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From a talk, we learned that

  • The symplectic volume of the space $M$ of gauge equivalence classes of flat G connections is given by the “Witten zeta function”: enter image description here

where we sum over irreducible representations $R$ of $G$. In the formulas, the symbol $∼$ means that the left hand side is proportional to the right hand side by a known proportionality constant.

In the special case of SU(2), we have enter image description here

where we sum over the irreducible representations of SU(2), which are parametrized by their dimensions n. In this case, the Witten zeta function reduces to the Riemann zeta function.

Question:

  1. Is it correct to say the known proportionality constant is given by Reidemeister–Ray–Singer torsion? Is it correct to say the symplectic volume forms is also given by Reidemeister–Ray–Singer torsion? How is that mathematically rigorous derived?

  2. For SU(2), $$ \text{Witten zeta function = Riemann zeta function} $$ What is the analogy for U(1)? What is the analogy for SU($N$)? $$ \text{Witten zeta function for U(1) or SU($N$) = ?} $$

  3. What is the simple intuition behind based on "the asymptotics of the Verlinde formula" to obtain Witten zeta function (in the Ref below)?

Ref: E. Witten, ‘On quantum gauge theories in two dimensions’, Commun. Math. Phys. 141 (1991) 153–209.

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    $\begingroup$ The article by Witten that you mention gives an exact formula. For instance for $\operatorname{SU}(2) $, the proportionality factor is just $2(2\pi ^2)^{1-g}$. $\endgroup$
    – abx
    Commented Oct 31, 2018 at 7:48
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    $\begingroup$ Regarding Q1: Have you checked to what extent Witten's paper answers this? Regarding Q2: $U(1)$ has infinitely many one-dimensional representations, so it doesn't have a Witten zeta function. As for $SU(n)$, its Witten zeta function is, as far as I know, unrelated to classical zeta/$L$ functions such as Dedekind zetas (or at least not related in an easy way). While some of its properties, like special values at even positive integers, mirror those of some classical zeta functions, other properties, like the apparent lack of a functional equation, break the analogy. $\endgroup$
    – A Stasinski
    Commented Oct 31, 2018 at 11:56

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