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}}.


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}}.



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

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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