Question about the Yangian - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:09:45Z http://mathoverflow.net/feeds/question/105399 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105399/question-about-the-yangian Question about the Yangian Kevin Costello 2012-08-24T17:12:00Z 2012-08-25T20:05:26Z <p>I've a slightly technical question about the Yangian which I'm hoping an expert out there can answer.</p> <p>Recall that the Yangian $Y(\mathfrak{g})$ is a Hopf algebra quantizing $U(\mathfrak{g}[z])$. Drinfeld, in his quantum groups paper, explains that the algebra of integrals of motion of certain integrable lattice models can be described in terms of the Yangian, as follows. </p> <p>Let $C(\mathfrak{g})$ be the algebra of linear maps $$l : Y(\mathfrak{g}) \to \mathbb{C}$$ with $l([a,b] ) = 0$. In other words, $C(\mathfrak{g})$ is the linear dual of the zeroth Hochschild homology of $Y(\mathfrak{g})$. The product on $C(\mathfrak{g})$ comes from the coproduct on the Yangian, and the existence of the $R$-matrix implies that $C(\mathfrak{g})$ is commutative.</p> <p><b>Question 1:</b> Is there an explicit description of the algebra $C(\mathfrak{g})$? </p> <p>Let $C'(\mathfrak{g})$ be the classical analog of $C(\mathfrak{g})$, defined by replacing $Y(\mathfrak{g})$ in the above discussion with $U(\mathfrak{g}[z])$. </p> <p>One can compute that $C'(\mathfrak{g})$ is the algebra of $\mathfrak{g}[z]$-invariant formal power series on $\mathfrak{g}[z]$ (invariant under the adjoint action). </p> <p><b>Question 2:</b> Is there a PBW theorem for $C(\mathfrak{g})$, stating that $C(\mathfrak{g})$ has a filtration whose associated graded is $C'(\mathfrak{g})$? (This is equivalent to asking if part of the spectral sequence computing Hochschild homology of the Yangian from Hochschild homology of $U(\mathfrak{g}[z])$ degenerates). </p> <p>Thanks, Kevin</p> http://mathoverflow.net/questions/105399/question-about-the-yangian/105460#105460 Answer by Alexander Chervov for Question about the Yangian Alexander Chervov 2012-08-25T12:27:35Z 2012-08-25T20:05:26Z <p>It seems that subalgebra in the question is the so-called "Bethe subalgebra" of the Yangian. </p> <p>It is not immediate for me to recognize the connection with definition given in the question and the definition I'll give below - but I am sure that should simple and well-known. Can someone clarify this ? However the sentence "One can compute that C′(g) is the algebra of g[z]-invariant formal power series on g[z] (invariant under the adjoint action)" makes me no doubt that this is Bethe subalgebra. (I somehow missed this point when I wrote the first version of answer). </p> <h2>Some references</h2> <p>The subalgebra originates from the work of Faddeev's school on quantum integrable systems, and many things were known and "obvious" for Leningrad's team and it is not always easy to provide appropriate references.</p> <p>In mathematical literature it was introduced in "Bethe Subalgebras in Twisted Yangians" Maxim Nazarov, Grigori Olshanski <a href="http://arxiv.org/abs/q-alg/9507003" rel="nofollow">http://arxiv.org/abs/q-alg/9507003</a> where the name "Bethe" subalgebra was proposed, Leningrad's constructions has been mathematically written up for Y(gl_n) and for twisted Yangian. (Twisted Yangians for semisimple g, were introduced by Olshanski ~1989)</p> <p>Surprisingly similar construction for Yangians(g) for classical $g$ (not twisted Yangian, not gl_n/sl_n) , is quite recent (as far as I understand): "Feigin-Frenkel center in types B, C and D" A. I. Molev <a href="http://arxiv.org/abs/1105.2341" rel="nofollow">http://arxiv.org/abs/1105.2341</a></p> <p>The moral of the papers is that these subalgebras are very similar to the center of U(g). They are free commutative algebras with generators which can be indexed by C_{i,k} k=1...inf (corresponds to loop variable z^k) and i=1...rank(g) - corresponds to generators of the center of U(g). I.e. center of U(g) can be "loopified" to get commutative subalgebra in Y(g). They degenerate to appropriate subalgebras in U(g[t]).