# Math 202C: Lecture 4

In Lecture 3, we introduced the Young-Jucys-Murphy elements $Y_1,\dots,Y_d$ in the group algebra $\in \mathcal{A}(d)$ of the symmetric group $\mathrm{S}(d)$. The YJM elements form a commuting family of elements in the group algebra of the symmetric group $\mathrm{S}(d)$, and have the amazing feature that symmetric polynomials in $Y_1,\dots,Y_d$ always belong to the class algebra $\mathcal{Z}(d)$ despite the non-centrality of the $Y_j$‘s themselves. This is quite remarkable, as there is no obvious reason why an arbitrary symmetric polynomial $f(Y_1,\dots,Y_d)$ should evaluate to a linear combination of conjugacy classes.

Since $f(Y_1,\dots,Y_d)$ is central, Schur’s Lemma implies that it acts in any given irreducible representation of $\mathcal{A}(d)$ as a scalar operator. More precisely, if $(\mathbf{V}^\lambda,R^\lambda)$ is the irreducible representation of $\mathcal{A}(d)$ corresponding to the partition $\lambda \vdash d$ (more precisely, a representative of the isomorphism class of all such representations), then the operator $R^\lambda(f(Y_1,\dots,Y_d) \in \mathrm{End}\mathbf{V}^\lambda$ is of the form

$R^\lambda(f(Y_1,\dots,Y_d)) = \omega_f(\lambda) I_{\mathbf{V}^\lambda},$

where $I_{\mathbf{V}^\lambda} \in \mathrm{End}\mathbf{V}^\lambda$ is the identity operator and $\omega_f(\lambda) \in \mathbb{R}$ is a scalar. Note that this scalar must in fact be real, because each $Y_i$, being a sum of transpositions, is a selfadjoint element of $\mathcal{A}(d).$ Indeed,

$R^\lambda(Y_1),\dots,R^\lambda(Y_d) \in \mathrm{End}\mathbf{V}^\lambda$

are commuting selfadjoint operators, and hence are simultaneously diagonalizable, i.e. admit a common eigenbasis. A natural question is then how to explicitly describe the eigenvalue $\omega_f(\lambda).$ This question has an amazing answer, which we now describe.

According to Schur’s Lemma, the class algebra $\mathcal{Z}(d)$ may be described as the subalgebra of $\mathcal{A}(d)$ consisting of elements which act as scalar operators in all irreducible representations. Note that we are including here a converse to Schur’s Lemma, namely the statement that if $R^\lambda(C)$ is a scalar operator in $\mathrm{End}\mathbf{V}^\lambda$ for all $\lambda \vdash d$, then $C \in \mathcal{Z}(d)$; this follows immediately from the fact that the map

$\mathcal{F} \colon \mathcal{A}(d) \longrightarrow \bigoplus\limits_{\lambda \vdash d} \mathrm{End}\mathbf{V}^\lambda$

defined by

$\mathcal{F}(A) = \bigoplus\limits_{\lambda \vdash d} R^\lambda(A)$

is an algebra isomorphism (often referred to as the noncommutative Fourier transform). We can generalize this as follows.

As you know from Math 202B, the branching of irreducible $\mathcal{A}(d)$-modules is multiplicity free. More precisely, we may consider the group algebra $\mathcal{A}(d-1)$ of the symmetric group $\mathcal{S}(d-1)$ as the subalgebra of $\mathcal{A}(d)$ spanned by permutations in $\mathcal{S}(d)$ satisfying $\pi(d)=d.$ The isotopic decomposition of an irreducible $\mathcal{A}(d)$-module $\mathbf{V}^\lambda$ as an $\mathcal{A}(d-1)$-module is

$\mathbf{V}^\lambda = \bigoplus\limits_{\mu \nearrow \lambda} \mathbf{V}^\mu,$

where the direct sum is over all irreducible $\mathcal{A}(d-1)$-modules $\mathbf{V}^\mu$ such that the Young diagram $\lambda\vdash d$ can be obtained from the Young diagram $\mu\vdash d-1$ by adding a single cell.

