\documentstyle[12pt]{article} \textwidth 6.1in \textheight 9in \oddsidemargin .3in \evensidemargin .3in \title{A METHOD FOR OBTAINING QUANTUM DOUBLES FROM THE YANG-BAXTER $R$-MATRICES} \author{A.A.VLADIMIROV\thanks{E-mail address: alvladim@theor.jinrc.dubna.su}} \date{Laboratory of Theoretical Physics, \\ Joint Institute for Nuclear Research, \\ Dubna, Moscow region, 141980, Russia} \begin{document} \setcounter{page}0 \def\thepage{} \maketitle \begin{abstract} We develop the approach of refs.~\cite{FRT} and~\cite{MA} that enables one to associate a quasitriangular Hopf algebra to every regular invertible constant solution of the quantum Yang-Baxter equations. We show that such a Hopf algebra is actually a quantum double. \end{abstract} \vspace{9cm} \pagebreak \def\thepage{\arabic{page}} {\bf 1}. It is well known~\cite{FRT} that any invertible constant matrix solution $R$ of the quantum Yang-Baxter equation (QYBE) \begin{equation} R_{12}R_{13}R_{23}=R_{23}R_{13}R_{12} \end{equation} naturally generates a bialgebra $A_R=\{1,t_{ij}\}$ defined by \[R_{12}T_1T_2=T_2T_1R_{12},\ \ \ \ \Delta (T_1)=T_1\otimes T_1, \ \ \ \ \varepsilon (T)={\bf 1},\] (generators $t_{ij}$ form a matrix $T$, \ $\Delta$ is a coproduct and $\varepsilon$ a counit) and also another bialgebra $U_R=\{1,l^+_{ij},l^-_{ij}\}$ with \begin{eqnarray} R_{12}L^{\pm}_2L^{\pm}_1 & = & L^{\pm}_1L^{\pm}_2R_{12}, \\ R_{12}L^+_2L^-_1 & = & L_1^-L^+_2R_{12}, \\ \Delta(L^{\pm}_1) = L^{\pm}_1\otimes L^{\pm}_1 & , & \ \ \varepsilon(L^{\pm}) = {\bf 1} , \end{eqnarray} which is paired to $A_R$. This pairing~\cite{FRT,MA} is established by the relations $$ =R_{12}, \ \ \ =R_{21}^{-1}, $$ obeys the duality conditions $$ <\alpha\beta,a>=<\alpha\otimes\beta,\Delta(a)>,\ \ \ <\Delta(\alpha),a\otimes b>=<\alpha,ab>,$$ and appears to be degenerate. With some additional effort (quotienting by appropriate null bi-ideals) these bialgebras can be made Hopf algebras $\check{A}_R$ and $\check{U}_R$, dual to each other. Their antipodes are defined by \begin{eqnarray*} &=&=R_{12}^{-1}, \\ ~&=&=R_{21}. \end{eqnarray*} With essential use of this duality Majid \cite{MA} showed that in fact, with a certain reservation, $\check{U}_R$ proves to be a quasitriangular Hopf algebra with the universal $\cal R$-matrix given by implicit formulas originated from $=R_{12}.$ By the way, Majid claims ~\cite{MA} that $\check{U}_R$ is `more or less' of the form of a quantum double. In the present note we argue that, modulo the same reservation, $\check{U}_R$ is actually a quantum double. \vspace{.5cm} {\bf 2}. Recall that a quantum double $ {\cal D}(A)$ is the Hopf algebra of the following type (\cite{DR}, see also~\cite{FRT,MA,BU}). Let $A\otimes A^{\circ}$ be the tensor product of the Hopf algebra $A$ and its antidual $A^{\circ}$. Antiduality (i.e.