## Polynomials and Representations XX

From now onwards, we will assume the base field K has characteristic 0.

## Example: d=3

Following the previous article, we examine the case of $S_3$. We get 3 partitions: $\lambda_1 = (3)$, $\lambda_2 = (2,1)$ and $\lambda_3 = (1,1,1).$ Let us compute $M_{ij} := \left < U_{\l_i}, U_{\l_j}\right>$ for all $1 \le i, j \le 3.$ From the previous article, we have:

Since $\left< U_{\lambda_1}, U_{\lambda_1}\right> = 1$, $U_{\lambda_1}$ is absolutely irreducible (i.e. irreducible even over the algebraic closure of K). Since

$\displaystyle \left< U_{\lambda_2}, U_{\lambda_1} \right> = \left< U_{\lambda_3}, U_{\lambda_1}\right> = 1$

each of $U_{\lambda_2}$ and $U_{\lambda_3}$ contains exactly one copy of $U_{\lambda_1}.$ Thus we write

$\displaystyle U_{\lambda_2} = U_{\lambda_1} \oplus U'_{\lambda_2}, \quad U_{\lambda_3} = U_{\lambda_1} \oplus U'_{\lambda_3},$

where $U'_{\lambda_i}$ does not contain $U_{\lambda_1}$, i.e. $\left< U_{\lambda_1}, U'_{\lambda_i}\right> = 0$. Hence

$\displaystyle \left< U_{\lambda_2}, U_{\lambda_2}\right> = \left< U_{\lambda_1}, U_{\lambda_1}\right>+ \left< U'_{\lambda_2}, U'_{\lambda_2}\right> \implies \left< U'_{\lambda_2}, U'_{\lambda_2}\right> = 1.$

Taking $\left< U'_{\lambda_i}, U'_{\lambda_j}\right>$, we obtain:

Thus $\left< U'_{\lambda_2}, U'_{\lambda_2}\right> = 1$ so $U'_{\lambda_2}$ is absolutely irreducible. Since $\left< U'_{\lambda_2}, U'_{\lambda_3}\right> = 2$, $U'_{\lambda_3}$ contains two copies of $U'_{\lambda_2}.$ Thus we write:

$U'_{\lambda_3}=(U'_{\lambda_2})^{\oplus 2}\oplus U''_{\lambda_3},$

where $U''_{\lambda_3}$ does not contain $U_{\lambda_1}$ or $U'_{\lambda_2}$. We have

$\displaystyle 5 = \left< U'_{\lambda_3}, U'_{\lambda_3}\right> = 4\left< U'_{\lambda_2}, U'_{\lambda_2}\right> +\left< U''_{\lambda_3}, U''_{\lambda_3}\right> \implies \left< U''_{\lambda_3}, U''_{\lambda_3}\right> = 1.$

Since $S_3$ has 3 irreducible representations, we have found all of them: $U_{\lambda_1}$, $U'_{\lambda_2}$, $U''_{\lambda_3}$ and:

\begin{aligned}U_{\lambda_1} &= U_{\lambda_1}, \\ U_{\lambda_2} &= U_{\lambda_1} \oplus U'_{\lambda_2}, \\ U_{\lambda_3} &= U_{\lambda_1} \oplus (U'_{\lambda_2})^{\oplus 2} \oplus U''_{\lambda_3}.\end{aligned}

Furthermore, we have shown that the 3 irreps of $S_3$ are in fact defined over $\mathbb{Q}$.

## General Case

We will show that the above computations can be generalised to any $S_n.$ Indeed, the representations $U_\lambda := K[X_\lambda]$ satisfy:

$\displaystyle \left< U_\lambda, U_\mu \right> = M_{\lambda\mu}.$

Recall that the matrix $\mathbf M = (M_{\lambda\mu})$ satisfies $\mathbf M = \mathbf K^t \mathbf K$ where $K_{\lambda\mu}$ is the number of SSYT of shape $\lambda$ and type $\mu$ and we have:

1. $K_{\lambda\lambda} = 1$ for all $\lambda$;
2. if $K_{\mu\lambda} \ne 0$, then $\mu \trianglerighteq \lambda$ and so $\mu \ge \lambda$ lexicographically.

