The Heisenberg Group

Abstract

In this chapter we prove the Stone-von Neumann Theorem, which gives a full characterization of the unitary dual of the Heisenberg group ${\cal H}$. We then apply the trace formula to describe the spectral decomposition of ${L^2}(\Lambda \backslash H)$, where π is the standard integer lattice in ${\cal H}$.

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In this chapter we prove the Stone-von Neumann Theorem, which gives a full characterization of the unitary dual of the Heisenberg group ${\cal H}$. We then apply the trace formula to describe the spectral decomposition of ${L^2}(\Lambda \backslash H)$, where π is the standard integer lattice in ${\cal H}$.

10.1 Definition

The Heisenberg group ${\cal H}$ is defined to be the group of real upper triangular 3 × 3 matrices with ones on the diagonal:

$$\cal H{\ \stackrel{{\rm def}}{=}} \left\{\left. \left(\begin{array}{ccc} 1 & x & z \\ & 1 & y \\ & & 1\end{array}\right) \right| x,y,z\in{\mathbb R}\right\}.$$

It can also be identified with ${\mathbb R}^3$, where the group law is given by

$$(a,b,c)(x,y,z){\ \stackrel{{\rm def}}{=}\,} (a+x,b+y,c+z+ay).$$

The inverse of $(a,b,c)$ is

$$(a,b,c)^{-1}= (\!\!-\!a,-b,ab-c).$$

The center of ${\cal H}$ is $Z({\cal H})=\{(0,0,z)\mid z\in{\mathbb R}\}$ , and the projection to the first two coordinates induces an isomorphism

$$\cal H/Z({\cal H})\ \cong\ {\mathbb R}^2.$$

An easy calculation ([Dei05] Chap. 12) shows that ${\cal H}$ is unimodular and that a Haar integral on ${\cal H}$ is given by

$$\int_{\cal H} f(h)\, dh{\ \stackrel{{\rm def}}{=}} \int_{\mathbb R}\int_{\mathbb R}\int_{\mathbb R} f(a,b,c)\, da\, db\, dc,\ \ f\in C_c({\cal H}).$$

We will use this Haar measure on ${\cal H}$ for all computations in the sequel.

10.2 The Unitary Dual

In this section we are going to describe the unitary dual $\widehat{\cal H}$ of the Heisenberg group ${\cal H}$.

Let $\widehat{\cal H}_0$ denote the subset of $\widehat{\cal H}$ consisting of all classes $\pi\in\widehat{\cal H}$ such that $\pi(h)=1$ whenever h lies in the center $Z({\cal H})$ of ${\cal H}$. Since ${\cal H}/Z ({\cal H})\cong{\mathbb R}^2$, it follows that

$$\widehat{\cal H}_0=\widehat{{\cal H}/Z({\cal H})}\ \cong\ \widehat{{\mathbb R}^2}\ \cong\ \left(\widehat{\mathbb R}\right)^2,$$

and the latter can be identified with ${\mathbb R}^2$ in the following explicit way. Let $(a,b)\in{\mathbb R}^2$ and define a character

$$\begin{array}{cccl} \chi_{a,b}: & H & \to & {\mathbb T},\\ & (x,y,z) &\mapsto & e^{2\pi\! i(ax+by)}.\end{array}$$

The identification is given by $(a,b)\mapsto \chi_{a,b}$. In particular, it follows that all representations in $\widehat{\cal H}_0$ are one-dimensional. This observation indicates the importance of the behavior of the center under a representation.

As a consequence of the Lemma of Schur 6.1.7, for each $\pi\in\widehat{\cal H}$ there is a character $\chi_\pi: Z({\cal H})\to{\mathbb T}$ with $\pi(z)=\chi_\pi(z){\rm Id}$ for every $z\in Z({\cal H})$. This character $\chi_\pi$ is called the central character of the representation π.

For every character $\chi\ne 1$ of $Z({\cal H})$, we will now construct an irreducible unitary representation of the Heisenberg group that has χ for its central character. So let $t\ne 0$ be a real number and consider the central character

$$\chi_t(0,0,c)= e^{2\pi\! i ct}.$$

For $(a,b,c)\in{\cal H}$ we define the operator $\pi_t(a,b,c)$ on $L^2({\mathbb R})$ by

$$\pi_t(a,b,c)\phi(x){\ \stackrel{{\rm def}}{=}\,} e^{2\pi\! i (bx+c)t} \phi(x+a).$$

It is straightforward that π _t is a unitary representation.