</p> <p>Small detail: you can work with either U(tg[t]) or U(g[t]). In the first case you get maximal commutative subalgebra, in the second it is not maximal and you can add arbitrary constant matrix to make it maximal. I mean morally it is maximal, but there are details.</p> <p>Similar subalgebras are in U_q(g[t]), elliptic algebras, many twisted versions - physics literature devoted to "Bethe ansatz" - the method to find joint eigenpairs is enourmous. Concerning the Yangian: there are surveys 1) "MNO" <a href="http://arxiv.org/abs/hep-th/9409025" rel="nofollow">http://arxiv.org/abs/hep-th/9409025</a>, 2) Molev <a href="http://arxiv.org/abs/math/0211288" rel="nofollow">http://arxiv.org/abs/math/0211288</a>, and 3) Molev's boook Yangians and Classical Lie Algebras. Mathematical Surveys and Monographs. Providence, RI: American Mathematical Society . They do not cover physics literature and some recent developments. </p> <h2>Quick facts. Everyone can understand.</h2> <p>Let me give some facts as abstract ones and later relate to the subalgebra in question.</p> <p>Fact 1. (AKS-lemma (Adler-Kostant-Symmes)).</p> <p>Consider any associative algebra A, and its two subalgebras $B,C$. Assume $A$ is isomorphic to $B\otimes C$ as a vector space. Then the projection of center of $A$ to $B$ gives commutative subalgebra in $B$. (Same for $C$). (Same for Poisson algebras).</p> <p>Proof - pleasant exercise. Importance: all natural commutative subalgebras and integrable systems come in this way or around:)</p> <p>Fact 2. (Poisson center of $S(g)$ vs. center of $S(g(t))$).</p> <p>Consider Lie algebra $g$ and its loop algebra $g(t)$. How the Poisson centers of $S(g)$ and $S(g(t))$ are related ? It is simple (if I am not mistaking:) - from any element $C$ of $Z(S(g))$ one can make a generating function $C(t)= \sum t^iC_i$ for elements of the center of $Z(S(g(t)))$. </p> <p>Explanation: think of the loop space for g^* as infinite product of copies of g^* indexed by points of the circle S^1. Structure of the center of the finite product is clear. We just consider infinite limit. </p> <p>Proof: Exercise or if anyone is interested I can try to write it.</p> <p>The same works not for only for g^*, but for any Poisson manifold. (Open question - what happens when quantizing ? What is general "anomaly cancellation = critical level"? (This MO question is related: <a href="http://mathoverflow.net/questions/95240/what-is-the-center-of-u-gt-for-g-not-reductive-lie-algebra" rel="nofollow">http://mathoverflow.net/questions/95240/what-is-the-center-of-u-gt-for-g-not-reductive-lie-algebra</a>).</p> <h2>Where all "these" commutative subalgebras (integrable systems) comes from ?.</h2> <p>Now it is pretty simple. Take Lie algebra $g$ (usually reductive).</p> <p>Consider loop algebra $g(t)$ and split it to $g[t], g[t^{-1}]$. From fact 2 above we have huge center in $S(g(t))$. Let us project it to $S(g[t])$ - we get commutative subalgebra there.</p> <p>Variation on this theme - take quantum-super generalization of $g(t)$ - same will work. E.g. Yangian case.</p> <p>That is all.</p> <h2>What is difficult ? Quantization, anomalies, Langlands correspondence</h2> <p>Everything works well when we consider the "classical world" i.e. work with Poisson brackets, not commutators.</p> <p>We might ask what happens for $U(g(t))$, not for $S(g(t))$. Here the story becomes interesting - moral is that everything will work, but not in the obvious way.</p> <p>In physics it is related to "anomalies" "renormalization", change of k->k+2 in Chern-Simons and WZ coupling constants.</p> <p>The miracle is that - if we will think properly how to construct the center of $U(g(t))$ then we will come to a form of Langlands correspondence. The breakthrough is due to Dmitry Talalaev: <a href="http://arxiv.org/abs/hep-th/0404153" rel="nofollow">http://arxiv.org/abs/hep-th/0404153</a></p> <p>The key point is that generators of the center of $Z(U(g(t))$ should be organized in the differential operator of order rank(g): $\sum_i C_i(z) (\partial_z)^i$.