# Characters of Symmetric and Antisymmetric Powers

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).

• 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
• math.stackexchange.com/questions/39751/… 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 C1 on this exercise sheet.

• 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 $\operatorname{Sym}^m V$, 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 first:

1. Let $X_1,X_2,X_3,\ldots$ be indeterminates (finitely or infinitely many). Let $e_i$ be the $i$-th elementary symmetric polynomial of these indeterminates for each $i \in \mathbb{N}$ (this polynomial is $1$ if $i = 0$). Let $p_i=X_1^i+X_2^i+X_3^i+\cdots$ be the $i$-th power sum for each $i\geq 1$. 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: see the solution to Exercise 2.9.13 (b) in Darij Grinberg and Victor Reiner, Hopf Algebras in Combinatorics (the numbering might change in the future, but search for "These are not novel for us" to find it). An equivalent version of the identity can be stated in the language of lambda-rings (lambda, solution to Exercise 9.6 (b), currently on pp. 203--205, but refers back to the solution to Exercise 9.3 (b) which is on pp. 196--198).

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

2. The symmetric power $\operatorname{Sym}^n V$ is analogous to $\wedge^n V$. Here is the counterpart of the above for symmetric powers:

Let $X_1,X_2,X_3,\ldots$ be indeterminates (finitely or infinitely many). Let $h_i$ be the $i$-th complete homogeneous symmetric polynomial of these indeterminates for each $i \in \mathbb{N}$ (this polynomial is $1$ if $i = 0$). Let $p_i=X_1^i+X_2^i+X_3^i+\cdots$ be the $i$-th power sum for each $i\geq 1$. For each $i \geq 1$, define $q_i = \left(-1\right)^{i-1} p_i$. Then, for every $n\in\mathbb N$, we have the formula $n!h_n = \det C_n$, where $C_n$ is the $n\times n$ matrix defined as follows:

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

Here is how this matrix looks like:

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

There is a quick way to derive this formula from the above formula $n!e_n = \det A_n$. Namely, let us restrict ourselves to the case where we have infinitely many indeterminates $X_1, X_2, X_3, \ldots$. Indeed, it suffices to prove the result in this case only, since the case of finitely many indeterminates can be obtained from it by setting the remaining indeterminates to $0$. Let $\Lambda$ be the ring of symmetric functions in the infinitely many indeterminates $X_1, X_2, X_3, \ldots$. Then, it is known that there exists an automorphism $\omega$ of $\Lambda$ sending each $e_i$ to $h_i$, and that this automorphism also sends each $p_i$ to $\left(-1\right)^{i-1} p_i = q_i$. (This automorphism $\omega$ is studied in Section 2.4 of Darij Grinberg and Victor Reiner, Hopf Algebras in Combinatorics, but the fact that it sends $p_i$ to $\left(-1\right)^{i-1} p_i$ is not spelt out explicitly there; instead, you can derive it from (2.4.7) and Proposition 2.4.1 (i).) Now, applying the automorphism $\omega$ to both sides of the equality $n!e_n = \det A_n$, we obtain the equality $n!h_n = \det C_n$ (since $\omega$ sends each entry of the matrix $A_n$ to the corresponding entry of $C_n$).

To get an explicit formula for characters of the symmetric powers $\operatorname{Sym}^n V$ of a representation $V$ of $G$, take $g\in G$, and throughout the above, replace $h_i$ by $\chi_{\operatorname{Sym}^i V}\left(g\right)$ and replace $q_i$ by $\left(-1\right)^{i-1} \chi_V\left(g^i\right)$.

• 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