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This is an equality that I am gleaning out of some papers that I have been reading. I am not sure I am reading it right. Hopefully people will correct it.

Let $U$ be a group element and $R$ be a representation of the group. Let $sym^n (R)$ be the $n^{th}$ symmetric power of the representation and similarly $anti^n (R)$ is the $n^{th}$ anti-symmetric power of the representation . And $\chi$ be the character of the representations. Then apparently the following hold,

$$ \sum _{n=0} ^{\infty} t^n \chi _{sym ^n (R)} (U) = e^{\sum _{m =1} ^\infty \frac{t^m}{m} \chi _R (U^m)} $$

$$ \sum _{n=0} ^{\infty} t^n \chi _{anti ^n (R)} (U) = e^{\sum _{m =1} ^\infty \frac{(-1)^{m+1}t^m}{m} \chi _R (U^m)} $$

Can someone tell me its derivation or let me know if it is known by some name which I can track down in the books?

Also does any special thing happen if $U$ is in the adjoint representation?

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Anirbit, this is precisely the identity I described in my answer to your last question. You can prove it by expanding everything out in terms of the eigenvalues of U. –  Qiaochu Yuan Jan 31 '11 at 18:35
    
@Qiachu I did notice the similarity between the first equation I wrote here and the one you referred to in your last answer. But I still put it up here because of the second formula I wrote here which doesn't seem to have an obvious parallel in the last discussion. I was wondering if there is a general framework in which both of these fit together. May be I was being naive. –  Anirbit Feb 1 '11 at 10:36

2 Answers 2

up vote 3 down vote accepted

I explain a combinatorial proof of this identity in this blog post and this one; I'll sketch the second proof. First, this is really an identity of symmetric functions in the eigenvalues: the first says that

$$\sum_{n \ge 0} h_n t^n = \exp \left( \sum_{n \ge 1} \frac{p_n}{n} t^n \right)$$

where $h_n$ are the complete homogeneous symmetric polynomials and $p_n$ the power symmetric functions in the eigenvalues of $U$. The second says that

$$\sum_{n \ge 0} e_n t^n = \exp \left( \sum_{n \ge 1} (-1)^{n+1} \frac{p_n}{n} t^n \right)$$

where $e_n$ are the elementary symmetric functions in the eigenvalues. It is not hard to see that $\sum h_n t^n = \frac{1}{\sum e_n t^n}$, so actually these identities are equivalent; we will thus concentrate on the first one, which has no signs.

First, here is the standard proof (with analytic details omitted, since this is an equality of formal power series anyway), which I find unsatisfactory. Consider the operator $\frac{1}{1 - Ut} = \sum_{n \ge 0} U^n t^n.$ A standard matrix identity asserts that $\det \exp M = \exp \text{tr } M$ for any matrix $M$. Applying this identity to the logarithm of the above, we conclude that

$$\det \frac{1}{1 - Ut} = \exp \text{tr } \log \frac{1}{1 - Ut} = \exp \left( \sum_{n \ge 1} \frac{p_n}{n} t^n \right).$$

On the other hand, $\det \frac{1}{1 - Ut} = \frac{1}{\prod (1 - \lambda_i t)}$ where $\lambda_i$ are the eigenvalues, and this is precisely $\sum h_n t^n$.


Now here is a sketch of the combinatorial proof. First we assume that $U$ is the adjacency matrix of a finite graph $G$. Then $\text{tr } U^n$ is the number of closed walks of length $n$ on $G$. On the other hand, $\text{tr } \text{Sym}^n(U)$ describes something more complicated: the graph $G$ has a symmetric product $\text{Sym}^n(G)$ whose vertices are the unordered $n$-tuples of vertices of $G$ and whose edges are chosen so that $\text{Sym}^n(U)$ is its adjacency matrix.

Now $\text{tr } \text{Sym}^n(U)$ describes the number of closed walks of length $1$ on the symmetric product; that is, the number of loops. A loop on $\text{Sym}^n(U)$ is (roughly speaking) an unordered collection of closed walks on $U$ with total length $n$. This turns out to imply, via the well-known exponential formula in combinatorics, the desired identity.

Now, since the adjacency matrices of finite graphs are Zariski dense in $\text{GL}_n(\mathbb{C})$, the result follows for all matrices.


This identity has hidden depths. For a finite graph $G$ with adjacency matrix $A$, the function $\frac{1}{\det(I - At)}$ is a kind of zeta function for $G$, and the argument above is closely related to the Euler product of this zeta function.

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Thanks for the help. The inverse relation between your first two displayed equations is not obvious to me since if I multiply the two RHS then it seems that the odd n terms remain. Also the second equality in your third displayed equation also seems non-obvious. I guess I have read your 3 blog posts fully to figure them out. –  Anirbit Feb 2 '11 at 9:40
    
@Anirbit: ah, sorry, you need to replace one of the t's with -t first. In the third equation use the Taylor series for the logarithm. –  Qiaochu Yuan Feb 2 '11 at 11:32

Notice that neither identity involves the whole group: you only need a vector space $R$ and a linea map $U:R\to R$. You can pick a basis of $R$ so that $U$ is in its Jordan canonical form, so that it is easy to compute its trace and that of its powers; if the field is not algebraically closed, you can actually extend scalars, for nothing breaks by doing it: your identities are invariant under field extensions.

Now compute the trace of the action of $U$ on the symmetric and the antysimmetric powers of $R$ explicitely, and you are left with an identity between formal series (they actually converge, in fact, over the complex numbers, say) Now prove that identity is true.

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