Skip to main content
Commonmark migration
Source Link

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][1]][1]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? [1]: https://i.sstatic.net/bhqrH.png

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][1]][1]

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? [1]: https://i.sstatic.net/bhqrH.png

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?

Source Link
D_S
  • 6.2k
  • 1
  • 15
  • 42

Decomposition of $L^2(\Gamma \backslash H)$ into irreducible representations using the spectral theorem

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][1]][1]

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? [1]: https://i.sstatic.net/bhqrH.png