The Selberg Trace Formula

Abstract

In this chapter we introduce the Selberg trace formula, which is a natural generalization of the Poisson summation formula to non-abelian groups. Applications of the trace formula will be given in the next two chapters.

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In this chapter we introduce the Selberg trace formula, which is a natural generalization of the Poisson summation formula to non-abelian groups. Applications of the trace formula will be given in the next two chapters.

9.1 Cocompact Groups and Lattices

We will study the following situation. G will be a locally compact group, and H will be a closed subgroup, which is unimodular. We say that a subgroup H of a topological group G is a cocompact subgroup if the quotient $G/H$ is a compact space.

Examples 9.1.1

• A classical example is $G={\mathbb R}$ and $H={\mathbb Z}$.

• For a unimodular group H and a compact group K together with a group homomorphism $\eta:K\to{\rm Aut}(H)$ such that the ensuing map $K\times H\to H$ given by $(k,h)\mapsto \eta(k)(h)$ is continuous, one can form the semi-direct product $G=H\rtimes K$. As a topological space, one has $G=H\times K$. The multiplication is $(h,k)(h_1,k_1)=(h\,\eta(k)(h_1),kk_1)$. Then H is a cocompact subgroup of G.

• For $G={\rm SL}_2({\mathbb R})$, one uses hyperbolic geometry [Bea95] to construct discrete subgroups ${\Upgamma}=H\subset G$, which provide examples of the situation considered here.

Proposition 9.1.2

Let G be a locally compact group. If G admits a unimodular closed cocompact subgroup, then G is unimodular itself.

Proof

Let $H\subset G$ be a unimodular closed cocompact subgroup. The inversion on G induces a homeomorphism $G/H\to H{\backslash} G$ by $gH\mapsto Hg^{-1}$. Therefore $H{\backslash} G$ is compact as well. We install a Radon measure μ on the compact Hausdorff space $H{\backslash} G$ using Riesz’s representation theorem as follows. For $f\in C(H{\backslash} G)$ choose $g\in C_c(G)$ with $^H\!g=f$, where $^H\!g(x){\ \stackrel {{\rm def}}{=}} \int_Hg(hx)\, dh.$ In this case define $\int_{H{\backslash} G} f(x)\,d\mu(x)$ by $\int_Gg(x)\,dx$. For this to be well-defined, we have to show that $^H\!g=0$ implies $\int_Gg (x)\,dx=0$. Note that, as H is unimodular, we have $\int_Hg(h^{-1}x)\,dh=\int_Hg(hx) \,dh=0$. Let $\phi\in C_c(G)$ with $^H\phi\equiv 1$. One gets

$$\begin{aligned} \int_G g(x)\,dx &= \int_G\int_H\phi(hx)g(x)\,dh\,dx\\ &= \int_H\int_G\phi(hx)g(x)\,dx\,dh\\ &= \int_H\int_G\phi(x)g(h^{-1}x)\,dx\,dh\\ &= \int_G\phi(x)\int_Hg(h^{-1}x)\,dh\,dx= 0.\end{aligned}$$

This measure μ on $H{\backslash} G$ satisfies

$$\begin{aligned} \int_{H{\backslash} G}f(xy)\,d\mu(x)&= \int_Gg(xy)\,dx=\Updelta(y^{-1})\int_Gg(x) \,dx\\ &= \Updelta(y^{-1})\int_{H{\backslash} G}f(x)\,d\mu(x).\end{aligned}$$

So in particular, for $f\equiv 1\in C(H{\backslash} G)$ one gets

$$\begin{aligned} 0 <\ \int_{H{\backslash} G}f(x)\,d\mu(x)&=\int_{H{\backslash} G}f(xy)\,d\mu(x) =\Updelta(y^{-1})\int_{H{\backslash} G}f(x)\,d\mu(x),\end{aligned}$$

which implies $\Updelta(y^{-1})=1.$ □

Definition

A subgroup ${\Upgamma}\subset G$ of a topological group is called a discrete subgroup if the subspace-topology on ${\Upgamma}$ is the discrete topology.

Lemma 9.1.3

• (a) A subgroup ${\Upgamma}\subset G$ is discrete if and only if there is a unit-neighborhood $U\subset G$ with ${\Upgamma}\cap U=\{1\}$.

• (b) A discrete subgroup is closed in G.

Proof

As ${\Upgamma}$ is a topological group, it is discrete if and only if $\{1\}$ is open, which is equivalent to the existence of an open set U with ${\Upgamma} \cap U=\{1\}$. This implies part (a). For part (b) let ${\Upgamma}$ be a discrete subgroup, and let ${\gamma}_j\to x$ be a net in ${\Upgamma}$, convergent in G. We have to show that $x\in{\Upgamma}$. Let U be a unit-neighborhood in G such that ${\Upgamma}\cap U=\{1\}$ and let V be a symmetric unit-neighborhood such that $V^2\subset U$. As xV is a neighborhood of x, there exists an index j ₀ such that for $i,j\ge j_0$ one has ${\gamma}_i,{\gamma}_j\in xV$, therefore ${\gamma}_i^{-1}{\gamma}_j\in V^2\subset U$, hence ${\gamma}_i={\gamma}_j$, so the net is eventually constant, hence $x={\gamma}_{j_0}\in{\Upgamma}$. □

Example 9.1.4

• The subgroup ${\mathbb Z}$ of ${\mathbb R}$ is a discrete subgroup.

• Let G be Hausdorff and compact. Then a subgroup ${\Upgamma}\subset G$ is discrete if and only if it is finite.

• Let $G={\rm SL}_2({\mathbb R})$ be the group of real 2 × 2 matrices of determinant one. Then ${\Upgamma}={\rm SL}_2({\mathbb Z})$ is a discrete subgroup of G.

Definition

Let G be a locally compact group. A discrete subgroup ${\Upgamma}$ such that $G/{\Upgamma}$ carries an invariant Radon measure μ with $\mu(G/{\Upgamma})\ <\ \infty$ is called a lattice in G. A cocompact lattice is also called a uniform lattice .

Proposition 9.1.5

Let G be a locally compact group. A discrete, cocompact subgroup ${\Upgamma}$ is a uniform lattice.

Proof

Let ${\Upgamma}\subset G$ as in the proposition. Then G is unimodular by Proposition 9.1.2. As ${\Upgamma}$ is unimodular, by Theorem 1.5.3 there exists an invariant Radon measure on $G/{\Upgamma}$, and, as the latter space is compact, its volume is finite, so ${\Upgamma}$ is a lattice. □

Remark

There are lattices, which are not uniform, like ${\Upgamma}={\rm SL}_2({\mathbb Z})$ in $G={\rm SL}_2({\mathbb R})$ (See [Ser73]).

Theorem 9.1.6

Let G be a locally compact group. If G admits a lattice, then G is unimodular.

