What is the Selberg Trace Formula?

What is the Selberg Trace Formula? August 19, 2010

Posted by Sarah in Uncategorized.
Tags: harmonic analysis, selberg trace formula
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Ngo Bau Chau was one of the winners of the Fields Medal for 2010, so I thought I’d try to understand what it is he studied. The laudation gives a clear explanation of his contribution to the Langlands program. Ngo’s proof applies the Selberg trace formula, and since that’s accessible, I thought I’d try to understand that.

If $G$ is a Lie group and $\Gamma$ is a cofinite discrete group, we want to understand the operator

$R(f) = \int_\Gamma f(y) R(y) dy$
on $L^2(G/\Gamma)$ .

$R(f)$ is an integral operator, with kernel
$K(f) = \sum_{\gamma \in \Gamma} f(x^{-1} \gamma y)$
And now we can express the trace of the operator as
$\int_G \sum_{\gamma \in \Gamma} f(x^{-1} \gamma x) dx$
The result is
$\sum_{[\gamma]} \int_{\Gamma_\gamma / G} f(x^{-1} \gamma x) dx = \sum_{\pi} tr(\pi(f))$
where $[\gamma]$ are the conjugacy classes of $\gamma$ , $\Gamma_\gamma$ is the centralizer of $\gamma$ in $\Gamma$ , and $\pi$ are the automorphic representations.

When $\Gamma = \mathbb{Z}$ and $G = \mathbb{R}$ , the Selberg Trace Formula is the Poisson Summation Formula

$\sum_{n = -\infty}^\infty f(t + nT) = \frac{1}{T}\sum_{k = -\infty}^\infty \hat{f}(k/T) exp(2 \pi i k/T t).$

The Poisson Summation formula can be seen to be a trace formula as follows:
The eigenvectors of the positive Laplacian on the unit circle are $m^2$ for integer $m$ , and the eigenfunctions are $\phi_m = (2\pi)^{1/2} e^{i mx}.$
Consider the linear operator
$(Lf)(x) = \int_0^{2\pi} k(x, y) f(y) dy$
with kernel
$k(x, y) = \sum h(m)\phi_m(x) \bar{\phi_m(y)}$
Then
$L \phi_m = h(m) \phi_m$ and so the Poisson Summation Formula says
$Tr L = \sum_m h(m) = \sum \int h(\rho) e^{2 \pi i n \rho} d\rho$

The Selberg Trace Formula is proven in a similar manner to the proof of the Poisson Summation Theorem. If $R$ is the regular representation of $G$ on $L^2(\Gamma/G)$
$[R(g) \phi](x) = \phi(xg), g\in G, x \in \Gamma / G$
then we can write R as
$R(f)\phi(x) = \int_G f(x^{-1}g) \phi(g) dg$
fixing a Haar measure on $G$ . Then, by splitting the integral
$R(f) \phi(x) = \int_{\Gamma / G} \sum_{\gamma \in \Gamma} f(x^{-1} \gamma g)\phi(g) dg = \int_{\Gamma / G} K_f(x, y) \phi(y) dy$ .

We now take the trace of this operator in two different ways:
on one side,
$tr R = \int_{\Gamma / G} K_f(x, x) dx$
We break the sum over $\gamma$ into conjugacy classes of $\gamma$ . Each conjugacy class contributes
$\int_{\Gamma / G} \sum_{\delta \in \Gamma_\gamma / \Gamma}f(x^{-1}\delta^{-1} \gamma \delta x) dx = \int_{\Gamma / G} f(x^{-1} \gamma x) dx$
On the other side, we can compute the trace alternatively by quoting a result that says $L^2(\Gamma / G)$ decomposes into a direct sum of irreducible representations of G:
$\sum_{\pi} tr(\pi(f))$ .

This completes the proof. Conjugacy classes and irreducible representations are dual — in the same way that integers (in the Fourier series) and functions $e^{2 \pi i nx}$ are dual in the Poisson summation formula.

So what can you do with the trace formula? One thing is to let $\Gamma$ be the fundamental group of a Riemann surface, and describe the spectrum of differential operators such as the Laplace-Beltrami operator using geometric data like the lengths of geodesics. (Way back in sophomore year I was trying to learn about this from Peter Buser’s book.) The Selberg Trace Formula is also useful in analyzing the Riemann zeta function.

See these notes for a bunch of background on the Selberg Trace Formula.

Comments»

1. Rahul Krishna - August 22, 2010: There is no way I would ever call the Selberg Trace formula accessible, although those notes you linked to were quite nice. Both sides of the equation are really hard to deal with in general–conjugacy classes v. automorphic representations–and using the trace formula to say anything at all always strikes me as some sort of magic.

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2. Sarah - August 22, 2010: Ok — more accessible than the bleeping Langlands program?

You’ll notice I (and the notes) conveniently black-boxed the part that explains why the trace of the operator is a weighted sum of unitary representations. Still have no idea about that.

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