useful and important by-products,\nobtain the FLRW $C = C(z)$ and $\\hat{M}_0 = \\hat{M}_0(z)$ closed-form functional\nrelationships for $\\Lambda \\neq 0$ universes, parallel to those which\nare well-known for $\\Lambda = 0$ FLRW models (Ellis and Stoeger 1987;\nStoeger, {\\it et al.} 1992), as well as a very simple observational\ncriterion for flatness in terms of $C_{max}$. Here $C(z)$, of course, is\nsimply the angular-diameter distance as a function of the redshift $z$, and \n$\\hat{M}_0(z)$ is the mass density per source counted as a function of\n$z$, which is closely related to the differential galaxy number counts\n$dN\/dz$ (see Stoeger, {et al.} 1992). To our knowledge, these more\ngeneral results, along with the closed-form expressions and the flatness\ncriterion are new. \\\\\n\nWe have already indicated that these measurements will be able to be\nimplemented once we have luminosity distances and redshifts for SN Ia, or\nfor other standard candles or standard rods, in the interval $1.5 < z < 1.8$.\nAs we shall show, it is precisely in this region that a flat FLRW universe will\nhave a maximum in its angular-diameter distance, if $0.59 \\leq \\Omega_{\\Lambda}\n< 0.82$. For the best fit FLRW given by Riess {\\it et al.} (2004) with\n$\\Omega_M = 0.29$ and $\\Omega_{\\Lambda} = 0.71$, $z_{max} = 1.62$. Another \npotential way of obtaining such precise measurements is -- following Krauss\nand Schramm's (1993) idea -- the use of VLBI to determine the\nangular-size\/redshift relation for ultra-compact (milliarcsecond) radio\nsources. These have been argued to be standard rods (Jackson and Dodgson\n1997; Jackson 2004).  If we actually do \nfind the maximum angular-diameter distance near this value of the redshift,\nthis would be independent confirmation of the concordance model. If we\ndo not, but find the maximum angular-diameter distance $C$ at some other value\nof $z$, this will require further work at reconciling the models, and\npossibly modifying them. In that case, either the universe may still be flat,\nbut the relative amounts of matter and dark energy would be quite different\nfrom that given by the concordance, or there is a significant deviation from\nflatness that must be taken into account, or possibly there are significant\ndeviations from FLRW on the largest scales which must be included -- or all\nthree! At the very least, this would be a good consistency check on  \nour cosmological fitting so far. Alternatively, as we have already mentioned,\nwe could simply include both $C_{max}$ and $z_{max}$ data in our over-all \nfitting scheme -- which would further improve the relibility of our results. \\\\\n\nIt is important to point out that this redshift range is already attracting \nspecial attention. That is because there have been preliminary indications\n(Gilliland {\\it et al.} 1999) from an SN Ia at $z \\approx 1.7$ that the\nuniverse was decelerating at that time! Further studies (Riess {\\it et al.}\n2001; Mortsell {\\it et al.} 2001; Ben\\'{i}tez {\\it et al.} 2002) have confirmed\nthe plausibility of that conclusion, but were unable to strengthen it without\nfurther SN Ia measurements in that interval. Thus, we now have two strong\nmotivations for pursuing precise SN Ia searches and measurements in this\nredshift range. \\\\\n \nFinally, one might wonder how measurements of the luminosity distances\nof SN Ia can reveal maxima in the angular-diameter (or observer-area)\ndistances. The luminosity distances themselves will not have such maxima. The\nanswer to this question is simple, though rarely adverted to. According the\nreciprocity theorem of Etherington (1933; see also Ellis 1971), the luminosity\ndistance $d_L$ is very generally related to the angular-diameter, or\nobserver-area, distance by\n\n\\begin{equation}\nd_L = (1+z)^2 C.  \\label{recth}\n\\end{equation}\n\n\\noindent\nThis simple but important relationship holds for all cosmologies, even very\ninhomogenous ones. Thus, with observed luminosity distances and redshifts in the\nabove mentioned crucial redshift range, we can very quickly convert to\nangular-diameter distances, and determine whether the maximum for those \ndistances lies within that range. \\\\\n\nNow we shall go on to work out the simple relationship between $z_{max}$\nand $\\Omega_{\\Lambda}$ for flat FLRW. \n\n\\section{The Maximum Angular-Diameter Distance in Flat FLRW with $\\Lambda \\neq\n0$}\n\nThe basic equations relating $z_{max}$ and $\\Omega_{\\Lambda}$ in  flat FLRW\nwith $\\Lambda \\neq 0$ are not difficult, but require some effort to obtain\nand check, because they involve elliptic integrals. As we have already \nmentioned, this represents the simplest and clearest example of a more general\nrelationship between the redshift of the maximum of the angular-diameter\ndistance (in LTB models this is often referred to as the ``areal radius'') and\nthe matter and vacuum-energy content of the universe for all FLRW and LTB\nmodels (Hellaby 2006). Furthermore, neither Krauss and Schramm (1993) nor\nHellaby (2006) illustrate the actual calculation. Their results were obtained\nnumerically, and presented in plotted or table form. \\\\\n\nIn flat FLRW, the angular-diameter (or observer-area) distance $C(\\eta, y)$ is\ngiven by \n\n\\begin{equation}\nC(\\eta, y) = R(\\eta)y = \\frac{R_0 y}{1 + z}\\>, \\label{oadef}\n\\end{equation}\n\n\\noindent\nwhere $R(\\eta)$ is the scale factor, $\\eta$ is the conformal time, $R_0$ is\nthe scale factor now, $y$ is the comoving radial coordinate, and $z$ is the\nredshift of signals from distant sources. Here we have used the important\nFLRW relationship\n\n\\begin{equation}\n1 + z = \\frac{R_0}{R(\\eta)}\\>. \\label{red}\n\\end{equation}\n\nClearly, if we differentiate equation (\\ref{oadef}) with respect to $y$ and set\nthe result equal to zero, we\nshall have the equation for the maximum of $C(\\eta, y)$, subject to the \nusual condition that $d^2C\/dy^2 < 0$ for $dC\/dy = 0$. We have then from\nequation  (\\ref{oadef})\n\n\\begin{equation}\ndC\/dy = \\frac{R_0}{1+z} - \\frac{R_0 y}{(1+z)^2}dz\/dy = 0\\,, \\label{dcdy}\n\\end{equation}\nwhich becomes\n\n\\begin{equation}\n\\frac{R_0}{1+z} - \\frac{R_0 y}{(1 + z)^2} R_0 H_0 \\sqrt{\\Omega_{\\Lambda} \n+ (1 - \\Omega_{\\Lambda})(1+z)^3} = 0\\,, \\label{dcdy2}\n\\end{equation}\n\\noindent\nsince the Friedmann equation in this case yields\n\n\\begin{equation}\ndz\/dy = R_0 H_0 \\sqrt{\\Omega_{\\Lambda} + (1 - \\Omega_{\\Lambda})(1+z)^3}\\>.\n                                 \\label{dzdy}\n\\end{equation}\n\nThus, from solving equation (\\ref{dcdy2}) for $y$, we obtain the equation for\n$y_{max}$, the comoving radial coordinate distance to the point down\nthe observer's past light cone at which the angular-diameter distance is a\nmaximum, as a function of $z_{max}$, the redshift there, and of\n$\\Omega_{\\Lambda}$:\n\n\\begin{equation}\ny_{max} = \\frac{1 + z_{max}}{R_0H_0 \\sqrt{\\Omega_{\\Lambda} + (1+z_{max})^3\n                                    (1-\\Omega_{\\Lambda})}} \\> . \\label{ymax}\n\\end{equation}\n\nThis is the first and most essential step in finding the equation we are\nlooking for. \\\\\n\nThe second step involves finding the explicit solution to the Friedmann\nequation, essentially equation (\\ref{dzdy}), to give us another expression for\n$y_{max}$ at $z_{max}.$ Substituting this expression into left-hand-side of\nequation (\\ref{ymax}) gives a unique implicit equation for $\\Omega_{\\Lambda}$ as a\nfunction simply of $z_{max}$. This is the relationship we have been looking\nfor. \\\\\n\nSo, what is the solution of equation (\\ref{dzdy})? Normally, we might want to simply\ndo a numerical integration. However, this would not be very useful in\nour case. It turns out, as is well known (Byrd \\& Friedman (1954), pp.\n8-10 and formula 260.00 (p. 135); see also Jeffrey (1995), pp. 225-226), that,\nsince this equation involves the square \nroot of a cubic polynomial, it has an analytic solution in terms of elliptic\nintegrals. In our case the most useful form of the solution is:\n\n\\begin{equation}\ny = \\frac{g}{R_0 H_0 \\Omega_{\\Lambda}^{1\/2}}\\biggl[F(\\phi, k) \\mid_{(1+z)^{-1}=1} -\n                      F(\\phi, k)\\mid_{(1+z)^{-1}}\\biggr] \\>, \\label{yeliptic}\n\\end{equation}\n\n\\noindent\nwhere the $F(\\phi, k)$ are standard elliptic integrals of the first kind,\nfor the angle $\\phi$, which is a function of $1+z$, and $k$ is the modulus.