Proof

Let Δ be the modular function of G, and let $H={\rm ker}(\Updelta)$. Then H is unimodular by Corollary 1.5.5. Let ${\Upgamma}\subset G$ be a lattice. As there is an invariant measure on $G/{\Upgamma}$, Theorem 1.5.3 implies that Δ is trivial on ${\Upgamma}$, so ${\Upgamma}\subset H$. On $G/H$ and on $H/{\Upgamma}$ there are invariant measures by Theorem 1.5.3, so by the uniqueness of these respective measures we conclude that

$$\infty>\ {\rm vol}(G/{\Upgamma})=\int_{G/{\Upgamma}} 1\,dx=\int_{G/H}\int_{H/{\Upgamma}} 1\,dx \,dy.$$

Therefore $G/H$ has finite volume. Being a locally compact group, it follows that $G/H$ is compact. By Proposition 9.1.2. the group G is unimodular. □

9.2 Discreteness of the Spectrum

Let G be a locally compact group, and let ${\Upgamma}\subset G$ be a cocompact lattice. It is a convention that one uses the right coset space ${\Upgamma}{\backslash} G$ instead of the left coset space. This quotient carries a Radon measure μ that is invariant under right translations by elements of G and has finite volume. On the ensuing Hilbert space $L^2({\Upgamma}{\backslash} G)$ the right translations give a unitary representation R of G by

$$R(x)\phi(y){\ \stackrel{{\rm def}}{=}\,} \phi(yx).$$

Example 9.2.1

Let $G={\mathbb R}$, and ${\Upgamma}={\mathbb Z}$ then ${\Upgamma}$ is a closed cocompact subgroup and the theory of Fourier series, [Dei05] Chap. 1, implies that the representation R decomposes into the direct sum $\widehat{\bigoplus}_{n\in{\mathbb Z}}\chi_n$, where $\chi_n:{\mathbb R}\to{\mathbb T}$ is the character given by $\chi_n(x)=e^{2\pi i n x}$. Thus R decomposes into a discrete sum of irreducible representations. This is a general phenomenon, as the next theorem shows.

For a representation π and a natural number N we write $N\pi$ for the N-fold direct sum $\pi\oplus\pi\dots\oplus\pi$.

Theorem 9.2.2

Let G be a locally compact group, and let ${\Upgamma}\subset G$ be a unimodular closed cocompact subgroup. The representation R on $L^2({\Upgamma}{\backslash} G)$ decomposes as a direct sum of irreducible representations with finite multiplicities, i.e.,

$$L^2({\Upgamma}{\backslash} G)\ \cong\ \bigoplus_{\pi\in \widehat G} N_{\Upgamma}(\pi)\pi,$$

where the sum runs over the unitary dual $\widehat G$ of G, and $N_{\Upgamma}(\pi)\in{\mathbb N}_0$ is a finite multiplicity for $\pi\in\widehat G$.

Proof

We need a lemma that tells us that for $f\in C_c(G)$ the operator R(f) is given by a continuous integral kernel. For later use we will extend this to a greater class of functions f. Let U be a compact unit-neighborhood in G. For a continuous function f on G let $f_U: G\to [0,\infty)$ be defined by

$$f_U(y){\ \stackrel{{\rm def}}{=}} \sup_{x,z\in U}|f(xyz)|.$$

Lemma 9.2.3

The function f _U is continuous.

Proof

It suffices to show that for $a\ge 0$ the sets $f_U^{-1}((a,\infty))$ and $f_U^{-1}([0,a))$ are open. For the former assume $f_U(x)>a$, then there exist $u_1,u_2\in U$ with $|f(u_1xu_2)|>a$. As the function $y\mapsto f(u_1yu_2)$ is continuous, there exists an open neighborhood V of x such that $|f(u_1vu_2)|>a$ for every $v\in V$. This implies $f_U(v)>a$ for every $v\in V$.

The second assertion is equivalent to saying that $f_U^{-1}([a,\infty))$ is closed. So let (x _j ) be a net in this set, convergent to some $x\in G$. This means that $f_U(x_j)\ge a$ and we have to show that $f_U(x)\ge a$. For each j fix some $u_j,v_j\in U$ such that $f_U(x_j)=|f(u_jx_jv_j)|$. By switching to a subnet, we may assume that the nets $(u_j),(v_j)$ are both convergent in the compact set U with limits u and v respectively. It follows that

$$\begin{aligned} f_U(x)\ge |f(uxv)|=\lim_j|f(u_jx_jv_j|\ge a. \end{aligned}$$

Definition

We say that a continuous function f is a uniformly integrable function if there exists a compact unit-neighborhood U such that f _U is in $L^1(G)$. If f is uniformly integrable, then $f\in L^1(G)$ as $|f|\le f_U$. Let $C_{\rm unif}(G)$ be the set of all continuous functions f on G that are uniformly integrable.

Example 9.2.4

• If $f\in C_c(G)$, then f is uniformly integrable, as $f_U\in C_c(G)$ again.

• Every Schwartz-function on ${\mathbb R}$ is uniformly integrable (Exercise 9.1).

Lemma 9.2.5

Let G be unimodular. Every uniformly integrable function vanishes at infinity, and the space $C_{\rm unif}(G)$ is an algebra under convolution.

Proof

Assume that f is uniformly integrable and let U be a compact symmetric unit-neighborhood in G such that f _U is integrable. If f does not vanish at ∞, there exists an ${\varepsilon}>0$ such that for every compact set $K\subset G$ there is $x\in G{\smallsetminus} K$ with $|f(x)|\ge{\varepsilon}$. Let $x_1\in G$ be any element with $|f(x_1)|\ge{\varepsilon}$. Then there exists $x_2\notin x_1U^2$ such that $|f(x_2)|\ge{\varepsilon}$. As U is symmetric, it follows that $x_1U\cap x_2U=\emptyset$. Next we pick an element x ₃ outside of $x_1U^2\cup x_2U^2$ with $|f(x_3)|\ge{\varepsilon}$. Repeating this argument, we find a sequence $\{x_n: n\in{\mathbb N}\}$ in G such that $x_nU\cap x_mU=\emptyset$ for all $n\neq m$ and $|f(x_n)|\geq{\varepsilon}$ for every n. But then $f_U\geq{\varepsilon}$ on x _nU for every n, which contradicts the integrability of f _U .

Since integrable functions in $C_0(G)$ are square integrable, we get

$$C_{\rm unif}(G)\ \subset\ L^2(G).$$

Let $f,g\in C_{\rm unif}(G)$. We can write $f*g(x)={\left\langle{f,L_xg^*}\right\rangle}$. The map $x\mapsto L_xg^*$ is continuous as a map from G to $L^2(G)$. The inner product is continuous, hence so is $f*g$. Finally, choose a unit-neighborhood U such that f _U and g _U are both integrable. Then

$$\begin{aligned} (f*g)_U(y) = & \sup_{x,z\in U}\left|\int_G f(\xi)g(\xi^{-1} xyz)\,d\xi \right| \\ = & \sup_{x,z\in U}\left|\int_G f(x\xi)g(\xi^{-1} yz)\,d\xi\right|\\ &\le \sup_{x,z\in U}\int_G |f(x\xi)g(\xi^{-1} yz)|\,d\xi\\ &\le \int_G f_U(\xi)g_U(\xi^{-1} y)\,d\xi = f_U *g_U(y).\end{aligned}$$

This implies that $(f*g)_U$ is integrable over G. □

Lemma 9.2.6

For $f\in C_{\rm unif}(G)$ and $\phi\in L^2({\Upgamma}{\backslash} G)$ one has

$$R(f)\phi(x)=\int_{{\Upgamma}{\backslash} G}k(x,y)\phi(y)\,dy,$$

where $k(x,y)=\sum_{{\gamma}\in{\Upgamma}}f(x^{-1}{\gamma} y)\,dh$. The kernel k is continuous on ${\Upgamma}{\backslash} G\times{\Upgamma}{\backslash} G$.

