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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?
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Where are all those minus signs coming from if $A$ is positive? –  Qiaochu Yuan Apr 19 '12 at 14:52

3 Answers 3

up vote 13 down vote accepted

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.

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I very much like this answer. Thank you! I'm more-or-less happy with Jacobi's identity popping up in this context, because it is how the Alexander polynomial (equivalent to Ray-Singer torsion for a knot complement) indeed arises as a quantum invariant (Aarhus integral gives rise to exp tr of the equivariant linking matrix, whatever that means), so it fits well with my preconceptions. –  Daniel Moskovich Apr 20 '12 at 1:26
    
@Tom: done. (More characters.) –  Qiaochu Yuan Apr 20 '12 at 14:45

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.

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Here are some suggestions, until somebody more knowledgeable appears.

  1. Why does this definition make sense: Analytic continuation is unique, and hence if there is analytic continuation to neighbourhood of $0$, and the formula is valid in finite dimension, then why not take it. Unfortunately, we have in general that $$ \det (AB) \neq \det(A) \det(B).$$

  2. Why does it possibly generalize the theory of the Riemann zeta function: If you take the circle with the Laplace operator $-\frac{\partial^2}{\partial^2 x}$ on $L^2[0,1]$ has eigenfunctions $\exp(2 \pi inx)$ for $n \in \mathbb{Z}$, you actually will get the Riemann zeta function as the spectral zeta function $\zeta_A(s) = 2 \zeta(2s)$.

  3. Why does it not: Certain properties are not shared by the general spectral zeta function. There is in general no functional equation relating $\zeta_A(s)$ to $\zeta_A( D-s)$ for some $D$, and there is no Euler product.

But here is also a quote from page 20, of "Ten physical applications of spectral zeta functions" from Elizalde, pg.20

Actually, a universal definition for the determinant [...] is still missing.

Also Singer and Ray refer in http://www.sciencedirect.com/science/article/pii/0001870871900454 to Seeley http://www.ams.org/mathscinet/search/publications.html?pg1=IID&s1=157950 , who has proven the regularity at $0$. He has a nice formula there:

$$A^s=i/2π\int_{Γ} λ^s(A−λ)^{−1}dλ$$

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It seems that this whole theory (of spectral zeta functions, etc) was developed by Bob Seeley, who is not quite getting all the credit he deserves. –  Igor Rivin Apr 19 '12 at 16:31
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It's not the first time, or is it? –  Marc Palm Apr 19 '12 at 17:24
    
$ det(A.B)=detA.detB $ in case its commutator is zero (i think) $ [A,B]=0 $ –  Mathman Apr 21 '12 at 13:56

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