[Math] Understanding zeta function regularization

analytic-number-theoryca.classical-analysis-and-odeslinear algebraregularization

I attended a talk this morning on Ray-Singer torsion, in which Rafael Siejakowski introduced zeta function regularization in a compelling way. The goal is to define the determinant of a positive self-adjoint operator $A$ with "pure point spectrum" $0>\lambda_1>\lambda_2>\cdots$. The definition of the determinant is $\exp(-\zeta_A^\prime(0))$ where $\zeta_A$ is the zeta function $\zeta_A(s)=\sum_{i=1}^\infty (-\lambda_i)^{-s}$. This sum diverges in general- but it converges for values of $s$ with large enough real part, and we define it for other values of $s$ (including zero) by analytic continuation.

Why should this be related to the determinant? Well, in the finite dimensional case (the motivating case is when $A$ is the `combinatorial Laplacian'), then $\zeta_A(s)=\sum_{i=1}^N (-\lambda_i)^{-s}$ is a finite sum. In this case:

$\zeta^\prime_A(s)=\sum_{i=1}^N -\ln (-\lambda_i)(-\lambda_i)^{-s}$

and

$\zeta^\prime_A(0)=-\ln \prod_{i=1}^N(-\lambda_i)=-\ln \det A$.

This looks to me like an ad-hoc trick, indicating that I don't understand what is actually going on.

The equation $\det(A)=\exp(-\zeta_A^\prime(0))$ (in the finite dimensional case an equation, in the infinite dimensional case a definition) equates two familiar mathematical quantities:

  1. The determinant, which I can think of as a volume, as an action on a highest exterior power, or maybe most evocatively as the signed sum of weights of non-intersecting paths in a graph between "source" vertexes $a_1,\ldots,a_n$ and "sink" vertices $b_1,\ldots,b_n$. See this blog post.
  2. The Riemann zeta function, which I don't understand conceptually almost as well, but which is heavily studied and so is clearly important and natural.

Question: Is there a conceptual (hand-wavy is fine) explanation for zeta function regularization, and for how this expression in the zeta function is capturing the idea of a "determinant"? How is the derivation which I wrote above more than an ad-hoc trick? Is there a sense in which the derivative of a zeta function at zero heuristically calculates a signed sum of weights of non-intersecting paths, or something like that?

Best Answer

Not a complete answer. First, here is an alternate derivation of the result in the finite-dimensional case which might be more enlightening. If $A$ is positive self-adjoint, we can write $A = \exp(L)$ for some self-adjoint $L$. This lets us define $$A^s = \exp(sL)$$

for all real $s$. The trace $$\text{tr}(A^s) = \sum_{i=1}^n \lambda_i^s = \zeta_A(s)$$

is then the zeta function associated to $A$ (I am getting rid of all of the minus signs). Now, for small $\epsilon$ we can write $$A^{s+\epsilon} = A^s A^{\epsilon} = A^s (1 + \epsilon L + O(\epsilon^2))$$

so it follows that $$\zeta_A(0)' = \text{tr}(L).$$

But Jacobi's identity $\det \exp M = \exp \text{tr } M$ gives $$\det A = \det \exp L = \exp \text{tr } L = \exp \zeta_A(0)'$$

and we conclude.


So what conceptual significance can we attach to the above? Well, it seems to me like we should think of the map $s \mapsto A^s$ as a representation of the Lie group $\mathbb{R}$, so the zeta function is the character of the corresponding representation. The derivative of the zeta function at zero gives the trace of the infinitesimal generator of this representation, $L$, which generates the abelian Lie algebra $\mathbb{R}$. And this is connected to the determinant of $A$ by Jacobi's identity.

So I think most of what needs explanation is Jacobi's identity. I freely admit that I do not have a good conceptual explanation of Jacobi's identity (beyond the fact that it's obvious for diagonalizable matrices). In these two blog posts I attempted to meander towards a combinatorial proof of Jacobi's identity in the form $$\det (I - At)^{-1} = \exp \text{tr } \log (I - At)^{-1}$$

(where $A$ was the adjacency matrix of a graph) but didn't quite succeed. There is a combinatorial proof of Jacobi's identity in the literature due to Foata but I haven't gone through it in detail.