Proof

Let $f\in L^1(G)$. For $\phi\in L^2({\Upgamma}{\backslash} G)$, one computes with the quotient integral formula

$$\begin{aligned} R(f)\phi(x) &= \int_G f(y) R(y)\phi(x)\,dy=\int_Gf(y)\phi(xy)\,dy\\ &= \int_Gf(x^{-1}y)\phi(y)\,dy\\ &= \int_{{\Upgamma}{\backslash} G}\sum_{{\gamma}\in{\Upgamma}}f(x^{-1}{\gamma} y)\phi({\gamma} y)\,dy = \int_{{\Upgamma}{\backslash} G}\sum_{{\gamma}\in{\Upgamma}}f(x^{-1}{\gamma} y)\phi(y)\,dy,\end{aligned}$$

as claimed, and the integral converges almost everywhere in x. In particular, for $f\in C_{\rm unif}(G)$ this argument works with f replaced by f _U for a suitable symmetric unit-neighborhood U to get a kernel k _U . We choose $g\in C_c(G)$ with $g\ge 0$ and $\sum_{{\gamma}\in{\Upgamma}}g({\gamma} x)= 1$ for all $x\in G$ and use the quotient integral formula to compute

$$\begin{aligned} \int_{{\Upgamma}{\backslash} G\times{\Upgamma}{\backslash} G}k_U(x,y)\,dx\,dy &= \int_{{\Upgamma}{\backslash} G}\int_{{\Upgamma}{\backslash} G}\sum_{{\gamma}\in{\Upgamma}}g({\gamma} x)\sum_{\tau\in{\Upgamma}}f_U(x^{-1} \tau y)\,dx\,dy\\ &= \int_{{\Upgamma}{\backslash} G}\int_{G}g(x)\sum_{\tau\in{\Upgamma}}f_U(x^{-1} \tau y)\,dx\,dy\\ &= \int_{G}\int_{G}g(x)f_U(x^{-1}y)\,dx\,dy=\Vert{g*f_U}\Vert_1<\infty\end{aligned}$$

By the quotient integral theorem, the sum $\sum_{{\gamma}\in{\Upgamma}} f_U(x^{-1}{\gamma} y)$ converges almost everywhere in $(x,y)$, so it converges on a dense set of $(x,y)$. Let $(x_0,y_0)$ be such a point of convergence. We show that $k(x,y)$ is continuous in the set $x_0U\times y_0U$. For given ${\varepsilon}>0$ there exists a finite set $S\subset{\Upgamma}$ such that $\sum_{{\gamma}\notin S}f_U(x_0^{-1}{\gamma} y_0)<{\varepsilon}/2$. This means that for $(x,y)\in x_0U\times y_0U$ one has $\sum_{{\gamma}\notin S}|f(x^{-1}{\gamma} y)|<{\varepsilon}/2$. This implies for $(x',y')\in x_0U\times y_0U$,

$$|k(x,y)-k(x',y')|\le\sum_{{\gamma}\in S}|f({x}^{-1}{\gamma}y)-f({x'}^{-1}{\gamma} y')|+{\varepsilon},$$

from which the continuity of k follows □

As $k\in C({\Upgamma}{\backslash} G\times{\Upgamma}{\backslash} G)\subset L^2({\Upgamma}{\backslash} G\times{\Upgamma}{\backslash} G)$, the operator R(f) is a Hilbert-Schmidt operator, hence compact. The theorem follows from the next lemma.

Lemma 9.2.7

Let A be a *-closed subspace of $C_c(G).$ Let $(\eta, V_\eta)$ be a unitary representation of G such that for every $f\in A,$ the operator $\eta(f)$ is compact and such that for every non-zero $v\in V_\eta$ the space $\eta(A)v$ is non-zero. Then η is a direct sum of irreducible representations with finite multiplicities.

Note that the condition $\eta(A)v\ne 0$ is always satisfied if A contains a Dirac-net $(\phi_U)_U$.

Proof

Zorn’s Lemma provides us with a subspace E, maximal with the property that it decomposes as a sum of irreducibles. The assumption of the Lemma also holds for the orthocomplement $E^\perp$ of E in $V=V_\eta$. This orthocomplement cannot contain any irreducible subspace. We have to show that it is zero. In other words, we have to show that a representation η as in the assumption, always contains an irreducible subspace.

The space A is generated by its self-adjoint elements. Let $f\in A$ be self-adjoint. Then $\eta(f)$ is self-adjoint and compact. By the spectral theorem for compact operators 5.2.2 the space $V=V_\eta$ decomposes,

$$V= V_{f,0}\ \oplus\ \bigoplus_{j=1}^\infty V_{f,j},$$

where $V_{f,0}$ is the kernel of $\eta(f)$ and $V_{f,j}$ is the eigenspace of $\eta(f)$ for an eigenvalue ${\lambda}_j\ne 0$. The sequence ${\lambda}_j$ tends to zero and each $V_{f,j}$ is finite dimensional for $j>0$. For every closed invariant subspace $E\subset V$ one has a similar decomposition

$$E= E_{f,0}'\ \oplus\ \bigoplus_{j=1}^\infty E_{f,j}',$$

and $E'_{f,j}\subset V_{f,j}$ for every $j\ge 0$. It follows that every non-zero closed invariant subspace E has a non-zero intersection with one of the $V_{f,i}$ for some $i>0$ and some $f\in A$, since otherwise $E\subset{\rm ker}\,\eta(f)$ for every $f\in A$. Fix f and j and consider the set of all non-zero intersections $V_{f,j} \cap E$, where E runs over all closed invariant subspaces. Among these intersections choose one $W=V_{f,j}\cap E$ of minimal dimension $\ne 0$, which is possible as $V_{f,j}$ is finite-dimensional. Let $E^1=\bigcap_{E:E\cap V_{f,j}=W}E,$ where the intersection runs over all closed invariant subspaces E with $E \cap V_{f,j}=W$. Then E ¹ is a closed invariant subspace. We claim that it is irreducible. For this assume that $E^1=F\oplus F'$ with closed invariant subspaces $F,F'$. Then by minimality, $W\subset F$ or $W\subset F'$, which implies that one of the spaces F or $F'$ is zero, so E ¹ is indeed irreducible. We have shown that V indeed has an irreducible subspace E ¹. As shown above this implies that V decomposes as a sum of irreducibles.

It remains to show that the multiplicities are finite. For this note first that if τ and σ are unitarily equivalent representations, then ${\lambda}$ is an eigenvalue of $\tau(f)$ if and only if it is one for $\sigma(f)$. Thus finiteness of the multiplicities follows from the fact that any collection of orthogonal subspaces, which give rise to unitarily equivalent representations, must all have non-trivial intersection with the same eigenspaces $V_{f,j}$. Since the $V_{f,j}$ are finite dimensional, there can only exist finitely many of them. □

9.3 The Trace Formula

Let X be a locally compact Hausdorff space, and let μ be a Radon measure on X. A continuous L ²-kernel k on X is called admissible kernel if there exists a function $g\in C(X)\cap L^2(X)$ such that $|k(x,y)|\le g(x)g(y)$. Note that if X is compact, then every continuous kernel is admissible.

