Why are Jucys-Murphy elements' eigenvalues whole numbers?

Question

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.

Answer 1

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).

Answer 2

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 $\lambda$.

Answer 3

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.

Answer 4

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).

Answer 5

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.

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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

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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

http://arxiv.org/abs/nlin/0306005