\nMore explicitly \n\n\\begin{eqnarray}\n \\phi &=&cos^{-1}\\Biggl[\\frac{-m(1+z)+ (\\sqrt 3 -1)}{-m(1+z)-(\\sqrt 3 + 1)}\\Biggr]\\>, \\nonumber  \\\\\n m &= &\\Biggl[\\frac{1 - \\Omega_{\\Lambda}}{\\Omega_{\\Lambda}}\\Biggr]^{1\/3}, \\nonumber \\\\\nk^2 &=&\\frac{1}{2} + \\frac{\\sqrt 3}{4}\\>, \\nonumber  \\\\\ng &=&\\frac{1}{3^{1\/4}}\\Biggl[\\frac{\\Omega_{\\Lambda}}{1 - \\Omega_{\\Lambda}}\\Biggr]^{1\/3}. \\nonumber \n\\end{eqnarray}\n\n\\noindent\nThis solution was obtained and checked using elliptic integral tables\nin Byrd \\& Friedman (1954) (entry 260.00, p. 135) in conjunction with\nMAPLE.\n\nWith equation  (\\ref{yeliptic}) being substituted for $y$, equation (\\ref{oadef}) is the characteristic\nFLRW relationship for the angular-diameter distance $C = C(z)$ in terms of\n$z$. It turns out (see below) that this same form of the relationship\nholds in the general (non-flat) FLRW cases -- with the parameters $\\phi$,\n$k$, and $g$ being more complicated functions, involving $\\Omega_{\\Lambda}$,\neither $\\Omega_M$ or $C_{max}$, and $H_0$. We shall explicitly write these\ndown in the next section. Similarly, we quickly can write down the\ncomplementary characteristic $\\Lambda \\neq 0$ mass density per source counted\nas a function of $z$ (see Ellis and Stoeger 1987 and Stoeger, {\\it et al.}\n1992):\n\n\\begin{equation}\n \\hat{M}_0(z) = \\frac{\\mu_{m_{0}}(1+z)^2}{R_0H_0\\sqrt{\\Omega_{\\Lambda} +\\Omega_{M} (1+z)^3 - (\\Omega_0 - 1)(1+z)^2}}, \\label{Mz}\n\\end{equation}\nwhere $\\mu_{m_{0}}$ is the mass-energy density now and\n$\\Omega_0\\equiv\\Omega_{\\Lambda} +\\Omega_M$, and the last term under the\nradical sign in the denominator is zero when the universe is flat (see\nbelow). These characteristic FLRW relationships for $C(z)$ and for\n$\\hat{M}_0(z)$ are very useful to know (Ellis and Stoeger 1987;\nStoeger, {\\it et al.}(1992). If the universe is FLRW and $\\Lambda = 0$,\nthen these relationships inevitably follow. If, on the other hand, the data\ncan be put into these functional forms, then it can be shown by solving\nthe field equations with this data (Stoeger, {\\it et al.} 1992;\nAra\\'{u}jo, Stoeger, {\\it et al.}, in preparation) that the universe\nmust be FLRW. Thus, being able to fit the data to these forms, assures us\nthat the universe is FLRW. Not being able to do so, assures us that it is\nnot FLRW.\n \nReturning to the main object of our derivation, substituting equation (\\ref{yeliptic}) into\nthe left-hand-side of equation  (\\ref{ymax}),\nwe have simply:\n\n\\begin{eqnarray}\n\\frac{g}{\\Omega_{\\Lambda}^{1\/2}}\\biggl[F(\\phi, k)\\mid_{(1+z)^{-1} = 1} &-& F(\\phi, k)\n  \\mid_{(1+z_{max})^{-1}}\\biggr]    \\nonumber \\\\\n& & = \\frac{1+z_{max}}{\\sqrt{\\Omega_{\\Lambda} + (1+z_{max})^3(1 - \\Omega_{\\Lambda})}} \\>.  \\label{elzmax}\n\\end{eqnarray}\n\n\nThis is a transcendental relationship for $\\Omega_{\\Lambda}$ as a\nfunction of $z_{max}$. It is worth noticing that it does not involve\nany other parameters! This is the relationship which represents the\nnumerical results obtained by Krauss and Schramm (1993). \\\\\n\nThe solutions to this implicit algebraic equation were obtained using\nMAPLE, and were checked by hand for values of $\\Omega_{\\Lambda}$ near\nthe concordance model value of $\\Omega_{\\Lambda} = 0.73$. They are given in \nTable 1 and Figure 1 below.\\footnote{There are alternative sequences of\nsteps for obtaining these results -- for instance using the solution of (6) to\nwrite down a general formula for $C$ as a function of $z+1$ and then\ndifferentiating this, setting the result to zero, and solving for\n$\\Omega_{\\Lambda}$ in terms of $z_{max}$. But they all involve explicitly or\nimplicitly the steps we have indicated -- solving the Friedmann equation to\nobtain the relationship between $y$ and the observable $z$ (redshift), and\ndetermining the equation for $C_{max}$ in terms of $y_{max}$ or, from the first\nstep, its observational equivalent $z_{max}$. Because of the complication of\nincluding a non-zero $\\Omega_{\\Lambda}$, at some point a numerical solution\nwill always be needed. See, for instance Carroll, {\\it et al.} (1992), pp.\n510-512. We have chosen to keep the solution of Friedman\nequation analytic, in terms of elliptic integrals, in order to derive the\ncharacteristic FLRW closed-form expression for $C(z)$ and to solve the resulting\nalgebraic equation numerically.} We can immediately see, that for the\nconcordance model we should find $z_{max} = 1.64$. For the nearby best fit\nmodel of Riess,\n{et al.} (2004) we have already mentioned, $z_{max} = 1.62$. Values of\n$z_{max}$ for many other values of $\\Omega_{\\Lambda}$ are given, as well. These\nverify the values presented in Krauss and Schramm (1992), and those\nevident in the plots of Refsdal, {\\it et al.} (1967), Carroll, {\\it et al.}\n(1992), and Hellaby (2006). \\\\\n\n\\section{Non-Flat FLRW Universes}\n\nIf the universe is not flat, a slight\ngeneralization of these same equations obtains, with the solution for $y$\ntaking the same general form as given in equation  (\\ref{yeliptic}). The generalization\nof equation (\\ref{elzmax}) in this case will, however, include -- as is intuitively\nclear -- a dependence on $\\Omega_M$ as well as on $\\Omega_{\\Lambda}$. Using the\ngeneral relationship emphasized by Hellaby (2006)\n\n\\begin{equation}\n \\Lambda C_{max}^3 - 3C_{max} + 6 M_{max} = 0,  \\label{lcmmax}\n\\end{equation}\n\n\\noindent\nwe can determine $\\Omega_M$ through $M_{max}$ in terms of $C_{max}$ and \n$\\Lambda$. It is important to stress that equation (\\ref{lcmmax}) holds for these \nquantities as measured at $z_{max}$, or $y_{max}$, down the observer's past\nlight cone. From Hellaby's (2006) results, we easily find that, for FLRW,\n\n\\begin{equation}\nM_{max} = \\frac{4}{3} \\pi \\rho_M C_{max}^3, \\label{masseq}\n\\end{equation}\nwhere $\\rho_M = \\rho(t_{max}) = \\rho_0(1+z_{max})^3.$ Here $\\rho_0$ is\nthe density at our time now, $t_0$.\nUsing this together with the definition of $\\Omega_M \\equiv 8\\pi\\rho_0\/3{H_0}^2$\nand equation (\\ref{lcmmax}), we easily obtain\\footnote{As in Hellaby (2006), we also use units such that $G=c=1$.}\n\n\\begin{equation}\n\\Omega_M = \\frac{1}{H_0^2(1+z_{max})^3}[C_{max}^{-2} - \\Omega_{\\Lambda}H_0^2]. \\label{omegam}\n\\end{equation}\n\nThis can be substituted into the non-flat versions of equations (\\ref{dzdy}) and (\\ref{ymax}),\n\n\\begin{equation}\ndz\/dy = R_0H_0 \\sqrt{\\Omega_{\\Lambda} + \\Omega_M(1+z)^3 - (\\Omega_0 -1)\n   (1+z)^2}, \\label{dzdy2}\n\\end{equation}\nand\n\\begin{equation}\ny_{max} = \\frac{1+z_{max}}{R_0H_0\\sqrt{\\Omega_{\\Lambda}+\\Omega_M(1+z_{max})^3\n  - (\\Omega_0 - 1)(1+z_{max})^2}}, \\label{ymax2}\n\\end{equation}\n\nIn passing, we immediately see from equation (\\ref{omegam}) that we have a \nuseful observational criterion for flatness of an FLRW universe:\n\n\\begin{equation}\n\\Omega_0 = 1 \\Rightarrow (1+z_{max})^{-3}\\Biggl[\\frac{1}{H_0^2C_{max}^2} - \\Omega_{\\Lambda}\\Biggr] + \n  \\Omega_{\\Lambda} - 1 = 0, \\label{flatness}\n\\end{equation}\nThus, if already know that the universe is flat, or nearly so, and we know both $z_{max}$ and $C_{max}$, we\ncan directly determine $\\Omega_{\\Lambda}$, and therefore $\\Omega_M$ itself\nfrom equation (\\ref{flatness}).\n\nProceeding on, then, equation (\\ref{omegam}) can therefore be substituted into the\nnon-flat version of equation (\\ref{elzmax}), which\nis the same as equation (\\ref{elzmax}), except that its right-hand-side is identical\nto right-hand-side of equation (\\ref{ymax2}) without the $R_0H_0$ factors\nin the denominator (these have cancelled out, as before). Thus, we have, finally,\nthe resulting algebraic relationship involving  $C_{max}$, $z_{max}$,\n$H_0$ and $\\Omega_{\\Lambda}$ as the general FLRW relationship corresponding to the flat case\ngiven in equation (\\ref{elzmax}):\n\n\\begin{eqnarray}\n&  &\\frac{g}{\\Omega_{\\Lambda}^{1\/2}}\\biggl[ F(\\phi, k) \\mid_{(1+z)^{-1} = 1} -F(\\phi, k) \\mid_{(1+z_{\\max})^{-1}}\\biggr]  \\nonumber \\\\\n&  & \\hfill{\\qquad}= \\frac{1+z_{\\max}}{\\sqrt{\\Omega_{\\Lambda} + \\Omega_{M}(1+z_{\\max})^3 -(\\Omega_0-1)(1+z_{\\max})^2}}.  \\label{omegaleq}\n\\end{eqnarray}\nHere and in the solution of the Friedmann  equation for the general\nFLRW case, the parameters associated with that solution are now given by:\n\n\\begin{eqnarray}\n\\phi_{(1+z)^{-1}} &=&  \\cos^{-1}\\Biggl[\\frac{(A-B) - (\\bar{A}+\\bar{B})A(1+z)}\n{(A+B)-(\\bar{A}+\\bar{B})A(1+z)}\\Biggr], \\nonumber \\\\\nk^2 &=& \\frac{(A+B)^2 - (a - b)^2}{4AB}, \\nonumber \\\\\ng &=& \\frac{1}{\\sqrt{AB}}, \\nonumber\n\\end{eqnarray}\nwith $a \\equiv -\\frac{\\Omega_0-1}{\\Omega_{\\Lambda}}$,\n$b \\equiv \\frac{\\Omega_M}{\\Omega_{\\Lambda}}$,\nand\n\n\n\\begin{eqnarray}\nA^2 &=& \\bar{A}^2 + \\bar{B}^2 - \\bar{A}\\bar{B}, \\nonumber \\\\\nB^2 &=& 3(\\bar{A}^2 + \\bar{B}^2) + 3\\bar{A}\\bar{B}. \\nonumber \n\\end{eqnarray}\n\nHere, further, \n\n\\begin{eqnarray}\n\\bar{A} = \\Biggl\\{\\frac{\\Omega_M}{2\\Omega_{\\Lambda}} +\n\\Biggl[\\frac{{\\Omega_M}^2}{4\\Omega_{\\Lambda}^2} -\n\\frac{(\\Omega_0-1)^3}{27 \\Omega_{\\Lambda}^3}\\Biggr]^{1\/2}\\Biggr\\}^{1\/3},  \\nonumber \\\\\n\\bar{B} = \\Biggl\\{\\frac{\\Omega_M}{2\\Omega_{\\Lambda}} -\n\\Biggl[\\frac{{\\Omega_M}^2}{4\\Omega_{\\Lambda}^2} -\n\\frac{(\\Omega_0-1)^3}{27 \\Omega_{\\Lambda}^3}\\Biggr]^{1\/2}\\Biggr\\}^{1\/3}. \\nonumber \n\\end{eqnarray}\nIn these equations, remember that $\\Omega_M$ is given by equation (\\ref{omegam}), so that relationship given by\nequation (\\ref{omegaleq}) is indeed an algebraic relationship involving $C_{max}$,\n$z_{max}$, $H_0$ and $ \\Omega_{\\Lambda}$.\nThus, if both $C_{max}$ and $z_{max}$, together with $H_0$, are all known from data, then\nequation (\\ref{omegaleq}) will determine $\\Omega_{\\Lambda}$, the only unknown. Using \nthat result in equation (\\ref{omegam}) will also determine $\\Omega_M$. Thus,\nobservational determination of both $C_{max}$ and $z_{max}$,\nwill determine both $\\Omega_M$ and $\\Omega_{\\Lambda}$, as long as we also\nknow  $H_0$. \\\\\n\n\\clearpage\n\\begin{table}\n\\begin{tabular} {c c c c c c c c } \\hline\n${\\bf \\Omega_{\\Lambda}}$&${\\bf z_{max}}$&${\\bf \\Omega_{\\Lambda}}$&${\\bf z_{max}}$&${\\bf \\Omega_{\\Lambda}}$&${\\bf z_{max}}$&${\\bf \\Omega_{\\Lambda}}$&${\\bf z_{max}}$ \\\\ \n0.59&1.50&0.65&1.55&{\\bf0.71}&{\\bf1.62}&0.77&1.71 \\\\ \n0.60&1.51&0.66&1.56&0.72&1.63&0.78&1.72 \\\\ \n0.61&1.51&0.67&1.57&{\\bf0.73}&{\\bf1.64}&0.79&1.74 \\\\ \n0.62&1.52&0.68&1.58&0.74&1.66&0.80&1.76 \\\\ \n0.63&1.53&0.69&1.59&0.75&1.67&0.81&1.78 \\\\ \n0.64&1.54&0.70&1.61&0.76&1.69&0.82&1.81 \\\\ \\hline\n\\end{tabular}\n\\caption{List of pairs ($\\Omega_{\\Lambda}$,$z_{max}$) for $0.59 \\leq \\Omega_{\\Lambda}\n\\leq 0.82$ and $1.5 \\leq z_{max} \\leq 1.81$.}\n\\end{table}\n\\clearpage\n\\begin{figure}\n\\begin{center}\n\\includegraphics {asfigure1}\n\\end{center}\n\\caption{Plot of $\\Omega_{\\Lambda}$ -- $z_{max}$, given by equation (\\ref{elzmax}), which\nis for a flat FLRW universe. Here $z_{max}$ is the\nredshift at which the maximum of the angular diameter distance, $C_{max}$\noccurs.}\n\\label{araujo}\n\\end{figure}\n\\clearpage\n\n\\section{Observational Prospects and Conclusion}\n\nWhat are the prospects for actually determining $C_{max}$ and $z_{max}$ from\nobservations? We would\ncertainly need precise SN Ia luminosity-distance, or ultra-compact radio-source\nangular-diameter distance, and redshift data out to\n$z \\approx 1.8$ or so. In the SN Ia case this would require careful,\nlong-range programs\nusing space-telescopes. However, as already mentioned, we already have\ndetected and measured SN Ia out to $z \\approx 1.7$, and in a recent\nassessment (Davis, Schmidt and Kim 2006), precision SN Ia measurements\nto $z \\approx 1.8$ are considered attainable. This is already considered\nan important goal, in order to confirm at what redshift (and cosmic epoch)\nthe universe made the transition from deceleration to acceleration. It is\ncertainly fortuitous that the same redshift range promises to provide a\nstrong independent test of the concordance FLRW model we have derived from\nCMB, SN Ia, and large-scale structure measurements. \\\\\n\nHere we have provided a brief presentation of the straightforward relationship\n(first found in numerical form by Krauss and Schramm (1992)) between the\npresent value of $\\Omega_{\\Lambda}$ and the redshift $z_{max}$ at\nwhich the angular-diameter (or observer area) distance $C$ occurs in a flat\nFLRW cosmology, like that which apparently models our universe. Furthermore,\nwe have generalized this to non-flat FLRW cases, adding the $C_{\\max}$\nmeasurements themselves. In doing this we have derived the characteristic\nFLRW observational relationships in closed form for $C(z)$ and $\\hat{M}_0\n(z)$ in the $\\Lambda \\neq 0$ case, and found a very simple and potentially\nuseful observational criterion for flatness.  These results\npromise to provide improved determination of the\nbest fit cosmological model, or a strong consistency test of it, (depending on\nhow the relationship and the data supporting it are used), once we have\nenough precise high-redshift luminosity-distance (or angular-diameter distance)\ndata. That should be possible in the near future with the rapid progress being\nmade in SN Ia measurements from space. If the concordance model --\na nearly flat universe with $\\Omega_M = 0.27$ and $\\Omega_{\\Lambda} = 0.73$ --\nis approximately correct, we should find observationally that $z_{max}\n\\approx 1.64$. \\\\ \n\nOur thanks to George Ellis and Charles Hellaby for discussions and comments,\nand to an anonymous referees for several helpful suggestions for clarification \nand for checking our results, and to one of them for pointing out the\nmuch earlier 1993 Krauss and Schramm paper.  \\\\\n  \n\n\\noindent\n{\\Large \\bf References } \\\\\n\n\\noindent\nAlbrecht, A., {\\it et al.}, 2006, Report of the Dark Energy Task Force, \nastro-ph\/0609591. \\\\\nBen\\'{i}tez, N., Riess, A. G., Nugent, P., Dickinson, M., Chornook, R., \\& \nFilippenko, A. V. 2002, ApJ, 577, L1. \\\\\nBennett, C. L., {\\it et al.} 2003, ApJS, 148, 1. \\\\\nByrd, P. F. \\& Friedman, M. D. 1954, {\\it Handbook of Elliptic\nIntegrals for Engineers and Physicists}, Springer Verlag. \\\\\nCarroll, S. M.,  Press, W. H., \\& Turner, E. L., 1992, ``The Cosmological\nConstant,'' Ann. Rev. Astron. \\& Astrophys. 30, 499-542.  \\\\\nEfstathiou, G., {\\it et al.} 2002, MNRAS, 330, L29 \\\\\nDavis, T. M., Schmidt, B. P. \\& Kim, A. G. 2006, PASP, 118, 205. \\\\\nEllis, G. F. R. 1971, ``Relativistic Cosmology,'' in {\\it General Relativity\nand Cosmology}, Proc. Int. School Phys. ``Enrico Fermi,'' R. K. Sachs,\neditor (New York: Academic Press), pp. 104-182 (see especially pp. 153-1540.\\\\\nEllis, G. F. R., Nel, S. D., Maartens, R., Stoeger, W. R., \\& Whitman, A. P.\n1985, Phys. Reports, 124 (No. 5 and 6), 315. \\\\\nEllis, G. F. R. \\& Tivon, G. 1985, Observatory, 105, 189.\\\\\nEllis, G. F. R. \\& Stoeger, W. R. 1987. Class. Quantum Grav., 4, 1697. \\\\\nEtherington, I. M. H. 1933, Phil. Mag., 15, 761. \\\\\nGilliland, R. L., Nugent, P. E., \\& Phillips, M. M. 1999, ApJ, 521, 30. \\\\\nHellaby, C. W. 2006, MNRAS, 370, 239 (astro-ph\/0603637). \\\\\nHoyle, F., 1961, in Moller, C., ed., Proc. Enrico Fermi School of Physics,\nCourse XX, Varenna, {\\it Evidence for Gravitational Theories}, Academic\nPress, New York, p. 141. \\\\\nJackson, J. C. \\& Doddgson, M. 1997, Mon. Not. R. Astron. Soc., 285, 806. \\\\\nJackson, J. C. 2004, JCAP, 11, 007. \\\\\nJeffrey, A., 1995, {\\it Handbook of Mathematical Formulas and Integrals},\nAcademic Press, Inc., pp. 225-234. \\\\\nKrauss, L. M., and Schramm, D. N. 1993, ApJ, 405, L43. \\\\\nMcCrea, W. H., 1935, Z. Astrophys., 9, 290. \\\\\nMortsell, E., Gunnarson, C., \\& Goobar, A. 2001, ApJ 561, 106. \\\\\nPeacock, J. A., {\\it et al.} 2001, Nature, 410, 169. \\\\\nPeebles, P. J. E., 1993, {\\it Principles of Physical Cosmology}, Princeton\nUniversity Press, Princeton, NJ, pp. 325-329. \\\\\nPercival, W. J., {\\it et al.} 2001, MNRAS, 327, 1297. \\\\\nPerlmutter, S., {\\it et al.} 1999, ApJ, 517, 565. \\\\\nRefsdal, S., Stabell, R., \\& de Lange, F. G. 1967, Mem. R. Astron. Soc., 71,\n143. \\\\\nRiess, A. G., {\\it et al.} 1998, AJ, 116, 1009. \\\\\nRiess, A. G., {\\it et al.} 2001, ApJ, 560, 49. \\\\\nRiess, A. G., {\\it et al.} 2004, ApJ, 607, 665. \\\\\nSpergel, D. N., {\\it et al.} 2003, ApJS, 148, 175.\\\\\nStoeger, W. R., Ellis, G. F. R. \\& Nel, S. D. 1992, Class. Quantum Grav., 9,\n509.  \n\\end{document}  \n","meta":{"redpajama_set_name":"RedPajamaArXiv"}}