Recall from Exercise 3.17 that the Schwartz space ${\cal S}({\mathbb R}^n)$ consists of all $C^\infty$-functions $f:{\mathbb R}^n\to{\mathbb C}$ such that $x^\alpha\partial^\beta f$ is bounded for all multi-indices $\alpha,\beta\in{\mathbb N}_0^n$. We have ${\cal S}({\mathbb R}^n)\subseteq L^p({\mathbb R}^n)$ for every $p\geq 1$ and ${\cal S}({\mathbb R}^n)$ is stable under Fourier transform.

Theorem 10.2.1

(Stone-von Neumann). For $t\ne 0$ the unitary representation π _t is irreducible. Every irreducible unitary representation of ${\cal H}$ with central character $\chi_t$ is isomorphic to π _t . It follows that

$$\widehat{\cal H}=\widehat{\mathbb R}^2\ \cup\ \{\pi_t: t\ne 0\}.$$

Proof

We show irreducibility first. Fix $t\ne 0$, and let $V\subset L^2({\mathbb R})$ be a closed non-zero subspace that is invariant under the set of operators $\pi_t({\cal H})$. If $\phi\in V$, then so is the function $\pi_t(\!\!-\!a,0,0)\phi(x)=\phi(x-a)$. As V is closed, it therefore contains $\psi *\phi(x)=\int_{\mathbb R}\psi(a)\phi(x- a)\,da$ for $\psi\in{\cal S}={\cal S}({\mathbb R})$. These convolution products are smooth functions, so V contains a smooth function $\phi\ne 0$. One has $\pi_t(0,-b,0)\phi(x)= e^{-2\pi\! ibtx}\phi(x)\ \in\ V.$ By integration it follows that for $\psi\in{\cal S}$ one has that $\hat\psi(tx) \phi(x)$ lies in V. The set of possible functions $\hat\psi(tx)$ contains all smooth functions of compact support, as the Fourier transform is a bijection on the space of Schwartz functions to itself. Choose an open interval I, in which φ has no zero. It follows that $C_c^\infty(I)\subset V$. Translating, taking sums, and using a partition of unity argument gives $C_c^\infty({\mathbb R})\subset V$. This space is dense in $L^2({\mathbb R})$, so $V=L^2({\mathbb R})$, which means that π _t is irreducible.

For the second assertion of the theorem, let π be an irreducible unitary representation with central character χ _t . We want to show that π is isomorphic to π _t . For notational ease, let’s first reduce to the case t = 1. Consider the map $\theta_t:{\cal H}\to{\cal H}$ given by $\theta_t(a,b,c)=(a,bt,ct)$. A calculation shows that θ _t is an automorphism of ${\cal H}$ with $\chi_1\circ\theta_t=\chi_t$. So we conclude that π has central character χ _t if and only if $\pi\circ\theta_t^{-1}$ has central character χ₁. It therefore suffices to show the uniqueness for t = 1.

Let $(\eta,V_\eta)$ be an irreducible unitary representation of ${\cal H}$ with central character χ₁. Let ${\Gamma}$ be the subgroup of ${\cal H}$ consisting of all $(0,0,k)$ with $k\in{\mathbb Z}$. Then ${\Gamma}$ lies in the center and $\chi_1({\Gamma})=\{1\}$. So the representation η factors over the quotient group $B{\ \stackrel{{\rm def}}{=}} {\cal H}/{\Gamma},$ which is homeomorphic to the space ${\mathbb R}^2\times{\mathbb R}/{\mathbb Z}$. For $\phi\in{\cal S}({\mathbb R}^2)$, let $\phi_B(a,b,c)=\phi(a,b)e^{-2\pi\! ic}$. Then $\phi(a,b)=\phi_B(a,b,0)$. For $\phi,\psi\in{\cal S}({\mathbb R}^2)$, one computes

$$\begin{aligned} \hspace{-20pt}\phi_B *\psi_B (a,b,c)&= \int_B\phi_B(h)\psi_B(h^{-1}(a,b,c)) \,dh\\ &= \int_{{\mathbb R}^2}\int_{{\mathbb R}/{\mathbb Z}}\phi(x,y)e^{-2\pi\! iz}\psi_B(a-x,b-y,c-z+xy-xb) \,dx\,dy\, dz\\ &= \int_{{\mathbb R}^2} \int_{{\mathbb R}/{\mathbb Z}}\phi(x,y)e^{-2\pi\! iz}\psi(a-x,b-y)e^{-2\pi\! i(c-z +xy-xb)}\,dx\,dy\,dz\\ &= e^{-2\pi\! ic} \int_{{\mathbb R}^2} \phi(x,y)\psi(a-x,b-y)e^{2\pi\! i x(b-y)}\,dx \,dy.\end{aligned}$$

