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The following is a historical question, but first some background:

Let $\psi$ be a linear operator from a vector space to itself. The following two expressions, viewed as formal power series, can be shown to be equal (even for an infinite dimensional vector space):

$$ \det(1 - \psi T) = \exp \left(-\sum_{s=1}^{\infty} \text{Tr}(\psi^{s})T^{s}/s\right) $$

Here is a simple example using the associated matrix in the finite dimensional case:

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For more on this topic (and how to ensure the definitions make sense for infinite dimensional vector spaces) see: N. Koblitz, p-adic Numbers, p-adic Analysis, and Zeta-Functions, Springer-Verlag, New York, 1984.

In particular, see Koblitz's book for how ($p$-adic versions of) this identity can be used in Bernard Dwork's proof of the rationality of the zeta function (resolving the first of the Weil Conjectures).

I would welcome any insight as to how one thinks up, observes, or conjectures such an identity in the first place, but my question is: what is the origin of this identity and where else has it appeared (perhaps in a somewhat modified form)?

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Its discussed on the wikipedia page for the determinant. It can be viewed as a form of the identity $\exp(\mathrm{tr}(M))=\det\exp(M)$. –  Chris Godsil Jan 13 '13 at 20:10
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@ChrisGodsil If you can write up something formal showing why this is the case, I will accept it as an answer. (Unless others can shed more light on where else this identity shows up and what its history is.) –  Benjamin Dickman Jan 13 '13 at 21:20
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Since you ask the question in relation to its use in Dwork's proof of the first Weil conjectures... I think this kind of identity was known to Weil, who first defined the zeta-function of a variety in fairly general terms. You might look at his 1949 paper where he stated his conjectures, just to see if it provides some inspiration for you. –  KCd Jan 14 '13 at 4:23

2 Answers 2

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+300

The development of this operator identity can be traced back till the very beginning of the 20th century.

I think the paper On the numerical evaluation of Fredholm determinants from Folkmar Bornemann (2008) could be useful for you. Bornemann presents in Section 3: Definition and Properties of Fredholm and Operator Determinants of this paper four different representations of your stated operator identity. He writes:

For a trace class operator $A \in \mathcal{J}_1(\mathcal{H})$ there are several equivalent constructions that all define one and the same entire function \begin{align*} d(z)=\text{det}(I+zA)\qquad(z\in\mathbb{R}) \end{align*} in fact, each construction has been chosen at least once, in different places of the literature, as the basic definition of the operator determinant:

$1.)$ Gohberg and Krein ($1969$, p. $157$) define the determinant by the locally uniformly convergent (infinite) product \begin{align*} \text{det}(I+zA)=\prod_{n=1}^{N(A)}(1+z\lambda_n(A)) \end{align*} which possesses zeros exactly at $z_n=-1/\lambda_n(A)$, counting multiplicity.

$2.)$ Gohberg et al. ($1990$, p. $115$) define the determinant as follows. Given any sequence of finite rank operators $A_n$ with $A_n\rightarrow A$ converging in trace class norm, the sequence of finite dimensional determinants \begin{align*} \text{det}\left(I+zA_n\upharpoonright_{\text{ran}(A_n)}\right) \end{align*} (which are polynomials in z) converges locally uniform to $\text{det}(I+zA)$, independently of the choice of the sequence $A_n$. The existence of at least one such sequence follows from the singular value representation at $(2.1)$.

$3.)$ Dunford and Schwartz ($1963$, p. $1029$) define the determinant by what is often called Plemelj’s formula \begin{align*} \text{det}(I+zA)=\exp(\text{tr}\ \text{log}(I+zA))=\exp\left(-\sum_{n=1}^{\infty}\frac{(-z)^n}{n}\text{tr}A^n\right) \end{align*} which converges for $|z|<1/|\lambda_1(A)|$ and can analytically be continued as an entire function to all $z\in\mathbb{C}$.

$4.)$ Grothendieck ($1956$, p. $347$) and Simon ($1977$, p. $254$) define the determinant most elegantly with a little exterior algebra (Greub $1967$). With $\bigwedge^n(A)\in\mathcal(J)_1\left(\bigwedge^n(\mathcal{H})\right)$ being the $n^\text{th}$ exterior product of $A$, the power series \begin{align*} \text{det}(I+zA)=\sum_{n=0}^{\infty}z^n\text{tr}\bigwedge^n(A) \end{align*} converges for all $z\in\mathbb{C}$. Note that $\text{tr}\bigwedge^{n}(A)=\sum_{i_1<\cdots<i_n}\lambda_{i_1}(A)\cdots\lambda_{i_n}(A)$ is just the $n^\text{th}$ symmetric function of the eigenvalues of $A$.

Proofs of the equivalence can be found in (Gohberg et al. $2000$, Chap. $2$) and (Simon $2005$, Chap. $3$).


In a footnote for $3.)$ above, Bornemann provides additional information to the Plemelj Formula, namely:

Plemelj ($1904$, Eq. $(62)$) had given a corresponding form of the Fredholm determinant for integral operators. However, it can already be found in Fredholm ($1903$, p. $384$).

So, the formula seems to go back till Fredholm $1903$. I suppose according to some references I've found, it's the following paper: I. Fredholm, Sur une classe d`équation fonctionelle, Acta Math. $27, 365-390 (1903)$.

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@Benjamin Dickman: Thanks for the bounty. Since you put the question long time ago, its really good to see the answer pleases you! :-) –  Markus Scheuer May 3 at 17:11

Using $\det\exp=\exp\mathrm{tr}$, we have $$ \det(I-tA) = \det \exp\log(I-tA) = \exp\mathrm{tr}\log(I-tA) $$ and as $$ \log(I-tA) = -\sum_{k\ge1} \frac1k t^k A^k, $$ we have $$ \exp\mathrm{tr}\log(I-tA) = \exp\left( -\sum_{k\ge1} \frac1k t^k \mathrm{tr}(A^k)\right) $$ This shows that your identity is a form of the $\det\exp$ identity.

The question about the origins is interesting and I do not have much to say (which was why I restricted myself to a comment in the first place). For complex matrices it is very easy to prove that $\det\exp(A)=\exp\mathrm{tr}(A)$, which might explain why I have never seen a name attached to it.

I have seen it used in connection with Lie groups (to prove that sl is the Lie algebra of SL), and (in forms like what you stated) in work on enumeration.

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