Is there some explication for the transformation of the eigenvalues in Selberg trace formula Selberg trace formula http://en.wikipedia.org/wiki/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.
 A: 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. 
A: 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
A: 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 $\Re s = 0$. 