Inspired by this, we define a new, non-commutative convolution product on ${\cal S}({\mathbb R}^2)$ by

$$\phi *_B\psi (a,b) {\ \stackrel{{\rm def}}{=}} \int_{{\mathbb R}^2} \phi(x,y)\psi(a-x,b-y)e^{2\pi\! i x(b-y)} \,dx\,dy.$$

For $\phi\in{\cal S}({\mathbb R}^2)$ we define $\eta(\phi)=\eta(\phi_B)$. This equals

$$\eta(\phi)=\int_{{\mathbb R}^2}\phi(a,b)\eta(a,b,0)\,da\,db.$$

By construction of the convolution product one gets $\eta(\phi *_B\psi)=\eta (\phi)\eta(\psi)$, so η is an algebra homomorphism of the algebra ${\cal A}= ({\cal S}({\mathbb R}^2),*_B)$. One computes that $(\phi_B)^*=(\phi^*)_B$, where the involution on the left is the usual involution of functions g on the group B, given by $g^*(b) ={\overline{g(b^{-1})}}$. The involution on the algebra ${\cal A}={\cal S}({\mathbb R}^2)$ on the other hand is defined by $\phi^*(a,b)={\overline{\phi(\!\!-\!a,-b)}}e^{2\pi\! iab}$. So η is a homomorphism of *-algebras.

Lemma 10.2.2

The *-algebra homomorphism $\eta: {\cal A}\to{\cal B}(V_\eta)$ is injective.

Proof

Assume $\eta(\phi)=0$. Then, for $h=(a,b,0)\in B$ and $v,w\in V_\eta$ we have

$$\begin{aligned} 0&= {\left\langle{\eta(h)\eta(\phi)\eta(h^{-1})v,w}\right\rangle}\\ &= \int_{{\mathbb R}^2}{\left\langle{\eta\, \textit{(x,\;y,\;0)\;v,\;w}}\right\rangle}\phi(x,y)e^{2\pi\! i(ay-bx)}\,dx\,dy\end{aligned}$$

The continuous, rapidly decreasing function ${\left\langle{\eta(x,y,0)v,w}\right\rangle}\phi(x,y)$ therefore has zero Fourier transform, hence is zero. Varying v and w, this leads to φ being zero. □

In the special case $\eta=\pi_1$ the lemma has the consequence that relations among the $\pi_1(\phi)$ for $\phi\in{\cal A}$ already hold in ${\cal A}$. We will make use of this principle for special elements as follows. For $F,G\in{\cal S}({\mathbb R})$ let $\tilde\phi_{F,G}(a,b)= {\overline{F(a+b)}}G(b),$ and $\phi_{F,G}(a,b)={\cal F}_2\tilde\phi_{F,G}(a,b)\ \in\ {\cal S}({\mathbb R}^2),$ where ${\cal F}_2$ denotes taking Fourier transform in the second variable. An application of the properties of the Fourier transform yields

$$\begin{aligned} \pi_1(\phi_{F,G})f(x) &= \int_{{\mathbb R}^2} \phi_{F,G}(a,b)\pi_1(a,b,0)f(x)\,da\,db\\ &= \int_{{\mathbb R}^2} \phi_{F,G}(a,b)f(x+a)e^{2\pi\! i bx}\,da\,db\\ &= \int_{{\mathbb R}^2} {\cal F}_2\tilde\phi_{F,G}(a-x,b)f(a)e^{2\pi\! i bx}\,da\,db\\ &=\int_{\mathbb R}\tilde\phi_{F,G}(a-x,x)f(a)\,da\\ &= \int_{\mathbb R} {\overline{F(a)}}G(x)f(a)\,da= {\left\langle{f,F}\right\rangle}G(x).\end{aligned}$$

In particular, if the norm of F is one, then $\pi_1(\phi_{F,F})$ is the orthogonal projection onto the one dimensional space ${\mathbb C} F$.

For $h\in B={\cal H}/{\Gamma}$ and $\phi\in{\cal S}({\mathbb R}^2)$ define $L_h\phi$ by the formula $\left(L_h\phi\right) _B=L_h(\phi_B)$. Explicitly, one computes that $L_{(x,y,z)}\phi(a,b)= \phi(a-x,b-y)e^{2\pi\! i(bx-xy+z)}.$ For $\phi,\psi\in{\cal A}$ we also write $\phi\psi$ for the convolution product $\phi *_B\psi$.

