# Trace of the exterior powers of linear operators

Given linear operators $K_1,\ldots,K_m$ on a Hilbert space $\mathcal H$, what can we say about the trace of their exterior product $Tr \,(K_1\wedge \cdots \wedge K_m)$ ?

More precisely:

1) If we assume assume $K=\sum_n f_n\otimes g_n$ (with $f_n\in\mathcal H$ and $g_n\in\mathcal H'$) is of finite rank (trace class would be enough), we have $$Tr\, (\underbrace{K\wedge \cdots \wedge K}_m)=\frac{1}{m!} \sum_{n_1<\cdots <n_m}\det\Big[\langle f_{n_i},g_{n_j}\rangle\Big]_{i,j=1}^m.$$

Does a similar kind of formula exists for $Tr \,(K_1\wedge \cdots \wedge K_m)$ when $K_j=\sum_n f_n^{(j)}\otimes g_n^{(j)}$ ?

2) Given any permutation $\sigma\in\frak S_m$ is there any relations between $Tr \,(K_{\sigma(1)}\wedge \cdots \wedge K_{\sigma(m)})$ and $Tr \,(K_1\wedge \cdots \wedge K_m)$ ? I have the feeling they should be equal. The reason why is the following argument (that you could transpose from $2$ to $m$):

Let $(e_i)_i$ be a basis for $\mathcal H$. Then, $(e_i\wedge e_j)_{i<j}$ is a basis for $\wedge^2\mathcal H$ and thus, by setting $A_{ij}=\langle A e_i,e_j\rangle$ and similarly for $B$, we obtain

$$Tr(A\wedge B)=\sum_{i<j}\langle (A\wedge B) e_{i}\wedge e_j,e_i\wedge e_j\rangle=\sum_{i<j}(A_{ii} B_{jj} -A_{ij}B_{ji}).$$ Next, since $(e_i\wedge e_j)_{i>j}$ is also a basis for $\wedge^2\mathcal H$, we have $$Tr(B\wedge A)=\sum_{i>j}\langle (B\wedge A) e_{i}\wedge e_j,e_i\wedge e_j\rangle=\sum_{i>j}(B_{ii} A_{jj} -B_{ij}A_{ji}).$$ But I feel there is something fishy, I am right ?

I don't see how to understand $A \wedge B$. Is it the operator that takes $x \wedge y$ to $Ax \wedge By$, or the operator that takes $x\wedge y = -y\wedge x$ to $-Ay\wedge Bx = Bx \wedge Ay$? So it only makes sense if $A = B$. Also, we don't have $A \wedge A = 0$, we only have $x\wedge x = 0$. And in fact some people prefer the notation $\Lambda^{(2)} (A)$ as it is the induced map on $\Lambda^{(2)}(\mathcal H)$.
If $\mathcal H$ is finite dimensional, and the eigenvalues of $A$ are $\lambda_1,\dots,\lambda_n$, then the eigenvalues of $\Lambda^{(2)} (A)$ are $(\lambda_i \lambda_j)_{i>j}$. Hence $$\text{Tr}(\Lambda^{(2)} (A)) = \sum_{i>j} \lambda_i \lambda_j = \tfrac12 \sum_i \lambda_i \sum_{j\ne i} \lambda_j = \tfrac12 \sum_i (\text{Tr}(A) - \lambda_i) = \tfrac12(\text{Tr}(A)^2 - \|A\|^2_{\text{Frobenius}}) .$$ and this equals $\tfrac12 \sum_{i,j} A_{ii} A_{jj} - A_{ij} A_{ij} = \sum_{i>j} A_{ii} A_{jj} - A_{ij} A_{ij}$, which is the same as the formula you obtained if you had written it in the case $A = B$.
• I realized later my question sounded weird indeed, but nobody seemed to care... I also realized there may be a proper definition for $A_1\wedge \cdots \wedge A_m$ : Take the operator $\mathcal Alt : \mathcal H^{\otimes m}\to \mathcal H^{\wedge m}$, $v_1\otimes \cdots \otimes v_m\mapsto \sum_{\sigma\in S_m}(-1)^\sigma v_{\sigma(1)}\otimes \cdots\otimes v_{\sigma(m)}$. Then, define $(A_1\wedge \cdots \wedge A_m)$ by $\mathcal Alt (A_1\otimes \cdots \otimes A_m)\mathcal Alt$ (with the identification/definition $\mathcal Alt(v_1\otimes \cdots\otimes v_m)=v_1\wedge\cdots\wedge v_m$). – Student Jul 27 '15 at 1:26