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I've a slightly technical question about the Yangian which I'm hoping an expert out there can answer.

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.

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.

Question 1: Is there an explicit description of the algebra $C(\mathfrak{g})$?

Let $C'(\mathfrak{g})$ be the classical analog of $C(\mathfrak{g})$, defined by replacing $Y(\mathfrak{g})$ in the above discussion with $U(\mathfrak{g}[z])$.

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).

Question 2: 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).

Thanks, Kevin

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  • $\begingroup$ I get used to think of integrable systems related to Y(g) in a little different fashion. Let me explain on the example of g[z]: U(g[z,z^-1]) has a huge center (modula tech. details), you project it to U(g[z]) and get a commutative subalgebra in U(g(z)) in suitable representation it gives Gaudin integrable spin chain (it also has many other names). Is there any relation between what you right and this construction ? ` $\endgroup$ Aug 24, 2012 at 20:11
  • $\begingroup$ What happens in the toy model case - consider just $g$ itself, not g[z], not Y(g). Is there any description of $C(g)$ ? I would simplify the question even more - better to consider the Poisson algebra $S(g)$. What happens in this case ? It seems the question is about Poisson (co?)-homology of g - it should be known... I think if understand this question it might clarify original, because the Y(g) has asymptotically faithful family of homomorhisms to $U(g)^{\otimes N}$ ? $\endgroup$ Aug 24, 2012 at 20:20
  • $\begingroup$ As a guess (suggested by connection with integrable systems) I would think that the answer as an algebra is free polynomial algebra with generators indexed by T_{i,k} k=1...\infty and i=1...n-1 for sl(n) , and "Coxeter" weights(?) in general. I mean same degrees as Z(U(g)) $\endgroup$ Aug 24, 2012 at 20:23
  • $\begingroup$ Thanks for your comments, Alexander. I guess that the algebra $C(g)$ must be the integrals of motion for the spin-chain (Drinfeld doesn't use the term spin-chain, but how many lattice models can there be controlled by the Yangian?) Drinfeld also says you can view $C(g)$ as a subalgebra of the Yangian, rather than as it's dual, using the $R$-matrix. So I'd guess this must be the same algebra as what you describe. It would be very helpful if you had a reference for the construction you describe of the integrals of motion for the spin-chain. $\endgroup$ Aug 25, 2012 at 4:29
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    $\begingroup$ I don't know much about this stuff. I have been told that one must be careful when comparing U(g[z,z^-1]) and U(g[z]), as for many situations the "correct" map between them is not z \mapsto z, but z \mapsto \exp(z). This probably doesn't effect the question at hand. $\endgroup$ Aug 25, 2012 at 13:14

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It seems that subalgebra in the question is the so-called "Bethe subalgebra" of the Yangian.

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 it should be 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)" leaves me with no doubt that this is Bethe subalgebra. (I somehow missed this point when I wrote the first version of this answer).

Some references

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.

In the mathematical literature it was introduced in "Bethe Subalgebras in Twisted Yangians", by Maxim Nazarov, Grigori Olshanski where the name "Bethe" subalgebra was proposed, Leningrad's constructions have been mathematically written up for $Y(gl_n)$ and for twisted Yangians. (Twisted Yangians for semisimple $g$, were introduced by Olshanski ~1989).

Surprisingly, a 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", by A. I. Molev.

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 $C_{i,k}$, $k=1,\dots,\infty$ (corresponds to loop variable $z^k$) and $i=1,\dots,rank(g)$ - corresponds to generators of the center of $U(g)$. I.e. the center of $U(g)$ can be "loopified" to get commutative subalgebra in $Y(g)$. They degenerate to appropriate subalgebras in $U(g[t])$.

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.

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" http://arxiv.org/abs/hep-th/9409025,
  2. Molev http://arxiv.org/abs/math/0211288, and
  3. Molev's boook "Yangians and Classical Lie Algebras." Mathematical Surveys and Monographs. Providence, RI: American Mathematical Society.

They do not cover the physics literature and some recent developments.

Quick facts. Everyone can understand.

Let me give some facts as abstract ones and later relate to the subalgebra in question.

Fact 1. (AKS-lemma (Adler-Kostant-Symmes)).

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 a commutative subalgebra in $B$. (Same for $C$). (Same for Poisson algebras).

Proof - pleasant exercise. Importance: all natural commutative subalgebras and integrable systems come in this way or around :)

Fact 2. (Poisson center of $S(g)$ vs. center of $S(g(t))$).

Consider Lie algebra $g$ and its loop algebra $g(t)$. How are the Poisson centers of $S(g)$ and $S(g(t))$ 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 $Z(S(g(t)))$.

Explanation: think of the loop space for $g^*$ as infinite product of copies of $g^*$ indexed by points of the circle $S^1$. The structure of the center of the finite product is clear. We just consider the infinite limit.