Lemma 10.2.3

In the algebra ${\cal A}$ , one has the relations

$$\phi_{F,G}\phi_{H,L}={\left\langle{L,F}\right\rangle}\phi_{H,G}\ {\rm and}\ \phi_{F,G}^*=\phi_{G,F}.$$

For $h\in B$ one has $L_h\phi_{F,G}=\phi_{F,\pi_1(h)G}$.

Proof

Using the formula $\pi_1(\phi_{F,G})f= {\left\langle{f,F}\right\rangle}G$ one computes for $f\in L^2({\mathbb R})$,

$$\begin{aligned} \pi_1(\phi_{F,G}\phi_{H,L})f &= \pi_1(\phi_{F,G})\pi_1(\phi_{H,L})f\\ &= {\langle{\pi_1(\phi_{H,L})f,F}\rangle}G\\ &= {\left\langle{{\left\langle{f,H}\right\rangle}L,F}\right\rangle}G= {\left\langle{L,F}\right\rangle}{\left\langle{f,H}\right\rangle}G.\end{aligned}$$

This implies the first claim as π₁ is injective. The second follows from

$$\langle{\pi_1(\phi_{F,G})f\!,h}\rangle={\left\langle{f,F}\right\rangle}{\left\langle{G,h}\right\rangle}={\langle{f,\pi_1(\phi_{G,F})h}\rangle}.$$

For the last claim compute

$$\begin{aligned} \pi_1(L_h\phi_{F,G})f &= \pi_1(h)\pi_1(\phi_{F,G})f= \pi_1(h){\left\langle{f\!,F}\right\rangle}G\\ &= {\left\langle{f\!,F}\right\rangle}\pi_1(h)G=\pi_1(\phi_{F,\pi_1(h)G})f.\end{aligned}$$

The lemma follows. □

As ${\cal S}({\mathbb R})$ is dense in $L^2({\mathbb R})$, by the orthonormalizing scheme one can find a sequence F _j in ${\cal S}({\mathbb R})$ that forms an orthonormal base of $L^2({\mathbb R})$. For $j,k\in{\mathbb N}$ write $\phi_{j,k}$ for $\phi_{F_j,F_k}$. Lemma 10.2.3 implies that $\phi_{j,k}\phi_{s,t}=\delta_{j,t}\phi_{s,k}$ (Kronecker-delta), and $\phi_{j,j}^2=\phi_{j,j}=\phi_{j,j}^*$.

Let now $(\eta,V_\eta)$ again denote an irreducible unitary representation with central character χ₁. By integration, η gives an algebra homomorphism ${\cal A}\to{\cal B}(V_\eta)$.

Lemma 10.2.4

The $\eta(\phi_{j,j})_{j\in{\mathbb N}}$ are non-zero projections. They are pairwise orthogonal. Write V _j for the image of $\eta(\phi_{j,j})$. Then $\eta(\phi_{j,k})$ is an isometry from V _j to V _k and it annihilates every V _l for $l\ne j$.

Proof

Since $\phi_{j,j}^2=\phi_{j,j}=\phi_{j,j}^*$, the same holds for $\eta(\phi_{j,j})$ hence the latter are projections. They are non-zero by Lemma 10.2.2. For $v\in V_j$ one computes ${\langle{\eta(\phi_{j,k})v,\eta(\phi_{j,k})v}\rangle}= {\langle{\eta(\phi_{k,j}\phi_{j,k})v,w}\rangle} = {\langle{\eta(\phi_{j,j})v,w}\rangle}={\langle{v,v}\rangle}.$ The claim follows. □

Now choose $v_1\in V_1$ of norm one and set $v_j=\eta(\phi_{1,j})v_1$. Then (v _j ) is an orthonormal system in $V_\eta$. Define an isometry $T: L^2({\mathbb R})\to V_\eta$ by mapping F _j to v _j . Then

$$\begin{aligned} \eta(h)T(F_j)&= \eta(h)v_j= \eta(h)\eta(\phi_{j,j})v_j\\ &= \eta(L_h\phi_{j,j})v_j=\eta(\phi_{F_j,\pi_1(h)F_j})v_j.\end{aligned}$$