An operator $S: L^2(X)\to L^2(X)$ is called an admissible operator if there exists an admissible kernel k such that

$$S\phi(x)=\int_X k(x,y)\phi(y)\,dy,$$

where we have written dy for $d\mu(y)$.

Proposition 9.3.1

Let X be a locally compact Hausdorff space equipped with a Radon measure. Assume that X is first countable or compact. Let $T$ be an integral operator with continuous L ²-kernel on X. Assume that there exists admissible operators $S_1,S_2$ with $T=S_1S_2.$

Then T is of trace class and

$$\hspace{1pt}{\rm tr}\hspace{2pt}(T)=\int_X k(x,x)\,dx.$$

Proof

Replacing S ₂ with $S_2^*$ we can assume $T=S_1S_2^*$ in the proposition. The map $\sigma:(S_1,S_2)\mapsto S_1S_2^*$ is sesquilinear, so it obeys the polarization rule,

$$\sigma(S,R)=\frac14\left(\sigma(S+R)-\sigma(S-R)+i\left(\sigma(S+iR)-\sigma(S-iR)\right)\right),$$

where we have written $\sigma(S+R)$ for $\sigma(S+R,S+R)$. This implies that, in order to prove the proposition, it suffices to assume that $T=SS^*$ for some admissible operator S. As S is a Hilbert-Schmidt operator, $T=SS^*$ is trace class and also ${\hspace{1pt}{\rm tr}\hspace{2pt}}(T)=\Vert S\Vert_{\rm HS}^2$. Let $l(x,y)$ be the admissible kernel of S. For $\phi\in C_c(X)$ we compute

$$\begin{aligned} SS^*\phi(x) &= \int_Xl(x,u)S^*\phi(u)\,du\\ &=\int_X\int_X l(x,u){\overline{l(y,u)}}\phi(y)\,dy\,du\\ &=\int_X\left(\int_X l(x,u){\overline{l(y,u)}}\,du\right)\phi(y)\,dy.\end{aligned}$$

So the operator $SS^*$ has the kernel $l*l^*(x,y)=\int_X l(x,z){\overline{l(y,z)}}\,dz.$ If l is admissible and X is first countable, it suffices to check convergence with sequences, and therefore the Theorem of Dominated Convergence implies that the kernel $l*l^*$ is continuous. If X is compact, the Radon measure is finite and the function l is uniformly continuous in the following sense: For every $x_0\in X$ and every ${\varepsilon}>0$ there exists a neighborhood U of x ₀ such that for all $x\in U$ and all $z\in X$ we have $|l(x,z)-l(x_0,z)|<{\varepsilon}.$ This implies the continuity of $l*l^*$. For $\phi,\psi\in C_c(X)$ we have

$$\begin{aligned} \int_X\int_X l*l^*(x,y)\phi(y)\psi(x)\,dy\,dx &={\left\langle{SS^*}\phi,{\overline{\psi}}\right\rangle}\\ &=\int_X\int_X k(x,y)\phi(y)\psi(x)\,dy\,dx.\end{aligned}$$

Varying φ and ψ we conclude that the continuous kernels k and $l*l^*$ coincide. Therefore,

$$\int_{X} k(x,x)\,dx {=}\int_{X}\int_{X} |l(x,u)|^{2}\, du \, dx{=}\Vert{S}\Vert_{\rm HS}={\hspace{1pt}{\rm tr}\hspace{2pt}}(T).$$

The proposition follows. □

Recall that for $\pi\in{\widehat G}$ the number $N_{\Upgamma}(\pi)\ge 0$ is the multiplicity of π as a subrepresentation of $(R,L^2({\Upgamma}{\backslash} G))$. Let $\widehat{G}_{\Upgamma}$ denote the set of all $\pi\in\widehat{G}$ with $N_{\Upgamma}(\pi)>~0$.

Definition

We write $C_{\rm unif}(G)^2$ for the space of all linear combinations of functions of the form $g*h$ with $g,h\in C_{\rm unif}(G)$. Moreover, if ν is a lattice in G and ${\gamma}\in{\Upgamma}$, then we denote by $[{\gamma}]$ the conjugacy class of ${\gamma}$ in ${\Upgamma}$, $G_{\gamma}{\ \stackrel{ {\rm def}}{=}\,}\{x\in G: x{\gamma} x^{-1}={\gamma}\}$ denotes the centralizer of ${\gamma}$ in G, and ${\Upgamma}_{\gamma}{\ \stackrel{{\rm def}}{=}\,} G_{\gamma}\cap{\Upgamma}$ denotes the centralizer of ${\gamma}$ in ${\Upgamma}$. Note that the map from ${\Upgamma}$ to $[{\gamma}]$, which sends ν to $\nu^{-1}{\gamma}\nu$, factors through a bijection ${\Upgamma}_{\gamma}{\backslash}{\Upgamma}\cong [{\gamma}]$.

Theorem 9.3.2

(Trace Formula) Let G be a locally compact group and ${\Upgamma}\subset G$ a uniform lattice. Let ${\widehat{G}}_{\Upgamma}$ denote the set of all $\pi\in \widehat{G}$ which appear as subrepresentations of the representation R on $L^2({\Upgamma}{\backslash} G)$ and let $f\in C_{\rm unif}(G)^2.$ For every $\pi\in\widehat G_{\Upgamma}$ the operator $\pi(f)$ is of trace class and

$$\sum_{\pi\in\widehat G_{\Upgamma}}N_{\Upgamma}(\pi)\,{\hspace{1pt}{\rm tr}\hspace{2pt}}\pi(f)=\sum_{[{\gamma}]}{\rm vol}({\Upgamma}_{\gamma}{\backslash} G_{\gamma})\,{\cal O}_{\gamma}(f),$$

where the summation on the right runs over all conjugacy classes $[{\gamma}]$ in the group ${\Upgamma},$ and ${\cal O}_{\gamma}(f)$ denotes the orbital integral ,

$$\cal O_{\gamma}(f)=\int_{G_{\gamma}{\backslash} G} f(x^{-1}{\gamma} x)\,dx.$$

We shall see in Lemma 9.3.3 below that the centralizer $G_{\gamma}$ is unimodular and that ${\Upgamma}_{\gamma}{\backslash} G_{\gamma}$ has finite measure for every ${\gamma}\in{\Upgamma}$. The expression ${\rm vol}({\Upgamma}_{\gamma}{\backslash} G_{\gamma}){\cal O}_{\gamma}(f)$ is therefore well-defined. It does not depend on the choice of Haar measure on $G_{\gamma}$.

