I want to compute the derivative of the Selberg $\zeta$-function:

$$ \mathcal{Z}(s)=\prod_{\gamma \; \text{primitive}} \prod_{n=0}^\infty (1-e^{-l(\gamma)(n+s)}); \qquad \Re(s)>1.$$

Where $\gamma$ are primitive closed geodesics on a given manifold and $l(\gamma)$ is their lenght. We can assume $s\in \mathbb{R}$. Observe that the given expression is justified by: $l(\gamma^n)= n l(\gamma)$.

A couple of reliable articles (Lou's "On the zeros of the derivative of the Selberg Zeta function" page 1143 for the statement and page 1141 for its definition of $\mathcal{Z}(s)$, and D'Hoker&Phong's "The geometry of String Perturbation theory, implicitly in the last passage of page 1005), and give, more or less explicitly, the following formula for the logarithmic derivative:

$$ \frac{\frac{d}{ds}\mathcal{Z}(s)}{\mathcal{Z}(s)} = \sum_{\gamma \; \text{primitive}} \sum_{n=1}^\infty\frac{l(\gamma) e^{-l(\gamma)ns}}{1-e^{-l(\gamma)n}}.$$

But my computation seems to give a slightly different result. Indeed, first we observe:

$$ \frac{d}{dx} \left(\prod_{n=0}^\infty f_n(x)\right)= \left(\prod_{n=0}^\infty f_n(x)\right)\left(\sum_{n=0}^\infty \frac{\frac{d}{dx}f_n(x)}{f_n(x)}\right).$$

Is it true? Wikipedia confirms it only in the finite case.

Then, applying it to $ \mathcal{Z}(s)$, I deduce:

$$ \frac{\frac{d}{ds}\mathcal{Z}(s)}{\mathcal{Z}(s)} = \sum_{\gamma \; \text{primitive}} \sum_{n=0}^\infty\frac{l(\gamma) e^{-l(\gamma)(n+s)}}{1-e^{-l(\gamma)(n+s)}}.$$

This expression seems to be close to the claimed one but I couldn't prove they are the same. So I guess I'm doing something wrong in the computation. Do you know how to do it?

Thank you very much!

  • $\begingroup$ I'm going to link them! $\endgroup$ – Giovanni De Gaetano Oct 9 '12 at 19:29

Maybe something like that?

Write $$ \frac{1}{1-e^{-l(\gamma)n}} = \sum_{k=0}^\infty e^{-l(\gamma)kn} \qquad\text{and}\qquad \frac{e^{-l(\gamma)(k+s)}}{1-e^{-l(\gamma)(k+s)}} = \sum_{n=1}^\infty e^{-l(\gamma)(k+s)n}. $$ Then $$ \begin{align*} \sum_{n=1}^\infty\frac{e^{-l(\gamma)ns}}{1-e^{-l(\gamma)n}} & =\sum_{n=1}^\infty \sum_{k=0}^\infty e^{-l(\gamma)ns} e^{-l(\gamma)kn} \\ & = \sum_{k=0}^\infty \sum_{n=1}^\infty e^{-l(\gamma)(k+s)n} \\& = \sum_{k=0}^\infty \frac{e^{-l(\gamma)(k+s)}}{1-e^{-l(\gamma)(k+s)}} \end{align*} $$ multiply by $l(\gamma)$ and sum over the primitives and you seem to get what you want.

  • $\begingroup$ Perfect! 'sorry for the delay in the answer and thank you very much! :) $\endgroup$ – Giovanni De Gaetano Oct 10 '12 at 7:52

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