By the fact that (F _j ) is an orthonormal basis, one concludes

$$\pi_1(h)F_j=\sum_k{\langle{\pi_1(h)F_j,F_k}\rangle}F_k.$$

So that by Lemma 10.2.3

$$\begin{aligned} \eta(h)T(F_j)&= \sum_k{\langle{\pi_1(h)F_j,F_k}\rangle}\eta(\phi_{F_j,F_k})v_j\\ &=\sum_k{\langle{\pi_1(h)F_j,F_k}\rangle}v_k= T(\pi_1(h)F_j).\end{aligned}$$

Hence T is an ${\cal H}$-homomorphism onto a non-zero closed subspace of $V_\eta$. As η is irreducible, T must be surjective, so T is unitary and $\eta$ is equivalent to π₁. This finishes the proof of the Theorem of Stone and von Neumann. □

10.3 The Plancherel Theorem for ${\cal H}$

We have an identification ${\cal H}\cong{\mathbb R}^3$. We interpret $f\in{\cal S}({\mathbb R}^3)$ as a function on ${\cal H}$, and we write ${\cal S}({\cal H})$ for this space of functions.

Theorem 10.3.1

(Plancherel Theorem). Let $f\in{\cal S}({\cal H})$. For every $t\in{\mathbb R}^\times$ the operator $\pi_t(f)$ is a Hilbert-Schmidt operator, and we have

$$\int_{{\mathbb R}^\times} {\left|\hspace{-1pt}\left| {\pi_t(f)}\right|\hspace{-1pt} \right|}_{HS}^2\, |t|\, dt=\int_{\cal H} |f(h)|^2\, dh.$$

It follows that the Plancherel measure of ${\cal H}$ in the sense of Theorem 8.5.3 equals $|t|\,dt$ and that the set of one dimensional representations of ${\cal H}$ has Plancherel measure zero.

Proof

This is proved in [Dei05], Chap. 12. □

10.4 The Standard Lattice

Let ${\Lambda}$ be the set of all $(a,b,c)\in{\cal H}$, where $a,b,c$ are integers. The multiplication and inversion keep this set stable, so ${\Lambda}$ is a discrete subgroup.

Lemma 10.4.1

${\Lambda}$ is a uniform lattice in ${\cal H}$.

Proof

Let $K=[0,1]\times[0,1]\times[0,1]\subset{\cal H}$. Then K is a compact subset. We claim that the projection map $K\to{\cal H}/{\Lambda}$ is surjective. This means we have to show that for every $h\in{\cal H}$ there exists ${\lambda}\in{\Lambda}$ such that $h{\lambda}\in K$. Let $h=(x,y,z)\in{\cal H}$, and let ${\lambda}=(a,b,c)\in{\Lambda}$. Then $h{\lambda}=(x+a,y+b,z+c+xb)$. So we can find $a,b\in{\mathbb Z}$ such that $x+a$ and $y+b$ lie in the unit interval. So we may assume as well that $x,y\in [0,1]$. Assuming that, we only consider such ${\lambda}$, which are central, i.e., of the form ${\lambda}=(0,0,c)$ and it becomes clear that we can find ${\lambda}$ such that $h {\lambda}\in K$. □

By Theorem 9.2.2 we conclude that as ${\cal H}$-representation one has the decomposition

$$L^2({\Lambda}{\backslash}{\cal H})\ \cong\ \bigoplus_{\pi\in\widehat{\cal H}}N(\pi)\pi,$$

with finite multiplicities $N(\pi)$.

We want to apply the trace formula to compute the numbers $N(\pi)$. For this recall the characters of ${\cal H}$ given by $\chi_{r,s}(a,b,c)=e^{2\pi\! i(ra+sb)}$, where $r,s\in{\mathbb R}$.

Theorem 10.4.2

For a character $\chi_{r,s}$ of ${\cal H}$ the multiplicity $N(\chi_{r,s})$ in $L^2({\Lambda}{\backslash} {\cal H})$ is equal to one if r and s are both integers. Otherwise it is zero. For $t\in{\mathbb R}$, $t\ne 0$, the multiplicity $N(\pi_t)$ is $|t|$ if t is an integer and zero otherwise. So the decomposition of the representation $\left(L^2({\Lambda}{\backslash} {\cal H}), R\right)$ reads

$$R\ \cong\ \bigoplus_{r,s\in{\mathbb Z}}\chi_{r,s}\ \oplus\ \bigoplus_{^{k\in{\mathbb Z}}_{k\ne 0}}|k|\pi_k.$$

