The Jucys-Murphy elements of the group algebra of a finite symmetric group (here's the definition in Wikipedia) are known to correspond to operators diagonal in the Young basis of an irreducible representaion of this group. As one can see from the Wikipedia entry, all of the elements of such a diagonal matrix (in other words the operator's eigenvalues) are integers.

I'm looking for a simple way of explaining this fact (that the eigenvalues are wholes). By simple I mean without going into more or less advanced representation theory of the symmetric group (tabloids, Specht modules etc.), so trying to prove the specific formula given in Wikipedia is not an option. (I'm considering the Young basis as the Gelfand-Tsetlin basis of the representation for the inductive chain of groups $S_1\subset S_2\subset \ldots\subset S_n$, which is uniquely defined thanks to this chain's spectrum's simplicity, not as a set of vectors in correspondence with the standard tableaux.)

In fact, I'm trying to prove the first statement ($a_i\in \mathbb{Z}$) of proposition 4.1 in this article.

  • 3
    $\begingroup$ I myself have asked this question to several people to no avail. Not only are the eigenvalues integers; we also have $\prod\limits_{i=-n+1}^{n-1}\left(X-i\right)=0$, where $X=\left(1,n\right)+\left(2,n\right)+...+\left(n-1,n\right)\in\mathbb Z\left[S_n\right]$ is the $n$-th YJM element. I am sure this has a combinatorial proof, probably even a smart elementary induction one - but I had no success whatsoever in finding one over several weeks. $\endgroup$ – darij grinberg Dec 10 '11 at 23:00
  • 2
    $\begingroup$ I'm sorry for the off topic, but I had to say this. Don't know if you remember, but I know you, me and my twin brother Anton visited you at your home with our parents some 8-10 years ago when we (and you) lived in Karlsruhe. I'm in my 5th year at MSU now =) Quite a coincidence that you were the first person to respond to my first question here. $\endgroup$ – imakhlin Dec 10 '11 at 23:23
  • $\begingroup$ Oh hi! So we meet again. I've just started studying for a PhD at MIT. As you see by the comment I'm doing some algebraic combinatorics, at least as a pastime. Are you, too, or do you need this for some kind of infinite symmetric groups / probability theory? $\endgroup$ – darij grinberg Dec 11 '11 at 15:03
  • $\begingroup$ Well, actually I have been reading this article simply with the purpose of educating myself on the subject of representation theory, which (I guess) is my main field of interest. But there's a lot of combinatorics to it, yes. $\endgroup$ – imakhlin Dec 11 '11 at 15:55

This can be shown using the following two facts:

  1. $X_n=(1,n)+(2,n)+\ldots+(n-1,n)$ commutes with any element of $\mathbb Z S_{n-1}$
  2. Any irreducible $\mathbb Q S_n$-module $V$ restricts to a multiplicity-free $\mathbb Q S_{n-1}$-module (this follows from the classical branching rule; of course you said you didn't want to use tableaux's etc., so I'm not entirely sure whether this is compatible with your idea of "elementary").

The first one implies that $$X_n\in End_{\mathbb Q S_{n-1}}(Res^n_{n-1}V)$$ for any irreducible $\mathbb Q S_n$-module $V$. But, due to the second point and the fact that $\mathbb Q$ is a splitting field for $S_{n-1}$, there is an isomorphism of algebras $$ End_{\mathbb Q S_{n-1}}(Res^n_{n-1}V) \cong \mathbb Q \oplus \ldots \oplus \mathbb Q $$ This shows that $X_n$ acts semisimply on $V$ with eigenvalues in $\mathbb Q$. That the eigenvalues lie in $\mathbb Z$ then actually follows from the fact that $\mathbb Z S_n$ is a $\mathbb Z$-order (elements of $\mathbb Z$-orders have integral characteristic polynomial and therefore their eigenvalues will be integral over $\mathbb Z$ no matter on what module we let them act).

The assertion for all JM elements reduces to this (hope this is clear).

