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Is there exists some simple formula for characters $$\chi_{\Lambda^{k}V}~~~~\text{and}~~~\chi_{\text{Sym}^{k}V}$$ for some representation $V$ of finite group?

Thanks.

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3 Answers 3

Suppose $V$ is a vector space over an field $k.^\dagger$ If an $A\in\mathrm{GL}(V)$ is diagonalizable, it induces an eigenbasis $\{v_i\}$ for $V$ with associated eigenvalues $\{\lambda_i\}$. Consider the inherited action of $A$ on the exterior and symmetric powers $\Lambda^k(V)$ and $S^k(V)$ respectively. In particular, we can characterize the eigenpairs or $\Lambda^k(A)$ and $S^k(A)$ respectively as

$$(\lambda_{i_1}\cdots\lambda_{i_k},~v_{i_1}\wedge\cdots\wedge v_{i_k});$$ $$(\lambda_{j_1}\cdots\lambda_{j_k},~v_{j_1}\cdots v_{j_k}),$$

where $(i_1,\cdots,i_k)$ and $(j_1,\cdots,j_k)$ are respectively strictly increasing and nondecreasing sequences of integers taken from the index set $\{1,\cdots,n\}$. We thereby obtain the trace formulas

$$\mathrm{tr}\,\Lambda^k(A)=e_k(\lambda_1,\cdots,\lambda_n);$$ $$\mathrm{tr}\,S^k(A)=h_k(\lambda_1,\cdots,\lambda_n),$$

where $e_k$ is the $k$th elementary symmetric polynomial and $h_k$ the $k$th complete homogeneous symmetric polynomial. Fortunately, the symmetric power sum polynomials

$$p_k(x_1,\cdots,x_n)=x_1^k+\cdots+x_n^k$$

also form a basis for the symmetric polynomials $k[x_1,\cdots,x_n]^{S_n}$, and $\mathrm{tr}(A^k)=p_k(\lambda_1,\cdots,\lambda_k)$, so we may recursively write the traces on $\Lambda^k$ and $S^k$ as polynomials in the traces of powers of $A$, using generating functions. Helpfully, Qiaochu writes an explicit formula I was not previously aware of, for the particular case of $k=n$, the dimension of $V$, we have

$$\mathrm{tr}\,\Lambda^n(A)=\frac{1}{n!}\sum_{\pi\in S_n}\mathrm{sgn}(\pi)\mathrm{tr}_\pi(A).$$

Note $\mathrm{sgn}(\pi)$ is the parity of the permutation $\pi$, and Qiaochu defines $\mathrm{tr}_\pi(A)$ to be $\mathrm{tr}(A^{f_1})\cdots\mathrm{tr}(A^{f_m})$, where $(f_1,\cdots,f_m)$ is the cycle type of $\pi$. See the references given in his answer for more info. Also, the symmetric trace $\mathrm{tr}\,S^k(A)$ has the same formula but without the $\mathrm{sgn}(\pi)$. (Representations act as invertible linear maps, so we may now choose to write $g$ instead of $A$. Assuming it isn't defective.)

Otherwise, we may use the family of Newton-Girard-type identites in order to recursively define trace formulae, as given by draks' answer on the previously linked question;

$$\mathrm{tr}\,\Lambda^k(g)=\frac{1}{k}\sum_{m=1}^k(-1)^{m-1}\mathrm{tr}(g^m)\mathrm{tr}\,\Lambda^{k-m}(g).$$

It's the same for $\mathrm{tr}\,S^k(g)$, except the summands are not alternating like above, i.e. no $(-1)^{m-1}$.

$^\dagger$The characteristic should not divide $n!$, where $n=\dim V$. The conclusions here can be extended to defective $A$ by considering e.g. continuity after taking care of the case of diagonalizable $A$ on the assumption of an underlying topology.


Explicitly, in addition to the $k=2$ case in rattle's answer, this gives us

$$\begin{array}{c l} \chi_{\Lambda^2(V)},~\chi_{S^2(V)} & =\frac{\chi(g)^2\mp\chi(g^2)}{2}, \\ \chi_{\Lambda^3(V)},~\chi_{S^3(V)} & =\frac{\chi(g)^3\mp 3\chi(g)\chi(g^2)+2\chi(g^3)}{6}, \\ \chi_{\Lambda^4(V)},~\chi_{S^4(V)}& =\frac{\chi(g)^4\mp6\chi(g)^2\chi(g^2)+3\chi(g^2)^2+8\chi(g)\chi(g^3)\mp6\chi(g^4)}{24}\end{array}$$

$$\cdots \cdots \cdots \cdots$$

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This is not quite an answer, but Fulton & Harris, §2.1 on page 13 gives a Formula for $k=2$:

$$\chi_{\bigwedge^2 V}(g) = \frac{1}{2}\cdot\left( \chi_V(g)^2 - \chi_V(g^2)\right)$$

as well as, in the Exercise below,

$$\chi_{\mathrm{Sym}^2(V)}(g) = \frac{1}{2}\cdot\left( \chi_V(g)^2 + \chi_V(g^2)\right)$$

Maybe you can look into the proof for the first equality and generalize.

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6  
$2\chi_{\bigwedge^2 V}(g)$ looks like the square of a standard deviation $\sigma = \sqrt{\operatorname E[X^2]-(\operatorname E[X])^2}. $ Is there a connection? –  draks ... Apr 30 '12 at 9:42

I ran into this question when trying to find a reference for some formulas that I think should be true. I will leave them here, in case they are of some help to someone. Here are the formulas, and I comment on the derivation below: $$ \sum_{n=0}^{\infty}t^n\chi_{S^n\alpha}(g)=\exp\left\{\sum_{p=1}^{\infty}t^p\frac{\chi_\alpha(g^p)}{p}\right\},\\ \sum_{n=0}^{\infty}t^n\chi_{\Lambda^n\alpha}(g)=\exp\left\{-\sum_{p=1}^{\infty}(-t)^p\frac{\chi_\alpha(g^p)}{p}\right\}. $$ And in particular $$ \sum_{n=0}^{\infty}t^n\chi_{S^n\alpha}=\frac{1}{\sum_{n=0}^{\infty}(-t)^n\chi_{\Lambda^n\alpha}}. $$ The latter formula also can be seen as a direct consequence of the facts described in anon's answer and the relation between generating functions for complete homogeneous symmetric polynomials and for elementary symmetric polynomials.

As for the first two formulas, the key observation is given in Qiaochu's answer $$ \chi_{\Lambda^n \alpha}(g)=\frac{1}{n!}\sum_{\pi\in S_n}\mathrm{sgn}{\pi} \,\mathrm{tr}_\pi \alpha(g), $$ where $\mathrm{tr}_\pi(A)$ is defined in Qiaochu's answer and in anon's answer above. Slight modification gives $$ \chi_{S^n \alpha}(g)=\frac{1}{n!}\sum_{\pi\in S_n}\mathrm{tr}_\pi \alpha(g). $$ Now these can be more or less straightforwardly made into the generating functions above by the use of Polya enumeration (Qiaochu gives the relevant corollary, $Z_G$ is defined here). One just has to express the sign of the permutation in terms of the cycle type.

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