## What is the Selberg Trace Formula?August 19, 2010

Posted by Sarah in Uncategorized.
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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.