Proof

We first consider the subspace H of $L^2({\Lambda}{\backslash} {\cal H})$, which is invariant under the action of the center $Z\subset{\cal H}$. This is isomorphic to $L^2\left({\mathbb Z}^2{\backslash} {\mathbb R}^2\right)$ as ${\Lambda}{\backslash} {\cal H}/Z\cong{\mathbb Z}^2{\backslash} {\mathbb R}^2$ and the representation of ${\cal H}$ on H factors through the representation of ${\mathbb R}^2$ on $L^2\left({\mathbb Z}^2{\backslash} {\mathbb R}^2\right)$. By the theory of Fourier series this gives us the value of $N\left(\chi_{r,s}\right)$ as in the theorem. The difficult part is to determine $N(\pi_t)$.

Let $h\in C_c^\infty({\cal H})$ and set $f=h*h^*$. Then the trace formula says,

$$\hspace{1pt}{\rm tr}\hspace{2pt} R(f)=\sum_{\pi\in\widehat{\cal H}}N(\pi){\hspace{1pt}{\rm tr}\hspace{2pt}}\pi(f)=\sum_{[{\lambda}]}{\rm vol}\left({\Lambda}_{\lambda}{\backslash}{\cal H}_{\lambda}\right)\, {\cal O}_{\lambda}(f),$$

where the sum is taken over all conjugacy classes $[{\lambda}]$ in ${\Lambda}$, ${\Lambda}_{\lambda}$ and ${\cal H}_{\lambda}$ denote the centralizers of ${\lambda}$ in ${\Lambda}$ and ${\cal H}$, respectively, and ${\cal O}_{\lambda}(f)=\int_{{\cal H}_{\lambda}{\backslash}{\cal H}} f(x^{-1}{\lambda} x)\, dx$. First consider $\pi\in\widehat{\cal H}$ with trivial central character, say $\pi=\chi_ r,s$. Then

$$\begin{aligned} {\hspace{1pt}{\rm tr}\hspace{2pt}}\pi(f)&=\chi_{r,s}(f)=\int_{\mathbb R}\int_{\mathbb R}\int_{\mathbb R} f(a,b,c) e^{2\pi\! i (ar+bs)}\,da\,db\,dc\\ &={\cal F}_1{\cal F}_2{\cal F}_3f(\!\!-\!\!r,-s,0),\end{aligned}$$

where ${\cal F}_i$ denotes Fourier transform on ${\mathbb R}$ applied to the i-th component of ${\cal H}$. Next the center $Z_{\Lambda}$ of ${\Lambda}$, the set of all $(0,0,k)$ for $k\in{\mathbb Z}$, is contained in the center of ${\cal H}$ and therefore acts trivially on $L^2({\Lambda} {\backslash}{\cal H})$. This implies that for $\pi\in\widehat{\cal H}$ one has that $N(\pi)\ne 0$ implies $\pi(Z_{\Lambda})=\{1\}$. Therefore, $N(\pi_t)\ne 0$ implies that $t\in{\mathbb Z}$. For $t=k\in{\mathbb Z}$ one computes

$$\begin{aligned} \pi_k(f)\phi(x) &= \int_{\mathbb R}\int_{\mathbb R}\left(\int_{\mathbb R} f(a,b,c)e^{2\pi\! ikc}\,dc \right) e^{2\pi\! ikbx}\phi(a+x)\, da\,db\\ &= \int_{\mathbb R} {\cal F}_2{\cal F}_3f(a-x,-kx,-k)\phi(a)\,da\end{aligned}$$

So $\pi_k(f)$ is an integral operator with kernel

$$k(x,y)={\cal F}_2{\cal F}_3f(y-x,-kx,-k).$$

Analogously, the kernel of $\pi_k(h)$ is ${\cal F}_2{\cal F}_3h(y-x,-kx,-k)$. The latter kernel is in ${\cal S}({\mathbb R}^2)$ and therefore is admissible. By Proposition 9.3.1 and Fourier inversion, we have

$$\hspace{1pt}{\rm tr}\hspace{2pt}\pi_k(f)=\int_{\mathbb R}{\cal F}_2{\cal F}_3f(0,-kx,-k)\,dx= \frac 1{|k|}{\cal F}_3f(0,0,-k).$$

Together we get

$$\hspace{1pt}{\rm tr}\hspace{2pt} R(f)= \sum_{r,s\in{\mathbb Z}}{\cal F} f(r,s,0)+\sum_{k\in{\mathbb Z}{\smallsetminus}\{0\}} \frac{N(\pi_k)}{|k|}{\cal F}_3f(0,0,k),$$

where ${\cal F} f={\cal F}_1{\cal F}_2{\cal F}_3 f$.

