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Let $S:V \rightarrow V$ be a linear operator on the function space $V$. It is possible to define a functional determinant for $S$ via the zeta function regularization process.

In specific we define first the associated $\zeta$-function $\zeta_S(z):=\text{Tr}(S^{-z})$ for $\Re(z)>>0$, then we extend $\zeta_S(z)$ by analytic continuation, and finally we define the functional determinant by $\det(S):=\text{exp}(-\zeta_s'(0))$.

The details of the definition are rather clear, but I'm puzzled about the meaning of the construction.

Question What does the functional determinant tell us about the operator $S$? To be more specific, what can we say about $S$ if $\det(S)=0$?

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