## Algebraic combinatorics Lecture 9: symmetric group representations and isomorphism of symmetric functions and class functions

Since I am not a superhuman, my blog lengths will significantly reduce in size from now on. Let ${L(G)}$ be the space of functions on a finite group ${G}$. We can define the usual inner product on it, and the convolution makes it into a group algebra; one can associate each function with an element of the group algebra in the obvious way. The center of ${L(G)}$ consists of class functions, as testified by the indicator functions of single elements. Thus a basis of the center is given by ${\delta_C}$, where ${C}$ ranges over conjugacy classes. For class functions on ${S_n}$,

$\displaystyle \langle f, h \rangle = \sum_\lambda \frac{f(\lambda) h(\lambda)}{z_\lambda}. \ \ \ \ \ (17)$

Definition 9 Let ${CL = \bigoplus_{n=0}^\infty CL_n}$ to be the graded algebra of class functions, where multiplication is defined by

$\displaystyle f \circ g = \rm{IND}_{S_n \times S_m}^{S_{n+m}}(f \times g) \in CL_{n+m}. \ \ \ \ \ (18)$

Further let ${\rm{ch}: CL \rightarrow \Lambda}$ called the characteristic map be defined by

$\displaystyle \rm{ch}(f) = \frac{1}{n!} \sum_{\omega \in S_n} f(\omega) p_{\rho(\omega)} = \sum_{\lambda\vdash n} f(\lambda) p_\lambda / z_\lambda. \ \ \ \ \ (19)$

Proposition 10 Ch is an isometry.

The induced map ${\rm{IND}_{S_n \times S_m}^{S_{n+m}}(f)}$ is defined by the Frobenius relation:

$\displaystyle \langle \rm{IND}_{S_n \times S_m}^{S_{n+m}}(f), \chi \rangle_G = \langle f, \rm{Res}_H^G \chi\rangle_G \ \ \ \ \ (20)$

or equivalently,

$\displaystyle \rm{IND}_{S_n \times S_m}^{S_{n+m}}(f)(s) = \sum_{i=0}^k \bar{f}(t_i^{-1} s t_i), \ \ \ \ \ (21)$

where ${t_i}$ are representative of right cosets of ${H}$ in ${G}$, and ${\bar{f}}$ equals f on H and 0 o.w.
The proposition above is easy, next we show

Proposition 11 Ch is an algebra isomorphism.

Proof: One can write

$\displaystyle Ch(f) = \langle f, \psi \rangle \ \ \ \ \ (22)$

where ${\psi: S_n \rightarrow \Lambda^n}$ given by ${\psi(\omega) = p_{\rho(\omega)}}$. Note ${\psi}$ is a function on ${S_n}$ with value in the space of symmetric functions. This coupled with Frobenius reciprocity easily prove the assertion. $\Box$

Need some representation theory of ${S_n}$. For a G-set X (G acts transitively), one has the permutation reprsentation on it. For set partitions of ${[n]}$ given by a partition ${\lambda}$, called partial rankings, ${S_N}$ acts on them in the natural way, i.e., permuting the labels. This representation is called Specht module.
Fact: ${m^\lambda = \rm{IND}_{S_\lambda}^{S_n} (1)}$, where ${S_\lambda = S_{\lambda_1} \times \ldots \times S_{\lambda_k} \subset S_n}$. But these representations are reducible, for instance:

$\displaystyle m^{n-1,1} = S^n \oplus S^{n-1,1}. \ \ \ \ \ (23)$

Let ${\eta^\lambda}$ be the character of ${m^\lambda}$.

Proposition 12

$\displaystyle \rm{ch}(\eta^\lambda) = h_\lambda = h_{\lambda_1} \ldots h_{\lambda_l} \ \ \ \ \ (24)$

Proof: First we establish it for ${f = 1 = \delta_{S_n}}$. So we basically need ${Ch(f) = \sum_\lambda \frac{p_\lambda}{z_\lambda} \overset{?}{=} h_n}$. Recall ${\prod_{i,j} (1-x_i y_j)^{-1} = \sum_\lambda \frac{p_\lambda(x) p_\lambda(y)}{z_\lambda}}$. Setting ${y_1 = 1}$, ${y_2 = \ldots =0}$, then ${p_\lambda(y) = t^{|\lambda|}}$. Then

$\displaystyle \prod_i (1-tx_i)^{-1} = \sum_\lambda \frac{p_\lambda(x)}{z_\lambda} t^{|\lambda|}\\ = \sum_n t^n \sum_{\lambda \vdash n} \frac{p_\lambda(x)}{z_\lambda} \ \ \ \ \ (25)$

But ${[t^n] \prod_i (1-x_i t)^{-1} = h_n}$.
Finally

$\displaystyle Ch(\eta^\lambda) = Ch(\delta_{S_{\lambda_1}} \circ \ldots \circ \delta_{S_{\lambda_l}}) = h_{\lambda_1} h_{\lambda_2} \ldots h_{\lambda_l} = h_\lambda \ \ \ \ \ (26)$

$\Box$

Next we find linear combinations of ${\{\eta_\lambda\}}$ giving an irreducible character. Recall Munagen-Nakayama:

$\displaystyle p_\mu = \sum_{\lambda \vdash n} \chi^\lambda(\mu) s_\lambda\\ s_\mu = \sum_{\lambda \vdash n} \frac{\chi^\lambda(\mu)}{z_\lambda} p_\lambda. \ \ \ \ \ (27)$

${\chi^\lambda(\mu)}$ is the number of tableaux of shape ${\lambda}$ and type ${\mu}$. Define the following linear combination of products of ${\eta_\lambda}$‘s:

$\displaystyle \psi^\lambda = \det (\eta^{\lambda_i + j -i}) \ \ \ \ \ (28)$

in analogy of Jacobi-Trudi formula for Schur functions. Then using isometry and algebra isomorphism of ${\rm{ch}}$, we see

$\displaystyle \langle \psi^\lambda, \psi^\nu \rangle = \delta_{\lambda \nu}. \ \ \ \ \ (29)$

Since the ${\eta^k}$‘s are characters of possibly reducible representations, which in turn are ${{\mathbb Z}}$-linear combinations of irreps, we know that ${\pm \psi^\lambda}$‘s must be irreducible characters. And since ${\psi^\lambda(1) = }$ the number of SSYT’s which is positive, we know that ${\psi^\lambda}$ is the irreducible characters (they evaluate to a positive trace on identity) and can be identified with ${\chi^\lambda}$.