Proof of the trace formula

The algebra $C_{\rm unif}(G)^2$ consists of all finite linear combinations of convolution products of the form $g*h^*$ with $g,h\in C_{\rm unif}(G)$. So it suffices to show the trace formula for $f=g*h^*$. According to Lemma 9.2.6, the operators R(g) and R(h) are integral operators with continuous kernels

$$k_g(x,y)=\sum_{{\gamma}\in{\Upgamma}}g(x^{-1}{\gamma} y)\qquad{\rm and}\qquad k_h(x,y)=\sum_{{\gamma}\in{\Upgamma}}h(x^{-1}{\gamma} y).$$

By Proposition 9.3.1 the operator R(f) is of trace class and ${\hspace{1pt}{\rm tr}\hspace{2pt}}(R(f))=\int_{{\Upgamma}{\backslash} G} k_f(x,x)\,dx.$ With R(f), all its restrictions to subrepresentations are of trace class. It follows that

$$\begin{aligned} \sum_{\pi\in\widehat G_{\Upgamma}}N_{\Upgamma}(\pi){\hspace{1pt}{\rm tr}\hspace{2pt}}\pi(f)&={\hspace{1pt}{\rm tr}\hspace{2pt}}(R(f)) =\int_{{\Upgamma}{\backslash} G}k_f(x,x)\,dx\\ &=\int_{{\Upgamma}{\backslash} G}\sum_{{\gamma}\in{\Upgamma}}f(x^{-1}{\gamma} x)\,dx.\end{aligned}$$

We order the sum according to conjugacy classes $[{\gamma}]$ in ${\Upgamma}$, interchange integration and summation and use the quotient integral formula to get

$$\begin{aligned} {\hspace{1pt}{\rm tr}\hspace{2pt}}(R(f))&=\int_{{\Upgamma}{\backslash} G}\sum_{[{\gamma}]}\sum_{\sigma\in \Upgamma_\gamma\backslash \Upgamma}f(x^{-1}\sigma^{-1} {\gamma} \sigma x)\,dx\\ &=\sum_{[{\gamma}]}\int_{{\Upgamma}{\backslash} G}\sum_{\sigma\in \Upgamma_\gamma\backslash \Upgamma} f((\sigma x)^{-1} {\gamma} \sigma x)\,dx\\ &=\sum_{[{\gamma}]}\int_{{\Upgamma}_{\gamma}{\backslash} G}f(x^{-1} {\gamma} x)\,dx.\end{aligned}$$

Lemma 9.3.3

For every ${\gamma}\in{\Upgamma},$ the centralizer $G_{\gamma}$ is unimodular and ${\Upgamma}_{\gamma}{\backslash} G_{\gamma}$ has finite invariant measure ${\rm vol}({\Upgamma}_{\gamma}{\backslash} G_{\gamma})$.

Proof

The above calculation shows that for every f as in the theorem, which is positive, i.e., for $f\ge 0$, one has $\int_{{\Upgamma}_{\gamma}{\backslash} G} f(x^{-1}{\gamma} x)\,dx < \infty$ for every ${\gamma}\in{\Upgamma}$. Consider the projection $p:{\Upgamma}_{\gamma}{\backslash} G\to G_{\gamma}{\backslash} G$. Since G is unimodular by Theorem 9.1.6, the space ${\Upgamma}_{\gamma}{\backslash} G$ carries an invariant Radon measure ν. Let μ be the image under p of ν, i.e., we define $\int_{G_{\gamma}{\backslash} G} f(x)\,d\mu(x)= \int_{{\Upgamma}_{\gamma}{\backslash} G} f(p(y))\,d\nu(y)$ for every $f\in C_c(G_{\gamma}{\backslash} G)$. We need to show that this is finite. For this let $0\le f\in C_c(G_{\gamma}{\backslash} G)$, and let $\Upphi: G_{\gamma}{\backslash} G\to G$ be given by $\Upphi(G_{\gamma} x)=x^{-1}{\gamma} x$. Then $K=\Upphi({\operatorname{supp}} f)\subseteq G$ is compact in G and by Tietze’s extension theorem we can find a function $0\le \tilde f\in C_c(G)$ such that $\tilde f(y^{-1}{\gamma} y)=f(G_{\gamma} y)$ for every $y\in G$ with $f(G_{\gamma} y)>0$. Choose $0\le F\in C_c(G)^2$ such that $\tilde f\le F$. To show the existence of such a function, let $g\ge 0$ in C _c (G) such that $g>0$ in a neighborhood of ${\operatorname{supp}}(f)$. There exists a unit-neighborhood U, such that $\phi_U*g$ is $>0$ on the set ${\operatorname{supp}}(f)$, where φ _U is a Dirac function of support in U. Set $F=c \phi_U*g$ for some sufficiently large $c>0$, then $f\in C_c(G)^2$ and $F\ge f$. Further F satisfies the conditions of the theorem, so we get

$$\begin{aligned} \int_{G_{\gamma}{\backslash} G} f(x)\,dx = & \int_{{\Upgamma}_{\gamma}{\backslash} G} f(p(y))\,d\nu(y) =\int_{{\Upgamma}_{\gamma}{\backslash} G} \tilde f(y^{-1}{\gamma} y)\,d\nu(y)\\ &\le \int_{{\Upgamma}_{\gamma}{\backslash} G} F(y^{-1}{\gamma} y)\, d\nu(y),\end{aligned}$$

which is finite by computations preceding the lemma. Thus μ is a well defined Radon measure. One easily checks that it is invariant. By Theorem 1.5.3 it follows that the modular function of $G_{\gamma}$ agrees with the one of G. As G is unimodular, so is $G_{\gamma}$. Finally, since for every function in $C_c(G_{\gamma}{\backslash} G)$ the function $f\circ p$ is invariant under $G_{\gamma}$, it follows from the finiteness of $\int_{{\Upgamma}_{\gamma}{\backslash} G} f(p(y))\,d\nu(y)$ that ${\Upgamma}_{\gamma}{\backslash} G_{\gamma}$ has finite invariant volume. □

By the lemma, we can continue the calculation above to arrive at

$$\begin{aligned} {\hspace{1pt}{\rm tr}\hspace{2pt}} R(f)&= \sum_{[{\gamma}]}\int_{G_{\gamma}{\backslash} G}\int_{{\Upgamma}_{\gamma}{\backslash} G_{\gamma}} f((\sigma x)^{-1}{\gamma} \sigma x)\,d\sigma\,dx\\ &=\sum_{[{\gamma}]}{\rm vol}({\Upgamma}_{\gamma}{\backslash} G_{\gamma})\,{\cal O}_{\gamma}(f).\end{aligned}$$

This proves the theorem. □

Example 9.3.4

Consider the case $G={\mathbb R}$ and ${\Upgamma}={\mathbb Z}$. Every $t\in{\mathbb R}$ gives a character $x\mapsto e^{2\pi itx}$, and in this way we identify $\widehat G$ with ${\mathbb R}$. The subset $\widehat G_{\Upgamma}$ is mapped to ${\mathbb Z}$ and the multiplicities $N_{\Upgamma} (\pi)$ are each equal to one. Therefore the spectral side of the trace formula equals $\sum_{k\in{\mathbb Z}} \hat f(k),$ and the geometric side equals $\sum_{k\in{\mathbb Z}} f(k).$ In other words, the trace formula is the same as the Poisson summation formula.

For applications of the trace formula, the following lemma will be important.

Lemma 9.3.5

Let G be a locally compact group, and let A be a *-subalgebra of C _c (G) which contains a Dirac net $(\phi_U)_U.$ Assume further that A is stable under left translations L _y , $y\in G.$ Let $(\pi, V)$ and $(\sigma,W)$ be unitary representations of G such that for every $f\in A$ the operators $\pi(f)$ and $\sigma(f)$ are trace class and that ${\hspace{1pt}{\rm tr}\hspace{2pt}}\pi(f)={\hspace{1pt}{\rm tr}\hspace{2pt}}\sigma(f).$ Then π and σ are both direct sums of irreducible representations with finite multiplicities, and they are equivalent.

Proof

The fact that π and σ are both direct sums of irreducible representations with finite multiplicities is Lemma 9.2.7. There is a subrepresentation of π, maximal with the property of being isomorphic to a subrepresentation of σ. Restricting to the orthogonal space we can assume that π and σ do not have isomorphic subrepresentations. We have to show that they both are zero.

