Notes on Microlocal Aspects of Representation Theory

Notes on microlocal aspects of representation theory

Paul Nelson March 4, 2018

Contents

1 Overview 2

2 Lecture summaries and homework 2 2.1 2/21 and 2/22: Kirillov formula for SU(2) ...... 2

3 Prerequisites 5

4 Some background 5 4.1 Peter–Weyl theorem ...... 5 4.2 Symplectic manifolds ...... 5 4.2.1 ...... 5 4.2.2 ...... 6 4.3 Unitary representations ...... 6

5 Kirillov formula: the case of SU(2) 7 5.1 ...... 7 5.2 ...... 8 5.3 ...... 9 5.4 ...... 10 5.5 ...... 11 5.6 ...... 11 5.7 ...... 12

6 Kirillov formula: the general statement 12

7 The Heisenberg group 15 7.1 ...... 15 7.2 ...... 15 7.3 ...... 16 7.4 ...... 17 7.5 ...... 18

1 1 Overview

These notes are recorded to complement the lectures. Material here has been rearranged from how it was presented in lecture to serve as a reference rather than an introduction. A principal aim of the lectures is that after attending them, you should be able to do the homework problems given below their summaries.

2 Lecture summaries and homework 2.1 2/21 and 2/22: Kirillov formula for SU(2) Objectives. You should be able to state the Kirillov formula for compact Lie groups and to prove it for SU(2). Summary. We proved the Kirillov formula for SU(2) by explicitly computing the Fourier transforms of the characters of the irreducible representations. We philosophized about the possibility of a “combinatorial proof” and indicated the aims of the course. We stated the general Kirillov formula in the nilpotent and compact cases. By computing the canonical symplectic forms on the coadjoint orbits of SU(2), we veriﬁed that the general formula specialized to what we had shown directly. Homework 1 (Due Mar 8). 1. Skim at least the ﬁrst few pages of Chapter 1 of Kirillov’s book (which should be available via mathscinet from the ETH network).

The remaining exercises concern G := SU(2) and the representation πn : n n−1 n G → GL(Vn), as in the lecture, with standard basis X ,X Y,...,Y ∼ 3 ∼ ∗ and character χn. We ﬁx coordinates g = R = g as in lecture. We ∗ denote by Oπn ⊆ g the coadjoint orbit of πn (i.e., the sphere of radius n+1) and ﬁx a G-invariant inner product h, i on Vn, with associated norm kvk = hv, vi1/2.

∞ 2. Let φ1, φ2 ∈ Cc (R) with sin(θ) φ2(θ) = φ1(θ) for all θ ∈ . θ R

Write dθ for Lebesgue measure on the real line. For k ∈ R, set φcj(k) = R ikθ e φj(θ) dθ. Let νj denote the compactly supported measure on g given θ R R in standard coordinates by g f dνj = θ f(0, 0, θ)φj(θ) dθ. (a) Using Kirillov’s formula and Archimedes’s theorem, verify that Z Z n+1 x p dk χn(e ) j(x) dν1(x) = φc1(k) , x∈g k=−n−1 2 where dk denotes Lebesgue measure.

2 R x (b) Deﬁne T ∈ End(Vn) by T := x∈g πn(e ) dν2(x). Verify that T is n+k n−k diagonalized by the basis X 2 Y 2 of πn, where k = −n, −n + 2, . . . , n, with eigenvalues φc2(k), hence that Z x X χn(e ) dν2(x) = trace(T ) = φc2(k). x∈g k=−n,−n+2,...,n √ (c) Verify that dν2 = j dν1. Conclude that Z n+1 dk X φc1(k) = φc2(k). 2 k=−n−1 k=−n,−n+2,...,n [This may be deduced more directly by noting that the Fourier transform of sin(x)/x is a multiple of the characteristic function of the interval [−1, 1] and that the Fourier transform transports multiplication to convolution. The point is to observe the relationship between this fact and the SU(2) case of the Kirillov formula.] 3. Following the steps outlined below, establish the following: there is a

