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 $\int\limits_H f(y)\phi(xy)dy$.

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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?

  • $\begingroup$ 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

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