  • 1
    $\begingroup$ A shrewd argument, but I'm not sure this is what Igor is searching for (it is, at least, not what I was searching for). When I asked Pavel Etingof for a proof, he came up with a slightly similar proof (he actually computed the eigenvalues with multiplicities using tableau combinatorics - and that's not more tableau combinatorics than needed to prove the branching rule). This is problem 4.52 (a) in his Introduction to representation theory ( arxiv.org/abs/0901.0827 ). What I am looking for (but may be unrealistic) is a completely elementary argument without tableaux or representations. $\endgroup$ – darij grinberg Dec 11 '11 at 15:08
  • $\begingroup$ Thank you very much, this is the kind of argument I was looking for. The restrictions being multiplicity-free are within my understanding of "simple", actually this is what I meant by "the chain's spectrum's simplicity" in my question. That fact is a corollary of the centralizer $Z(\mathbb{C}[S_n],\mathbb{C}[S_{n-1}])$ being commutative. $\endgroup$ – imakhlin Dec 11 '11 at 15:41
  • $\begingroup$ Anyway, it would be very nice to have an elementary combinatoric proof of the identity provided by Darij in his first comment. $\endgroup$ – imakhlin Dec 11 '11 at 15:44
  • 4
    $\begingroup$ Alternate proof that the eigenvalues are integers: They are eigenvalues of a matrix with integer entries, so they are algebraic integers, and you just showed that they are rational. $\endgroup$ – David E Speyer Dec 19 '11 at 15:09

I came up with a more or less elementary proof of the identity from the top comment to my question. It involves nothing more advanced than some basic linear algebra.

Namely, let us denote $X_k=\sum\limits_{i=1}^{k-1} (i,k) \in\mathbb{C}[S_n]$, then we are to prove $$\prod\limits_{i=-k+1}^{k-1}(X_k-i)=0$$ for all $1\le k\le n$. For convenience sake we will also use $X_k$ to denote the linear operator on $\mathbb{C}[S_n]$ of left multiplication by $X_k \in \mathbb{C}[S_n]$.

First of all, let us show that $X_k$ is a diagonalizable operator. There are many ways to prove this fact, for example it is easy to see that $X_k$'s matrix is symmetric in the standard basis consisting of all the elements of $S_n$ (since the matrix corresponding to any $(i,k)$ is obviously such).

With the diagonalizability taken into account it is now sufficient to prove that $X_k$'s spectrum is a subset of $\{-k+1,-k+2,\ldots,k-1\}$. Starting with $X_1=0$ we conduct by induction on $k$.

Suppose that $\lambda\not\in\{-k,\ldots,k\}$ is an eigenvalue of $X_{k+1}$. $X_{k+1}$ commutes with all of $\mathbb{C}[S_k]$ including $X_k$, which implies that $X_k$ and $X_{k+1}$ are simultaneously diagonalizable. Thus exists such a nonzero $v\in\mathbb{C}[S_n]$ that $X_{k+1}v=\lambda v$ and $X_kv=\mu v$. Our choice of $\lambda$ together with the inductive hypothesis provides $(\lambda-\mu)\not\in\{-1,0,1\}$ which lets us consider the element $$u=\left(s_k-\frac1{\lambda-\mu}\right)v$$ where $s_k=(k,k+1)$. $u\neq0$, otherwise we would have $s_kv=\frac1{\lambda-\mu}v\implies \lambda-\mu=\pm1$ since $s_k^2v=v$. Finally $$X_ku=X_ks_kv-\frac1{\lambda-\mu}X_kv=(s_kX_{k+1}-1)v-\frac\mu{\lambda-\mu}v=\lambda s_kv-v-\frac\mu{\lambda-\mu} v=\lambda u$$ where we employ the easily obtainable $s_kX_{k+1}=X_ks_k+1$. However $\lambda$ being an eigenvalue of $X_k$ contradicts the inductive hypothesis due to our choice of $\lambda$.

  • $\begingroup$ "is a an eigenvalue" is a typo. This argument is very much in the spirit of Okounkov-Vershik, but way simpler than what they are doing (then again I might be misunderstanding them; the paper is not among the most readable...). For a representation theorist, it is a very natural argument. $\endgroup$ – darij grinberg Dec 16 '11 at 3:12
  • 1
    $\begingroup$ Thanks for the typo. Yes, this proof was definitely inspired by their paper. And yes, as one can see from this question, they made an unexpected choice of statements to be left unproven. $\endgroup$ – imakhlin Dec 16 '11 at 7:27

I will show that the eigenvalues of $X_{k+1}$ lie in the interval $[-k, k]$. Since Florian has already given a nice proof that the eigenvalues are integers, this answers your question.