constant C > 0 so that for all n ∈ Z≥0 and each ξ ∈ Oπn , there is a unit vector v ∈ πn so that for all x ∈ g, ihx,ξi √ 1/2 kπn(exp(x))v − e vk ≤ C( n|x| + |x| ). (1) The moral is that the vector v is an approximate eigenvector, with eigenvalue√ corresponding to ξ, of group elements g = exp(x) satisfying |x| < ε/ n for some small ε > 0. ∗ (a) Using that Ad (G) acts transitively on Oπn , reduce to the case ξ = (0, 0, n + 1). [Hint: if (1) holds for a given pair (ξ, v), then a modiﬁed form of (1) ∗ holds for (Ad (g)ξ, πn(g)v).] (b) Choose v ∈ πn to be a unit vector that is a multiple of the monomial Xn. Using the G-invariance of the inner product h, i, verify that v is orthogonal to the vectors Xn−1Y,...,Y n and that a b hπ (g)v, vi = an g = ∈ G. n for all c d (2)

(c) Using the triangle inequality, reduce to the case √ |x| ≤ 1/ n. (3) ix ix + x For x = 3 2 1 ∈ g satisfying (3), write exp(x) = ix2 − x1 −ix3 a b c d , and verify that

2 ix3 2 a = 1 + ix3 + O(|x| ) = e (1 + O(|x| )) (4)

3 and

an = einx3 (1 + O(n|x|2)) = ei(n+1)x3 + O(n|x|2 + |x|), (5)

where O(A) denotes a quantity bounded in magnitude by C|A| for some constant C > 0. (d) Square both sides of (1), expand the norm squared as a sum of four inner products, and apply (2) and (5) to conclude.

∗ 4. For λ > 0, let Oλ := {ξ ∈ g : |ξ| = λ} denote the corresponding coadjoint orbit and ωλ its canonical symplectic volume measure, of total volume λ. ∗ ∗ ∗ Let p : g × g → g denote the addition map. For λ1, λ2 > 0, verify the measure disintegration

Z λ1+λ2

p∗(ωλ1 ⊗ ωλ2 ) = ωλ dλ, |λ1−λ2|

∗ i.e., that for f ∈ Cc(g ),

Z Z Z λ1+λ2 Z

f(x+y) dωλ1 (x) dωλ2 (y) = ( f(z) dωλ(z)) dλ, x∈Oλ1 y∈Oλ2 |λ1−λ2| z∈Oλ

where dλ denotes Lebesgue measure. Compare with the decomposition ∼ m+n πm ⊗ πn = ⊕k=|m−n|πk.

4 3 Prerequisites

• basics on Lie groups, e.g., deﬁnitions of Lie algebra, exponential map, adjoint action

• basic language of representation theory (e.g., of finite groups) • definition of Haar measure; reading the Wikipedia article should suffice

4 Some background 4.1 Peter–Weyl theorem

Let G be a group. Recall that f : G → C is a class function if it is constant on conjugacy classes: f(ghg−1) = f(h). Assume henceforth that G is compact. Then it has a Haar measure dg, which we may normalize to have total volume one. The square-integrable class functions on G form a Hilbert space with inner product Z hf1, f2i := f1(g)f2(g) dg. g∈G

Let π : G → GL(V ) be a finite-dimensional representation. Its character χ = χπ = χV : G → C is the class function given by χ(g) := trace(π(g)). Theorem 1 (Peter–Weyl theorem). Every finite-dimensional representation of G decomposes as a finite direct sum of irreducible subrepresentations. The characters of the irreducible finite-dimensional representations of G, taken up to isomorphism, form an orthonormal basis for the space of class functions.

In the special case that G is finite, this is proved in a first course on representation theory. The case of compact G may be proved similarly assuming the existence of Haar measure. We note also that for a finite-dimensional representation V , V is irreducible iff hχV , χV i = 1, as follows from (the easy part of) Theorem 1. If π : G → GL(V ) is any finite-dimensional representation, then Maschke’s theorem (proved as in the case of finite groups by averaging with respect to Haar measure) shows that there is a G-invariant inner product on V , which is moreover unique if V is irreducible.

