Frobenius characteristic map

Properties

Inner product and isometry

Hall inner product

Source:

The inner product on the ring of symmetric functions is the Hall inner product. It is required that $⟨h_{\mu },m_{\lambda }⟩={\delta }_{\mu \lambda }$ . Here, ${\displaystyle m_{\lambda }}$ is a monomial symmetric function and ${\displaystyle h_{\mu }}$ is a product of completely homogeneous symmetric functions. To be precise, let ${\displaystyle \mu =({\mu }_1,{\mu }_2,\cdots )}$ be a partition of integer, then$${\displaystyle h_{\mu }=h_{{\mu }_1}{h}_{{\mu }_2}\cdots .}$$In particular, with respect to this inner product, ${\displaystyle \{p_{\lambda }\}}$ form a orthogonal basis: $⟨p_{\lambda },p_{\mu }⟩={\delta }_{\lambda \mu }{z}_{\lambda }$, and the Schur polynomials ${\displaystyle \{s_{\lambda }\}}$ form a orthonormal basis: $⟨s_{\lambda },s_{\mu }⟩={\delta }_{\lambda \mu }$, where ${\displaystyle {\delta }_{\lambda \mu }}$ is the Kronecker delta.

Inner product of characters

Let ${\displaystyle f,g\in R^n}$, their inner product is defined to be

$${\displaystyle ⟨f,g{⟩}_n=\frac{1}{n!}\sum _{w\in S_n}f(w)g(w)=\sum _{\mu ⊢n}{z}_{\mu }^{-1}f(\mu )g(\mu )}$$If ${\displaystyle f=\sum _n{f}_n,g=\sum _n{g}_n}$, then

$${\displaystyle ⟨f,g⟩=\sum _n⟨f_n,g_n{⟩}_n}$$

Frobenius characteristic map as an isometry

One can prove that the Frobenius characteristic map is an isometry by explicit computation. To show this, it suffices to assume that ${\displaystyle f,g\in R^n}$:$${\displaystyle \begin{array}{cc}⟨\mathrm{ch}(f),\mathrm{ch}(g)⟩& =⟨\sum _{\mu ⊢n}{z}_{\mu }^{-1}f(\mu )p_{\mu },\sum _{\lambda ⊢n}{z}_{\lambda }^{-1}g(\lambda )p_{\lambda }⟩\\ & =\sum _{\mu ,\lambda ⊢n}{z}_{\mu }^{-1}{z}_{\lambda }^{-1}f(\mu )g(\mu )⟨p_{\mu },p_{\lambda }⟩\\ & =\sum _{\mu ,\lambda ⊢n}{z}_{\mu }^{-1}{z}_{\lambda }^{-1}f(\mu )g(\mu )z_{\mu }{\delta }_{\mu \lambda }\\ & =\sum _{\mu ⊢n}{z}_{\mu }^{-1}f(\mu )g(\mu )\\ & =⟨f,g⟩\end{array}}$$

Ring isomorphism

The map ${\displaystyle \mathrm{ch}}$ is an isomorphism between ${\displaystyle R}$ and the ${\displaystyle \mathbb{Z}}$-ring ${\displaystyle \mathrm{\Lambda }}$. The fact that this map is a ring homomorphism can be shown by Frobenius reciprocity. For ${\displaystyle f\in R^n}$ and ${\displaystyle g\in R^m}$,$${\displaystyle \begin{array}{cc}\mathrm{ch}(f\cdot g)& =⟨{\mathrm{ind}}_{S_n\times S_m}^{S_{m+n}}(f\times g),\psi {⟩}_{m+n}\\ & =⟨f\times g,{\mathrm{res}}_{S_n\times S_m}^{S_{m+n}}\psi ⟩\\ & =\frac{1}{n!m!}\sum _{\pi \sigma \in S_n\times S_m}(f\times g)(\pi \sigma )p_{\rho (\pi \sigma )}\\ & =\frac{1}{n!m!}\sum _{\pi \in S_n,\sigma \in S_m}f(\pi )g(\sigma )p_{\rho (\pi )}{p}_{\rho (\sigma )}\\ & =\left[\frac{1}{n!}\sum _{\pi \in S_n}f(\pi )p_{\rho (\pi )}\right]\left[\frac{1}{m!}\sum _{\sigma \in S_m}g(\sigma )p_{\rho (\sigma )}\right]\\ & =\mathrm{ch}(f)\mathrm{ch}(g)\end{array}}$$

Defining ${\displaystyle \psi :S_n\to {\mathrm{\Lambda }}^n}$ by ${\displaystyle \psi (w)=p_{\rho (w)}}$, the Frobenius characteristic map can be written in a shorter form:

$${\displaystyle \mathrm{ch}(f)=⟨f,\psi {⟩}_n,f\in R^n.}$$

In particular, if ${\displaystyle f}$ is an irreducible representation, then ${\displaystyle \mathrm{ch}(f)}$ is a Schur polynomial of ${\displaystyle n}$ variables. It follows that ${\displaystyle \mathrm{ch}}$ maps an orthonormal basis of ${\displaystyle R}$ to an orthonormal basis of ${\displaystyle \mathrm{\Lambda }}$. Therefore it is an isomorphism.

Example

Computing the Frobenius image

Let ${\displaystyle f}$ be the alternating representation of ${\displaystyle S_3}$, which is defined by ${\displaystyle f(\sigma )v=\mathrm{sgn}(\sigma )v}$, where ${\displaystyle \mathrm{sgn}(\sigma )}$ is the sign of the permutation ${\displaystyle \sigma }$. There are three conjugacy classes of ${\displaystyle S_3}$, which can be represented by ${\displaystyle e}$ (identity or the product of three 1-cycles), ${\displaystyle (12)}$(transpositions or the products of one 2-cycle and one 1-cycle) and ${\displaystyle (123)}$ (3-cycles). These three conjugacy classes therefore correspond to three partitions of ${\displaystyle 3}$ given by ${\displaystyle (1,1,1)}$, ${\displaystyle (2,1)}$, ${\displaystyle (3)}$. The values of ${\displaystyle f}$ on these three classes are ${\displaystyle 1,-1,1}$ respectively. Therefore:$${\displaystyle \begin{array}{cc}\mathrm{ch}(f)& =z_{(1,1,1)}^{-1}f((1,1,1))p_{(1,1,1)}+z_{(2,1)}f((2,1))p_{(2,1)}+z_{(3)}^{-1}f((3))p_{(3)}\\ & =\frac{1}{6}(x_1+x_2+x_3{)}^3-\frac{1}{2}(x_1^2+x_2^2+x_3^2)(x_1+x_2+x_3)+\frac{1}{3}(x_1^3+x_2^3+x_3^3)\\ & =x_1{x}_2{x}_3\end{array}}$$Since ${\displaystyle f}$ is an irreducible representation (which can be shown by computing its characters), the computation above gives the Schur polynomial of three variables corresponding to the partition ${\displaystyle 3=1+1+1}$.