Let $V=\bigoplus_{{\alpha}\in I}V_{\alpha}$ be a decomposition in pairwise orthogonal subrepresentations and let $v_{{\alpha},\mu}\in V_{\alpha}$ be vectors, such that

$$\sum_{\alpha}\sum_\mu\Vert{\pi(f) v_{{\alpha},\mu}}\Vert^2\ <\ \infty$$

for every $f\in A$. Choose a vector $0\ne w\in W$. We claim that for every ${\varepsilon}>0$ there exists some $f\in A$, such that

$$\sum_{\alpha}\sum_\mu\Vert{\pi (f)v_{{\alpha},\mu}}\Vert^2<{\varepsilon}\Vert{\sigma(f)w}\Vert^2.$$

If this were not the case, the map

$$\sum_{\alpha}\sum_\mu\pi(f)v_{{\alpha},\mu}\mapsto\sigma(f)w, \qquad f\in A,$$

would be well-defined and could be extended to a non-trivial intertwining map from the closure L of the space

$$\left\{\sum_{\alpha}\sum_\mu\pi(f)v_{{\alpha},\mu}: f\in A\right\}\subset \bigoplus_{\alpha}\bigoplus_\mu V_{{\alpha},\mu}$$

to W. Here every $V_{{\alpha},\mu}$ is a new copy of the space $V_{\alpha}$. The space L decomposes as a direct sum of irreducibles and it only contains isotypes, which don’t occur in W, so such an intertwiner cannot exist.

Now assume $\sigma\ne 0$. As A contains Dirac functions of arbitrary small supports, there is a function h in A, such that $\sigma(h)\ne 0$. Let $f=h*h^*$. Then $\sigma(f)$ is of trace class and positive. Therefore $\sigma(f)$ possesses a largest eigenvalue and we can scale h in such a way that this eigenvalue is equal to 1. Let $w\in W$ of norm one with $\sigma(f)w=w$. Let ${\lambda}>0$ be the largest eigenvalue of $\pi(f)$. For every ${\alpha}$, let $(v_{{\alpha},\mu})$ be an orthonormal basis of $V_{\alpha}$ consisting of eigenvectors of $\pi(f)$, say $\pi(f)v_{{\alpha},\mu} ={\lambda}_{{\alpha},\mu} v_{{\alpha},\mu}$. For every $g\in A$, the sum $\sum_{\alpha}\sum_\mu\Vert{\pi(g)v_{{\alpha},\mu}}\Vert^2$ equals the square of the Hilbert-Schmidt norm of $\pi(g)$ and therefore is finite. By the above remark there is a $g\in A$ with

$$\sum_{\alpha} \sum_\mu\Vert{\pi(g)v_{{\alpha},\mu}}\Vert^2\ <\ \frac 1{{\lambda}^2} \Vert{\sigma(g) w}\Vert^2.$$

The trace of $\pi(g*f)^*\pi(g*f)$ equals

$$\begin{aligned} \sum_{\alpha}\sum_\mu\Vert{\pi(g*f)v_{{\alpha},\mu}}\Vert^2 = & \sum_{\alpha}\sum_\mu{\lambda}_{{\alpha},\mu}^2\Vert{\pi(g)v_{{\alpha}\mu}}\Vert^2\\ &\le{\lambda}^2\sum_{\alpha}\sum_\mu\Vert{\pi(g)v_{{\alpha},\mu}}\Vert^2.\end{aligned}$$

The right hand side is strictly smaller than $\Vert{\sigma(g)w}\Vert^2=\Vert{\sigma(g*f)w}\Vert^2,$ which again is smaller than

$$\hspace{1pt}{\rm tr}\hspace{2pt}\left(\sigma(g*f)^*\sigma(g*f)\right)={\hspace{1pt}{\rm tr}\hspace{2pt}}\left(\pi(g*f)^*\pi(g*f)\right),$$

a contradiction!

This implies $\sigma=0$ and, by symmetry, also $\pi=0$. □

9.4 Locally Constant Functions

Let G be a locally compact group with a cocompact lattice ${\Upgamma}\subset G$. A function f on G is called locally constant , if for every $x\in G$ there exists a neighborhood U _x of x such that f is constant on U _x . There are not many locally constant functions if G is connected, but there are many if G is totally disconnected.

The function f is called uniformly locally constant , if there exists a unit-neighborhood U such that f is constant on every set of the form UxU, $x\in G$. For example, if f is locally constant and of compact support, then f is uniformly locally constant.

Proposition 9.4.1

Let G be totally disconnected and f a uniformly locally constant and integrable function on G. Then $f\in C_{\rm unif}(G)^2.$ Hence the trace formula is valid for f.

Proof

Let U denote a compact open subgroup such that f is constant on every set of the form UxU. Then $f_U=|f|$ and therefore $f\in C_{\rm unif}(G)$. The same holds for the function $g=\frac{1}{{\rm vol}(U)}1_U$ and one sees that $f=g*f$, so $f\in C_ {\rm unif}(G)^2$ as claimed.

9.5 Lie Groups

In this section, we shall present a simple class of functions which satisfy the trace formula in the case of a Lie-group, i.e., a group which is a smooth manifold such that the group operations are smooth maps. We shall freely make use of the notion of a smooth manifold as in [War83].

Theorem 9.5.1

Let G be a Lie group of dimension n and let ${\Upgamma}\subset G$ be a cocompact lattice. Let f be a continuous integrable function on G, such that the sum

$$k(x,y)=\sum_{{\gamma}\in{\Upgamma}}f(x^{-1}{\gamma} y)$$

converges uniformly and the kernel k is 2r-times continuously differentiable, where r is the smallest integer with $r>n/2.$ Then the trace formula is valid for f.

In particular, the trace formula holds for every $f\in C_c^r(G)$.

Proof

To show the theorem, we need a partition of unity with a smooth square-root. The easiest way to achieve that is to adapt the classical construction of a partition of unity as is done in the following proposition.

Lemma 9.5.2

Let M be a smooth manifold and let $(U_i)_{i\in I}$ be an open covering of M. Then there are smooth functions $u_i: M\to [0,1],$ such that the support of u _i in contained in U _i and that

$$\sum_{i\in I} u_i\ \equiv 1,$$

where the sum is locally finite, i.e., for every $p\in M$ there is a neighborhood U, such that the set

$$\{i\in I: u_i|_U\ne 0\}$$

is finite. The family (u _i ) is called a smooth partition of unity subordinate to the covering $(U_i).$ One can choose the u _i in a way that for each $i\in I$ the function $\sqrt{u_i}$ is smooth as well.

Proof

Except for the smoothness of the square-root, this is Theorem 1.11 of [War83], the proof of which is adapted to give the lemma. In loc.cit., one constructs a sequence $(\psi_j)_{j\ge 1}$ of smooth functions, which are $\ge 0$ and such that the sets $\{p:\psi_j(p)>0\}$ form a locally-finite covering of M, subordinate to the covering (U _i ). So the function

$$\psi=\sum_{j\ge 1}\psi_j^2$$

is a smooth function with $\psi(p)>0$ for every $p\in M$. The functions

$$u_j=\frac{\psi_j^2}{\psi}$$

are smooth with smooth square root and satisfy $\sum_ju_j=1$. □

A Borel measure ν on ${\mathbb R}^n$ is called a smooth measure , if it has a smooth, strictly positive density with respect to the Lebesgue measure ${\lambda}$, i.e., if there is a smooth function

$$d:{\mathbb R}^n\to (0,\infty)$$

such that

$$\nu(A)=\int_A d(x)\,d{\lambda}(x)$$

holds for every Borel set $A\subset{\mathbb R}^n$.