Thus: $M_{\lambda\mu} = \sum_{\nu \trianglerighteq\lambda,\, \nu\trianglerighteq\mu} K_{\nu\lambda} K_{\nu\mu}.$

Theorem. Suppose $(\Sigma, \le)$ is a finite partially ordered set and $\{U_\lambda\}_{\lambda\in\Sigma}$ is a collection of representations. And suppose there are non-negative integers $\{K_{\lambda\mu}\}_{\lambda, \mu\in\Sigma}$ satisfying $K_{\lambda\lambda} = 1$ and:

$\displaystyle \left< U_\lambda, U_\mu\right> = \sum_{\nu\ge \lambda, \nu\ge \mu} K_{\nu\lambda} K_{\nu\mu}.$

Then there are non-isomorphic absolutely irreducible $\{V_\lambda\}_{\lambda\in\Sigma}$ defined over K such that:

$U_\lambda = \bigoplus_{\mu\ge\lambda} V_\mu^{\oplus K_{\mu\lambda}}.$

Proof

Choose a maximal element $\lambda\in \Sigma$; then

$\left< U_\lambda, U_\lambda\right> = \sum_{\nu\ge\lambda} K_{\nu\lambda}^2 = K_{\lambda\lambda}^2 = 1.$

So $V_\lambda := U_\lambda$ is absolutely irreducible. The number of copies of $V_\lambda$ contained in each $U_\mu$ is:

$\left< U_\mu, U_\lambda\right> = \sum_{\nu\ge \lambda,\, \nu\ge \mu} K_{\nu\lambda} K_{\nu\mu} =K_{\lambda\lambda} K_{\lambda\mu} = K_{\lambda\mu}.$

so we can write: $U_\mu = V_\lambda^{\oplus K_{\lambda\mu}} \oplus U_\mu'$ where $U_\mu'$ does not contain $V_\lambda.$ This then gives us:

\displaystyle\begin{aligned}\left < U_\mu', U_\nu'\right> = \left < U_\mu, U_\nu\right> - K_{\lambda\mu} K_{\lambda\nu} = \sum_{\substack{\lambda' \ne \lambda, \\ \lambda' \ge \mu,\ \lambda' \ge \nu}} K_{\lambda'\mu} K_{\lambda'\nu}\end{aligned}

so the collection of $U_\mu'$ for $\mu \in \Sigma - \{\lambda\}$ with $K_{\mu\nu}$ satisfy the given conditions and we may proceed recursively with this reduced set. ♦

Applying the above theorem to the set $\Sigma = \{\lambda: \lambda\vdash d\}$ and the representations $U_\lambda$ of $S_d,$ we get:

Corollary. There are absolutely irreducible representations $\{V_\lambda\}_{\lambda\vdash d}$ of $S_d$ defined over $\mathbb{Q}$ such that:

$\displaystyle U_\lambda = \bigoplus_{\mu\trianglerighteq \lambda} V_\mu^{\oplus K_{\mu\lambda}}.$

## Consequences

If we apply this to the case d=3, we get irreps $V_3, V_{21}, V_{111}$ such that:

$U_3 = V_3, \qquad U_{21} = V_{21} \oplus V_3, \qquad U_{111} = V_{111} \oplus V_{21}^{\oplus 2} \oplus V_3$

as we had computed earlier.

Note 1

Recall that $f_\lambda$ is the number of SYT of shape $\lambda$, i.e. $f_\lambda = K_{\lambda\mu}$ where $\mu = (1,1,\ldots,1).$ Now $U_\mu$ is the regular representation and $K_{\lambda\mu}$ is the number of copies of $V_\lambda$ contained in it. From character theory we thus have

• $f_\lambda = \dim V_\lambda$ and
• $|G| = \sum_{\lambda\vdash d} K_{\lambda\mu}^2 = \sum_{\lambda} f_\lambda^2 \implies \sum_{\lambda\vdash d} f_\lambda^2 = d!$ which we had proven earlier with the RSK correspondence.

Note 2: Frobenius Map

If we let $\chi_\lambda$ be the character for the irrep $V_\lambda$, then:

$\chi(U_\lambda) = \sum_{\mu} K_{\mu\lambda} \chi_\mu \implies \vec\chi_U = \mathbf K^t \vec \chi$

where the second equation is the vectorial form of the first.

Hence, if each $U_\lambda$ corresponds to the complete symmetric polynomial $h_\lambda \in \Lambda^{(d)},$ then $V_\lambda$ corresponds to the Schur polynomial $s_\lambda$ and the Hall inner product $\Lambda^{(d)}\times \Lambda^{(d)} \to \mathbb{Z}$ corresponds to $(V, W) \mapsto \dim_K \text{Hom}_{K[G]}(V, W).$

Thus we have a group isomorphism between $\Lambda^{(d)}$ and the group of virtual characters of $S_d,$ called the Frobenius characteristic map. [Recall that virtual characters are formal differences of the form $\chi_1 - \chi_2$ where $\chi_1, \chi_2$ are group characters. The set of virtual characters of a finite group G forms a ring R(G), under addition and multiplication of characters as functions. R(G) is a finitely generated free abelian group, with a basis given by the irreps of G.]

Taken over the rational field, we obtain a linear isomorphism between:

• $\Lambda^{(d)} \otimes_{\mathbb Z} \mathbb Q$, the space of symmetric polynomials of degree d with rational coefficients;
• the space of class functions $S_d \to \mathbb{Q}$.

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