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


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



Can an operator have Exp(z) as its characteristic "polynomial"?
Here's a proof that $\exp(z)$ is not a characteristic function using the product expansion for the determinant, which is essentially equivalent to Lidskii's theorem stating that the trace of a trace class operator is the sum of its eigenvalues.



> 
> If $T$ is a trace class operator on a Hilbert space $\mathcal{H}$ with eigenvalues $\lambda\_n$ (counted up to algebraic multiplicity), then $\sum\_n\vert\lambda\_n\vert < \infty$ and
>  $$
> {\rm det}(1+zT)\equiv\sum\_{k=0}^\infty{\rm Tr}(\Lambda^kT)z^k=\prod\_n(1+z\lambda\_n)
> $$
>  for all $z\in\mathbb{C}$.
> 
> 
> 


See [*Trace Ideals and Their Applications*, Theorem 3.7](http://books.google.co.uk/books?id=-1snx2pn59cC&lpg=PA32&ots=tzxSfOUAhs&dq=%252B%2522trace%2520class%2522%2520%252Beigenvalues%2520%252Blidskii&pg=PA35#v=onepage&q=Theorem%25203.7&f=false). Actually, they use this result to prove Lidskii's theorem, that ${\rm Tr}(T)=\sum\_n\lambda\_n$.


Then, ${\rm det}(1+zT)$ cannot be equal to $\exp(z)$ everywhere. This is because $\exp$ has no zeros, so $\lambda\_n$ would have to be all zero, giving ${\rm det}(1+zT)=1$.


