Frobenius characteristic map

Source:{{Cite book|last=MacDonald|first=Ian Grant|title=Symmetric functions and Hall polynomials|publisher=Oxford University Press; 2nd edition|year=2015|isbn=9780198739128|pages=112}}

Let $R^n$ be the $\backslash mathbb\{Z\}$-module generated by all irreducible characters of $S\_n$ over $\backslash mathbb\{C\}$. In particular $S\_0=\backslash \{1\backslash \}$ and therefore $R^0=\backslash mathbb\{Z\}$. The ring of characters is defined to be the direct sum$$R=\backslash bigoplus\_\{n=0\}^\{\backslash infty\}R^n$$with the following multiplication to make $R$ a graded commutative ring. Given $f\; \backslash in\; R^n$ and $g\; \backslash in\; R^m$, the product is defined to be$$f\; \backslash cdot\; g\; =\; \backslash operatorname\{ind\}\_\{S\_m\; \backslash times\; S\_n\}^\{S\_\{m+n\}\}(f\; \backslash times\; g)$$with the understanding that $S\_m\; \backslash times\; S\_n$ is embedded into $S\_\{m+n\}$ and $$

\operatorname{ind} denotes the induced character.

For $f\; \backslash in\; R^n$, the value of the Frobenius characteristic map $\backslash operatorname\{ch\}$ at $f$, which is also called the Frobenius image of $f$, is defined to be the polynomial

$$\backslash operatorname\{ch\}(f)=\backslash frac\{1\}\{n!\}\backslash sum\_\{w\; \backslash in\; S\_n\}f(w)p\_\{\backslash rho(w)\}=\backslash sum\_\{\backslash mu\; \backslash vdash\; n\}z\_\backslash mu^\{-1\}f(\backslash mu)p\_\backslash mu.$$

Here, $\backslash rho(w)$ is the integer partition determined by $w$. For example, when $n=3$ and $w=(12)(3)$, $\backslash rho(w)=(2,1)$ corresponds to the partition $3=2+1$. Conversely, a partition $\backslash mu$ of $n$ (written as $\backslash mu\; \backslash vdash\; n$) determines a conjugacy class $K\_\backslash mu$ in $S\_n$. For example, given $\backslash mu=(2,1)\backslash vdash\; 3$, $K\_\backslash mu=\backslash \{(12)(3),(13)(2),(23)(1)\backslash \}$ is a conjugacy class. Hence by abuse of notation $f(\backslash mu)$ can be used to denote the value of $f$ on the conjugacy class determined by $\backslash mu$. Note this always makes sense because $f$ is a class function.

Let $\backslash mu$ be a partition of $n$, then $p\_\backslash mu$ is the product of power sum symmetric polynomials determined by $\backslash mu$ of $n$ variables. For example, given $\backslash mu=(3,2)$, a partition of $5$,

$\backslash begin\{aligned\}$

p_\mu(x_1,x_2,x_3,x_4,x_5)&=p_3(x_1,x_2,x_3,x_4,x_5)p_2(x_1,x_2,x_3,x_4,x_5) \\

&=(x_1^3+x_2^3+x_3^3+x_4^3+x_5^3)(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2)

\end{aligned}

Finally, $z\_\backslash lambda$ is defined to be $\backslash frac\{n!\}\{k\_\backslash lambda\}$, where $k\_\backslash lambda$ is the cardinality of the conjugacy class $K\_\backslash lambda$. For example, when $\backslash lambda\; =\; (2,1)\backslash vdash\; 3$, $z\_\backslash lambda\; =\; \backslash frac\{3!\}\{3\}=2$. The second definition of $\backslash operatorname\{ch\}(f)$ can therefore be justified directly:$$\backslash frac\{1\}\{n!\}\backslash sum\_\{w\; \backslash in\; S\_n\}f(w)p\_\{\backslash rho(w)\}\; =\; \backslash sum\_\{\backslash mu\; \backslash vdash\; n\}\backslash frac\{k\_\backslash mu\}\{n!\}f(\backslash mu)p\_\backslash mu$$

= \sum_{\mu \vdash n}z_\mu^{-1}f(\mu)p_\mu

Source:{{Cite book|last=Macdonald|first=Ian Grant|title=Symmetric functions and Hall polynomials|publisher=Oxford University Press; 2nd edition|year=2015|isbn=9780198739128|pages=63|language=English}}