4.2 Symplectic manifolds 4.2.1 Let W be a vector space of ﬁnite-dimension over the reals. A symplectic form on W is a map σ : W × W → R that is

5 1. bilinear, 2. alternating (σ(v, v) = 0, hence σ(u, v) = −σ(v, u)), and 3. non-degenerate: the map W → W ∗ given by v 7→ σ(v, ·) is an isomorphism.

The ﬁrst two properties say that

σ ∈ Λ2(W )∗ =∼ Λ2(W ∗).

Lemma 2. If W admits a symplectic form σ, then W is even-dimensional, say dim(W ) = 2n, and there is a basis e1, . . . , en, f1, . . . , fn of W so that σ(ei, ej) = 0 = σ(fi, fj) and σ(ei, fj) = δij = −σ(fi, ej).

0 Proof sketch. Choose any nonzero e1, choose f1 with he1, f1i = 1, set W := {v ∈ W : hv, e1i = hv, f1i = 0}, and continue.

∗ ∗ ∗ Denoting by e1, . . . , fn the basis of W dual to that furnished by the lemma, we have ∗ ∗ ∗ ∗ σ = e1 ∧ f1 + ··· + en ∧ fn. We check readily that

σn := σ ∧ · · · ∧ σ = n!e∗ ∧ f ∗ ∧ · · · ∧ e∗ ∧ f ∗ ∈ Λ2n(W ∗). | {z } 1 1 n n n times

In particular, σn 6= 0.

4.2.2 Let M be a manifold. Recall that a 2-form σ on M is the smooth assignment 2 ∗ M 3 ξ 7→ σξ ∈ Λ (Tξ M), thus σξ is a bilinear alternating pairing on the tangent space TξM. Its exterior derivative dσ is a 3-form; recall that σ is closed if dσ = 0. We call σ non-degenerate if each σξ is non-degenerate in the sense of §4.2.1. A symplectic form σ on M is a closed, non-degenerate 2-form. A symplectic form exists only if M has even dimension, say 2n. In that case the n-fold wedge product ωn deﬁnes a volume form (i.e., a nonvanishing top degree form), hence an orientation and a volume measure on M.

4.3 Unitary representations

Let V be a Hilbert space. Recall that a sequence vj ∈ V converges to v ∈ V

• strongly if kvj − vk → 0, and

• weakly if hvj, wi → hv, wi for each w ∈ V . Let U(V ) denote the space of unitary operators on V . The topologies on this space include

6 • the norm topology: Tj → T if kTj − T k → 0, where k.k denotes the operator norm.

• the strong topology: Tj → T if Tjv → T v strongly for each v ∈ V .

• the weak topology: Tj → T if Tjv → T v weakly for each v ∈ V .

Exercise 1. Verify that if vj → v weakly and kvjk ≤ C, then vj → v strongly. Let G be a topological group. Exercise 2. Let π : G → U(V ) be a homomorphism. Verify that if π is continuous for the weak topology on the target, then it is continuous for the strong topology on the target.

Exercise 3. Verify that the map π : R → U(L2(R)) given by π(x)f(y) = f(y + x) is not continuous for the norm topology on the target. A unitary representation of a group G is a pair (π, V ), where V is a Hilbert space and π : G → U(V ) is a homomorphism. If G is a topological group, then we require π to be continuous for the weak (and hence the strong) operator topology on the target; in other words, the map G × V → V is continuous. It is irreducible if it is nonzero and if {0} and V are the only closed invariant subspaces. Schur’s lemma: if (π, V ) is a unitary representation of a group G, then the following are equivalent: 1. π is irreducible. 2. Each bounded operator T on V that commutes with π is a scalar.

5 Kirillov formula: the case of SU(2)

Set ∗ G := SU(2) := {g ∈ SL2(C): gg = 1}, where g∗ denotes conjugate transpose. In this section we will compute explicitly the characters of the irreducible representations of G and show that, after pulling back to the Lie algebra and “twisting” a bit, they are given by Fourier transforms of surface measures on spheres. We then philosophize a bit.