Next we consider the geometric side of the trace formula. Let ${\lambda}\in{\Lambda}$. First assume that ${\lambda}$ lies in the center of ${\Lambda}$. Then ${\lambda}=(0,0,k)$ for some $k\in{\mathbb Z}$, and the centralizer ${\cal H}_{\lambda}$ equals ${\cal H}$, so that the orbital integral equals ${\cal O}_{\lambda}(f)= f({\lambda})= f(0,0,k).$ Next let ${\lambda}=(r,s,t)\in{\Lambda}={\mathbb Z}^3$ with $(r,s)\ne (0,0)$. Let $h=(a,b,c)\in{\cal H}$, then $h{\lambda} h^{-1}=(r,s,t+as-br).$ So h lies in the centralizer ${\cal H}_{\lambda}$ if and only if as = br. It follows that ${\cal H}_{\lambda}$ is the subspace generated by the vectors $(r,s,0)$ and $(0,0,c)$ as a subspace of ${\mathbb R}^3$. On ${\cal H}_{\lambda}$ we pick the Haar measure that comes from the euclidean length in ${\mathbb R}^3$; then the volume of ${\Lambda}_{\lambda}{\backslash}{\cal H}_{\lambda}$ is the minimal value $\sqrt{a^2+b^2}$ where $a,b\in{\mathbb Z}$, not both zero, and as = br. This minimum is taken in $a=\frac{r}{{\rm gcd}(r,s)}$ and $b=\frac{s}{{\rm gcd}(r,s)}$, where ${\rm gcd}(r,s)$ denotes the greatest common divisor of r and s. Therefore,

$$\rm vol({\Lambda}_{\lambda}{\backslash}{\cal H}_{\lambda})=\frac{\sqrt{r^2+s^2}}{{\rm gcd}(r,s)}.$$

The orbital integral equals

$$\cal O_{\lambda}(f)=\int_{U^\perp}f(r,s,t+xs-yr)\,d(x,y),$$

where U is the subspace of all $\left(\begin{smallmatrix}a\\ b\end{smallmatrix} \right)\in{\mathbb R}^2$ with as-br = 0. Then $U={\mathbb R}\left(\begin{smallmatrix}r \\ s\end{smallmatrix}\right)$ and the orthogonal space is spanned by the norm one vector $\frac 1{\sqrt{r^2+s^2}}\left(\begin{smallmatrix}s\\ -r\end{smallmatrix}\right).$ So we get

$$\cal O_{\lambda}(f) = \int_{\mathbb R} f\left(r,s,t+x\frac{s^2+r^2}{\sqrt{r^2+s^2}}\right)\,dx= \frac 1{\sqrt{r^2+s^2}}\,{\cal F}_3f(r,s,0).$$

The number of conjugacy classes with given $(r,s,*)$ equals ${\rm gcd}(r,s)$, hence

$$\hspace{1pt}{\rm tr}\hspace{2pt} R(f)=\sum_{k\in{\mathbb Z}}f(0,0,k) +\sum_{(r,s)\ne (0,0)}{\cal F}_3f(r,s,0).$$

Using the Poisson summation formula in the first two variables we get

$$\sum_{(r,s)\in{\mathbb Z}^2}{\cal F}_3f(r,s,0)=\sum_{(r,s)\in{\mathbb Z}^2}{\cal F} f(r,s,0)$$

with ${\cal F}={\cal F}_1{\cal F}_2{\cal F}_3$ and using the formula in the third variable gives

$$\sum_{k\in{\mathbb Z}}f(0,0,k)=\sum_{k\in{\mathbb Z}}{\cal F}_3 f(0,0,k).$$

Combining this with the two expressions for ${\hspace{1pt}{\rm tr}\hspace{2pt}} R(f)$ we conclude

$$\sum_{k\ne 0}\frac{N(\pi_k)}{|k|}{\cal F}_3f(0,0,k)=\sum_{k\ne 0}{\cal F}_3 f(0,0,k).$$

By Lemma 9.3.5 this implies $N(\pi_k)=|k|$. The theorem is proven. □

10.5 Exercises and Notes

Exercise 10.1

Let ${\cal H}$ be the Heisenberg group, and let $\cal Z$ be its center. Show that every normal subgroup of ${\cal H}$ lies in the center ${\cal Z}$ or contains the center.

