MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Selberg trace formula says $$\sum_nh(r_n)=...$$ where $1/4+r_n^2$ is the eigenvalue of the Laplacian.

My question is "what is the geometric exlication for the transformation $r_n\to r_n^2+1/4$?"

Or it just makes the formula beautiful.

share|cite|improve this question
The question is not appropriate. Because Selberg's trace formula is just a link between geometry and analysis. – Denis Serre Sep 13 '11 at 14:51
up vote 1 down vote accepted

This transformation only plays a role in the "trivial K-type" formulation of the trace formula, which is only special case of the whole story. One considers test-functions in the functional calculus of the Laplace-Beltrami operator. The trace formula describes the trace of a convolution operator on the group level, so the spectral side is parametrized by irreducible representations of the group ${\rm SL}_2({\mathbb R})$, in this case principal series representations, which are parametrized by the parameter $r$. The corresponding Laplace-Belytrami-eigenvalue is $\lambda=r^2+1/4$. Note that the Laplace-Beltrami operator is induced by the Casimir-operator on the group.

The best way to view the Selberg trace formula is on the group level, where such difficulties disappear. For details see the book "Principles of Harmonic Analysis" (Springer) by Siegfried Echterhoff and myself.

share|cite|improve this answer

the 'Selberg zeros' appear as momenta of the Laplacian $ p= \sqrt (E-1/4) $ NOT over the Eigenvalues.. a smilar thing happens with Poisson sum formula

$ \sum_{n\in Z}\delta (x-n)=\sum_{n\in Z}exp(2i\pi nx)$

but the Energies are $ E= n^{2}\pi^{2}$ this is the analogy between Selberg and Poisson

share|cite|improve this answer

This normalization originates from representation theory. Let $B$ be upper triangular matrices.

The induced representation $Ind_{B(\mathbb{R})}^{SL_2(\mathbb{R}} \left| \cdotp \right|^s$ is precisely unitarizable, if and only if $\Re s = 0$ or $-1/2< s <1/2$.

To translate to you picture $s = r -1/2$.

Note that the Selberg eigenvalue conjecture asserts that only $\Re s = 0$ should occur in certain cases. The difference is that $Ind_{B(\mathbb{R})}^{SL_2(\mathbb{R}} \left| \cdotp \right|^s$ is tempered, if and only if $\Re s = 0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.