# Can an operator have Exp(z) as its characteristic “polynomial”?

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.

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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

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

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Since $e^z$ has no zeros, it seems that such a $T$ either cannot exist or must be extremely weird to begin with. –  ARupinski Oct 11 '11 at 0:45
Well, it does have zeros... –  András Bátkai Oct 11 '11 at 5:26
@András Bátkai What do you mean? –  John Wiltshire-Gordon Oct 11 '11 at 13:58
@ARupinski: wouldn't the Fredholm determinant of a quasinilpotent trace-class operator be an entire, nowhere-vanishing function? –  Yemon Choi Oct 11 '11 at 23:25
I think it's true that $f(z)={\rm Det}(I+zT)$ can be written as $\prod\_n(1+z\lambda\_n)$ where $\lambda\_n$ are the eigenvalues of $T$ (I just say 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. –  George Lowther Oct 13 '11 at 0:36

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. 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 ${\rm det}(1+zT)=1$.

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This is not an answer, but just a sketch of a possible answer; I think the answer is no. For $T$ as desired, consider $Det(I-z^2T^2)=Det(I-zT)\cdot Det(I+zT)=e^{-z}\cdot e^z\equiv 1$. Considering the Taylor expansion of this function, one can conclude that $Trace(T^{2n})=0$ for any natural $n$. I think this yields $T^2=0$, and then the claim follows.

Actually, I am not sure if such determinant may have positive exponential type - this could be another approach.

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If $T^2$ is a quasinilpotent trace class operator, then it has trace zero (Lidskii). So all powers of $T^2$ have trace zero. Does this imply that $\Tr(\wedge^{2k} T) \neq 1/(2k!)$? –  Yemon Choi Oct 12 '11 at 22:07
If so, maybe my first argument that $T^2=0$ is wrong; I got it from the finite-dimensional situation. –  kap44 Oct 12 '11 at 22:19
Don't you mean ${\rm Trace}(\Lambda^nT^2)=0$? –  George Lowther Oct 12 '11 at 22:21
or are these the same thing? –  George Lowther Oct 12 '11 at 22:24
Ok. That proves that $T$ is quasinilpotent (all eigenvalues are zero). Then, I think, $\Lambda^nT$ is also quasinilpotent and ${\rm Det}(I+zT)=1$, giving a negative result to the question. –  George Lowther Oct 12 '11 at 22:53