Lemma: Let $A$ be a symmetric matrix with nonnegative entries whose rows and columns all add up to $k$. Then, for any real vector $v$, we have $-k \langle v,v \rangle \leq \langle v, Av \rangle \leq k \langle v, v \rangle$.

Proof: For any vector $v$, we have $$\langle v, Av \rangle = k \sum v_i^2 - \sum_{i<j} A_{ij} (v_i-v_j)^2 \leq k \sum v_i^2 = k \langle v,v \rangle$$ as desired. The equality $\langle v, Av \rangle = -k \sum v_i^2 + \sum_{i<j} A_{ij} (v_i+v_j)^2$ proves the other direction. $\square$

Now, the matrix of $X_{k+1}$ acting on the regular representation clearly obeys the conditions of the lemma. Taking $v$ to be an eigenvector with eigenvalue $\lambda$, we deduce that $-k \leq \lambda \leq k$, as desired.

  • 2
    $\begingroup$ Nice proof, but I think your $X_k$ is Igor's $X_{k+1}$. $\endgroup$ – darij grinberg Dec 16 '11 at 3:05
  • $\begingroup$ Good catch! Fixed now. $\endgroup$ – David E Speyer Dec 16 '11 at 13:46

Awfully sophisticated proof for the fact :) Just to relate it with the question about Knizhnik-Zamolodchikov equation:

Find polynom p(z) with values in C[S_n] such that p'(z) = \sum_i (Id+(1i))/(z-i) p(z). [Knizhnik-Zamolodchikov equation for S_n]

Consider the following KZ ODE:

$ p'(z) = \sum_{i=2...n} \frac{ Id + \pi( (1i) )}{z-z_i} p (z) $

As it is discussed in MO-question above it is known to have polynomial solution.

The reside at infinity is equal to $Res=-\sum_{i=2...n} { Id + \pi( (1i) )}$. Which is our beloved JM-element up to sign and n*Id.

Hence its eigenvalues must be non-positive integers (this is obvious since at infinity the solution looks like $(1/z)^{Res}, so in order to be polynomial in z they must be non-positive ints). Hence we are done.

Moreover we got that eigs are greater or equal -n (as David Speyer proved directly above).


It is not an answer, rather long comment...

1) I am sorry: my previous posts were incorrect, I will correct below.

2) I would suggest you guys insert the statement and may be proof to Wiki article, it is quite worth and since it was mainly written by me, imho I might give such a suggestion.

The main message is that there is "certain relation" (described below) between standard Gelfand-Tsetlin maximal commutative subalgebra in $U(gl_M)$ and the maximal commutative subalgebra in $C[S_M]$ generated by Jucys-Murphy elements. The relation consists of two steps which can be seen as generalized Schur-Weyl duality and generalized $gl_M - gl_N$ duality. Both steps involve an intermediate object - "bending flow" commutative subalgebra in $U(gl_N \oplus ... \oplus gl_N)$ (sum contains $M$ terms). Briefly speaking these generalized dualities say that: images in certain representations of these commutative subalgebras coincide.

Since I forget some details I would NOT make again the claim that "JM elements go to "quadratic Casimirs"", which might give another (but very long) way to answer Igor's question. Just simply describe the relation which might be interesting on its own.

Step 1. Generalized Schur-Weyl from JM to "bending flows". (Rather trivial step).

Consider $V=C^N \otimes ... \otimes C^N$ ($M$ terms in tensor product). $C[S_M]$ acts here in a natural way. $U(gl_N \oplus ... \oplus gl_N)$ surjects on $End(V)$. Since it surjects we can find certain elements in $U(gl_N \oplus ... \oplus gl_N)$ which are mapped to JM elements, moreover we require such elements to be quadratic in generators of $U(gl_N \oplus ... \oplus gl_N)$, and it would fix these elements. The basic idea is that the permutation operator (12) acting in $C^N\otimes C^N$ is OBVIOUSLY an image of $\sum_{ij} E_{ij}\otimes E_{ji} \in U(gl_N)\otimes U(gl_N)=U(gl_N\oplus gl_N)$ and nothing more than that.

By $E_{ij}$ denoted the matrix with $1$ at position $(ij)$ and zeros everywhere else.

So we get certain commutative subalgebra in $U(gl_N \oplus ... \oplus gl_N)$ such that it is "Schur-Weyl dual" to JM subalgebra, meaning that the images of these subalgebras in $End(V)$ coincide. Such a commutative subalgebra is called "generalized bended flows" or just "bending flows", by reason commented below.