\ the duality with opposite coproduct and inverse antipode) means $=\delta^i_j$ and \begin{equation} <\alpha \beta,a> = <\alpha \otimes \beta,\Delta(a)>, \ \ <\Delta(\alpha), a\otimes b>=<\alpha,ba>, \end{equation} $$ \varepsilon(a)=<1,a>,\ \ \varepsilon(\alpha)=<\alpha,1>,\ \ =<\alpha,S^{-1}(a)> ,$$ where $a,b\in A,\ \alpha,\beta\in A^{\circ}$, and $\{e_j\},\{e^i\}$ are the corresponding bases. To equip $A\otimes A^{\circ}$ with the Hopf algebra structure of the quantum double, one must define a very specific cross-multiplication recipe. If $$ e_ie_j=c^k_{ij}e_k,\ \ \Delta(e_i)=f_i^{jk}(e_j\otimes e_k),\ \ S(e^i)=\sigma_j^ie^j, $$ it reads $$ e^ie_j={\cal O}_{jq}^{ip}\,e_pe^q, \ \ \ \ \mbox{where}\ {\cal O}_{jq}^{ip}=c^t_{nq}c^i_{ts}\sigma_r^sf_j^{rl}f_l^{pn}. $$ In invariant form this looks like \begin{equation} \alpha a=\sum \sum<\alpha_{(3)},a_{(3)}> a_{(2)}\alpha_{(2)}, \end{equation} where $$\Delta^2(\alpha)=\sum \alpha_{(1)}\otimes \alpha_{(2)}\otimes\alpha_{(3)}, \ \ \ \Delta^2(a)=\sum a_{(1)}\otimes a_{(2)}\otimes a_{(3)}.$$ Here the usual notation for coproducts (cf. $ \Delta(a)=\sum a_{(1)}\otimes a_{(2)} $ ) is used. The resulting Hopf algebra proves to be quasitriangular with the universal $\cal R$-matrix $$ {\cal R}=\sum_{i}^{}(e_i\otimes1)\otimes(1\otimes e^i). $$ One easily finds that $gl_q(2)$ and other simple examples of quantum universal enveloping algebras are both the $\check{U}_R$-type algebras and quantum doubles. Can it happen that $\check{U}_R$ would be a quantum double for any $R$? Majid's approach based on the $\check{U}_R \leftrightarrow \check{A}_R$ duality does not readily answer this question. That is why we choose another way: not to use $A_R$ at all. The key observation is that there exists an inherent antiduality between $U^+_R$ and $U^-_R$ which is precisely of the form required for the quantum-double construction. \vspace{.5cm} {\bf 3}. Let us define bialgebras $U^+_R=\{1,l^+_{ij}\}$ and $U^-_R=\{1,l^-_{ij}\}$ by eqs. (2),(4). Note that the cross-multiplication relation (3) is not yet imposed, so $U^+_R$ and $U_R^-$ are considered to be independent so far. However, the very natural pairing between them can be introduced. It is generated by \begin{equation} =R^{-1}_{12}, ~~~=<{\bf 1},L^+>=<{\bf 1},{\bf 1}>=1 \end{equation} and in the general case looks like \begin{equation} \left\langle L^-_{i_1}\ldots L^-_{i_m},L^+_{j_1}\ldots L^+_{j_n}\right\rangle =R^{-1}_{i_1j_n}\ldots R^{-1}_{i_qj_p}\ldots R^{-1}_{i_mj_1} \ , \end{equation} where the r.h.s.\ is a product of $mn \ \ R^{-1}$-matrices corresponding to all pairs of indices $i_qj_p$ with $j$-indices ordered from right to left. The consistency of (8) and (5) with (4) is evident, while the proof of the consistency with (2) reduces to manipulations like \begin{eqnarray*} &=& \\ &=& R_{12}R^{-1}_{01}R^{-1}_{02}-R^{-1}_{02}R^{-1}_{01}R_{12}=0 \end{eqnarray*} and repeated use of QYBE (1). For general $R$, this pairing is degenerate. To remove the degeneracy, i.e.