Quantization of conjugacy classes in a Lie group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:49:12Z http://mathoverflow.net/feeds/question/67328 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67328/quantization-of-conjugacy-classes-in-a-lie-group Quantization of conjugacy classes in a Lie group unknown (google) 2011-06-09T00:41:06Z 2011-06-10T14:49:12Z <p>Let $G$ be a Lie group (and to be safe, let's assume it is semisimple). Consider the action of $G$ on itself by conjugation, and form the GIT (algebro-geometric) quotient $G/\!/G$. Then let $\pi:G\to G/\!/G$ denote the quotient map. For example, if $G=SL(2,\mathbb C)$, then $G=\operatorname{Spec}\mathbb C[a,b,c,d]/(ad-bc-1)$ and $G/\!/G=\operatorname{Spec}\mathbb C[t]$, and the projection $\pi$ is induced from the map of rings which sends $t$ to $a+d$ (all I'm saying is that the trace determines the conjugacy class in $SL(2,\mathbb C)$, at least for semisimple ones).</p> <p>There is a well-known closed $2$-form on $G$ whose symplectic leaves are exactly the fibers of $\pi$ (i.e. the conjugacy classes). It can be defined, for example, as follows. Let $g\in G$, and denote its conjugacy class by $G_g$. Then $T_gG_g$ is exactly the image of the map $R:\mathfrak g\to\mathfrak g$ given by $\alpha\mapsto\alpha-\operatorname{ad}_g\alpha$. Now the map $\mathfrak g\otimes\mathfrak g\to\mathbb C$ given by $\alpha\otimes\beta\mapsto\operatorname{tr}(g[\alpha,\beta])$ has exactly the same kernel as $R$, and thus it descends to give an alternating bilinear form on $T_gG_g$ (by $\operatorname{tr}$ I mean trace in the adjoint representation). Some additional arguments show that this form is closed.</p> <p>This construction is sometimes done instead with $\mathfrak g^\ast\to\mathfrak g^\ast/\!G$, though I'm more interested in the case $G\to G/\!/G$.</p> <p>There are many papers which study the deformation quantization (or geometric quantization) of (the leaves of) this Poisson structure on $G$, but everything is very VERY abstract. Is there any place where I can find explicit formulae for things like:</p> <p>1) the Poisson bracket of standard functions on $G$ (e.g. the coordinates $a,b,c,d$ in the case $G=SL(2,\mathbb C)$).</p> <p>or even better</p> <p>2) the deformed product (star product) on $\mathcal O(G)$.</p> <p>?</p> <p>I feel like these should be known or easy to calculate, but an answer has eluded my search.</p> <p>I've tried doing some explicit calculations myself, but everything seems to hinge on getting a succinct formula for the $2$-form on $G$, and I always end up with something horrendous.</p> http://mathoverflow.net/questions/67328/quantization-of-conjugacy-classes-in-a-lie-group/67349#67349 Answer by Theo Johnson-Freyd for Quantization of conjugacy classes in a Lie group Theo Johnson-Freyd 2011-06-09T14:14:44Z 2011-06-10T14:49:12Z <p>I won't really answer your questions, but I'd like to remind other readers that much of what you say is somewhat special to the semisimple case (well, semisimplicity is a little stronger than necessary).</p> <p>There is a very general notion of <em>Poisson Lie group</em> $G$, which is a Lie group and a Poisson manifold such that the multiplication map $G \times G \to G$ is Poisson (i.e. the map $\mathcal C^\infty(G) \to \mathcal C^\infty(G\times G)$ is a Lie algebra homomorphism for the Poisson bracket). Poisson Lie groups (at least, the simply-connected ones) are in natural bijection with <em>Lie bialgebras</em>. Then something special happens in the semisimple case. Namely, there is a "standard Lie bialgebra structure", which I think requires a choice of upper- and lower- triangular decomposition — these choices are all related by conjugation in the group, so in particular are all isomorphic, but the assignment of a Poisson structure to a semisimple Lie group is not functorial, I think. </p> <p>For example, here's the Poisson structure for the matrix group $\mathrm{SL}(2)$. In the standard coordinates $a,b,c,d$ you give, the brackets are (up to a normalization): $$\lbrace a, b\rbrace = ab, \quad \lbrace a, c\rbrace = -ac,\quad \lbrace a, d\rbrace = 2bc, \quad \lbrace b,c\rbrace = 0, \quad \lbrace b,d\rbrace = -bd, \quad \lbrace c, d\rbrace = cd.$$ See for example Lecture 7 from my <em>very unedited</em> notes on Reshetikhin's class on quantum groups at Berkeley, Spring 2009 (<a href="http://math.berkeley.edu/~theojf/QuantumGroups.pdf" rel="nofollow">PDF</a>).</p> <p>But this is not, I think, the Poisson structure that you're after. Here is another general fact about Poisson Lie groups. Since Lie bialgebras are completely symmetric definition, to every Poisson Lie group $G$ there is a <em>dual Lie group</em> $G^\ast$, which is the simply-connected Poisson Lie group with the dual Lie bialgebra. Then it is always true that $G$ acts on $G^\ast$ and $G^\ast$ acts on $G$, although these actions can be complicated in general (in particular, in general $G$ does not act by automorphisms of $G^\ast$). The actions are called the <em>Dressing action</em>, which is a name from the 1970s and the theory of soliton waves. A theorem of Lu, Semenov-Tian-Shansky, and Weinstein holds that the symplectic leaves in $G$ are precisely the orbits under the Dressing action of $G^\ast$ on $G$. In the case of $G = \mathrm{SL}(2)$, the dual group is </p> <p>$$G^\ast = \left\lbrace \left( \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix}, \begin{pmatrix} a^{-1} & 0 \\ c & a \end{pmatrix} \right) \right\rbrace$$</p> <p>the group of pairs of matrices with the above shapes, i.e. it is a fibered product of the two Borel subgroups of $\mathrm{SL}(2)$ over the Cartan.