Step 2. $GL_M-GL_N$-duality from "bending flows" to Gelfand-Tsetlin. (This step is not so trivial). It is mainly due to Flaschka and Millson - section 8 of http://arxiv.org/abs/math.SG/0108191

Consider the vector space $W = S(C^N\otimes C^M) = S(C^N \oplus ... \oplus C^N)$ (M-terms in sum) and $S$ denotes symmetric algebra of the vector space. Lie algebras $gl_M$ and $U(gl_N \oplus ... \oplus gl_N)$ acts on $W$ in a natural way.

Theorem: the images in $End(W)$ of GT and "bending flows" coincide.

In such a form it is Theorem 2 page 9 in our paper: http://arxiv.org/abs/0710.4971

Why the name "bending flows" ? If we make similar considerations for $U(so_3 \oplus ... \oplus so_3)$ or more precisely its associated grade Poisson algebra $S(so_3 \oplus ... \oplus so_3)$ we get a (Poisson) commutative subalgebra there.

The beautiful fact is that "JM" type generators have very nice geometric interpretation. We can identify $so_3=R^3$ and so elements of $(so_3 \oplus ... \oplus so_3)$ can be seen as $M$-gons in $R^3$. The statement is that if we "bend" polygon along the non-intersecting diagonals then such flows will be hamiltonian and will be defined by JM-type generators in $S(so_3 \oplus ... \oplus so_3)$. Well, I omitted some details and may be comment is not so clear, one should draw simple pictures in order to see what is going on.

Bending flows were proposed for $S(so_3 \oplus ... \oplus so_3)$ in paper M. Kapovich, J. Millson, The symplectic geometry of polygons in Euclidean space,J. Differ. Geom. 44, 479–513 (1996)

Generalized further in several papers in particular in Gregorio Falqui, Fabio Musso, Gaudin Models and Bending Flows: a Geometrical Point of View, J. Phys. A 36 (2003), no. 46,11655–11676. nlin.SI/0306005


  • $\begingroup$ "JM elements will be mapped into quadratic Casimir operators". By what? Schur-Weyl duality? How exactly? $\endgroup$ – darij grinberg Dec 16 '11 at 21:47
  • $\begingroup$ Sorry to say, but I don't understand any of the sentences of your post. I am not used to physicists' terminology at all, but not even the maths part is clear to me. Feel free to explain in Russian if it is easier to you; both Igor and me understand it. $\endgroup$ – darij grinberg Dec 16 '11 at 21:48
  • $\begingroup$ Well, I'm trying to wrap my mind around fact 1. Saying that the images of the GT subalgebras in $\mathbb{C}[S_n]$ and $U(\mathfrak{gl}_k)$ in $\mathrm{End}(\mathbb{C}^k\otimes\mathbb{C}^n)$ coincide, you're surely considering some isomorphism between $\mathbb{C}^k\otimes\mathbb{C}^n$ and $\mathbb{C}^k\oplus\ldots\oplus\mathbb{C}^k$. Which one is it? Since I'm pretty sure that the natural $e_i\otimes e_j\to e_{j,i}$ wouldn't do the trick: in this basis the matrices corresponding to $\mathfrak{gl}_k$ will all be block diagonal ($n$ blocks $k\times k$), while $X_k$'s matrix won't be such. $\endgroup$ – imakhlin Dec 17 '11 at 14:18
  • $\begingroup$ Less formally, fact 1 seems pretty strange to me, since $S_n$ acts only on the second component of a tensor from $\mathbb{C}^k\otimes\mathbb{C}^n$, while $\mathfrak{gl}_k$ only on the first one. This is if we're applying the natural isomorphism mentioned above. $\endgroup$ – imakhlin Dec 17 '11 at 14:33
  • $\begingroup$ AFAIU, for fact 1 to hold it will have to be an isomorphism mapping (any two) GT bases in $\mathbb{C}^k\otimes\mathbb{C}^n$ and $\mathbb{C}^k\oplus\ldots\oplus\mathbb{C}^k$ into each other because our images in $\mathrm{End}$ are the endomorphisms diagonal is these bases. What I'm asking is whether this is the way you initially define your isomorphism. $\endgroup$ – imakhlin Dec 17 '11 at 14:48

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.