+{"text":"\\section{Preliminaries}\\label{sec:pre}\nIn this section we recall the definitions of Yang--Baxter operators and\nassociated link invariants.\nIf an invertible linear map $R:\\mathbb{C}^N\\otimes\\mathbb{C}^N\\to\\mathbb{C}^N\\otimes\\mathbb{C}^N$ satisfies the\nfollowing Yang--Baxter equation, it is called a Yang--Baxter operator.\n\\begin{equation*}\n  (R\\otimes{id})({id}\\otimes{R})(R\\otimes{id})\n  =\n  ({id}\\otimes{R})(R\\otimes{id})({id}\\otimes{R}),\n\\end{equation*}\nwhere $id:\\mathbb{C}^N\\to\\mathbb{C}^N$ is the identity.\nIf there exists a homomorphism $\\mu:\\mathbb{C}^N\\to\\mathbb{C}^N$ and scalars $\\alpha,\\beta$\nsatisfying the following two equations, the quadruple $S=(R,\\mu,\\alpha,\\beta)$\nis called an enhanced Yang-Baxer operator \\cite{Turaev:INVEM88}.\n\\begin{gather*}\n  (\\mu\\otimes\\mu)R=R(\\mu\\otimes\\mu),\n  \\\\\n  \\Sp_2\\left(R^{\\pm1}({id}\\otimes\\mu)\\right)=\\alpha^{\\pm1}\\beta\\,{id},\n\\end{gather*}\nwhere\n$\\Sp_k:\\operatorname{End}(\\mathbb{C}^{\\otimes{k}})\\to\\operatorname{End}(\\mathbb{C}^{\\otimes{k-1}})$ is the\noperator trace defined as\n\\begin{multline*}\n  \\Sp_k(f)(v_{i_1}\\otimes{v_{i_2}}\\otimes\\dots\\otimes{v_{i_{k-1}}})\n  \\\\=\n  \\sum_{j_1,j_2,\\dots,j_{k-1},j=0}^{N-1}\n  f_{i_1,i_2,\\dots,i_{k-1},j}^{j_1,j_2,\\dots,j_{k-1},j}\n  (v_{j_1}\\otimes{v_{j_2}}\\otimes\\dots\\otimes{v_{j_{k-1}}}\\otimes{v_{j}}),\n\\end{multline*}\nwhere\n\\begin{equation*}\n  f(v_{i_1}\\otimes{v_{i_2}}\\otimes\\dots\\otimes{v_{i_k}})\n  =\n  \\sum_{j_1,j_2,\\dots,j_k=0}^{N-1}\n  f_{i_1,i_2,\\dots,i_k}^{j_1,j_2,\\dots,j_k}\n  \\left({v_{j_1}}\\otimes{v_{j_2}}\\otimes\\dots\\otimes{v_{j_k}}\\right)\n\\end{equation*}\nfor a basis $\\{v_0,v_1,\\dots,v_{N-1}\\}$ of $\\mathbb{C}^N$.\n\\par\nFor an enhanced Yang--Baxter operator one can define a link invariant as follows\n\\cite{Turaev:INVEM88}.\nFirst we represent a given link $L$ as the closure of a braid $\\xi$ with\n$n$ strings.\nConsider the $n$-fold tensor product $\\left(\\mathbb{C}^N\\right)^{\\otimes{n}}$ and\nassociate the homomorphism\n$b_R(B):\\left(\\mathbb{C}^N\\right)^{\\otimes{n}}\\to\\left(\\mathbb{C}^N\\right)^{\\otimes{n}}$ by\nreplacing $\\sigma_i^{\\pm1}$\n(the usual $i$-th generator of the braid group) in $\\xi$ with\n\\begin{equation*}\n  \\underset{i-1}{\\underbrace{{id}\\otimes\\dots\\otimes{id}}}\n  \\otimes{R^{\\pm1}}\\otimes\n  \\underset{N-i-1}{\\underbrace{{id}\\otimes\\dots\\otimes{id}}}.\n\\end{equation*}\nThen taking the operator trace $n$ times we define\n\\begin{equation*}\n  T_S(\\xi)\n  =\n  \\alpha^{-w(\\xi)}\\beta^{-n}\n  \\Sp_1\\left(\\Sp_2\n         \\left(\\cdots\n           \\left(\\Sp_n\n             \\left(b_R(\\xi)\\mu^{\\otimes{n}}\n             \\right)\n           \\right)\n         \\right)\n       \\right),\n\\end{equation*}\nwhere $w(\\xi)$ is the sum of the exponents.\nThen $T_S(\\xi)$ defines a link invariant and denoted by $T_S(L)$.\n\\par\nTo define the (generalized) Alexander polynomial from an enhanced Yang--Baxter\noperator we have to be more careful, since $T_S$ always vanishes in this case.\nIf the following homomorphism\n\\begin{equation*}\n  T_{S,1}(\\xi)\n  =\n  \\alpha^{-w(\\xi)}\\beta^{-n}\n  \\Sp_2\\left(\\Sp_3\n         \\left(\\cdots\n           \\left(\\Sp_n\n             \\left(b_S(\\xi)({id}\\otimes\\mu^{\\otimes(n-1)})\n             \\right)\n           \\right)\n         \\right)\n       \\right)\n   \\in\\operatorname{End}(\\mathbb{C}^N)\n\\end{equation*}\nis a scalar multiple and\n\\begin{equation*}\n  \\Sp_1(\\mu)T_{S,1}(\\xi)=T_S(\\xi)\n\\end{equation*}\nfor any $\\xi$, then the scalar defined by $T_{S,1}(\\xi)$ becomes a link\ninvariant (even if $\\Sp_1(\\mu)=0$) and is denoted by $T_{S,1}(L)$.\nNote that this invariant can be regarded as an invariant for $(1,1)$-tangles,\nwhere a $(1,1)$-tangle is a link minus an open interval.\n\\section{The intersection of the generalized Alexander polynomials and the\ncolored Jones polynomials}\\label{sec:Alexander}\nIn \\cite{Akutsu\/Deguchi\/Ohtsuki:JKNOT92} Akutsu, Deguchi and Ohtsuki\ndefined a generalization of the multivariable Alexander polynomial for colored\nlinks.\nFirst we will briefly describe their construction only for the case where\nall the colors are the same according to \\cite{Deguchi:1994}.\n\\par\nFix an integer $N\\ge2$ and a complex number $p$.\nPut $s=\\exp(\\pi\\sqrt{-1}\/N)$ and $[k]=(s^k-s^{-k})\/(s-s^{-1})$ for a complex\nnumber $k$.\nNote that $[N]=0$ and $[N-k]=[k]$.\n\\par\nLet $U_q(sl(2,\\mathbb{C}))$ be the quantum group generated by $X,Y,K$ with the following\nrelations.\n\\begin{equation*}\n  KX=sXK,\\quad KY=s^{-1}YK,\\quad XY-YX=\\frac{K^2-K^{-2}}{s-s^{-1}}.\n\\end{equation*}\nLet $F(p)$ be the $N$-dimensional vector space over $\\mathbb{C}$ with basis\n$\\{f_0,f_1,\\dots,f_{N-1}\\}$.\nWe give an action of $U_q(sl(2,\\mathbb{C}))$ on $F(p)$ by\n\\begin{align*}\n  X(f_i)&=\\sqrt{[2p-i+1][i]}f_{i-1},\n  \\\\\n  Y(f_i)&=\\sqrt{[2p-i][i+1]}f_{i+1},\n  \\\\\n  K(f_i)&=s^{p-i}f_{i}.\n\\end{align*}\nUsing Drinfeld's universal $R$-matrix given in \\cite{Drinfeld:ICM86},\nwe can define a set of enhanced Yang--Baxter operators $S_A(p)$\nwith complex parameter $p$.\nThen Akutsu--Deguchi--Ohtsuki's generalized Alexander\npolynomial is defined to be $T_{S_A(p),1}$ by using the notation in the previous\nsection.\nWe denote it by $\\Phi_N(L,p)$ for a link $L$.\nNote that if $N=2$ the invariant $\\Phi_2(L,p)$ is the same as the\nmultivariable Alexander polynomial \\cite{Jun:PACJM93}.\n\\par\nNext we review the colored Jones polynomial at the $N$-th root of unity.\nThere is another $N$-dimensional representation of $U_q(sl(2,\\mathbb{C}))$,\ncorresponding to the usual $N$-dimensional irreducible representation of\n$sl(2,\\mathbb{C})$.\nLet $E$ be the $N$-dimensional complex vector space with basis\n$\\{e_0,e_1,\\dots,e_{N-1}\\}$ and we define the action of $U_q(sl(2,\\mathbb{C}))$\nby\n\\begin{align*}\n  X(e_i)&=[i+1]e_{i+1},\n  \\\\\n  Y(e_i)&=[i]e_{i-1},\n  \\\\\n  K(e_i)&=s^{i-(N-1)\/2}e_{i}.\n\\end{align*}\n(See for example \\cite[(2.8)]{Kirby\/Melvin:INVEM91}.)\nBy using Drinfeld's universal $R$-matrix again we have another enhanced\nYang--Baxter operator $S_J$.\nThen the invariant $T_{S_J,1}$ coincides with the colored Jones polynomial\nof a link each of whose component decorated by the $N$-dimensional irreducible\nrepresentation evaluated at $t=s^2=\\exp(2\\pi\\sqrt{-1}\/N)$.\nNote that before evaluating at $s^2$, we have to normalize the colored Jones\npolynomial so that its value of the trivial knot is one for otherwise\nthe invariant would be identically zero.\nThis is well-defined since the colored Jones polynomial defines a well-defined\n$(1,1)$-tangle invariant (\\cite[(3.9) Lemma]{Kirby\/Melvin:INVEM91}).\nWe will denote $T_{S_J,1}$ by $J_{N}$.\n\\par\nNow we put $p=(N-1)\/2$ in the Akutsu--Deguchi--Ohtsuki invariant.\nThen since $[N-k]=[k]$, we have\n\\begin{align*}\n  X(f_i)&=[i]f_{i-1},\n  \\\\\n  Y(f_i)&=[i+1]f_{i+1},\n  \\\\\n  K(f_i)&=s^{(N-1)\/2-i}f_{i}\n\\end{align*}\nand so the two representation $F((N-1)\/2)$ and $E$ are quite similar.\nIn fact if we exchange $X$ and $Y$, and replace $K$ with $K^{-1}$ then these two\ncoincide.\n(This automorphism is known as the Cartan automorphism\n\\cite[p.~123, Lemma VI.1.2]{Kassel:quantum_groups}.)\nTherefore they determine the same Yang--Baxter operator and the same\nlink invariant, that is, we have the following theorem.\n\\begin{thm}\\label{thm:Alexander=Jones}\n  The Akutsu--Deguchi--Ohtsuki invariant with all the colors $p=(N-1)\/2$\n  coincides with the colored Jones polynomial corresponding to the\n  $N$-dimensional irreducible representation evaluated at\n  $\\exp(2\\pi\\sqrt{-1}\/N)$.\n  More precisely, we have $\\Phi_{N}(L,(N-1)\/2)=J_{N}(L)$ for every link $L$.\n\\end{thm}\n\\begin{rem}\nAfter finishing this work we were informed by Deguchi that it has already\nobserved \\cite{Deguchi:JPHYS191} that the $R$-matrices given by $F((N-1)\/2)$\nand $E$ coincide.