Let $\mathcal{H}$ be a Hilbert space, and let $T: \mathcal{H} \rightarrow \mathcal{H}$ be a trace-class operator. Define $$ f_T(z) = \sum_{i=0}^\infty \mbox{Tr}(\wedge^k T) \cdot z^k, $$

the ordinary generating function for traces of exterior powers of $T$. Expressed another way,

$$ f_T(z) = \mbox{Det}(I + zT) $$

where $\mbox{Det}$ is the Fredholm determinant. This function is entire, and can be considered a generalization of the characteristic polynomial.

I am wondering if $$f_T(z) = e^z $$ for some natural choice of $T$ on some nice incarnation of Hilbert space.

The motivation for this problem comes from representation theory, where finite minors of such a $T$ provide a formula for dimensions of irreps of $S_n$, the symmetric group on $n$ letters.

--edit--

In order to provide slightly more background on the sort of application I have in mind, here is an attractive identity which might pique your interest:

If $\beta : \mathbb{C} \rightarrow \mathbb{C}$ is any function, define $$\gamma(z) = \frac{1}{\Gamma(z+1)}, \hspace{.4in} \delta(z) = \frac{\beta(z)}{\Gamma(z+1)}$$ Now

$$ \left| \begin{pmatrix} \gamma(3) & \gamma(4) & \gamma(5) \\\\ \gamma(-1) & \gamma(0) & \gamma(1) \\\\ \gamma(-2) & \gamma(-1) & \gamma(0) \end{pmatrix} \begin{pmatrix} \delta(3) & \delta(4) & \delta(5)\\\\ \delta(-1) & \delta(0) & \delta(1)\\\\ \delta(-2) & \delta(-1) & \delta(0) \end{pmatrix} \right| + $$ $$ \left| \begin{pmatrix} \gamma(2) & \gamma(3) & \gamma(4) \\\\ \gamma(0) & \gamma(1) & \gamma(2) \\\\ \gamma(-2) & \gamma(-1) & \gamma(0) \end{pmatrix} \begin{pmatrix} \delta(2) & \delta(3) & \delta(4)\\\\ \delta(0) & \delta(1) & \delta(2)\\\\ \delta(-2) & \delta(-1) & \delta(0) \end{pmatrix} \right| + $$ $$ \left| \begin{pmatrix} \gamma(1) & \gamma(2) & \gamma(3) \\\\ \gamma(0) & \gamma(1) & \gamma(2) \\\\ \gamma(-1) & \gamma(0) & \gamma(1) \end{pmatrix} \begin{pmatrix} \delta(1) & \delta(2) & \delta(3)\\\\ \delta(0) & \delta(1) & \delta(2)\\\\ \delta(-1) & \delta(0) & \delta(1) \end{pmatrix} \right| = \frac{\delta(1)^3}{3!} $$

think, because I haven't really looked at Fredholm determinants before). Then, if $f$ is nowhere zero, we have $\lambda\_n=0$, so $f(z)=1$. Therefore, the answer to the question is no. $\endgroup$ – George Lowther Oct 13 '11 at 0:36