\ to transform pairing into antiduality, one should factor out appropriate bi-ideals \cite{MA}. In simple cases this procedure is explicitly carried out and works well. For general $R$ it is of course not under our control. The situation is quite similar to \cite{MA}: we are to rely on that the factorization procedure is ``soft'' in a sense that it does not destroy the whole construction. \vspace{.5cm} {\bf 4}. Keeping this in mind, we observe that, being antidual, $U_R^{\pm}$ admit the Hopf algebra structure. Let us introduce an antipode $S$ in $U_R^-$ and an inverse antipode $S^{-1}$ in $U^+_R$ by the relations \begin{equation} =<{\bf 1},S^{-1}(L^+)>={\bf 1},\ ==R_{12}, \end{equation} extending them on the whole of $U^+_R$ (or $U_R^-$) as antihomomorphisms of algebras and coalgebras. The definition is correct due to \begin{eqnarray*} &=&\\ =&=&R_{12}R_{12}^{-1} ={\bf 1}=<\varepsilon(L^-_1),L^+_2>, \end{eqnarray*} $$ = <\Delta\circ S(L_0^-),R_{12}(L^+_1\otimes L^+_2)-(L^+_2 \otimes L^+_1)R_{12}> $$ $$ ==R_{12}R_{02}R_{01}-R_{01}R_{02}R_{12}=0,$$ $$ = $$ $$ =<\!R_{12}(S(L_1^-)\otimes S(L_2^-))-(S(L_2^-)\otimes S(L_1^-))R_{12}, L^+_0\otimes L^+_0\!>=R_{12}R_{10}R_{20}-R_{20}R_{10}R_{12}=0, $$ but there is no such a formula for $S(L^+)$ or $S^{-1}(L^-)$. Once again the factorization is hoped to be soft enough to allow the antipodes to be invertible. If so, our bialgebras $U_R^{\pm}$ become the mutually antidual Hopf algebras $\check{U}_R^{\pm}$, and it is possible to define the multiplicative structure of the quantum double upon $\check{U}_R^+\otimes\check{U}_R^-$. The cross-multiplication rule is deduced from (6): \begin{eqnarray} L_1^-L^+_2R_{12}&=&L^+_2L_1^-R_{12}\nonumber\\ &=& R_{12}L^+_2L_1^-R^{-1}_{12}R_{12}=R_{12}L^+_2L_1^-. \end{eqnarray} Thus we regain eq.(3) as the quantum-double cross-multiplication condition! Our conclusion is that $R$-matrices obeying QYBE generate the algebraic structures of quantum double in quite a natural way. \vspace{.5cm} {\bf 5}. To illustrate the proposed scheme, consider the $sl_q(2)\ \ R$-matrix $$R_q=q^{-1/2}\left( \begin{array}{cccc} q&0&0&0\\0&1&q-q^{-1}&0\\0&0&1&0\\0&0&0&q\end{array} \right),\ \ \ R_q^{-1} =R_{q^{-1}}.$$ Here the bialgebras $U^{\pm}_{R_q}$ have 8 generators $l_{ij}^{\pm}$. The bi-ideals to be factored out are generated by the relations $$l^-_{21}=0,\ \ l^+_{12}=0,\ \ l^-_{11}l^-_{22}=l^-_{22}l^-_{11}=1,\ \ l^+_{11}l^+_{22}=l^+_{22}l^+_{11}=1.$$ After factorization the number of independent generators is reduced to 4. We denote them $X^{\pm},H,H^{\prime}$ (note that $H^{\prime}\neq H$ so far): $$L^+=\left( \begin{array}{cc} q^{H/2}&0\\(q^{1/2}-q^{-3/2})X^+&q^{-H/2} \end{array} \right), \ \ \ \ L^-=\left( \begin{array}{cc} q^{-H^{\prime}/2}&(q^{-1/2}-q^{3/2})X^-\\0&q^{H^{\prime}/2} \end{array} \right).$$ The multiplication rules (inside each algebra), coproducts and antipodes are: \begin{eqnarray*} [H,X^+]=2X^+&,&[H^{\prime},X^-]=-2X^-,\\ \Delta(H)=H\otimes1+1\otimes H&,&\Delta(H')=H'\otimes1+1\otimes H',\\ \Delta(X^+)=X^+\otimes q^{H/2}+q^{-H/2}\otimes X^+&,& \Delta(X^-)=X^-\otimes q^{H'/2}+q^{-H'/2}\otimes X^-,\\ S(X^{\pm})=-q^{\pm 1}X^{\pm}&,&S(H)=-H,\ S(H^{\prime})=-H^{\prime}. \end{eqnarray*} The quantum-double cross-multiplication rules (6) take the form \begin{eqnarray*} [H^{\prime},X^+]=2X^+&,& [H,X^-]=-2X^-,\ \ [H,H^{\prime}]=0, \\ \ [ X^+,X^-] &=& \left( q^{(H+H^{\prime})/2} -q^{-(H+H^{\prime})/2} \right)/(q-q^{-1}). \end{eqnarray*} The