</p> <p>Of course, a special case of the Dressing action is when the Poisson structure on $G$ is $0$. Then the dual Poisson group $G^\ast$ is abelian, and it is precisely the vector space $\mathfrak g^\ast$ with the Kirillov-Kostant-Souriau Poisson structure (the algebra of functions on $\mathfrak g^\ast$ is $\operatorname{Sym}(\mathfrak g)$, or some completion thereof, and the Poisson bracket is what you get by extending the bracket on $\mathfrak g$ as a biderivation). Then the Dressing action is precisely the coadjoint action.</p> <p>So it seems that I have been spending a while not answering any of your questions. To get closer to your question, construct the dual group $G^\ast$. Construct also the <em>double</em> group $D = G^\ast \bowtie G$. It is the simply-connected Lie group with Lie algebra the double of $\mathfrak g$; there are inclusions of Poisson Lie groups $G^\ast \hookrightarrow D \hookleftarrow G$, and as a manifold $G = G^\ast \times G$ (this should remind you of the "KAN" factorization of matrices).<br> Now I'm going to again specialize to the semisimple case: $G$ is a semisimple Lie group with its standard Poisson structure. Then $D$ is isomorphic as a Lie group to $G\times G$ — this is not true for general Poisson Lie groups — and the inclusion $G \hookrightarrow D$ is the diagonal inclusion (see Lecture 17 of <em>op cit</em>). Moreover, there is a projection of manifold $D = G\times G \to G$. <strong>Edit:</strong> Thus, the composition $$G^\ast \to D = G\times G \to G$$ <strike>is an isomorphism</strike> exists as a map of manifolds. In fact, more can be said: the Dressing action of $G$ on $G^\ast$ is defined by the adjoint action of $G$ on $D$, and the projection is also compatible. So this <strike>isomorphism</strike> map $G^\ast \to G$ of manifolds intertwines the two $G$-actions.</p> <p>So take the Poisson structure on $G^\ast$, which is the thing whose leaves are given by a $G$-action, and push it through the isomorphism to a Poisson structure on $G$. This is the one you're after.</p> <p>I don't understand the map $G^\ast \to G$, so I can't work it out. But I can tell you the Poisson structure on $G^\ast$, in the case that $G = \mathrm{SL}(2)$ with its standard structure. I've already defined the group $G^\ast$ above. In those coordinates, the Poisson brackets are (up to a normalization): $$\lbrace a,b\rbrace = ab, \quad \lbrace a,c\rbrace = -ac, \quad \lbrace b,c \rbrace = -2(a^2 - a^{-2})$$</p> <p>So this gets close to answering your question (1). <strike>At least, I've given some coordinates on $\mathrm{SL}(2)$ and a Poisson structure therein for which the adjoint action picks out the symplectic leaves.</strike> I'm guessing someone who spends more time with simple groups (I generally think about general Lie groups and Lie algebras, and not the simple ones, because there are too many coincidences there for me to see the general structure) will be able to give you a better formula in the standard coordinates, but I cannot.</p> <hr> <p>Maybe I'll make one more comment. From the point of view of $\mathrm{SL}(2)$, this Poisson structure doesn't play very well with the group structure (it doesn't have a lot to do with the "standard" Poisson-Lie structure, for example; the relationship is analogous to the different group structures on $G$ given by the KAN decomposition, or the PBW theorem that $\mathrm U \mathfrak g = \mathrm U \mathfrak n_- \otimes \mathrm U \mathfrak b_+$ as vector spaces). So if you treat it just as a Poisson manifold and ask for quantizations thereof, you can do this with Kontsevich star product, which is at least (locally) given by completely explicit formulas. But star product is not particularly functorial, and in particular does not preserve group objects (or at least isn't known to do so). In some sense you can do better by adopting some version of Etingof-Kazhdan, and quantizing the Poisson Lie group $\mathrm{SL}(2)^\ast$ to a Hopf algebra. But the formulas are much worse (I don't know if they're ever really been written down). <strong>Edit:</strong> Or, as pointed out in the comments, both $\mathrm{SL}(2)$ and $\mathrm{SL}(2)^\ast$ have quantizations that can be essentially read off from the Drinfel'd-Jimbo quantum group, which is completely explicit.</p> http://mathoverflow.net/questions/67328/quantization-of-conjugacy-classes-in-a-lie-group/67362#67362 Answer by DamienC for Quantization of conjugacy classes in a Lie group DamienC 2011-06-09T18:06:54Z 2011-06-09T18:06:54Z <p>Joseph Donin and Andrei Mudrov have worked a lot on this question... the quantization of conjugacy classes seems to be related to dynamical r-matrices and and the so-called reflection equation. See e.g. <a href="http://webdoc.sub.gwdg.de/ebook/serien/e/mpi_mathematik/2005/3.pdf" rel="nofollow">this paper</a> of Mudrov (<em>Quantum conjugacy classes of simple matrix groups</em>)... there are explicit formulae in the end of the paper, but I am not sure this answers your question (other explicit formulae for $GL_n$ can also be found in <a href="http://arxiv.org/abs/math/0206049" rel="nofollow">this older paper</a> by Donin and Mudrov). </p>