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Let $V$ be a representation with character $\chi$. I would like to have a formula for the characters of the representations $\mathrm{Sym}^m[V]$ and $\wedge ^m[V]$ in terms of $\chi$. Fulton and Harris presents such a formula for $m=2$, but not for general $m$. Is there a "nice" such expression?

I found a paper by Zhou and Pulay, but this doesn't seem to be what I'm looking for (it doesn't seem to give a general formula in terms of $\chi$).

In case the link does not work, here is the reference:

X. Zhou, P. Pulay. Characters for Symmetric and Antisymmetric Higher Powers of Representations: Application to the Number of Anharmonic Force Constants in Symmetrical Molecules. Journal of Computational Chemistry, Vol 10, Issue 7, (1989).

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X. Zhou, P. Pulay. Characters for Symmetric and Antisymmetric Higher Powers of Representations: Application to the Number of Anharmonic Force Constants in Symmetrical Molecules. Journal of Computational Chemistry, Vol 10, Issue 7, (1989). – Jonathan Gleason Nov 21 '11 at 3:23
@Srivatsan Is this what you meant by "reference"? – Jonathan Gleason Nov 21 '11 at 3:24
2… is very similar, and gives a textbook reference – Jack Schmidt Nov 21 '11 at 5:54

For a given small $m$, this is a good exercise and I recommend that you do it yourself. I am not aware of a formula for general $m$.

Step 1: show that if $v_1,\ldots,v_n$ is a basis for $V$ (which you might as well take to consist of eigenvectors under $g\in G$), then $$ \left\{\sum_{\sigma\in S_m}v_{k_{\sigma(1)}}\otimes\cdots\otimes v_{k_{\sigma(m)}}: 1\leq k_1\leq k_2\leq\ldots\leq k_m\leq n\right\} $$ is a basis for $\text{Sym}^mV$, while $$ \left\{\sum_{\sigma\in S_m}\text{sign}(\sigma)v_{k_{\sigma(1)}}\otimes\cdots\otimes v_{k_{\sigma(m)}}: 1\leq k_1< k_2<\ldots< k_m\leq n\right\} $$ is a basis for $\Lambda^mV$. You need to prove linear independence and spanning (the latter being very easy).

Step 2: Assuming that you did take the $v_i$ to all be eigenvectors under $g\in G$, show that each of the above basis vectors is again a $g$-eigenvector and determine the eigenvalue (trivial!). This will give you the characters of $\text{Sym}^mV$ and $\Lambda^mV$ not quite in terms of $\chi$, but in terms of the eigenvalues of $g$ on $V$.

Given this expression, you can now express the character of any given $\text{Sym}^mV$ or $\Lambda^mV$ in terms of $\chi$. E.g. by simply multiplying out, you can show that $$ \chi_{\text{Sym}^3V}(g) = \frac{\chi(g)^3 + 3\chi(g^2)\chi(g) + 2\chi(g^3)}{6} $$ and $$ \chi_{\Lambda^3V}(g) = \frac{\chi(g)^3 - 3\chi(g^2)\chi(g) + 2\chi(g^3)}{6}. $$

For an approach via generating functions, see also question 5 on this exercise sheet.

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Let $\lambda _i$ be the eigenvalue corresponding to $v_i$. I then find that $\chi _{\mathrm{Sym}^m[V]}(g)=\sum _{1\leq k_1\leq \cdots \leq k_m\leq n}\lambda _{k_1}\cdots \lambda _{k_m}$. Similarly for $\wedge ^m[V]$. I don't see any way to write this in terms of the character of $V$, however. How do I do this step? – Jonathan Gleason Nov 21 '11 at 5:40
Dear Jonathan, I just made a small edit. I am not aware of a general formula for all $m$. For any given $m$, this is an exercise in symmetric functions, and I wrote down the answer for $m=3$ to get you started. In general, your character will be a linear combination of $\left\{\prod \chi(g^{r_i})^{e_i}:\sum e_ir_i = m\right\}$ (check that each of these has the right common degree in the $\lambda_j$). – Alex B. Nov 21 '11 at 5:47

The "exercise in symmetric functions" referred to by Alex in the above comment is actually the following exercise: Write down the $n$-th elementary symmetric polynomial in terms of the first $n$ power-sum symmetric polynomials. (This is for $\wedge^m V$. For $\mathrm{Sym}^m V$, we should replace "elementary" by "complete" here.) This can be done in a closed form if you consider determinants as "closed forms". Let me do it for the elementary symmetric polynomials:

If $X_1,X_2,X_3,...$ are indeterminates (finitely or infinitely many), $e_i$ is the $i$-th elementary symmetric polynomial of these indeterminates, and $p_i=X_1^i+X_2^i+X_3^i+...$ is the $i$-th power sum (for $i\geq 1$), then for every $n\in\mathbb N$, we have the formula $n!e_n = \det A_n$, where $A_n$ is the $n\times n$ matrix defined as follows:

The $\left(i,j\right)$-th entry of $A_n$ is $\left\lbrace \begin{array}{c} p_{i-j+1},\ \text{ if }i\geq j;\\ i,\ \text{ if }i=j-1;\\ 0,\ \text{ if }i < j-1 \end{array}\right.$ for all $1\leq i\leq n$ and $1\leq j\leq n$.

Here is how this matrix looks like:

$A_{n}=\left( \begin{array} {cccccc} p_{1} & 1 & 0 & \cdots & 0 & 0\\ p_{2} & p_{1} & 2 & \cdots & 0 & 0\\ p_{3} & p_{2} & p_{1} & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ p_{n-1} & p_{n-2} & p_{n-3} & \cdots & p_{1} & n-1\\ p_{n} & p_{n-1} & p_{n-2} & \cdots & p_{2} & p_{1} \end{array} \right) $.

The proof of this identity isn't as monstrous as the identity itself and procees by induction. I wrote it up some time ago (lambda, solution to Exercise 9.6 (b), currently on pp. 161-162, but refers back to the solution to Exercise 9.3 (b) which is on pp. 155-156), but as usual it is easier to prove things than to understand the mess of a proof I have written up.

To get an explicit formula for characters, take $g\in G$, and throughout the above, replace $e_i$ by $\chi_{\wedge^i V}\left(g\right)$ and replace $p_i$ by $\chi_V\left(g^i\right)$.

I wish I had the time to comment on why $\mathrm{Sym}^n V$ is analogous to $\wedge^n V$, but I don't.

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Why do you replace $e_i$ with $\chi _{\wedge ^iV}(g)$ instead of with $\chi _{\mathrm{Sym}^iV}(g)$? – Jonathan Gleason Dec 4 '11 at 19:47
Because the eigenvalues of $g$ acting on $\wedge^iV$ are the $i$-wise products of DISTINCT eigenvalues of $g$ acting on $V$, so the sum of the former eigenvalues (and that's $\chi_{\wedge^i V}\left(g\right)$) is the $i$-th elementary symmetric function of the latter. For $\mathrm{Sym}^iV$, it would be the $i$-th complete symmetric function, because there is no "DISTINCT" anymore. – darij grinberg Dec 4 '11 at 20:01

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