\n\\end{rem}\n\\section{$\\check{R}$-matrix for the colored Jones polynomial at the $N$th root of\n         unity}\\label{sec:Jones}\nLet $R_J$ be the $\\check{R}$-matrix shown in\n\\cite[Corollary~2.32]{Kirby\/Melvin:INVEM91}, which is the $N^2\\times N^2$ matrix\nwith\n$((i,j),(k,l))$th entry\n\\begin{align*}\n  \\left(R_J\\right)_{kl}^{ij}=\n  \\sum_{n=0}^{\\min{(N-1-i,j)}}\n  &\\delta_{l,i+n}\\delta_{k,j-n}\n  \\frac{(s-s^{-1})^n}{[n]!}\n  \\frac{[i+n]!}{[i]!}\\frac{[N-1+n-j]!}{[N-1-j]!}\n  \\\\\n  &\\times\n  s^{2(i-(N-1)\/2)(j-(N-1)\/2)-n(i-j)-n(n+1)\/2},\n\\end{align*}\nwhere $[k]!=[k][k-1]\\cdots[2][1]$.\nNote that our matrix $R_J$ corresponds to $\\check{R}$ in\n\\cite[Definition~2.35]{Kirby\/Melvin:INVEM91}.\nThis matrix is used to define an enhanced Yang--Baxter operator and\nthe link invariant $J_{N}$ described in the previous section.\n\\par\nThe aim of this section is to transform it to a matrix similar to Kashaev's\n$\\check{R}$-matrix.\nLet $W$ and $D$ be the $N\\times N$ matrices with $(i,j)$th entry\n$W_{j}^{i}=s^{2ij}$ and $D_{j}^{i}=\\delta_{i,j}s^{(N-1)i}$ respectively,\nwhere $\\delta_{i,j}$ is Kronecker's delta.\nWe will calculate the product\n$\\tilde{R}_J\n=(W\\otimes W)({id}\\otimes{D}){R_J}\n({id}\\otimes{D^{-1}})(W^{-1}\\otimes{W^{-1}})$\nwith $id$ the $N\\times{N}$ identity matrix and show the following proposition.\n\\begin{prop}\\label{prop:J}\n\\begin{equation*}\n  \\left(\\tilde{R}_J\\right)_{ab}^{cd}\n  =\n  \\begin{cases}\n    \\rho(a,b,c,d)(-1)^{a+b+1}\\dfrac{[d-c-1]![N-1+c-a]!}{[d-b]![b-a-1]!}\n    \\quad&\\text{if $d\\ge b>a\\ge c$},\n    \\\\[5mm]\n    \\rho(a,b,c,d)(-1)^{a+c}\\dfrac{[b-d-1]![N-1+c-a]!}{[c-d]![b-a-1]!}\n    \\quad&\\text{if $b>a\\ge c\\ge d$},\n    \\\\[5mm]\n    \\rho(a,b,c,d)(-1)^{b+d}\\dfrac{[N-1+b-d]![c-a-1]!}{[c-d]![b-a-1]!}\n    \\quad&\\text{if $c\\ge d\\ge b>a$},\n    \\\\[5mm]\n    \\rho(a,b,c,d)(-1)^{c+d}\\dfrac{[N-1+b-d]![a-b]!}{[c-d]![a-c]!}\n    \\quad&\\text{if $a\\ge c\\ge d\\ge b$},\n    \\\\[5mm]\n    0\\quad&\\text{otherwise},\n  \\end{cases}\n\\end{equation*}\nwhere\n$\\rho(a,b,c,d)=s^{-N^2\/2+1\/2+c+d-2b+(a-d)(c-b)}[N-1]!(s-s^{-1})^{2(N-1)}\/N^2$.\n\\end{prop}\n\\begin{proof}\nSince\n$(W\\otimes{W})_{ab}^{ef}=s^{2ae+2bf}$,\n$({id}\\otimes{D})_{ef}^{kl}=\\delta_{e,k}\\delta_{f,l}s^{(N-1)l}$,\n$\\left({id}\\otimes{D^{-1}}\\right)_{ij}^{gh}\n  =\\delta_{g,i}\\delta_{h,j}s^{-(N-1)j}$,\n$\\left(W^{-1}\\otimes{W^{-1}}\\right)_{gh}^{cd}=s^{-2cg-2dh}\/N^2$,\nwe have\n\\begin{align*}\n  &N^2\\left(\\tilde{R}_J\\right)_{ab}^{cd}\n  \\\\\n  &=\n  \\sum_{i,j,k,l,e,f,g,h=0}^{N-1}\\sum_{n=0}^{\\min{(N-1-i,j)}}\n  \\delta_{e,k}\\delta_{f,l}\\delta_{g,i}\\delta_{h,j}\\delta_{l,i+n}\\delta_{k,j-n}\n  s^{2ae+2bf-2cg-2dh+(N-1)(l-j)}\n  \\\\\n  &\\quad\\times\n  \\frac{(s-s^{-1})^n}{[n]!}\\frac{[i+n]!}{[i]!}\\frac{[N-1+n-j]!}{[N-1-j]!}\n  s^{2(i-(N-1)\/2)(j-(N-1)\/2)-n(i-j)-n(n+1)\/2}\n  \\\\\n  &=\n  \\sum_{i,j=0}^{N-1}\\sum_{n=0}^{\\min{(N-1-i,j)}}\n \n  s^{(N-1)^2\/2}s^{(2b-2a+N)n-n^2\/2-3n\/2+(2a-2d+n+2)j+(2b-2c+2j-n)i}\n  \\\\\n  &\\quad\\times\n  (s-s^{-1})^n\\frac{[N-1+n-j]!}{[N-1-j]!}\\qbinom{n+i}{i},\n\\end{align*}\nwhere\n\\begin{equation*}\n  \\qbinom{x}{y}=\\frac{[x]!}{[y]![x-y]!}.\n\\end{equation*}\nSince the summation $\\sum_{i,j=0}^{N-1}\\sum_{n=0}^{\\min{(N-1-i,j)}}$ is the same\nas $\\sum_{n=0}^{N-1}\\sum_{j=n}^{N-1}\\sum_{i=0}^{N-1-n}$, we have\n\\begin{align*}\n  N^2\\left(\\tilde{R}_J\\right)_{ab}^{cd}\n  &=\n  s^{(N-1)^2\/2}\\sum_{n=0}^{N-1}\\sum_{j=n}^{N-1}\n  s^{(2b-2a+N)n-n^2\/2-3n\/2+(2a-2d+n+2)j}(s-s^{-1})^n\n  \\\\\n  &\\quad\\times\n  \\frac{[N-1+n-j]!}{[N-1-j]!}S(n,2(b-c+j)-n)\n\\end{align*}\nwith $S(\\alpha,\\beta)=\\sum_{i=0}^{N-1}s^{\\beta i}\\qbinom{\\alpha+i}{i}$.\n\\par\nReplacing $j-n$ with $k$, the summation turns out be\n$\\sum_{k=0}^{N-1}\\sum_{n=0}^{N-1-k}$ and we have\n\\begin{equation*}\n  N^2\\left(\\tilde{R}_J\\right)_{ab}^{cd}\n  =\n  s^{(N-1)^2\/2}\\sum_{k=0}^{N-1}\n  s^{2(a-d+1)k}[N-1-k]!X(k),\n\\end{equation*}\nwhere\n\\begin{equation*}\n  X(k)=\n  \\sum_{n=0}^{N-1-k}\n  (-1)^n s^{2(b-d)n+kn+n(n+1)\/2}\n  \\frac{(s-s^{-1})^n}{[N-1-k-n]!}\n  S(n,2(b-c+k)+n).\n\\end{equation*}\nNote that from Lemma~\\ref{lem:ST}, we have\n\\begin{multline}\\label{eq:S}\n  S(n,2(b-c+k)+n)\n  =\n  \\\\\n  (1-s^{2(b-c+k-1)})(1-s^{2(b-c+k-2)})\\cdots(1-s^{2(b-c+k+n-N+1)}).\n\\end{multline}\n\\par\nAs easily seen from Lemma~\\ref{lem:ST}, $S(\\alpha,\\beta)$ and $T(\\alpha,\\beta)$\nvanishes if $(\\beta-\\alpha-2)\\ge0\\ge(\\beta+\\alpha-2N+2)$ and if\n$(\\beta+\\alpha-1)\\ge0\\ge(\\beta-\\alpha+1)$ respectively.\nWe will use this fact repeatedly from now on and divide the proof into\nsome cases according to the order of $a,b,c,d$.\n\\par\nFirst we divide the proof into two cases; $b>c$ and $c\\ge b$.\n\\medskip\\par\\noindent\n{\\bf Case 1 ($b>c$).}\\qquad\nIn this case $b-c+k-1\\ge 0$ and so from \\eqref{eq:S}, $S(n,2(b-c+k)+n)=0$ if\n$n\\le N-1-k-b+c$.\nIf $n>N-1-k-b+c$, we see that\n\\begin{multline*}\n  S(n,2(b-c+k)+n)=\n  \\\\\n  (-1)^{N-1-n}s^{(N-1-n)(2(b-c+k)+n-N)\/2}(s-s^{-1})^{N-1-n}\n  \\frac{[b-c+k-1]!}{[b-c+k+n-N]!}.\n\\end{multline*}\nTherefore\n\\begin{align*}\n  X(k)&=\n  (-1)^{N+1}(s-s^{-1})^{N-1}s^{(b-c+k)(N-1)-N(N-1)\/2}\n  \\frac{[b-c+k-1]!}{[b-c-1]!}\n  \\\\\n  &\\quad\\times\n  \\sum_{n=N-1-k-b+c}^{N-1-k}\n  (-1)^{n}s^{(b+c-2d)n}\n  \\qbinom{b-c-1}{N-1-k-n}\n  \\\\[5mm]\n  &\\text{(putting $i=N-1-k-n$)}\n  \\\\\n  &=\n  (-1)^{k}(s-s^{-1})^{N-1}s^{(2b-2d+k)(N-1)-(b+c-2d)k-N(N-1)\/2}\n  \\frac{[b-c+k-1]!}{[b-c-1]!}\n  \\\\\n  &\\quad\\times\n  \\sum_{i=0}^{b-c-1}\n  (-1)^{i}s^{(2d-b-c)i}\n  \\qbinom{b-c-1}{i}\n  \\\\\n  &=\n  (s-s^{-1})^{N-1}s^{(2d-2b)-(b+c-2d+1)k-N(N-1)\/2}\n  \\frac{[b-c+k-1]!}{[b-c-1]!}\n  \\\\\n  &\\quad\\times\n  T(b-c-1,2d-b-c),\n\\end{align*}\nwith $T(\\alpha,\\beta)=\\sum_{i=0}^{\\alpha}(-1)^{i}s^{\\beta i}\\qbinom{\\alpha}{i}$.\nFrom Lemma~\\ref{lem:ST} we have\n\\begin{align*}\\label{eq:X}\n  X(k)\n  &=(s-s^{-1})^{N-1}s^{2(d-b)-(b+c-2d+1)k-N(N-1)\/2}\n  \\frac{[b-c+k-1]!}{[b-c-1]!}\n  \\\\\n  &\\quad\\times\n  (1-s^{2(d-c-1)})(1-s^{2(d-c-2)})\\cdots(1-s^{2(d-b+1)}).\n\\end{align*}\nWe divide the case into two subcases; $d\\ge b$ and $b>d$.\n\\smallskip\\par\\noindent\n{\\bf Subcase 1.1 ($d\\ge b$).}\\quad\nSince\n\\begin{multline*}\n  (1-s^{2(d-c-1)})(1-s^{2(d-c-2)})\\cdots(1-s^{2(d-b+1)})\n  \\\\\n  =\n  (-1)^{b-c-1}s^{(2d-b-c)(b-c-1)\/2}(s-s^{-1})^{b-c-1}\n  \\frac{[d-c-1]!}{[d-b]!},\n\\end{multline*}\nwe have\n\\begin{align*}\n  X(k)\n  &=(-1)^{b+c+1}(s-s^{-1})^{N+b-c-2}\n  s^{2(d-b)-N(N-1)\/2+(2d-b-c)(b-c-1)\/2}\n  \\\\\n  &\\quad\\times\n  \\frac{[d-c-1]!}{[b-c-1]![d-b]!}\n  s^{(2d-b-c-1)k}[b-c+k-1]!\n\\end{align*}\nand so\n\\begin{align*}\n  &N^2\\left(\\tilde{R}_J\\right)_{ab}^{cd}\n  \\\\\n  &=\n  s^{(N^2+1)\/2}(-1)^{b+c}(s-s^{-1})^{N+b-c-2}\n  s^{2(d-b)-N(N-1)\/2+(2d-b-c)(b-c-1)\/2}\n  \\\\\n  &\\quad\\times\n  \\frac{[d-c-1]!}{[d-b]!}\n  \\sum_{k=0}^{N-1}s^{(2a-b-c+1)k}\\frac{[N-1-k]![b-c+k-1]!}{[b-c-1]!}\n  \\\\\n  &=\n  s^{(N^2+1)\/2}(-1)^{b+c}(s-s^{-1})^{N+b-c-2}\n  s^{2(d-b)-N(N-1)\/2+(2d-b-c)(b-c-1)\/2}\n  \\\\\n  &\\quad\\times\n  \\frac{[d-c-1]![N-1]!}{[d-b]!}\n  S(b-c-1,2a-b-c+1)\n  \\\\\n  &=\n  s^{(N^2+1)\/2}(-1)^{b+c}(s-s^{-1})^{N+b-c-2}\n  s^{2(d-b)-N(N-1)\/2+(2d-b-c)(b-c-1)\/2}\n  \\\\\n  &\\quad\\times\n  \\frac{[d-c-1]![N-1]!}{[d-b]!}\n  (1-s^{2(a-b)})(1-s^{2(a-b-1)})\\dots(1-s^{2(a-c-N+1)}),\n\\end{align*}\nwhere the second equality follows from $[N-1-k]!=[N-1]!\/[k]!$.\n\\par\nNoting that $a-c-N+1\\le 0$, we see that\n\\begin{multline*}\n  (1-s^{2(a-b)})(1-s^{2(a-b-1)})\\dots(1-s^{2(a-c-N+1)})\n  \\\\\n  =\n  \\begin{cases}\n    0 &\\quad\\text{if $a\\ge b$},\n    \\\\\n    (s-s^{-1})^{N+c-b}s^{(2a-b-c-N+1)(N+c-b)\/2}\\frac{[N-1+c-a]!}{[b-a-1]!}&\\quad\n    \\text{if $b>a$}.\n  \\end{cases}\n\\end{multline*}\nTherefore we have\n\\begin{equation*}\n  \\left(\\tilde{R}_J\\right)_{ab}^{cd}\n  =\\rho(a,b,c,d)(-1)^{a+b+1}\\frac{[d-c-1]![N-1+c-a]!}{[d-b]![b-a-1]!}\n\\end{equation*}\nif $b>a$ and zero otherwise.\nThus the proof is complete for the case where $d\\ge b>c$.\n(Note that $[N-1+c-a]=0$ if $c>a$.)\n\\smallskip\\par\\noindent\n{\\bf Subcase 1.2 ($b>d$).}\\quad\nIn this case we have\n\\begin{multline*}\n  (1-s^{2(d-c-1)})(1-s^{2(d-c-2)})\\cdots(1-s^{2(d-b+1)})\n  \\\\\n  =\n  s^{(2d-b-c)(b-c-1)\/2}(s-s^{-1})^{b-c-1}\n  \\frac{[b-d-1]!}{[c-d]!