Proof: Exercise or if anyone is interested I can try to write it.

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).

Where do all "these" commutative subalgebras (integrable systems) come from?

Now it is pretty simple. Take Lie algebra $g$ (usually reductive).

Consider the loop algebra $g(t)$ and split it to $g[t], g[t^{-1}]$. From fact 2 above we have a huge center in $S(g(t))$. Let us project it to $S(g[t])$ - we get a commutative subalgebra there.

Variation on this theme - take quantum-super generalization of $g(t)$ - the same thing will work. E.g. Yangian case.

That is all.

What is difficult? Quantization, anomalies, Langlands correspondence

Everything works well when we consider the "classical world" i.e. work with Poisson brackets, not commutators.

We might ask what happens for $U(g(t))$, not for $S(g(t))$. Here the story becomes interesting - the moral is that everything will work, but not in the obvious way.

In physics it is related to "anomalies" "renormalization", change of $k\rightarrow k+2$ in Chern-Simons and WZ coupling constants.

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 the Langlands correspondence. The breakthrough is due to Dmitry Talalaev: http://arxiv.org/abs/hep-th/0404153

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 $.

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".

Well, over the complex numbers there is not a Galois group, but we should think about differential operators as a kind of representations of Galois group.

Partly thing like these are described in our paper http://arxiv.org/abs/hep-th/0604128 .

Talking about this "Langlands related" stuff one should mention works by Feigin and E. Frenkel. There are several surveys by E. Frenkel on the arXiv. I like the old one: http://arxiv.org/abs/q-alg/9506003 from which I learnt a lot.

Back to question. 1) Explicit description

Question 1: Is there an explicit description of the algebra $C(g)$?

Yes there is for classical $g$. For $Y(gl_n)$ I think we know it pretty well. For $Y(g)$ for classical $g$ this is recent work by Molev mentioned above, which is the starting point, hopefully there will be further works. For exceptional I do not know anything.

Let me write something on an explicit description. It is important to keep in mind two levels: Poisson and quantum. In the Poisson case everything is obvious for any $g$ (even exceptional).

To describe this explicitly let me remind you of the "matrix notations" - the Yangian for $gl_n$ can be described as $RTT=TTR$ where "$T(z)$" is a matrix which contains all generators of $Y(gl_n)$. It is called the "Lax matrix" or "transfer matrix", sometimes the "monodromy matrix".

In the Poisson case:

$Trace( T^k(z)) = \sum_i T_{k,i} z^i$

or

$det(l-T(z)) = \sum_{k,i} l^kz^i C_{k,i} $

$T_{k,i}$ and $C_{k,i}$ always (any $g$) Poisson commute among themselves (easy exercise for R-matrix calculations). Any set $T_{k,i}$ or $C_{k,i}$ provides a set of free generators of $C(Y(g))$ (well I forget pffaffian in $so(2n)$ case).

Quantum case

It is similar to quantum groups - instead of the determinant, one should use the q-determinant. In the Yangian case we do not have "q" but we must insert the shift in "z" e.g.

$Trace( T(z) T(z+1) T(z+2) ... T(z+k-1) ) = \sum_i T_{k,i} z^i$

This will give the generators of Bethe subalgebra - surprisingly it is recent ( http://arxiv.org/abs/0711.2236 ).

Talalaev's determinant:

$det^{column} ( 1- exp(-d/dz) T(z) ) = \sum_{k,i} exp(-l d/dz) z^i C_{k,i}$

This will give another set of generators of the Bethe subalgebra. The generators themselves are old (going back to the works of Faddeev's team in the late 1980's or maybe even earlier). Talalaev's insight is to introduce $\exp(-d/dz)$ and to obtain this difference GL-oper.

This allowed him to make a degeneration to $U(g[z])$.

He proved that $\det(d/dz - L(z))$ will give generators of the Bethe subalgebra in $U(g[z])$, they were not known before in a nice form.

The situation might be strange - in the quantum group case we've known how to describe the Bethe subalgebra for 20 years, but in the seemingly simpler case of $U(g[z])$ it was not known. However this is true - in some sense quantum groups easier to deal with than classical ones.

Let me mention that $e^{-d/dz} T(z)$ is a Manin matrix.

Back to question 2): Is there a PBW theorem for $C(g)$, stating that $C(g)$ has a filtration whose associated graded is $C'(g)$?

For $gl_n$ this should be contained in Nazarov-Olshanski and the survey MNO mentioned above.

For classical $g$ see the recent paper of Molev mentioned above. (I am not sure about $so(2n)$ and pffafian).

For exceptional I think it is not known. Maybe the Feigin-Frenkel technique works here - they do not need explicit formulas.

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