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

Let $f,g\; \backslash in\; R^n$, their inner product is defined to be{{Cite book|last=Stanley|first=Richard|title=Enumerative Combinatorics: Volume 2 (Cambridge Studies in Advanced Mathematics Book 62)|publisher=Cambridge University Press|year=1999|isbn=9780521789875|pages=349}}

$$\backslash langle\; f,\; g\; \backslash rangle\_n\; =\; \backslash frac\{1\}\{n!\}\backslash sum\_\{w\; \backslash in\; S\_n\}f(w)g(w)\; =\; \backslash sum\_\{\backslash mu\; \backslash vdash\; n\}z\_\backslash mu^\{-1\}f(\backslash mu)g(\backslash mu)$$If $f\; =\; \backslash sum\_\{n\}f\_n,g\; =\; \backslash sum\_\{n\}g\_n$, then

$$\backslash langle\; f,g\; \backslash rangle\; =\; \backslash sum\_n\; \backslash langle\; f\_n,\; g\_n\; \backslash rangle\_n$$

One can prove that the Frobenius characteristic map is an isometry by explicit computation. To show this, it suffices to assume that $f,g\; \backslash in\; R^n$:$$\backslash begin\{aligned\}$$

\langle \operatorname{ch}(f),\operatorname{ch}(g) \rangle &=

\left\langle \sum_{\mu\vdash n}z_\mu^{-1}f(\mu)p_\mu,

\sum_{\lambda\vdash n}z_\lambda^{-1}g(\lambda)p_\lambda\right\rangle \\

&= \sum_{\mu,\lambda\vdash n}z_\mu^{-1}z_\lambda^{-1}

f(\mu)g(\mu)\langle p_\mu,p_\lambda \rangle \\

&= \sum_{\mu,\lambda\vdash n}z_\mu^{-1}z_\lambda^{-1}

f(\mu)g(\mu)z_\mu\delta_{\mu\lambda} \\

&= \sum_{\mu\vdash n}z_{\mu}^{-1}f(\mu)g(\mu) \\

&= \langle f,g \rangle

\end{aligned}

The map $\backslash operatorname\{ch\}$ is an isomorphism between $R$ and the $\backslash mathbb\{Z\}$-ring $\backslash Lambda$. The fact that this map is a ring homomorphism can be shown by Frobenius reciprocity.{{Cite book|last=Stanley|first=Richard|title=Enumerative Combinatorics: Volume 2 (Cambridge Studies in Advanced Mathematics Book 62)|publisher=Cambridge University Press|year=1999|isbn=9780521789875|pages=352|language=English}} For $f\; \backslash in\; R^n$ and $g\; \backslash in\; R^m$,$$\backslash begin\{aligned\}$$

\operatorname{ch}(f \cdot g) &= \langle \operatorname{ind}_{S_n \times S_m}^{S_{m+n}}(f \times g),\psi \rangle_{m+n} \\

&= \langle f \times g, \operatorname{res}_{S_n \times S_m}^{S_{m+n}}\psi \rangle \\

&= \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] \\

&= \operatorname{ch}(f)\operatorname{ch}(g)

\end{aligned}

Defining $\backslash psi:S\_n\; \backslash to\; \backslash Lambda^n$ by $\backslash psi(w)\; =\; p\_\{\backslash rho(w)\}$, the Frobenius characteristic map can be written in a shorter form:

$$\backslash operatorname\{ch\}(f)=\backslash langle\; f,\; \backslash psi\; \backslash rangle\_n,\; \backslash quad\; f\; \backslash in\; R^n.$$

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

Let $f$ be the alternating representation of $S\_3$, which is defined by $f(\backslash sigma)v=\backslash sgn(\backslash sigma)v$, where $\backslash sgn(\backslash sigma)$ is the sign of the permutation $\backslash sigma$. There are three conjugacy classes of $S\_3$, which can be represented by $e$ (identity or the product of three 1-cycles), $(12)$(transpositions or the products of one 2-cycle and one 1-cycle) and $(123)$ (3-cycles). These three conjugacy classes therefore correspond to three partitions of $3$ given by $(1,1,1)$, $(2,1)$, $(3)$. The values of $f$ on these three classes are $1,-1,1$ respectively. Therefore:$$\backslash begin\{aligned\}$$

\operatorname{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_1x_2x_3

\end{aligned}Since $f$ 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 $3=1+1+1$.

{{reflist}}

{{DEFAULTSORT:Frobenius Manifold}}

Category:Representation_theory

Category:Combinatorics