}\n\\end{multline*}\nif $c\\ge d$ and $0$ otherwise.\nTherefore $\\left(\\tilde{R}_J\\right)_{ab}^{cd}=0$ if $d>c$.\nIf $c\\ge d$, we have\n\\begin{align*}\n  &N^{2}\\left(\\tilde{R}_J\\right)_{ab}^{cd}\n  \\\\\n  &\\quad=\n  s^{(N-1)^2\/2}(s-s^{-1})^{N+b-c-2}\n  s^{2(d-b)-N(N-1)\/2+(2d-b-c)(b-c-1)\/2}\n  \\\\\n  &\\qquad\\times\n  \\frac{[N-1]![b-d-1]!}{[c-d]!}\n  S(b-c-1,2a-b-c+1)\n  \\\\\n  &\\quad=\n  s^{-(N-1)\/2}(s-s^{-1})^{N+b-c-2}\n  s^{2(d-b)+(2d-b-c)(b-c-1)\/2}\n  \\\\\n  &\\qquad\\times\n  \\frac{[N-1]![b-d-1]!}{[c-d]!}\n  (1-s^{2(a-b)})(1-s^{2(a-b-1)})\\cdots(1-s^{2(a-c-N+1)}).\n\\end{align*}\nThis vanishes if $a\\ge b$ and equals\n$N^2\\rho(a,b,c,d)(-1)^{a+c}\\frac{[b-d-1]![N-1+c-a]!}{[c-d]![b-a-1]!}$\notherwise, completing the proof for the case $b>d$ and $b>c$ since\n$[N-1+c-a]=0$ if $c>a$.\n\\medskip\\par\\noindent\n{\\bf Case 2 ($c\\ge b$)}.\\quad\nFirst note from \\eqref{eq:S}, $X(k)=0$ if $k\\ge c-b+1$.\nIf $k<c-b+1$, we have\n\\begin{multline*}\n  S(n,2(b-c+k)+n)\n  \\\\\n  =s^{(2(b-c+k)+n-N)(N-1-n)\/2}(s-s^{-1})^{N-1-n}\n  \\frac{[N-1-n+c-b-k]!}{[c-b-k]!}\n\\end{multline*}\nand so\n\\begin{align*}\n  X(k)\n  &=\n  (s-s^{-1})^{N-1}s^{(2(b-c+k)-N)(N-1)\/2}\\frac{[c-b]!}{[c-b-k]!}\n  \\\\\n  &\\quad\\times\n  \\sum_{n=0}^{N-1-k}s^{(b+c-2d)n}\\qbinom{N-1-k-n+c-b}{c-b}\n  \\\\[5mm]\n  &\\text{(putting $i=N-1-k-n$)}\n  \\\\\n  &=\n  (s-s^{-1})^{N-1}s^{(2(b-c+k)-N)(N-1)\/2+(b+c-2d)(N-1-k)}\\frac{[c-b]!}{[c-b-k]!}\n  \\\\\n  &\\quad\\times\n  S(c-b,2d-b-c)\n  \\\\\n  &=\n  (s-s^{-1})^{N-1}s^{(2(b-c+k)-N)(N-1)\/2+(b+c-2d)(N-1-k)}\\frac{[c-b]!}{[c-b-k]!}\n  \\\\\n  &\\quad\\times\n  (1-s^{2(d-c-1)})(1-s^{2(d-c-2)})\\cdots(1-s^{2(d-b-N+1)}).\n\\end{align*}\nSince $d-b-N+1\\le0$, $X(k)$ vanishes and so does\n$\\left(\\tilde{R}_J\\right)_{ab}^{cd}$ if $d>c$.\nTherefore we assume that $c\\ge d$.\n\\par\nIn this case since\n\\begin{multline*}\n  (1-s^{2(d-c-1)})(1-s^{2(d-c-2)})\\dots(1-s^{2(d-b-N+1)})\n  \\\\\n  =s^{(2d-b-c-N)(N-1+b-c)\/2}(s-s^{-1})^{N-1+b-c}\n  \\frac{[N-1+b-d]!}{[c-d]!},\n\\end{multline*}\nwe have\n\\begin{align*}\n  X(k)\n  &=\n  (s-s^{-1})^{2N-2+b-c}\n  \\\\\n  &\\quad\\times\n  s^{(2b-2c-N)(N-1)\/2+(b+c-2d)(N-1)+(2d-b-c-N)(N-1+b-c)\/2}\n  \\\\\n  &\\quad\\times\n  \\frac{[N-1+b-d]![c-b]!}{[c-d]!}\n  \\\\\n  &\\quad\\times\n  s^{(N-1+2d-b-c)k}\\frac{1}{[c-b-k]!}.\n\\end{align*}\nTherefore\n\\begin{align*}\n  N^{2}\\left(\\tilde{R}_J\\right)_{ab}^{cd}\n  &=\n  s^{(N-1)^2\/2}(s-s^{-1})^{2N-2+b-c}\n  \\\\\n  &\\quad\\times\n  s^{(2b-2c-N)(N-1)\/2+(b+c-2d)(N-1)+(2d-b-c-N)(N-1+b-c)\/2}\n  \\\\\n  &\\quad\\times\n  \\frac{[N-1]![N-1+b-d]!}{[c-d]!}T(c-b,2a-b-c+1)\n  \\\\\n  &=\n  s^{(N-1)^2\/2}(s-s^{-1})^{2N-2+b-c}\n  \\\\\n  &\\quad\\times\n  s^{(2b-2c-N)(N-1)\/2+(b+c-2d)(N-1)+(2d-b-c-N)(N-1+b-c)\/2}\n  \\\\\n  &\\quad\\times\n  (1-s^{2(a-b)})(1-s^{2(a-b-1)})\\cdots(1-s^{2(a-c+1)}).\n\\end{align*}\nThere are two subcases; $b>a$ and $a\\ge b$.\n\\smallskip\\par\\noindent\n{\\bf Subcase 2.1 ($b>a$)}.\\quad\nSince\n\\begin{multline*}\n  (1-s^{2(a-b)})(1-s^{2(a-b-1)})\\cdots(1-s^{2(a-c+1)})\n  =\n  \\\\\n  s^{(2a-b-c+1)(c-b)}(s-s^{-1})^{c-b}\\frac{[c-a-1]!}{[b-a-1]!},\n\\end{multline*}\nwe have\n\\begin{equation*}\n  \\left(\\tilde{R}_J\\right)_{ab}^{cd}\n  =\n  \\rho(a,b,c,d)(-1)^{b+d}\\frac{[N-1+b-d]![c-a-1]!}{[c-d]![b-a-1]!}\n\\end{equation*}\nand the proof for the case where $c\\ge b$ and $b>a$ is complete.\n(Note that $\\left(\\tilde{R}_J\\right)_{ab}^{cd}$ vanishes unless $c\\ge d\\ge b$.)\n\\smallskip\\par\\noindent\n{\\bf Subcase 2.2 ($a\\ge b$)}.\\quad\nIn this case we only have to consider the case where $a\\ge c$ for\notherwise $\\left(\\tilde{R}_J\\right)_{ab}^{cd}=0$.\nNow\n\\begin{multline*}\n  (1-s^{2(a-b)})(1-s^{2(a-b-1)})\\cdots(1-s^{2(a-c+1)})\n  =\n  \\\\\n  (-1)^{b+c}s^{(2a-b-c+1)(c-b)}(s-s^{-1})^{c-b}\\frac{[a-b]!}{[a-c]!}\n\\end{multline*}\nand so we have\n\\begin{equation*}\n  \\left(\\tilde{R}_J\\right)_{ab}^{cd}\n  =\n  \\rho(a,b,c,d)(-1)^{c+d}\\frac{[N-1+b-d]![a-b]!}{[c-d]![a-c]!}.\n\\end{equation*}\nThe proof for the case where $c\\ge b$ and $a\\ge b$ is now complete.\n(Note again that $\\left(\\tilde{R}_J\\right)_{ab}^{cd}=0$ unless $c\\ge d\\ge b$.)\n\\end{proof}\n\\section{Kashaev's R-matrix and his invariant}\\label{sec:Kashaev}\nIn this section we will calculate Kashaev's $\\check{R}$-matrix given in\n\\cite{Kashaev:MODPLA95} and prove that it coincides with the matrix $\\tilde{R}_J$\nup to a constant given in the previous section.\n\\par\nWe prepare notations following \\cite{Kashaev:MODPLA95}.\nFix an integer $N\\ge2$.\nPut $(x)_n=\\prod_{i=1}^{n}(1-x^i)$ for $n\\ge 0$.\nDefine $\\theta:\\mathbb{Z}\\to\\{0,1\\}$ by\n\\begin{equation*}\n  \\theta(n)\n  =\n  \\begin{cases}\n    1&\\quad\\text{if $N>n\\ge0$},\n    \\\\\n    0&\\quad\\text{otherwise}.\n  \\end{cases}\n\\end{equation*}\nFor an integer $x$, we denote by $\\operatorname{res}(x)\\in\\{0,1,2,\\dots,N-1\\}$ the\nresidue modulo $N$.\n\\par\nNow Kashaev's $\\check{R}$-matrix $R_K$ is given by\n\\begin{multline*}\n  {\\left(R_K\\right)}_{ab}^{cd}\n  \\\\\n  =\n  Nq^{1+c-b+(a-d)(c-b)}\n  \\frac{\\theta(\\operatorname{res}(b-a-1)+\\operatorname{res}(c-d))\\theta(\\operatorname{res}(a-c)+\\operatorname{res}(d-b))}\n  {(q)_{\\operatorname{res}(b-a-1)}(q^{-1})_{\\operatorname{res}(a-c)}(q)_{\\operatorname{res}(c-d)}(q^{-1})_{\\operatorname{res}(d-b)}}\n\\end{multline*}\nwith $q=s^2$.\nNote that we are using $P\\circ{R}$ with $R$ defined in\n\\cite[2.12]{Kashaev:MODPLA95} rather than $R$ itself where $P$ is the\nhomomorphism from $\\mathbb{C}^N\\otimes\\mathbb{C}^N$ to $\\mathbb{C}^N\\otimes\\mathbb{C}^N$ sending\n$x\\otimes{y}$ to $y\\otimes{x}$.\n\\par\nWe will show the following proposition.\n\\begin{prop}\\label{prop:K}\n\\begin{equation*}\n  {\\left({R}_K\\right)}_{ab}^{cd}\n    =\n    \\begin{cases}\n    \\lambda(a,b,c,d)(-1)^{a+b+1}\\dfrac{[d-c-1]![N-1+c-a]!}{[d-b]![b-a-1]!}\n    \\quad&\\text{if $d\\ge b>a\\ge c$},\n    \\\\[5mm]\n    \\lambda(a,b,c,d)(-1)^{a+c}\\dfrac{[b-d-1]![N-1+c-a]!}{[c-d]![b-a-1]!}\n    \\quad&\\text{if $b>a\\ge c\\ge d$},\n    \\\\[5mm]\n    \\lambda(a,b,c,d)(-1)^{b+d}\\dfrac{[N-1+b-d]![c-a-1]!}{[c-d]![b-a-1]!}\n    \\quad&\\text{if $c\\ge d\\ge b>a$},\n    \\\\[5mm]\n    \\lambda(a,b,c,d)(-1)^{c+d}\\dfrac{[N-1+b-d]![a-b]!}{[c-d]![a-c]!}\n    \\quad&\\text{if $a\\ge c\\ge d\\ge b$},\n    \\\\[5mm]\n    0\\quad&\\text{otherwise},\n  \\end{cases}\n\\end{equation*}\nwhere\n$\\lambda(a,b,c,d)=\ns^{-N^2\/2+N\/2+2+c+d-2b+(a-d)(c-b)}(s-s^{-1})^{1-N}N\/([N-1]!)^2$.\n\\end{prop}\n\\begin{proof}\nSince $N-1\\ge b-a-1\\ge-N$, $N-1\\ge c-d\\ge -N+1$, $N-1\\ge a-c\\ge -N+1$ and\n$N-1\\ge d-b\\ge -N+1$, we see that ${({R}_K)}_{ab}^{cd}$ vanishes except for the\nfollowing four cases, which have already appeared in Proposition~\\ref{prop:J}:\n(i) $d\\ge b>a\\ge c$, (ii) $b>a\\ge c\\ge d$, (iii) $c\\ge d\\ge b\\ge a$ and\n(iv) $a\\ge c\\ge d\\ge b$.\n\\par\nWe will only prove the first case because the other cases are similar.\nNoting that\n\\begin{align*}\n  (q     )_n&=(-1)^{n}s^{n(n+1)\/2}(s-s^{-1})^{n}[n]!,\n  \\\\\n  (q^{-1})_n&=s^{-n(n+1)\/2}(s-s^{-1})^{n}[n]!,\n\\end{align*}\nwe have\n\\begin{align*}\n  {\\left({R}_K\\right)}_{ab}^{cd}\n  &=\n  (-1)^{a+b+c+d}Ns^{2+2c-2b+2(a-d)(c-b)+a-d+N\/2+1\/2}(s-s^{-1})^{-N+1}\n  \\\\\n  &\\qquad\\times\n  \\frac{1}{[d-b]![b-a-1]![N+c-d]![a-c]!}\n\\end{align*}\nsince $\\operatorname{res}(b-a-1)=b-a-1$, $\\operatorname{res}(a-c)=a-c$, $\\operatorname{res}(c-d)=N+c-d$ and\n$\\operatorname{res}(d-b)=d-b$.\nNow since $[N-n]=[n]$, we see that\n\\begin{equation*}\n  \\frac{1}{[d-b]![b-a-1]![N+c-d]![a-c]!}\n  =\n  \\frac{1}{([N-1])^2}\\frac{[d-c-1]![N+c-a-1]!}{[b-a-1]![d-b]!}.\n\\end{equation*}\nTherefore\n\\begin{equation*}\n  {\\left({R}_K\\right)}_{ab}^{cd}\n  =\\lambda(a,b,c,d)(-1)^{a+b+1}\\frac{[d-c-1]![N-1+c-a]!}{[d-b]![b-a-1]!}\n\\end{equation*}\nas required.\n\\end{proof}\n\\par\nTherefore $R_K$ and $R_J$ are equal up to a constant depending\nonly on $N$.\nMore precisely we have\n\\begin{prop}\\label{prop:Kashaev=Jones}\nLet $R_K$ and $R_J$ be the $\\check{R}$-matrices defined by as above.\nThen we have\n\\begin{equation*}\n  {R}_K\n  =\n  s^{-(N+1)(N-3)\/2}(W\\otimes{W})({id}\\otimes{D})\n  R_J({id}\\otimes{D^{-1}})({W^{-1}}\\otimes{W^{-1}})\n\\end{equation*}\nfor any $N\\ge2$.