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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?

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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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$2\chi_{\bigwedge^2 V}(g)$ 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
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