3
$\begingroup$

I'm reading the introduction of An Introduction to the Trace Formula by James Arthur and wanted to understand something in the introduction.

Let $H$ be a unimodular locally compact Hausdorff group, and $\Gamma$ a discrete subgroup of $H$. Let $\mathscr H = L^2(\Gamma \backslash H)$ be the Hilbert space of measurable functions $\phi: \Gamma \backslash H \rightarrow \mathbb C$ satisfying $||\phi||^2 = \int\limits_{\Gamma \backslash H} |\phi(h)|dh < \infty$.

Assume $\Gamma \backslash H$ is compact. Fix $f \in C_c^{\infty}(H)$, and let $K \in L^2(\Gamma \backslash H \times \Gamma \backslash H)$ be the function defined by

$$K(x,y) = \sum\limits_{\gamma \in \Gamma} f(x^{-1}\gamma y)$$ which is a finite sum. To this kernel we can associate a compact operator $R(f)$ on $\mathscr H$ defined by

$$[R(f)\phi](x) = \int\limits_{\Gamma \backslash H} K(x,y)\phi(y)dy$$

It can be shown that this integral is equal to just $\int\limits_H f(y)\phi(xy)dy$.

enter image description here

Arthur arranges that $R(f)$ is a compact self adjoint operator, and claims the decomposition of $\mathscr H$ into a Hilbert space direct sum of irreducible subrepresentations (under the action of $H$ by right translation) follows from the spectral theorem for self adjoint operators. How does this follow?

$\endgroup$
  • $\begingroup$ Are you familiar with what the spectral theorem for compact self-adjoint operators says? $\endgroup$ – Yemon Choi Feb 10 at 3:22
  • $\begingroup$ Yes, you can find an orthonormal basis of eigenvectors for real eigenvalues which tend to zero. $\endgroup$ – D_S Feb 10 at 3:57
  • $\begingroup$ OK, I must confess I've never checked how the proof of the result mentioned by Arthur goes (something like this is in an introductory book on harmonic analysis by Deitmar, IIRC?) but I think this is the same idea as in the usual proof of the Peter-Weyl theorem, see e.g. Prop. 7 in terrytao.wordpress.com/2011/01/23/… $\endgroup$ – Yemon Choi Feb 10 at 4:10
  • $\begingroup$ I haven't checked carefully, but I think this is essentially in Tamagawa's paper "On Selberg's trace formula" (see Theorem 2). $\endgroup$ – Kimball Feb 11 at 19:28

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.