Exercise 10.2

On ${\mathbb R}^2$ consider the bilinear form

$$\begin{aligned} b(v,w) \stackrel{\rm def}{=}\, v^t\left(\begin{array}{@{}rr@{}}0 & 1 \\ -1 & 0\end{array} \right)\,w.\end{aligned}$$

Let L be the group ${\mathbb R}^2\times{\mathbb R}$ with the multiplication $(v,t)(w,s){\ \stackrel{{\rm def}}{=}\,} (v+w,s+t+\frac 12 b(v,w)).$ Show: The map $\psi:{\cal H}\to L$, defined by $\psi(a,b,c)= ((a,b)^t,c-\frac 12 ab)$ is a continuous group isomorphism with continuous inverse.

Exercise 10.3

Let L be defined as in Exercise 10.2. The group $G={\rm SL}_2({\mathbb R})$ acts on L via $g(v,t)=(gv,t)$. Show that G acts by group homomorphisms, which fix the center of L point-wise. Conclude that for every $g\in G$ one has $\pi_1\circ g\cong\pi_1$. Conclude that for every $g\in G$ there is a $T(g)\in U(L^2({\mathbb R}))$, which is unique up to scalar multiplication, such that $\pi_1(gl)= T(g)\pi_1(l) T(g)^{-1},$ where we consider π₁ as a representation of L via Exercise 10.2. Show that the map $g\mapsto T(g)$ is a group homomorphism $G\to U(L^2({\mathbb R}))/{\mathbb T}$.

Exercise 10.4

Let ${\rm Aut}({\cal H})$ be the group of all continuous group automorphisms of ${\cal H}$. Every $\phi\in{\rm Aut}({\cal H})$ preserves the center ${\cal Z}$, hence induces an element of ${\rm Aut}({\cal H}/{\cal Z})={\rm Aut}({\mathbb R}^2)={\rm GL}_2({\mathbb R})$. Show that the ensuing map ${\rm Aut}({\cal H})\to{\rm GL}_2({\mathbb R})$ is surjective.

Exercise 10.5

For $h\in{\cal H}$ let $\phi_h\in{\rm Aut}({\cal H})$ be defined by $\phi_h(x)=hxh^{-1}$. Every such automorphism is called an inner automorphism . Let ${\rm Inn}({\cal H})$ be the group of all inner automorphisms. Show that ${\rm Inn}({\cal H})\cong{\mathbb R}^2$, and that, together with the last exercise, one gets an exact sequence

$$1\to{\mathbb R}^2\to{\rm Aut}({\cal H})\to{\rm GL}_2({\mathbb R})\to 1.$$

Exercise 10.6

Let $N\subset{\rm GL}_n({\mathbb R})$ be the group of all upper triangular matrices with ones on the diagonal. Show that ${\Gamma}=N\cap{\rm GL}_n({\mathbb Z})$ is a uniform lattice in N.

Exercise 10.7

The $(2n+1)$-dimensional Heisenberg group ${\cal H}_n$ is defined as the group of all matrices of the form $\left(\begin{smallmatrix}1 & x^t & z \\ & 1 & y \\ & & 1\end{smallmatrix}\right)$ in ${\rm GL}_{n+2}({\mathbb R})$ where $x,y\in{\mathbb R}^n$ and $z\in{\mathbb R}$. Determine its unitary dual along the lines of the Stone-von Neumann Theorem.

Exercise 10.8

Let L be as in Exercise 10.2. Let $\Lambda\subset{\mathbb R}^2$ be a lattice with $b({\Lambda}\times{\Lambda})\subset 2{\mathbb Z}$. Show that ${\Lambda}\times{\Lambda}\times{\mathbb Z}$ is a uniform lattice in L. Use the proof of Theorem 10.4.2 to find the decomposition of $L^2({\Lambda}{\backslash} L)$.

10.5.1 Notes

As shown in Exercise 10.3, the Stone-von Neumann Theorem yields a group homomorphism ${\rm SL}_2({\mathbb R})\to U(L^2({\mathbb R}))/{\mathbb T}$. There is a unique non-trivial covering group ${\rm SL}_2^2({\mathbb R})$ of degree 2. On this group the homomorphism induces a proper unitary representation ${\rm SL}_2^2({\mathbb R})\to U(L^2({\mathbb R}))$. This representation is known as the Weil representation . It is used to explain the behavior of theta-series, in particular with respect to lifting of automorphic forms [How79, LV80, Wei64].