\n\\end{prop}\n\\begin{proof}\nFrom Propositions~\\ref{prop:J} and \\ref{prop:K}, we only have to check that\n$\\rho(a,b,c,d)\/\\lambda(a,b,c,d)=s^{(N+1)(N-3)\/2}$.\nWe have\n\\begin{equation*}\n  \\rho(a,b,c,d)\/\\lambda(a,b,c,d)\n  =(-1)^{N}s^{(N-3)\/2}\\left(\\frac{(s-s^{-1})^{N-1}[N-1]!}{N}\\right)^3\n\\end{equation*}\nbut this coincides with $s^{(N+1)(N-3)\/2}$ as shown below.\n\\par\nWe have\n\\begin{align*}\n  (s-s^{-1})^{N-1}[N-1]!\n  &=\\prod_{k=1}^{N-1}\\left(2\\sqrt{-1}\\sin(k\\pi\/N)\\right)\n  \\\\\n  &=\\sqrt{-1}^{N-1}\\prod_{k=1}^{N-1}\\left(2\\sin(k\\pi\/N)\\right).\n\\end{align*}\nOn the other hand from \\cite[I.392-1, p.~33]{Gradshteyn\/Ryzhik:1980}, we have\n\\begin{equation*}\n  \\sin(Nx)=2^{N-1}\\prod_{k=0}^{N-1}\\sin(x+k\\pi\/N).\n\\end{equation*}\nDivided by $\\sin x$ and taking the limit $x\\to 0$, we have\n\\begin{equation*}\n  N=\\prod_{k=1}^{N-1}\\left(2\\sin(k\\pi\/N)\\right).\n\\end{equation*}\nTherefore we have\n\\begin{align*}\n  (-1)^{N}s^{(N-3)\/2}\\left(\\frac{(s-s^{-1})^{N-1}[N-1]!}{N}\\right)^3\n  &=\n  (-1)^{N}s^{(N-3)\/2}\\sqrt{-1}^{3(N-1)}\n  \\\\\n  &=\n  s^{N^2+(N-3)\/2+3(N-1)N\/2}\n  =\n  s^{(N+1)(N-3)\/2},\n\\end{align*}\ncompleting the proof.\n\\end{proof}\n\\par\nWe will show that the matrix ${R}_K$ also satisfies the Yang--Baxter\nequation.\nTo do that we prepare a lemma.\n\\begin{lem}\\label{lem:D_through_J}\nThe matrices $D$ and $D^{-1}$ can go through ${R_J}$ in pair, that is, the\nfollowing equality holds.\n\\begin{equation*}\n  ({id}\\otimes{D}){R_J}({id}\\otimes{D^{-1}})\n  =\n  (D^{-1}\\otimes{id}){R_J}({D}\\otimes{id}).\n\\end{equation*}\n\\end{lem}\n\\begin{proof}\nIt is sufficient to show that $(D\\otimes{D})R_J=R_J(D\\otimes{D})$.\nSince $D_{j}^{i}=\\delta_{i,j}s^{(N-1)j}$,\n\\begin{align*}\n  \\left((D\\otimes{D})R_J\\right)_{kl}^{ij}\n  &=\\sum_{a,b}\\delta_{a,k}\\delta_{b,l}s^{(N-1)k}s^{(N-1)l}{(R_J)}_{ab}^{ij}\n  \\\\\n  &=s^{(N-1)(k+l)}{(R_J)}_{kl}^{ij}\n\\end{align*}\nand\n\\begin{align*}\n  \\left(R_J(D\\otimes{D})\\right)_{kl}^{ij}\n  &=\\sum_{a,b}\\delta_{a,i}\\delta_{b,j}s^{(N-1)i}s^{(N-1)j}{(R_J)}_{kl}^{ab}\n  \\\\\n  &=s^{(N-1)(i+j)}{(R_J)}_{kl}^{ij}.\n\\end{align*}\nBut these two coincide since $(R_J)_{kl}^{ij}$ vanishes unless $k+l=i+j$\n(the charge conservation law), completing the proof.\n\\end{proof}\nUsing Lemma~\\ref{lem:D_through_J} we can give another proof of the following\nproposition.\n\\begin{prop}[Kashaev]\\label{prop:YBE_K}\nKashaev's $\\check{R}$-matrix ${R}_K$ satisfies the Yang--Baxter equation, that is,\n\\begin{equation*}\n  ({R}_K\\otimes{id})({id}\\otimes{{R}_K})({R}_K\\otimes{id})\n  =\n  ({id}\\otimes{{R}_K})({R}_K\\otimes{id})({id}\\otimes{{R}_K}).\n\\end{equation*}\n\\end{prop}\n\\begin{proof}\nFrom Proposition~\\ref{prop:Kashaev=Jones}, we have\n\\begin{align*}\n  &(R_K\\otimes{id})({id}\\otimes{R_K})({R}_K\\otimes{id})\n  \\\\\n  &\\quad=\n    (W\\otimes{W}\\otimes{id})({id}\\otimes{D}\\otimes{id})(R_J\\otimes{id})\n    ({id}\\otimes{D^{-1}}\\otimes{id})(W^{-1}\\otimes{W^{-1}}\\otimes{id})\n  \\\\\n  &\\qquad\\times\n   ({id}\\otimes{W}\\otimes{W})({id}\\otimes{id}\\otimes{D})({id}\\otimes{R_J})\n   ({id}\\otimes{id}\\otimes{D^{-1}})({id}\\otimes{W^{-1}}\\otimes{W^{-1}})\n  \\\\\n  &\\qquad\\times\n    (W\\otimes{W}\\otimes{id})({id}\\otimes{D}\\otimes{id})(R_J\\otimes{id})\n    ({id}\\otimes{D^{-1}}\\otimes{id})(W^{-1}\\otimes{W^{-1}}\\otimes{id})\n  \\\\\n  &\\quad=\n    (W\\otimes{W}\\otimes{W})({id}\\otimes{D}\\otimes{D})(R_J\\otimes{id})\n  \\\\\n  &\\qquad\\times\n    ({id}\\otimes{D^{-1}}\\otimes{id})({id}\\otimes{R_J})\n    ({id}\\otimes{D}\\otimes{id})\n  \\\\\n  &\\qquad\\times\n    (R_J\\otimes{id})({id}\\otimes{D^{-1}}\\otimes{D^{-1}})\n    (W^{-1}\\otimes{W^{-1}}\\otimes{W^{-1}})\n  \\\\\n  &\\quad=\n    (W\\otimes{W}\\otimes{W})({id}\\otimes{D}\\otimes{D})(R_J\\times{id})\n  \\\\\n  &\\qquad\\times\n    (id\\otimes{id}\\otimes{D})({id}\\otimes{R_J})\n    (id\\otimes{id}\\otimes{D^{-1}})\n  \\\\\n  &\\qquad\\times\n    (R_J\\otimes{id})({id}\\otimes{D^{-1}}\\otimes{D^{-1}})\n    (W^{-1}\\otimes{W^{-1}}\\otimes{W^{-1}})\n  \\\\\n  &\\quad=\n    (W\\otimes{W}\\otimes{W})(id\\otimes{D}\\otimes{D^2})\n  \\\\\n  &\\qquad\\times\n    (R_J\\otimes{id})(id\\otimes{R_J})(R_J\\otimes{id})\n  \\\\\n  &\\qquad\\times\n    (id\\otimes{D^{-1}}\\otimes{D^{-2}})(W^{-1}\\otimes{W^{-1}}\\otimes{W^{-1}}).\n  \\\\\n  \\intertext{Similar calculation shows}\n  &({id}\\otimes{R_K})(R_K\\otimes{id})({id}\\otimes{R_K})\n  \\\\\n  &\\quad=\n    (W\\otimes{W}\\otimes{W})(id\\otimes{D}\\otimes{D^2})\n  \\\\\n  &\\qquad\\times\n    ({id}\\otimes{R_J})(R_J\\otimes{id})(id\\otimes{R_J})\n  \\\\\n  &\\qquad\\times\n    (id\\otimes{D^{-1}}\\otimes{D^{-2}})(W^{-1}\\otimes{W^{-1}}\\otimes{W^{-1}}).\n\\end{align*}\nFrom the Yang--Baxter equation for $R_J$ these two coincide, completing\nthe proof.\n\\end{proof}\n\\par\nTo show that $R_J$ and ${R}_K$ define the same link invariant,\nwe will construct enhanced Yang--Baxter operators precisely by using them.\n\\par\nLet $\\mu_J$ be the $N\\times N$-matrix with $(i,j)$-entry\n$\\left(\\mu_J\\right)_{j}^{i}=\\delta_{i,j}s^{2i-N+1}$.\nThen the quadruple $S_J=(R_J,\\mu_{J},s^{N^2-1},1)$\nis a Yang--Baxter operator and the following lemma holds.\n\\begin{lem}\\label{lem:EYB_J}\n\\begin{gather*}\n  (\\mu_J\\otimes\\mu_J)R_J=R_J(\\mu_J\\otimes\\mu_J),\n  \\\\\n  \\sum_{j=0}^{N-1}\\left((R_J)^{\\pm1}(id\\otimes\\mu_J)\\right)_{kj}^{ij}\n  =(s^{N^2-1})^{\\pm1}id.\n\\end{gather*}\n\\end{lem}\n\\par\nNext we will give a Yang--Baxter operator using $R_K$.\nLet $\\mu_K$ be the $N\\times N$-matrix with $(i,j)$-entry\n$\\left(\\mu_K\\right)_{j}^{i}=-s\\delta_{i,j+1}$.\nThen we have\n\\begin{lem}\\label{lem:mu_nu}\n\\begin{equation*}\n  W\\,D\\,\\mu_J\\,D^{-1}\\,W^{-1}=\\mu_K.\n\\end{equation*}\n\\end{lem}\n\\begin{proof}\nSince $W_{j}^{a}=s^{2aj}$, $D_{a}^{b}=\\delta_{a,b}s^{(N-1)b}$,\n$\\left(\\mu_J\\right)_{b}^{c}=\\delta_{b,c}s^{2c-N+1}$,\n$\\left(D^{-1}\\right)_{c}^{d}=\\delta_{c,d}s^{-(N-1)d}$\nand $\\left(W^{-1}\\right)_{d}^{i}=\\delta_{d,i}s^{-2di}\/N$, we have\n\\begin{align*}\n  \\left(W\\,D\\,\\mu_J\\,D^{-1}\\,W^{-1}\\right)_{j}^{i}\n  &=\n  \\frac{1}{N}(-s)\\sum_{a=0}^{N-1}s^{2(j-i+1)a}\n  \\\\\n  &=\n  -s\\delta_{i,j+1},\n\\end{align*}\ncompleting the proof.\n\\end{proof}\nCombining Lemmas~\\ref{lem:EYB_J} and \\ref{lem:mu_nu}, we show that\n$S_K=({R}_K,\\mu_K,-s,1)$ is also an enhanced Yang--Baxter\noperator.\n\\begin{lem}\\label{lem:EYB_K}\n\\begin{gather*}\n  (\\mu_K\\otimes\\mu_K){R}_K={R}_K(\\mu_K\\otimes\\mu_K),\n  \\\\\n  \\sum_{j=0}^{N-1}\\left((R_K)^{\\pm1}(id\\otimes\\mu_K)\\right)_{kj}^{ij}\n  =(-s)^{\\pm1}id.\n\\end{gather*}\n\\end{lem}\n\\begin{proof}\nNoting that $\\mu_J$ and $D$ commutes since they are diagonal, the first equality\nfollows immediately from that in Lemma~\\ref{lem:EYB_J}.\n\\par\nThe second equality follows from\n\\begin{multline*}\n  ({R}_K)^{\\pm1}({id}\\otimes\\mu_K)\n  \\\\\n  =\n  s^{\\mp(N+1)(N-3)\/2}(W\\otimes{W})(id\\otimes{D})R_J(id\\otimes{\\mu_J})\n  (id\\otimes{D^{-1}})(W^{-1}\\otimes{W^{-1}}),\n\\end{multline*}\ncompleting the proof.\n\\par\nNote that the lemma can also be proved by using\n\\cite[(2.8) and (2.17)]{Kashaev:MODPLA95}.\n\\end{proof}\nNow we see that $S_J$ and $S_K$ define the same link invariant by using\nthe following lemma.\n\\begin{lem}\\label{lem:b_K}\nLet $\\xi$ be an $n$-braid.\nThen\n\\begin{multline*}\n  b_{R_K}(\\xi)\n  =\n  \\left(W^{\\otimes{n}}\\right)\n  \\left(D^{k_1}\\otimes{D^{k_2}}\\otimes\\dots\\otimes{D^{k_n}}\\right)\n  b_{R_J}(\\xi)\n  \\\\\n  \\times\n  \\left(D^{-k_1}\\otimes{D^{-k_2}}\\otimes\\dots\\otimes{D^{-k_n}}\\right)\n  \\left(\\left(W^{-1}\\right)^{\\otimes{n}}\\right)\n\\end{multline*}\nfor some non-negative integers $k_1,k_2,\\dots,k_n$.\n\\end{lem}\n\\begin{proof}\nIn fact we can show that $(k_1,k_2,\\dots,k_n)$ is of the form\n$$(0,1,2,\\dots,d_1,0,1,2,\\dots,d_2,\\dots,0,1,2,\\dots,d_h)$$\nby using Lemma~\\ref{lem:D_through_J} repeatedly\nto `push' $D$ and $D^{-1}$ from left to right\n(See the proof of Proposition~\\ref{prop:YBE_K}).\nDetails are omitted.\n\\end{proof}\n\\par\nSince we know that $J_N=T_{S_J,1}$ is well-defined as described in\n\\S\\ref{sec:pre}, from the previous lemma $T_{S_K,1}(L)$ is also a link\ninvariant, which we denote by $\\langle{L}\\rangle_{N}$.