# Why do zeta regularization and path integrals agree on functional determinants?

When looking up the functional determinant on Wikipedia, a reader is treated to two possible definitions of the functional determinant, and their agreement is trivial in finite dimensions.

The first definition is based on zeta function regularization. If an operator $S$ has the spectrum of eigenvalues $\{\lambda_i\}$ then the associated zeta function is formally the operator trace

$$\zeta_S(s)=\mathrm{tr}(S^{-s})=\sum_{i=1}^\infty \lambda_i^{-s}.$$

The sum converges only when the $\mathrm{Re}(s)$ is sufficiently large, so $\zeta_S$ is defined by analytic continuation elsewhere on $\mathbb{C}$. Formally, this means that (in symbolic appearance at least)

$$\det S =e^{-\zeta_S'(0)}=\prod_{i=1}^\infty\lambda_i.$$

Though this isn't literally convergent, it does establish an intuitive basis for why the quantity may be called a determinant through analogy with the case in finite dimensions.

On the other hand, the following path integral quantity is a second possible avenue to defining the determinant for suitable operators:

$$\frac{1}{\sqrt{\det S}} \propto\int e^{-\pi\langle \phi,S\phi\rangle}\, \mathcal D\phi.$$

In finite-dimensional Euclidean space, the proportion is an actual equality, which can be seen by writing the inner product as $\langle x, Sx\rangle=\lambda_1x_1^2+\cdots+\lambda_nx_n^2$, separating the integral and then observing that each factor is either $\int dx_i=\infty$ when $\lambda_i=0$ or a rescaled Gaussian integral otherwise. However, in the infinite-dimensional case we can only compare determinants of operators in relative proportion to each other, so that divergent constants cancel appropriately.

It is stated that the results of these two definitions agree with each other, and Wikipedia cites the paper Extremals of Determinants of Laplacians as having established this fact. However, of what little in the paper that I can genuinely follow, I don't see any demonstration that the zeta regularization and path integral formulation agree with each other, so either I'm so out of my depth I can't even recognize the proof let alone understand it, or the Wikipedia article is misguided.

The former is very much a possibility - I understand what manifolds are and can do some basic tensor manipulations to, say, derive the geodesic equation, but other than this I'm not educated in differential geometry, and I am likewise ignorant to all but the basic construction of a path integral. I'd be appreciative if someone could shine a light on the underlying theory in play here at a level I can understand, if possible.

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I retagged from (path-integrals) to (functional-integration) as the former can be misinterpreted to mean line-integrals in the context of multivariable calculus and complex analysis. –  Willie Wong Jul 30 '11 at 13:58
I scanned through that paper and also couldn't find anything relevant. You might want to ask this on theoreticalphysics.stackexchange.com, where people will probably be more familiar with functional determinants. –  joriki Nov 23 '11 at 22:38
Tom left a comment about an arXiv paper underneath his answer. –  joriki Dec 11 '11 at 12:11