# How to generalise $(\wedge^2 \chi)(g) = \frac{1}{2}(\chi(g)^2-\chi(g^2))$?

One can decompose $\bigotimes^2 V = \bigvee^2 V \oplus \bigwedge^2 V$, getting a corresponding decomposition for representations, say when $V$ is a module for some finite group $G$. One then has the relation between the characters $\chi$, $\vee^2 \chi$ and $\wedge^2 \chi$ afforded by the $G$-modules $V$, $\bigvee^2 V$ and $\bigwedge^2 V$, given by $(\wedge^2 \chi)(g) = \frac{1}{2} (\chi(g)^2-\chi(g^2))$ and $(\vee^2 \chi)(g) = \frac{1}{2} (\chi(g)^2 + \chi(g^2))$.

More generally, one can decompose $\bigotimes^r V$ by Schur-Weyl duality in terms of the irreducible representations of the symmetric group $S_r$. Can we use this to give similar formulas for the representations coming up in this decomposition?

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From the GAP manual:

The symmetrization $\chi^{[\lambda]}$ of the character $\chi$ with the character $\lambda$ of the symmetric group $S_n$ of degree $n$ is defined by $$\chi^{[\lambda]}(g) = \frac{1}{n!}\left( \sum_{{\rho \in S_n}} \lambda(\rho) \prod_{{k=1}}^n \chi(g^k)^{{a_k(\rho)}} \right)$$
where $a_k(\rho)$ is the number of cycles of length $k$ in $\rho$.

You might also be interested in Murnaghan's refinement of these ideas which often can give smaller constituents. All of these ideas are important as they only rely on the power map of the character table, and so allow one to pretty quickly find a lot of nearly irreducible characters from very little information (which is then followed by LLL to magically make them irreducible).

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That's a great formula! Could you provide a good reference that explains it? – Will May 17 '11 at 23:21
Sure, this is theorem 2.7.9 on page 145 of Klaus Lux and Herbert Pahlings textbook on the Representations Of Groups. The formulas for n=3 are on the next page, and an application to the character table of M12 is on page 147, 150, and 157 (where the first only uses symmetrizations, the second uses a naive reduction, and the third uses LLL). – Jack Schmidt May 17 '11 at 23:25

Well, yes, but I'm not sure what Schur-Weyl duality has to do with it. The eigenvalues of $g \in G$ acting on $\Lambda^n V$ are the elementary symmetric polynomials of the eigenvalues of $g$ acting on $V$, so the character value can be read off the characteristic polynomial of $g$ as a linear endomorphism of $V$. Similarly, the eigenvalues of $g$ acting on $\text{Sym}^n V$ are the complete homogeneous symmetric polynomials of the eigenvalues of $g$, and can be related to the character of $\Lambda^n V$. With some work one can rewrite these as linear combinations of $\chi(g^m)$; for example $\displaystyle \Lambda^3 \chi(g) = \frac{1}{6} \left( \chi(g)^3 - 3 \chi(g^2) \chi(g) + 2 \chi(g^3) \right)$.

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The point with Schur-Weyl duality is that you find other interesting representations in $\bigotimes^r V$ by taking $\mathrm{Hom}_{K[S_r]}(V_\lambda,\bigotimes^r V)$ for some irreducible representation $V_\lambda$ of $S_r$. Taking the trivial and alternating representations gives $\bigvee^r V$ and $\bigwedge^r V$, but there are others too that one might be interested in. – Will May 17 '11 at 23:36
Also, your formula should be $\frac{1}{6} (\chi(g)^3 - 3\chi(g^2) \chi(g) +2\chi(g^3))$ as the conjugacy classes of $S_3$ are given by partitions $1+1+1$, $2+1$, $3$ with respective cardinalities 1,3,2. As you say, this is easy for $e_n$ and $h_n$ where you can read it off the cycle type. – Will May 18 '11 at 1:13