A measure μ on a smooth manifold M is called a smooth measure, if for every smooth chart $\phi:U\to{\mathbb R}^n$ the induced measure $\nu=\phi_*\mu$ on ${\mathbb R}^n$, given by $\mu(A)=\nu(\phi^{-1}(A))$ is smooth.

Proposition 9.5.3

Let M be a compact smooth manifold of dimension n with a smooth measure μ. Let $k:M\times M\to{\mathbb C}$ be continuous and 2r-times continuously differentiable in the first argument, where r is the smallest integer $r>n/2.$ Then the induced integral operator $T_k:L^2(M)\to L^2(M)$,

$$T_k(\phi)(x)=\int_Mk(x,y)\phi(y)\,dy,$$

is of trace class and

$$\hspace{1pt}{\rm tr}\hspace{2pt}(T_k)=\int_Mk(x,x)\,dx.$$

Proof

We first prove the assertion in the case $M={\mathbb R}^n/{\mathbb Z}^n$ and μ being the Lebesgue measure ${\lambda}$. In this case we define

$$l(x,y)=\sum_{k\in{\mathbb Z}^n}\left(\frac1{1+4\pi^2\Vert k\Vert^2}\right)^{r}e_k(x){\overline{e_k(y)}},$$

where $e_k(x)=e^{2\pi i(x_1k_1+x_2k_2+\dots+x_nk_n)}$ and $\Vert k\Vert^2=k_1^2+\dots+k_n^2.$ As the corresponding integral converges, this sum converges absolutely uniformly and the kernel $l(x,y)$ therefore is continuous. Let ∆ be the Laplace operator,

$$\Updelta=-\frac{\partial^2}{\partial x_1^2}-\dots-\frac{\partial^2}{\partial x_n^2}.$$

Lemma 9.5.4

For $\phi\in C^{2r}({\mathbb R}^n/{\mathbb Z}^n)$ one has

$$T_l(1+\Updelta)^r\phi=\phi.$$

Proof

Both sides are continuous functions, so it suffices to show that they agree as L ²-functions. But for all e _k , $k\in{\mathbb Z}^n$, we have $(1+\Updelta)^re_k=(1+4\pi^2\Vert k\Vert^2)^re_k$ and $T_le_k=(1+4\pi^2\Vert k\Vert^2)^{-r}e_k$, hence $T_l(1+\Updelta)^re_k=e_k$. Since $\{e_k: k\in{\mathbb Z}^n\}$ is an orthonormal basis of $L^2({\mathbb R}^n/{\mathbb Z}^n)$ we see that $T_l(1+\Updelta)^r={\rm Id}_{L^2({\mathbb R}^n/{\mathbb Z}^n)}$, and the result follows. □

We now show the proposition in the case $M={\mathbb R}^n/{\mathbb Z}^n$ equipped with the Lebesgue measure. Let k be as in the theorem. Then for $\phi\in L^2(M)$ the function $T_k\phi$ is in $C^{2r}(M)$, so by the lemma,

$$T_l(1+\Updelta)^rT_k\phi=T_k\phi.$$

One has $(1+\Updelta)^rT_k=T_{k'}$, where $k'$ is the continuous kernel

$$k'(x,y)=(1+\Updelta_x)^rk(x,y).$$

So $T_k=T_lT_{k'}$ is a product of two Hilbert-Schmidt operators, hence of trace class. As both, T _l and $T_{k'}$ are admissible, the claim follows from Lemma 9.3.1.

Next let M be an arbitrary smooth compact manifold of dimension n. Let $(U_i)_{i=1}^s$ be an open covering by chart sets, where we choose the chart maps ψ _i not as maps to ${\mathbb R}^n$ but instead to ${\mathbb R}^n/{\mathbb Z}^n$, which is possible as well. So for every i the map ψ _i is a homeomorphism of U _i to some open set $V_i\subset{\mathbb R}^n/{\mathbb Z}^n$. The sets $V_1,\dots, V_s\subset{\mathbb R}^n/{\mathbb Z}^n$ can be chosen pairwise disjoint. Let (u _i ) be a smooth partition of unity with smooth square root subordinate to the covering (U _i ). For $1\le i,j\le s$ let

$$k_{i,j}(x,y)=\sqrt{u_i(x)}k(x,y)\sqrt{u_j(y)}.$$

Then $k_{i,j}$ is a continuous kernel on $M\times M$. For given $1\le i,j\le s$, define a continuous kernel $\tilde k_{i,j}$ on ${\mathbb R}^n/{\mathbb Z}^n\times{\mathbb R}^n/{\mathbb Z}^n$ by

$$\tilde k_{i,j}(x,y)=\begin{cases}\sqrt{d_i(x)}k_{i,j}(\psi_i^{-1}(x),\psi_j^{-1}(y))\sqrt{d_j(y)}& (x,y)\in V_i\times V_j,\\ 0&\text{otherwise.}\end{cases}$$

Here d _j denotes the density of the measure $(\psi_j)_*\mu$ with respect to the Lebesgue measure, i.e. $\int_{V_i} f(x) d_i(x)\,d\lambda(x)=\int_{U_i} f(\psi_i(y))d\mu(y)$ for all $f\in C_c(V_i)$. We define $\tilde k=\sum_{i,j}\tilde k_{i,j}$ and for $\phi\in L^2(M)$ we set

$$\phi_j(x)=\phi(x)\sqrt{u_j(x)}.$$

We define $\tilde\phi_j\in L^2({\mathbb R}^n/{\mathbb Z}^n)$ by

$$\tilde\phi_j(x)=\begin{cases}\phi_j(\psi_j^{-1}(x))\sqrt{d_j(x)}& x\in V_j,\\ 0 &\text{otherwise.}\end{cases}$$

Finally set $\tilde\phi=\sum_j\tilde\phi_j$.

Lemma 9.5.5

The map $\Uppsi:\phi\mapsto\tilde\phi$ is a linear isometry $L^2(M)\hookrightarrow L^2({\mathbb R}^n/{\mathbb Z}^n)$ and one has

$$\Uppsi(T_k\phi)=T_{\tilde k}\Uppsi(\phi)$$

for every $\phi\in L^2(M).$ The operator $T_{\tilde k}$ equals $PT_{\tilde k}P,$ where P is the orthogonal projection $L^2({\mathbb R}^n/{\mathbb Z}^n)\to{\rm Im}(\Uppsi)$. Finally we have