5.1 Every element of G is conjugate to a diagonal element eiθ g := , θ e−iθ for some unique θ ∈ [0, π], so class functions f : G → C are determined by the values f(gθ).

7 Theorem 3 (Weyl integral formula). Let f be an integrable class function on G. Then Z Z π iθ −iθ 2 dθ f(g) dg = f(gθ)|e − e | . (6) g∈G 0 2π

Proof sketch. Let T denote the diagonal subgroup of G, thus T = {gθ : θ ∈ R/2πZ}. The map G/T × T → G −1 (hT, gθ) 7→ hgθh is generically 2-to-1. Computing its Jacobian leads to the required formula. Exercise 4. Verify directly that (6) holds in the special case f = 1.

5.2

The group G acts on the space C[X,Y ] of polynomials in two variables: g · f(X,Y ) := f((X,Y )g) := f(aX + cY, bX + dY )

a b g = ∈ G, f ∈ [X,Y ]. for c d C

For each n ≥ 0, the space Vn of homogeneous f ∈ C[X,Y ] of degree n is a G-invariant subspace of dimension n + 1, with basis Xn,Xn−1Y,...,Y n. Let πn : G → GL(Vn) denote the corresponding representation and χn : G → C its character. We compute χn. Since it is a class function, we need only compute χn(gθ). We have j k (j−k)θ j k gθ · X Y = e X Y , n n−1 n so πn(gθ) has the basis of eigenvectors X ,X Y,...,Y with eigenvalues einθ, ei(n−2)θ, . . . , e−inθ; thus

inθ i(n−2)θ i(n−4)θ −inθ χn(gθ) = e + e + e + ··· + e . (7)

Summing the geometric series, we obtain

ei(n+1)θ − e−i(n+1)θ χn(gθ) = . (8) eiθ − e−iθ Exercise 5.

1. Using Theorem 3 and the formula (8), verify that hχm, χni = δmn. Deduce in particular that the Vn are irreducible.

2. Verify that the χn span a dense subspace of the space of class functions.

3. Using Theorem 1, deduce that the Vn give a complete collection of ﬁnite- dimensional irreducible representations of G, up to isomorphism.

8 5.3 We record some preliminaries concerning the Lie algebra

g = Lie(G) = {x : 1 + εx ∈ G (mod ε2)}.

We have 1 + εx ∈ G = SU(2) iﬀ (1 + εx)(1 + εx)∗ ≡ 1 (mod ε2) iﬀ x + x∗ = 0, so ∗ g = su(2) = {x ∈ Mat2(C): x + x = 0}. Explicitly, ix3 ix2 + x1 3 g = x = :(x1, x2, x3) ∈ R . ix2 − x1 −ix3

3 Using these coordinates, we identify g with R . We may write x = x1J1 +x2J2 + x3J3 with respect to the basis elements 1 i i J = J = ,J = . 1 −1 2 i 3 −i whose Lie brackets are given by [Ji,Jj] = 2Jk for cyclic permutations {i, j, k} of {1, 2, 3}. The exponential map exp : g → G is the matrix exponential function X xn exp(x) = ex = . n! n≥0

We note that gθ = exp(θJ3). G acts on its Lie algebra g via the adjoint action,

Ad : G → GL(g),

Ad(g)x = gxg−1. For example, we may compute that with respect to the above coordinates, 2iθ Ad(gθ) ﬁxes x3 and sends x1 + ix2 to e (x1 + ix2), thus

Ad(gθ) = (rotation by 2θ radians clockwise around the x3-axis). (9)

A similar description applies to Ad(exp(θJ1)), Ad(exp(θJ2)). Among other general properties, one has

exp(Ad(g)x) = g exp(x)g−1. (10)

The kernel of Ad is always the center of G, which in the present case of SU(2) is {±1}. Using the above coordinates, we may identify GL(g) with GL3(R). We equip ∼ 3 2 2 2 2 g = R with the standard quadratic form |x| := x1 + x2 + x3. The following is proved in a ﬁrst course on Lie groups:

9 Lemma 4. The image Ad(G) ⊆ GL3(R) is the special orthogonal group SO(3) corresponding to the standard quadratic form on g =∼ R3. The map Ad : G → SO(3) is thus surjective, with kernel {±1}.