identification $H^{\prime}\equiv H$ leads to the ordinary $sl_q(2)$. \vspace{.5cm} {\bf 6}. To give one more illustration, let us consider a bialgebra introduced in \cite{LU}. In a slightly simplified form it has generators $\{1,t^i_j,u^i_j,E_j,F^i\}$ which obey the following relations (here we prefer to display all the indices): \begin{eqnarray} R^{ij}_{mn}\,t^m_p\,t^n_q=R^{mn}_{pq}\,t^j_n\,t^i_m &,& E_p\,t^j_q=R^{mn}_{pq}\,t^j_n\,E_m,\\ \Delta(t^i_j)=t^i_k\otimes t^k_j,\ \ \varepsilon(t^i_j)=\delta^i_j&,& \Delta(E_j)=E_i\otimes t^i_j+1\otimes E_j,\ \ \varepsilon(E_j)=0,\\ R^{ij}_{mn}\,u^m_p\,u^n_q=R^{mn}_{pq}\,u^j_n\,u^i_m &,& F^i\,u^j_p=R^{ji}_{mn}\,u^m_p\,F^n,\\ \Delta(u^i_j)=u^i_k\otimes u^k_j,\ \ \varepsilon(u^i_j)=\delta^i_j&,& \Delta(F^i)=F^i\otimes1+u^i_j\otimes F^j,\ \ \varepsilon(F^i)=0,\\ R^{ij}_{mn}\,u^m_p\,t^n_q=R^{mn}_{pq}\,t^j_n\,u^i_m &,& E_jF^i-F^iE_j=t^i_j-u^i_j,\\ u^i_p\,E_q=R^{mn}_{pq}\,E_n\,u^i_m&,& t^i_p\,F^j=R^{ji}_{mn}\,F^m\,t^n_p, \end{eqnarray} with $R$ obeying QYBE~(1). This is not a bialgebra of the form (2)-(4). Rather it is of the `inhomogeneous quantum group' type ~\cite{SC}. Let us make sure that it is a quantum double as well. Consider $T,E$-bialgebra (11),(12) and $U,F$-bialgebra (13),(14) firstly as being independent and fix nonzero pairings on the generators by \begin{equation} =R^{ip}_{jq},\ \ \ \ =<1,t^i_j>==\delta^i_j, \end{equation} extending them to the whole bialgebras with the help of~(5). The definition is correct due to \begin{eqnarray*} &=& \\ &=&R^{ij}_{kq}\,\delta^k_p-\delta^i_mR^{mn}_{pq}\,\delta^j_n=0,\\ &=& \\ &=&\delta^i_kR^{jk}_{pq}-R^{ji}_{mn}\,\delta^m_p\,\delta^n_q=0. \end{eqnarray*} After factoring out the corresponding null bi-ideals, we may define antipodes on the generators as follows: $$ =={(R^{-1})}^{ip}_{jq}, $$ $$ =<1,S^{-1}(t^i_j)>=\delta^i_j,\ \ S(F^i)=-S(u^i_j)F^j,\ \ S(E_j)=-E_iS(t^i_j). $$ The proof of correctness is in complete analogy with the $\check{U}_R$-case. Now a direct application of the recipe (6) exactly reproduces the cross-multiplication relations (15),(16). For example, $$ F^iE_j=<1,t^n_j>t^m_n+<1,t^m_j>E_mF^n $$ $$ +u^n_m=-t^i_j+E_jF^i+u^i_j, $$ because of $$ =-=-\delta^i_m. $$ Therefore, bialgebras of the type (11)-(14) are also transformed into the quantum double using our method. \vspace{.5cm} {\bf7}. Consider at last a bialgebra~\cite{BE} that is known to be related~\cite{CA} to bicovariant differential calculus on quantum groups. Its coalgebra structure is given by (12), whereas the multiplication relations (11) are to be supplemented by \begin{eqnarray} t^i_p\,E_q+f^i_{nm}\,t^n_p\,t^m_q&=&R^{nm}_{pq}\,E_m\,t^i_n+ f^n_{pq}\,t^i_n,\\ E_iE_j-R^{mn}_{ij}\,E_nE_m&=&f^m_{ij}\,E_m, \end{eqnarray} $f^i_{jk}$ being new structure constants. This bialgebra, unlike its ancestor (11),(12), exhibits the $R$-matrix-type representation \begin{equation} {\bf R}_{12}{\bf T}_1{\bf T}_2={\bf T}_2{\bf T}_1{\bf R}_{12},\ \ \Delta({\bf T})={\bf T}\otimes{\bf T}, \end{equation} where, in terms of multi-indices like $I=\{0,i\}$, $$ {\bf T}^I_J=\left( \begin{array}{cc}1&E_j\\0&t^i_j\end{array} \right),\ \ \ {\bf R}^{IJ}_{MN}=\left( \begin{array}{cccc} 1&0&0&0\\0&\delta^j_n&0&0\\0&0&\delta^i_m&f^i_{mn}\\0&0&0&R^{ij}_{mn} \end{array} \right). $$ Of course, ${\bf R}$ must satisfy the QYBE (1) which now involves the structure constants $f^i_{mn}$ as well as $R^{ij}_{mn}$. Note that, due to (18),(19), the bialgebra (11),(12) is not restored from (20) by mere setting $f^i_{mn}\equiv0$. Now let us try to develop a quantum double from the bialgebra (20). However, it seems to be quite uneasy task. A natural Ansatz for the candidate antidual bialgebra is $$ {\bf U}^I_J=\left( \begin{array}{cc} 1&0\\F^i&u^i_j \end{array} \right), $$ which causes the corresponding $R$-matrix to be $$ {\bf\overline{R}}^{IJ}_{MN}=\left( \begin{array}{cccc} 1&0&0&0\\0&\delta^j_n&0&0\\0&0&\delta^i_m&0\\0&\bar{f}_n^{ij}&0& R^{ij}_{mn} \end{array} \right) $$ with different structure constants $\bar{f}$ and another QYBE system involving $R$ and $\bar{f}$. Now, attempting to fix a pairing in the form \begin{equation} <{\bf U}_1,{\bf T}_2>={\bf Q}_{12} \end{equation} with a certain numerical matrix ${\bf Q}$, we immediately arrive at the following general statement: Let ${\bf R}$ and ${\bf\overline{R}}$ be invertible solutions of QYBE. If there exists an invertible solution ${\bf Q}$ of the equations \begin{eqnarray*} {\bf Q}_{12}{\bf Q}_{13}{\bf R}_{23}& =&{\bf R}_{23}{\bf Q}_{13}{\bf Q}_{12},\\ {\bf\overline{R}}_{12}{\bf Q}_{13}{\bf Q}_{23}& =&{\bf Q}_{23}{\bf Q}_{13}{\bf\overline{R}}_{12}, \end{eqnarray*} then (21) is a correct pairing between the ${\bf T}$-~and ${\bf U}$-bialgebras generated by ${\bf R}$ and $\bf\overline{R}$, respectively, and, assuming a proper quotienting procedure to be performed, the antipodes can be defined by the relations $$=<{\bf U}_1,S^{-1}({\bf T}_2)>= {\bf Q}^{-1}_{12} $$ and the quantum-double structure can be established on the tensor product of these bialgebras by the cross-multiplication formula $$ {\bf Q}_{12}{\bf U}_1{\bf T}_2={\bf T}_2{\bf U}_1 {\bf Q}_{12}.$$ Whether such a program can really be carried through in interesting cases (e.g. for $\bf R$ and $\bf\overline{R}$ given above) is the subject of further investigation. \vspace{.5cm} I wish to thank A.Isaev, R.Kashaev, A.Kempf and P.Pyatov for stimulating discussions. \begin{thebibliography}{99} \bibitem{FRT} L.D.\,Faddeev, N.Yu.\,Reshetikhin and L.A.\,Takhtajan, Algebra i Analiz\, 1 (1989) 178 (in Russian); English transl. in: Leningrad Math.\,J. 1 (1990) 193. \bibitem{MA} S.\,Majid, Int.\,J.\,Mod.\,Phys.\,A 5 (1990) 1. \bibitem{DR} V.G.\,Drinfeld, Quantum groups, in: ICM Berkeley-86 Proceedings, vol.\,1, p.\,798. \bibitem{BU} T.\,Tjin, Int.\,J.\,Mod.\,Phys.\,A 7 (1992) 6175. \bibitem{LU} M.\,L\"{u}dde, preprint BONN-HE-92-9. \bibitem{SC} M.\,Schlieker, W.\,Weich and R.\,Weixler, Z.\,Phys.\,C 53 (1992) 79. \bibitem{BE} D.\,Bernard, Phys.\,Lett.\,B 260 (1991) 389. \bibitem{CA} P.\,Aschieri and L.\,Castellani, preprint CERN-TH 6565\,/\,92, DFTT-22\,/\,92. \end{thebibliography} \end{document}