Example 1: For the “staircase” diagram $\lambda =(3,2,1),$ the isotypic decomposition of the irreducible $\mathcal{A}(6)$-module as an $\mathcal{A}(5)$-module is

$\mathbf{V}^{(3,2,1)} = \mathbf{V}^{(2,2,1)} \oplus \mathbf{V}^{(3,1,1)} \oplus \mathbf{V}^{(3,2)}.$

Iterating this construction, we obtain a decomposition of $\mathbf{V}^\lambda$ into a sum of $1-$dimensional spaces $\mathbf{V}^T$ indexed by “growth sequences”

$T=\mu^1 \nearrow \mu^2 \nearrow \dots \nearrow \lambda,$

where $\mu^i$ is a diagram with $i$-cells, which encode the possible life histories of the multicellular organism $\lambda \vdash d$ starting from the single-celled organism $\mu^1=\Box.$ Obviously, each such growth history corresponds uniquely to a standard Young tableau of shape $\lambda$ — the entry of a given cell encodes the time instant at which that cell was added. Thus, multiplicity-free branching gives us a basis $\{\mathbf{e}_T \colon T \in \mathrm{SYT}(\lambda)\}$ of each irreducible representation $\mathbf{V}^\lambda$ canonically determined up to scaling; this basis is known as the Young basis (or sometimes the Gelfand-Tsetlin basis). By Schur’s lemma, the subalgebra

$\mathcal{Y}(d) = \mathrm{Span}(\mathcal{Z}(1),\mathcal{Z}(2),\dots,\mathcal{Z}(d)$

coincides with the maximal commutative subalgebra of $\mathcal{A}(d)$ of elements which act diagonally on the Young basis of every irreducible representation $\mathbf{V}^\lambda$; this algebra is known as the Young subalgebra (or Gelfand-Tsetlin subalgebra) of the group algebra $\mathcal{A}(d)$, and its construction depends crucially on the fact that we have a natural inclusion $\mathcal{A}(d-1) \hookrightarrow \mathcal{A}(d)$. Observe that the YJM elements $Y_1,\dots,Y_d$ belong to the algebra $\mathcal{Y}(d)$ (explaining why is a conceptually optimal way to answer one of the homework problems). The following Theorem is an enormously useful tool, both theoretically and computationally.

Theorem 1 (Diaconis-Greene,Murphy,Okounkov-Vershik): The commutative algebra $\mathcal{Y}(d)$ is exactly the algebra of polynomials in the YJM elements $Y_1,\dots,Y_d.$ Moreover, for each irreducible representation $(\mathbf{V}^\lambda,R^\lambda)$ of $\mathcal{A}(d)$, the action of $Y_k$ on the corresponding Young basis $\{\mathbf{e}_T \colon T \in \mathrm{SYT}(d)\}$ is given by

$R^\lambda(Y_k)\mathbf{e}_T = c_T(k) \mathbf{e}_T,$

where $c_T(k)$ is the content (i.e. column index minus row index) of the cell in $T$ containing the number $k$.

Example 2: Consider the group algebra of the symmetric group of degree $d=6$ acting in the irreducible representation $\mathbf{V}^\lambda$ labeled by the staircase partition $\lambda=(3,2,1),$ and let $\mathbf{e}_T \in \mathbf{V}^\lambda$ be the Young basis vector corresponding to the standard tableau

$T = \begin{matrix} 1 & 3 & 6 \\ 2 & 5 & {} \\ 4 & {} & {} \end{matrix}.$

Theorem 1 says that $\mathbf{e}_T$ is an eigenvector of each of the operators $R^\lambda(Y_1),R^\lambda(Y_2),R^\lambda(Y_3),R^\lambda(Y_4),R^\lambda(Y_5),R^\lambda(Y_6),$ and that the corresponding eigenvalues are, respectively,

$0,-1,1,0,-2,2.$

Corollary 1: For any $f \in \mathbb{C}[X_1,\dots,X_d]^{\mathrm{S}(d)}$ and any $\lambda \vdash d$, we have

