Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Are there any estimates for the eigenvalues of the Laplace operator for $\Gamma \backslash SL(2, \mathbb{C})/SU(2)$ known beyond the main term? Here, $\Gamma$ should be congruence subgroup in $SL(2,o)$ for $o$ the ring of integers in an imaginary quadratic field.

I have checked Mathscinet and couldn't find anything except for the main term.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

For general compact manifolds (of dimension $n$), the error term (on the number of eigenvalues less than $T^2$, counted with multiplicity) is $O(T^{n-1})$. So for $\Gamma$ co-compact, the error term is $O(T^2)$.

For merely co-finite $\Gamma$, it should be possible to bound the error term, but I am unaware of any results. In Section 4 of this survey by Müller (by the way, equation 1.3 is Weyl's Law for compact manifolds with the error term, and Section 2 sketches a different approach for compact locally symmetric spaces), he sketches a proof of a strong form of the law (using the Selberg Zeta function) and implies that the argument would work for other rank-one groups.

The book Groups Acting on Hyperbolic Space seems to (using Google Preview) prove the strong form of the law in the co-compact case (Section 5.5), but only seems to prove the weak form in the co-finite case (Section 8.9).

share|improve this answer
    
... and $T^{n-1}$ is essentially optimal right, perhaps rather $T^{n-1}/log T$? I was aware of all this, but thanks already. –  plusepsilon.de Jul 25 '12 at 16:00
1  
Yeah, $O(T^{n-1})$ is optimal for compact spaces because of spheres, but I think conjecturally tighter bounds exist for certain hyperbolic spaces (and you are correct that for $SL_2(\mathbb Z)\backslash \mathfrak h$, Selberg obtained $O(T/{\rm log}\ T)$, so something better is certainly possible). I think it might be something like GRH (though more tractable), where getting nonvanishing of an $L$-function at the edge of the strip is hard, and making any improvement on that virtually impossible, even though we expect much more. –  B R Jul 25 '12 at 17:14

Your Answer

 
discard

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.