$$\int_Mk(x,x)\,d\mu(x)=\int_{{\mathbb R}^n/{\mathbb Z}^n}\tilde k(x,x)\,dx.$$

Proof

The map Ψ is linear. For $\phi\in L^2(M)$ we compute

$$\begin{aligned} \Vert{\Uppsi(\phi)}\Vert^2 &= \int_{{\mathbb R}^n/{\mathbb Z}^n}|\tilde\phi(x)|^2\,d{\lambda}(x)=\sum_j\int_{V_j}|\tilde\phi_j(x)|^2\,d{\lambda}(x)\\ &= \sum_j\int_{V_j}|\phi_j(\overbrace{\psi_j^{-1}(x)\!}^{=z})|^2\underbrace{d_j(x)\,d{\lambda}(x)}_{d(\psi_j)_*\mu(x)}\\ &=\sum_j\int_{U_j}|\phi_j(z)|^2\,d\mu(z)=\sum_j\int_M|\phi(z)|^2u_j(x)\,d\mu(z)\\ &=\int_M|\phi(z)|^2\,d\mu(x)=\Vert\phi\Vert^2,\end{aligned}$$

so Ψ is an isometry. In order to show $\Uppsi T_k=T_{\tilde k}\Uppsi$, we compute

$$\begin{aligned} T_{\tilde k}\Uppsi(\phi)(x)&=T_{\tilde k}\tilde\phi(x)=\sum_j T_{\tilde k}\tilde\phi_j(x)= \sum_j \int_{V_j}\tilde k(x,y)\tilde\phi_j(y)\,d{\lambda}(y)\\ &=\sum_{i,j}\sqrt{d_i(x)}\int_{V_j}k_{i,j}(\psi_i^{-1}(x),\psi_j^{-1}(y))\phi_j(\psi_j^{-1}(y))\underbrace{d_j(y)\,d{\lambda}(y)}_{=d(\psi_j)_*\mu(y)}\\ &=\sum_{i,j}\sqrt{d_i(x)}\int_{U_j}k_{i,j}(\psi_i^{-1}(x),z)\phi_j(z)\,d\mu(z)\\ &=\sum_{i,j}\sqrt{d_i(x)}\int_{U_j}\sqrt{u_i(\psi_i^{-1}(x))}k(\psi_i^{-1}(x),z)u_j(z)\phi(z)\,d\mu(z)\\ &=\sum_{i}\sqrt{d_i(x)}\sqrt{u_i(\psi_i^{-1}(x))}\int_{M}k(\psi_i^{-1}(x),z)\phi(z)\,d\mu(z)\\ &=\sum_{i}\sqrt{d_i(x)}\sqrt{u_i(\psi_i^{-1}(x))}T_k\phi(\psi_i^{-1}(x)) =\Uppsi T_k\phi(x).\end{aligned}$$

To show $T_{\tilde k}=PT_{\tilde k}P$ one has to show that for $g\in{\rm Im}(\Uppsi)^\perp$,

$$T_{\tilde k}(g)=0 \quad\text{and}\quad{\left\langle{T_{\tilde k}h,g}\right\rangle}=0$$

holds for every $h\in L^2({\mathbb R}^n/{\mathbb Z}^n)$. The first of these assertions follows from the fact that for fixed x, the map $y\mapsto\tilde k(x,y)$ lies in the image of Ψ. The second assertion follows similarly from the observation, that for fixed y the map $x\mapsto \tilde k(x,y)$ lies in the image of □.

We finish the proof of the lemma with

$$\begin{aligned} \int_{{\mathbb R}^n/{\mathbb Z}^n}\tilde k(x,x)\,d{\lambda}(x) &= \sum_{i,j}\int_{{\mathbb R}^n/{\mathbb Z}^n} \underbrace{\tilde k_{i,j}(x,x)}_{=0\,\text{if}\,i\ne i}\,d{\lambda}(x)\\ &=\sum_j\int_{V_j} k_{j,j}(\psi_j^{-1}(x),\psi_j^{-1}(x))d_j(x)\,d{\lambda}(x)\\ &=\sum_j\int_{U_j}k_{j,j}(x,x)\,d\mu(x)\\ &=\sum_j\int_{U_j}k(x,x)u_j(x)\,d\mu(x)\\ &=\int_{M}k(x,x)\,d\mu(x). \end{aligned}$$

The lemma implies

$$\begin{aligned} {\hspace{1pt}{\rm tr}\hspace{2pt}} T_k={\hspace{1pt}{\rm tr}\hspace{2pt}} T_{\tilde k}=\int_{{\mathbb R}^n/{\mathbb Z}^n}\tilde k(x,x)\,dx=\int_Mk(x,x)\,dx\end{aligned}$$

and the proposition follows. □

Now Theorem 9.5.1 follows from Proposition 9.5.3 in the same way as Theorem 9.3.2, i.e., the trace formula, follows from Proposition 9.3.1. □

9.6 Exercises

Exercise 9.1

Show that every Schwartz function on ${\mathbb R}$ is uniformly integrable.

Exercise 9.2

Let B be the group of upper triangular matrices in ${\rm SL}_2({\mathbb R})$. Show that B does not contain a lattice.

Exercise 9.3

Let ${\Upgamma}$ be a subgroup of the locally compact group G. Show that ${\Upgamma}$ is discrete if and only if for every compact subset $C \subset G$ the intersection ${\Upgamma}\cap C$ is finite.

Exercise 9.4

A group ${\Upgamma}$ acts discontinuously on a topological space X if for every $x\in X$ there is a neighborhood U of x, such that the set of all ${\gamma}\in{\Upgamma}$ with $U\cap{\gamma} U\ne\emptyset$ is finite.

Let $K\subset G$ be a compact subgroup of the locally compact group G. Show that a subgroup ${\Upgamma}\subset G$ is discrete if and only if it acts discontinuously on the space $G/K$.

Exercise 9.5

Let G be a finite group, and let H be a subgroup. For each choose the counting measure as Haar measure. For $x\in H$ let $H_x, G_x$ be the centralizers of x in H and G, respectively. For $\pi\in\widehat G$ let $\chi_\pi: G\to{\mathbb C}$ be defined by $\chi_\pi(x)={\hspace{1pt}{\rm tr}\hspace{2pt}}\pi (x)$. For $x\in H$ let $[x]_H$ be the H-conjugacy class of x, i.e., $[x]_H=\{hxh^{-1}: h\in H\}.$ Show:

$$\sum_{\pi\in\widehat G}({\rm dim} V_\pi^H)\,\chi_\pi=\sum_{[h]_H}\frac{|G_h|}{| H_h|}\,{\bf 1}_{[h]_G},$$

where $V_\pi^H$ denotes the space of H-invariant vectors in $V_\pi$.

(Hint: Let g be one of the sides of the equation. For $f\in L^2(G)$ consider the inner product ${\left\langle{f,g}\right\rangle}$ and use the trace formula.)

Exercise 9.6

Let $f\in{\cal S}({\mathbb R})$ be a Schwartz function. Apply the trace formula to f with $G={\mathbb R}$ and ${\Upgamma}={\mathbb Z}$ to give a proof of the classical Poisson summation formula:

$$\sum_{k\in{\mathbb Z}} f(k)=\sum_{k\in{\mathbb Z}} \hat f(k),$$

where $\hat f(x)=\int_{\mathbb R} f(y)e^{-2\pi ixy}\, dy$.

Exercise 9.7

Let G be a first countable compact group and $H\subset G$ a closed subgroup. Let $f\in C(G)^2$ with ${\operatorname{supp}}(f)\cap G.H=\emptyset$, where G.H is the union of all conjugates gHg ^-1 of H. Show that

$$\sum_{\pi\in\widehat G} {\rm dim}(V_\pi^H)\,{\hspace{1pt}{\rm tr}\hspace{2pt}}\pi(f)=0.$$