A bridge between two worlds: the Frobenius map

There is a reason I’ve been building up the theory of symmetric functions in the last few posts, one gemstone at a time: all this theory is needed for the proof of the beautiful Murnaghan-Nakayama rule for computing the characters of the symmetric group.

What do symmetric functions have to do with representation theory? The answer lies in the Frobenius map, the keystone that completes the bridge between these two worlds.

The Frobenius map essentially takes a character of a representation of the symmetric group and assigns it a symmetric function. To define it, recall that characters are constant across conjugacy classes, and in $S_n$, the conjugacy classes correspond to the partitions of $n$ by associating a permutation $\pi$ with its cycle type $c(\pi)$. For instance, the permutations $\pi=(12)(345)$ and $\sigma=(35)(142)$ both have cycle type $\lambda=(3,2)$. So, any character $\chi$ satisfies $\chi(\pi)=\chi(\sigma)$.

We can now define the Frobenius map $F$. For any character $\chi$ of a representation of $S_n$, define
$$F\left(\chi\right)=\frac{1}{n!}\sum_{\pi\in S_n}\chi(\pi)p_{c(\pi)}$$ where $p_\lambda=p_{\lambda_1}p_{\lambda_2}\cdots p_{\lambda_k}$ is the $\lambda$th power sum symmetric function. (Recall that $p_{\lambda_i}=x_1^{\lambda_i}+x_2^{\lambda_i}+\cdots+x_k^{\lambda_i}$.)

Then, combining the permutations with the same cycle type in the sum, we can rewrite the definition as a sum over partitions of size n:
$$F\left(\chi\right)=\sum_{|\lambda|=n}\frac{1}{z_\lambda}\chi(\lambda)p_{\lambda}$$ for some constants $z_\lambda$. (It’s a fun little combinatorial problem to show that if $\lambda$ has $m_1$ 1’s, $m_2$ 2’s, and so on, then $z_\lambda=1^{m_1}m_1!2^{m_2}m_2!\cdots.$)

As an example, let’s take a look at the character table of $S_3$:
$$\begin{array}{c|ccc} & [(1)(2)(3)] & [(12)(3)] & [(123)] \\\hline \chi^{(3)} & 1 & 1 & 1 \\ \chi^{(1,1,1)} & 1 & -1 & 1 \\ \chi^{(2,1)} & 2 & 0 & -1 \end{array}$$ Consider the third row, $\chi^{(2,1)}$, and let us work over three variables $x,y,z$. Then the Frobenius map sends $\chi^{(2,1)}$ to
$$F\chi^{(2,1)}=\frac{1}{6}(2p_{(1,1,1)}-2p_3)$$ by our definition above. This simplifies to:
$$\frac{1}{3}((x+y+z)^3-(x^3+y^3+z^3))=x^2y+y^2x+x^2z+z^2x+y^2z+z^2y+2xyz$$ which can be written as $m_{2,1}+2m_{1,1,1}$. Notice that, by the combinatorial definition of the Schur functions, this is precisely the Schur function $s_{2,1}$! In fact:

The Frobenius map sends the irreducible character $\chi^\lambda$ to the Schur function $s_\lambda$ for all $\lambda$.

And therein lies the bridge.

Why is this the case? Read on to find out…

3 thoughts on “A bridge between two worlds: the Frobenius map”

1. brent on said:

Thank you for a wonderful blog. I wish to be able to write mathematics expressions on my website as well. What software do you use to write your mathematics symbolism? Is it LaTex?

Thank you so much

• Yes, I use LaTeX here, with the MathJax plugin.

Best wishes,
Maria

2. Aritra on said:

Thanks for this wonderful post. In the inner product in the group side, why is there no conjugation on the second factor?
Also, I think the sentence “…since any transition matrix to another basis must be integral and orthogonal with respect to the Hall inner product…” is a little bit confusing, because the transition matrix is orthogonal as a matrix, and it does not need any reference to inner product.