\nNote that it is implicitly stated in \\cite{Kashaev:MODPLA95} that the invariant\ncan be regarded as an invariant for $(1,1)$-tangles.\nNote also that though the invariant was defined only up to a multiple of $s$ in\n\\cite{Kashaev:MODPLA95}, we can now define it without ambiguity.\n\\par\nSince\n\\begin{align*}\n  &b_{R_K}(\\xi)({id}\\otimes{\\mu_K}^{\\otimes{(n-1)}})\n  \\\\\n  &\\quad=\n  \\left(W^{\\otimes{n}}\\right)\n  \\left(D^{k_1}\\otimes{D^{k_2}}\\otimes\\dots\\otimes{D^{k_n}}\\right)\n  b_{R_J}(\\xi)\n  \\\\\n  &\\qquad\\times\n  \\left({id}\\otimes{\\mu_J}^{\\otimes(n-1)}\\right)\n  \\left(D^{-k_1}\\otimes{D^{-k_2}}\\otimes\\dots\\otimes{D^{-k_n}}\\right)\n  \\left(\\left(W^{-1}\\right)^{\\otimes{n}}\\right)\n\\end{align*}\nfrom Lemma~\\ref{lem:b_K}\nwe conclude that $S_J$ and $S_K$ define the same link\ninvariant.\n\\begin{thm}\n  For any link $L$ and any integer $N\\ge2$, $\\langle{L}\\rangle_{N}$ and\n  $J_{N}(L)$ coincide.\n\\end{thm}\n\\section{Relation between the simplicial volume and the colored Jones\npolynomials.}\nLet $K$ be one of the three simplest hyperbolic knots $4_1$, $5_2$ and $6_1$.\nKashaev found in \\cite{Kashaev:LETMP97} that the hyperbolic volume of\n$S^3\\setminus K$, denoted by $\\operatorname{Vol}(K)$, coincides numerically with the growth\nrate of the absolute value of $\\langle K\\rangle_{N}$ with respect to $N$.\nMore precisely,\n\\begin{equation*}\n  \\operatorname{Vol}(K)\n  =2\\pi \\, \\lim_{N\\to \\infty}\\,\n  \\frac{\\log\\left|\\langle{K}\\rangle_{N}\\right|}{N}.\n\\end{equation*}\n\\par\nWe would like to modify his conjecture taking Gromov's simplicial\nvolume (or Gromov norm) \\cite{Gromov:INSHE82} into account.\nLet us consider the torus decomposition of the complement of a knot $K$\n\\cite{Jaco\/Shalen:MEMAM79,Johannson:1979}.\nThen the simplicial volume of $K$, denoted by $\\|K\\|$ is equal to the sum\nof the hyperbolic volumes of hyperbolic pieces of the decomposition divided\nby $v_3$, the volume of the ideal regular tetrahedron in $\\mathbb{H}^3$, the\nthree-dimensional hyperbolic space.\nRecall that it is additive under the connect sum \\cite{Soma:INVEM81}\n\\begin{equation*}\n  \\|K_1 \\sharp K_2\\|\n  =\n  \\|K_1\\| + \\|K_2\\|,\n\\end{equation*}\nand that it does not alter by mutation \\cite[Theorem~1.5]{Ruberman:INVEM87}.\n\\par\nNoting that $J_{N}$ is multiplicative under the connect sum, that is,\n\\begin{equation*}\n  J_{N}(K_1 \\sharp K_2)\n  =\n  J_{N}(K) \\,J_{N}(K_2)\n\\end{equation*}\nand that it does not alter by mutation \\cite[Corollary~6.2.5]{Jun:ORSJM89},\nwe propose the following conjecture.\n\\begin{conj}[Volume conjecture]\nFor any knot $K$,\n\\begin{equation}\\label{eq:volume}\n  \\|K\\|\n  =\n  \\frac{2\\pi}{v_3}\\lim_{N \\to \\infty}\n  \\frac{\\log\\left|J_{N}(K)\\right|}{N}.\n\\end{equation}\n\\end{conj}\n\\begin{rem}\nFirst note that if Kashaev's conjecture is true then our conjecture holds\nfor hyperbolic knots and their connect sums.\nIt is also true for torus knots since Kashaev and O.~Tirkkonen\n[private communication] showed that the right hand side of\n\\eqref{eq:volume} vanishes in this case by using H.~Morton's formula\n\\cite{Morton:MATPC95} (see also \\cite{Rosso\/Jones:JKNOT93}).\n\\end{rem}\n\\begin{rem}\nNote however that the volume conjecture does not hold for links since $J_{N}$ of\nthe split union of two links vanishes.\n\\end{rem}\n\\par\nAs a consequence of the volume conjecture, we anticipate the following\nsimplest case of V.~Vassiliev's conjecture\n\\cite[6.1 Stabilization conjecture]{Vassiliev:1990}\n(see also \\cite[Chapter~1, Part V (L), Conjecture]{Kirby:problems}).\n\\begin{conj}[V.~Vassiliev]\\label{conj:Vassiliev}\nAssume that every Vassiliev (finite-type) invariant of a knot is\nidentical to that of the trivial knot.\nThen it is unknotted.\n\\end{conj}\n\\par\nWe show that the volume conjecture implies Conjecture~\\ref{conj:Vassiliev} by\nusing the following two lemmas.\n\\begin{lem}[{\\cite[Corollary~4.2]{Gordon:TRAAM83}}]\\label{lem:zero_volume}\nIf $\\|K\\| = 0$  then $K$ is obtained from the trivial knot\nby applying a finite number (possibly zero) of the following two operations:\n\\begin{enumerate}\n\\item[1)] making a connect sum,\n\\item[2)] making a cable.\n\\end{enumerate}\n\\end{lem}\n\\begin{lem}\\label{lem:Alexander_of_cable}\nIf $\\|K\\| = 0$, then the Alexander polynomial $\\Delta(K)$ of $K$ is trivial\nif and only if $K$ is the trivial knot.\n\\end{lem}\n\\begin{proof}\nThis lemma comes from Lemma~\\ref{lem:zero_volume} and the following three facts\n\\cite[\\S2]{Burau:ABHMS32}\n(see also \\cite[8.23~Proposition]{Burde\/Zieschang:1985}).\n\\begin{enumerate}\n  \\item[i)]\n    the Alexander polynomial of a non-trivial torus knot is not trivial.\n  \\item[ii)]\n    the Alexander polynomial is multiplicative under the connect sum.\n    Therefore if $\\Delta(K_1)$ and $\\Delta(K_2)$ are non-trivial, then\n    $\\Delta(K_1\\sharp K_2)$  is also non-trivial.\n  \\item[iii)]\n    if $K^\\prime$ is a knot obtained from $K$ by a cabling operation, then\n    $\\Delta(K^\\prime)$ is $\\Delta(K)f(t)$ with some Laurent polynomial $f(t)$.\n    Hence if $\\Delta(K)$ is non-trivial, so is $\\Delta(K^\\prime)$ .\n\\end{enumerate}\n\\end{proof}\n\\par\n\\begin{proof}\n[Proof that the volume conjecture implies Conjecture~\\ref{conj:Vassiliev}]\nFirst note that every coefficient of both colored Jones polynomial and Alexander\npolynomial as a power series in $h=\\log t$ is a Vassiliev invariant.\nSo a knot $K$ with every Vassiliev invariant trivial has the trivial colored\nJones polynomial for any color and the trivial Alexander polynomial.\nIn particular $J_{N}(K)=1$ for any $N$.\nTherefore assuming the volume conjecture, $\\|K\\|$ vanishes.\nFrom Lemma~\\ref{lem:Alexander_of_cable}, $K$ should be trivial, completing the\nproof.\n\\end{proof}\n\\begin{rem}\nIt was pointed out by Vaintrob and Bar-Natan that using the\nMelvin--Morton--Rozansky conjecture\n\\cite{Melvin\/Morton:COMMP95,Rozansky:COMMP96} proved by Bar-Natan and\nS.~Garoufalidis \\cite{BarNatan\/Garoufalidis:INVEM96}, we\ncan also show that a knot is trivial if and only if all of\nits colored Jones polynomials are trivial since the Melvin--Morton--Rozansky\nconjecture says that the Alexander polynomial can be determined by the colored\nJones polynomials.\n\\end{rem}\n\\renewcommand{\\thesection}{\\Alph{section}}\n\\setcounter{section}{0}\n\\section{appendix}\nIn this appendix we prove some technical formulas used in the paper.\nPut\n\\begin{align*}\n  S(\\alpha,\\beta)&=\\sum_{i=0}^{N-1}s^{\\beta i}\\qbinom{\\alpha+i}{i},\n  \\\\\n  T(\\alpha,\\beta)&=\\sum_{i=0}^{\\alpha}(-1)^{i}s^{\\beta i}\\qbinom{\\alpha}{i}.\n\\end{align*}\nNote that the summation in $S(\\alpha,\\beta)$ is essentially from $0$ to\n$N-1-\\alpha$.\nThen we have\n\\begin{lem}\\label{lem:ST}\nThe following formulas hold.\n\\begin{align*}\n  S(\\alpha,\\beta)&=\n    \\prod_{j=1}^{N-\\alpha-1}(1-s^{\\beta-\\alpha-2j})\n   =(1-s^{\\beta-\\alpha-2})(1-s^{\\beta-\\alpha-4})\\cdots(1-s^{\\beta+\\alpha-2N+2}),\n  \\\\\n  T(\\alpha,\\beta)&=\n    \\prod_{j=1}^{\\alpha}(1-s^{\\beta+\\alpha+1-2j})\n    =(1-s^{\\beta+\\alpha-1})(1-s^{\\beta+\\alpha-3})\\cdots(1-s^{\\beta-\\alpha+1}).\n\\end{align*}\n\\end{lem}\n\\begin{proof}\nWe only prove the equality for $S(\\alpha,\\beta)$ since the other case is\nsimilar.\nWe use the following quantized Pascal relation.\n\\begin{equation*}\n  \\qbinom{\\alpha+i}{i}\n  =s^{-\\alpha}\\qbinom{\\alpha+i-1}{i-1}+s^{i}\\qbinom{\\alpha+i-1}{i}.\n\\end{equation*}\nThen since\n\\begin{align*}\n  S(\\alpha,\\beta)\n  &=\n  s^{-\\alpha}\\sum_{i=0}^{N-1}s^{\\beta i}\\qbinom{\\alpha+i-1}{i-1}\n  +\n  \\sum_{i=0}^{N-1}s^{(\\beta+1)i}\\qbinom{\\alpha+i-1}{i}\n  \\\\[5mm]\n  &\\text{(putting $k=i-1$ in the first term)}\n  \\\\\n  &=\n  s^{\\beta-\\alpha}\\sum_{k=-1}^{N-2}s^{k}\\qbinom{\\alpha+k}{k}\n  +\n  S(\\alpha-1,\\beta+1)\n  \\\\\n  &=\n  s^{\\beta-\\alpha}S(\\alpha,\\beta)+S(\\alpha-1,\\beta+1),\n\\end{align*}\nwe have the following recursive formula.\n\\begin{equation*}\n  S(\\alpha-1,\\beta+1)=(1-s^{\\beta-\\alpha})S(\\alpha,\\beta).\n\\end{equation*}\nNow the required formula follows since $S(N-1,\\gamma)=1$ for any integer\n$\\gamma$.\n\\end{proof}\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}}