Given the Selberg trace formula, and the fact that the eigenvalues of the operator $\Delta -1/4 =T$ are the zeros of the Selberg zeta function, then would it be correct to say the number of eigenvalues of the operator $T$ below a certain quantity $E$ is equal to

$$\frac{1}{\pi}\arg Z(1/2+i \sqrt E ) ? $$

Here $Z$ is the Selberg zeta function defined as

$$ Z(s)= \prod_{p} \prod_{n=0}^{\infty}(1-p^{-(s+m)})$$

with the $p$'s representing the length of the closed geodesics.

This is a simple analogue to the math for the number of Riemann zeta zeros on the critical line.


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