TITLE: Understanding zeta function regularization
QUESTION [24 upvotes]: 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:

 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.
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?

REPLY [20 votes]: This is a variant of Qiaochu Yuan's answer. Suppose $A$ is some sort of differential operator. Then it is hard to think of things like the trace and determinant of $A$. But there is one quantity which is well behaved: it is the "heat trace" $Tr(e^{tA})$. Of course the heat trace is connected to the solution of the heat equation 
$$\left(\frac{\partial}{\partial t} - A\right) u = 0.$$
So how can we get at the eigenvalues of $A$ if we know $Tr(e^{tA})$? Suppose first that $A$ is finite dimensional. Then, 
$$Tr(e^{tA}) = \sum_i e^{t\lambda_i},$$ 
but it is somewhat inconvenient to extract $\sum \log \lambda_i$ directly from this. But if you plug in the identity
$$\lambda^s = \frac{1}{\Gamma(s)} \int_0^\infty e^{t \lambda} t^s \frac{dt}{t}$$
you get
$$\zeta_A(s) = \frac{1}{\Gamma(s)} \int_0^\infty Tr(e^{tA}) t^s \frac{dt}{t} \qquad \qquad (*) $$
and then $\log \det A = \sum \log \lambda_i = -\zeta_A'(0)$. So in some sense the zeta function is almost forced on you if you want to extract the determinant from the heat trace. 
Now if $A$ is a differential operator, then the equation (*) makes sense, but you cannot directly set $s=0$ because the integral diverges at the limit $t=0$. However, using the well known asymptotics of the heat trace as $t \to 0$, one can subtract the leading order terms, and then the rest will converge. This allows one to define the determinant of $A$, and one can write an expression which does not involve the zeta function at all, but only the heat trace. In some sense this is the real mathematical definition of det A.  Otherwise how do we even know that the zeta function has an analytic continuation to $s=0$?
So perhaps the zeta function regularization is just some shorthand for this argument. One can say "analytic continuation of the zeta function" instead of specifying exactly which terms to subtract from the heat trace.