For $f \in L^1(\mathbb R)$, the Fourier transform $\hat{f}: \mathbb R \rightarrow \mathbb C$ is defined by
$$\hat{f}(x) = \int_{\mathbb R} f(y) e^{2\pi i xy}dy.$$
The Poisson summation formula asserts that for $f$ smooth and compactly supported,
$$\sum\limits_{n\in \mathbb Z} f(n) = \sum\limits_{n \in \mathbb Z} \hat{f}(n).$$
I've heard that this formula arises as a special case of a trace formula for an integral operator of trace class on the Hilbert space $L^2(\mathbb R/\mathbb Z)$, or that Selberg or Arthur's trace formula is a "nonabelian Poisson summation formula." How can the Poisson summation formula be seen as arising in this way?
Generally, if $v_1, v_2, ... $ are an orthonormal basis of a given Hilbert space, and $T$ is a sufficiently nice operator on the given Hilbert space (specifically of "trace-class"), then the trace of $T$ can be defined by the formula
$$\operatorname{Tr}(T) = \sum\limits_{n\in \mathbb Z} \langle Tv_i, v_i \rangle$$
and this is independent of the choice of orthonormal basis. The Poisson summation formula comes from looking at the trace of a certain operator on the Hilbert space $V = L^2(\mathbb R/\mathbb Z)$ in two ways.
For $f$ smooth and compactly supported on $\mathbb R$, we set $K(x,y) = \sum\limits_{n \in \mathbb Z} f(-x+n+y)$, which is a finite sum for all pairs of real numbers $(x,y)$, and lies in the Hilbert space space $L^2(\mathbb R/\mathbb Z \times \mathbb R/\mathbb Z)$.
We define a bounded linear operator $T$ on $V$ by the formula
$$T(\varphi)(x) = \int\limits_{\mathbb R/\mathbb Z} K(x,y) \varphi(y)dy. \tag{$\varphi \in V$}$$
This operator turns out to be of trace class, so we may compute its trace $\operatorname{Tr}(T)$ in two ways.
First way (spectral expansion)
By the Peter-Weyl theorem, the characters $\phi_m(x) = e^{2\pi i mx} : m \in \mathbb Z$ form an orthonormal basis for $V$. Each vector $\phi_m$ is actually an eigenvector for the operator $T$, with eigenvalue $\hat{f}(n)$:
$$T(\phi_m)(x) = \int\limits_{\mathbb R/\mathbb Z} \sum\limits_{n \in \mathbb Z} f(-x+n+y) \phi_m(y)dy = \int\limits_{\mathbb R/\mathbb Z} \sum\limits_{n \in \mathbb Z} f(-x+n+y) \phi_m(n+y)dy = \int\limits_{\mathbb R} f(-x+y)\phi_m(y)dy $$
$$ = \int\limits_{\mathbb R} f(y) \phi_m(x+y)dy = \int\limits_{\mathbb R} f(y) e^{2 \pi i m(x+y)}dy = e^{2\pi i mx} \int\limits_{\mathbb R} f(y) e^{2\pi i my}dy = \hat{f}(m)e^{2\pi i mx}.$$
Therefore, we can compute $\operatorname{Tr}(T)$ as a sum of the eigenvalues $\hat{f}(n)$:
$$\operatorname{Tr}(T) = \sum\limits_{n\in \mathbb Z} \langle T(\phi_n), \phi_n \rangle = \sum\limits_{n\in \mathbb Z} \hat{f}(n) \langle \phi_n, \phi_n \rangle = \sum\limits_{n \in \mathbb Z} \hat{f}(n).$$
Second way (the trace as an integral along the diagonal)
On the other hand, we claim that the trace of $T$ can be calculated by integrating $K(x,y)$ along the diagonal $(x,x)$. Again by the Peter-Weyl theorem, the characters $\phi_{n,m}(x,y) = \phi_n(x) \overline{\phi_m(y)}$ of $(\mathbb R / \mathbb Z) \times(\mathbb R / \mathbb Z)$ form an orthonormal basis for $L^2(\mathbb R / \mathbb Z \times \mathbb R / \mathbb Z)$, and we may therefore write
$$K(x,y) = \sum\limits_{i,j} c_{ij} \phi_{i,j}(x,y)$$
for some complex numbers $c_{ij}$. We may compute
$$\int\limits_{\mathbb R/\mathbb Z} K(x,x)dx = \int\limits_{\mathbb R/\mathbb Z} \sum\limits_{i,j} c_{ij} \phi_{i,j}(x,x) dx = \sum\limits_{i,j} \int\limits_{\mathbb R/\mathbb Z} \phi_i(x) \overline{\phi_j(x)}dx = \sum\limits_{i,j} c_{i,j} \delta_{ij} = \sum\limits_i c_{i,i}$$
and at the same time,
$$\operatorname{Tr}(T) = \sum\limits_{n \in \mathbb Z} \int\limits_{\mathbb R/\mathbb Z} T(\phi_n)(x) \overline{\phi_n(x)}dx = \sum\limits_n \int\limits_{\mathbb R/\mathbb Z} \space \int\limits_{\mathbb R/\mathbb Z} K(x,y) \phi_n(y) \overline{\phi_n(x)} dx = \sum\limits_n \int\limits_{\mathbb R/\mathbb Z} \space \int\limits_{\mathbb R/\mathbb Z} \sum\limits_{i,j} c_{i,j} \phi_i(x) \overline{\phi_j(y)} \phi_n(y) \overline{\phi_n(x)}dydx$$
$$ = \sum\limits_{n,i,j} c_{i,j} \Bigg(\int\limits_{\mathbb R/\mathbb Z} \phi_i(x) \overline{\phi_n(x)}dx \Bigg) \Bigg( \int\limits_{\mathbb R/\mathbb Z} \phi_n(y) \overline{\phi_j(y)}dy \Bigg) = \sum\limits_{n,i,j} c_{i,j} \delta_{i,n} \delta_{n,j} = \sum\limits_{i,i} c_{i,i}.$$
Therefore,
$$\operatorname{Tr}(T) = \int\limits_{\mathbb R/\mathbb Z} K(x,x)dx.$$
Third way (geometric expansion):
Now that we have established that the trace of $T$ can be gotten by integrating $K(x,y)$ along the diagonal, we can compute the trace in one more way:
$$\operatorname{Tr}(T) = \int\limits_{\mathbb R/\mathbb Z} K(x,x) dx = \int\limits_{\mathbb R/\mathbb Z} \sum\limits_{n\in \mathbb Z} f(-x+n+x)dx = \int\limits_{\mathbb R/\mathbb Z} \sum\limits_{n\in \mathbb Z} f(n) dx = \sum\limits_{n \in \mathbb Z}f(n).$$
Thus the Poisson summation formula
$$\sum\limits_{n \in \mathbb Z}\hat{f}(n) = \sum\limits_{n \in \mathbb Z}f(n)$$
holds by looking at the trace of $T$ in two ways, first as a sum of eigenvalues, and second as an integral along the diagonal.