$\omega_f(\lambda) = f(c(\Box) \colon \Box \in \lambda),$

where the right hand side denotes the evaluation of the symmetric polynomial $f(X_1,\dots,X_d)$ on the numbers $c(\Box_1),\dots,c(\Box_d)$, with $\Box_1,\dots,\Box_d$ being an arbitrary enumeration of the cells of $\lambda.$

Example 2: Corollary 1 tells us that the eigenvalue of the symmetric polynomial

$f(Y_1,Y_2,Y_3,Y_4,Y_5,Y_6) = Y_1^2+Y_2^2+Y_3^2+Y_4^2+Y_5^2+Y_6^2$

acting in $\mathbf{V}^{(3,2,1)}$ is

$\omega_f(\lambda) = f(0,0,-1,1,-2,2) = 10.$

We will black-box Theorem 1, and (possibly) return to its proof later, though I will post a resource where you can look it up if you wish. Our goal will be to mine some rather remarkable information from Theorem 1. There are two natural directions in which one might want to go.

First, there is the question of whether evaluation on the YJM elements in fact defines a surjective morphism

$\mathbb{C}[X_1,\dots,X_d]^{\mathrm{S}(d)} \longrightarrow \mathcal{Z}(d).$

If this were so, then for each $\lambda \vdash d$ there would exist a symmetric polynomial $f_\alpha(X_1,\dots,X_d) \in \mathbb{C}[X_1,\dots,X_d]^{\mathrm{S}(d)}$ such that

$C_\alpha = f_\alpha(Y_1,\dots,Y_d,),$

i.e. we could express each conjugacy class as a symmetric polynomial function of the YJM elements. This would in turn have the following character-theoretic consequence. One one hand, the eigenvalue $\omega_\alpha(\lambda)$ of the conjugacy class $C_\alpha$ acting in the irreducible representation $\mathbf{V}^\lambda$ is given by

$\omega_\alpha(\lambda) = |C_\alpha| \frac{\chi^\lambda_\alpha}{\dim \mathbf{V}^\lambda},$

where as always the character $\chi^\lambda_\alpha$ is by definition the trace of the operator $R^\lambda(\pi) \in \mathrm{End}\mathbf{V}^\lambda$, with $\pi$ an arbitrary permutation in the conjugacy class $C_\alpha.$ Indeed, on one hand we have

$\mathrm{Tr} R^\lambda(C_\alpha) = \omega_\alpha(\lambda) \dim \mathbf{V}^\lambda,$

since the trace of a scalar operator is simply its unique eigenvalue summed dimension-many times, and on the other hand

$\mathrm{Tr} R^\lambda(C_\alpha) = \sum\limits_{\pi \in C_\alpha} \mathrm{Tr} R^\lambda(\pi) = |C_\alpha| \chi^\lambda_\alpha.$

Corollary 1 would then give us the formula

$\omega_\alpha(\lambda) = f_\alpha(c(\Box) \colon \Box \in \lambda),$

which in turn gives us a formula for the irreducible character $\chi^\lambda_\alpha$ as a function of the content alphabet of the Young diagram $\lambda$. In fact, this is all true, and is a consequence of combining Corollary 1 with the following classical theorem on the class algebra.

Theorem 2 (Farahat-Higman): Every conjugacy class $C_\alpha \in \mathcal{Z}(d)$ can be expressed as a polynomial in the levels $L_0,\dots,L_{d-1}$ of the Cayley graph of $\mathrm{S}(d).$

A good treatment of this approach to the irreducible characters of the symmetric groups has been given by our colleague Adriano Garsia, and I will post his lecture notes to our Piazza page. There is an inescapable downside to this seemingly miraculous result: the symmetric polynomials $f_\alpha(X_1,\dots,X_d)$ are not given by any simple formula. This is in a sense inevitable, as the computation of the irreducible characters of the symmetric groups is a $P^\sharp$-complete problem, and consequently the quest for an explicit closed form is quixotic.

Instead, one can take the opposite approach, and start from the fact that there are many interesting (symmetric) polynomials $f$ which we can write down explicitly, and Theorem 1 gives us a way to explicitly calculate the action of $f(Y_1,\dots,Y_d)$ in irreducible representations. It is this direction that we shall follow in the coming lectures.