</p> <p>So we have a form of Langlands correspondence: take an irrep $V$ (automorphic side); center acts on $V$ by scalars so we get $\sum_i C_i(z)|_{V} (d_z)^i$ - scalar differential operator. Differential operators should be thought as "Galois side" - their monodromy gives the "representations of the Galois group". </p> <p>Well, over complex numbers there is not Galois group, but should thought about differential operators as a kind of representations of Galois group.</p> <p>Partly thing like these are described in our paper <a href="http://arxiv.org/abs/hep-th/0604128" rel="nofollow">http://arxiv.org/abs/hep-th/0604128</a> .</p> <p>Talking about this "Langlands related" stuff one should mention works by Feigin and E. Frenkel. There are several surveys by E. Frenkel in arXiv. I like the old one: <a href="http://arxiv.org/abs/q-alg/9506003" rel="nofollow">http://arxiv.org/abs/q-alg/9506003</a> from which I learnt a lot.</p> <h2>Back to question. 1) Explicit description</h2> <p>Question 1: Is there an explicit description of the algebra C(g)?</p> <p>Yes there is for classical $g$. For Y(gl_n) I think we know pretty well. For Y(g) for classical $g$ this is recent work by Molev mentioned above, which is starting point, hopefully there will be further works. For exceptional I do not know anything.</p> <p>Let me write something on explicit description. It is important to keep in mind to levels: Poisson and quantum. In Poisson case everything is obvious for any $g$ (even exceptional).</p> <p>To describe this explicitly let me remind the "matrix notations" - Yangian for gl_n can be described as $RTT=TTR$ where "T(z)" is a matrix which contain all generators of Y(gl_n). It is called "Lax matrix" or "transfer matrix" sometimes "monodromy matrix".</p> <blockquote> <p>In Poisson case: </p> <p>$Trace( T^k(z)) = \sum_i T_{k,i} z^i$</p> </blockquote> <p>or </p> <blockquote> <p>$det(l-T(z)) = \sum_{k,i} l^kz^i C_{k,i}$</p> </blockquote> <p>T_{k,i} and C_{k,i} always (any $g$) Poisson commute among themselves (easy exercise for R-matrix calculations). Any set T_{k,i} or C_{k,i} provide set of free generators of C(Y(g)) (well I forget pffaffian in so(2n) case). </p> <blockquote> <p>Quantum case</p> </blockquote> <p>It is similar to quantum groups - instead of determinant one should use q-determinat. In Yangian case we do not have "q" but we must insert the shift in "z" e.g.</p> <blockquote> <p>Trace( T(z) T(z+1) T(z+2) ... T(z+k-1) ) = \sum_i T_{k,i} z^i</p> </blockquote> <p>This will give generators of Bethe subalgebra - surprisingly it is recent ( <a href="http://arxiv.org/abs/0711.2236" rel="nofollow">http://arxiv.org/abs/0711.2236</a> )</p> <p>Talalaev's determinat: </p> <blockquote> <p>$det^{column} ( 1- exp(-d/dz) T(z) ) = \sum_{k,i} exp(-l d/dz) z^i C_{k,i}$ </p> </blockquote> <p>This will give another set of generators of Bethe subalgebra. The generators themselves are old (goes back to Faddev's team works in late 198* or may be even earlier). Talalaev's insight is to introduce $exp(-d/dz)$ and to obtain this difference GL-oper. </p> <p>This allowed him to make degeneration to U(g[z]).</p> <p>He proved that det(d/dz - L(z)) will give generators of the Bethe subalgebra in U(g[z]), they were not know before in the nice form.</p> <p>Situation might be strange - in quatum group case we know how to describe the Bethe subalgebra for 20 years, but in seemingly more simple case of U(g[z]) it was not known. However this is true - in some sense quantum groups more easy to deal than classical ones.</p> <p>Let me mention that $e^{-d/dz} T(z)$ is a Manin matrix ( <a href="http://en.wikipedia.org/wiki/Manin_matrices" rel="nofollow">http://en.wikipedia.org/wiki/Manin_matrices</a> ) </p> <h2>Back to question. 2) Is there a PBW theorem for C(g), stating that C(g) has a filtration whose associated graded is C′(g)?</h2> <p>For gl_n this should be contained in Nazarov Olshanski and survey MNO mentioned above.</p> <p>For classical $g$ is recent Molev's paper mentioned above. ( I am not sure about so(2n) and pffafian).</p> <p>For exceptional I think it is not known. May be Feigin-Frenkel technique works here - they do not need explicit formulas. </p>