Proof sketch. Since SO(3) is connected, it suﬃces to verify that ad(g) = so(3). Indeed, it follows from (9) and its analogues that ad(J1)/2, ad(J2)/2, ad(J3)/2 give the standard basis of so(3). In particular, Ad(G) acts transitively on each sphere

{x ∈ g : |x| = r}, (11) so an Ad(G)-invariant function f : g → C is constant on such spheres, or equivalently, depends only upon |x|.

5.4

We now study the pullback of χn to g. Since χn is a class function on G, it x follows from (10) that its pullback g 3 x 7→ χn(e ) is Ad(G)-invariant; noting that each x ∈ g is in the Ad(G)-orbit of (0, 0, |x|) = |x|J3, we see that the formula (8) translates to

i(n+1)|x| −i(n+1)|x| x e − e χn(e ) = . (12) ei|x| − e−i|x|

x It turns out that the best object to study now is not χn(e ) but instead its x p twist χn(e ) j(x) by the Jacobian j of the exponential map, deﬁned by writing

dg = j(x) dx for g = ex, dx = Haar on g, (13) normalized by j(0) = 1. In the present case, one may compute that

2 ei|x| − e−i|x| j(x) = . (14) 2i|x|

Remark 5. Recall that the χn give an orthonormal basis for class functions in p L2(G, dg). To motivate the factor j(x), note that the exponential map is not a bijection, but “if it were,” then it would follow from (13) that the functions x p 2 χn(e ) j(x) give an orthonormal basis for L (g, dx). Combining (12) with (14) gives (after some cancellation) that

i(n+1)|x| −i(n+1)|x| x p e − e χn(e ) j(x) = . (15) 2i|x|

10 5.5 ∼ 3 3 For each x ∈ g = R and ξ ∈ R , let hx, ξi := x1ξ1 + x2ξ2 + x3ξ3 denote the standard pairing. (Note that to work in a coordinate-free manner, we would take ξ in the dual space g∗, which we implicitly identify here with R3.) 3 For λ > 0, let Sλ := {ξ ∈ R : |ξ| = λ} denote the sphere of radius λ. Let 2 µ denote the standard area measure on Sλ, of total volume 4πλ . The Fourier transform of this measure is deﬁned by Z µˆ(x) := eihx,ξi dµ(ξ). ξ∈Sλ

Let’s evaluate this explicitly. Since µ is rotation-invariant, we may replace x with (0, 0, |x|), giving Z µˆ(x) = eiξ3|x| dµ(ξ). ξ∈Sλ Lemma 6 (“Archimedes’s theorem”). Under the map

Sλ → [−λ, λ]

ξ 7→ ξ3, the measure µ on Sλ projects to the measure 2πλ dt on [−λ, λ].

Thus Z λ eiλ|x| − e−iλ|x| µˆ(x) = 2πλ eit|x| dt = 2πλ . (16) t=−λ i|x|

5.6 Combining (15) with the specialization of (16) to λ = n + 1, we arrive at the following:

Theorem 7 (Kirillov formula for SU(2)). Let n ∈ Z≥0. Set O := Sn+1 and µ ω = 4π(n+1) , with µ the standard surface measure on O. Then

Z x p ihx,ξi χn(e ) j(x) = e dω(ξ). (17) ξ∈O

This completes the rigorous part of today’s discussion. Some natural questions now arise: 1. Why should something like this be true? 2. Why care?

We address those brieﬂy below.

11 5.7 We may specialize (17) to x = 0. We have j(0) = 1 and χ(e0) = χ(1) = ih0,ξi dim(Vn) = n + 1 and e = 1, so

vol(O, dω) = dim(Vn) = n + 1.

Given any such identity of integers, it is natural to ask for a “combinatorial proof.” A vague idea in this direction is the following: given a nice partition F O = k Pk with vol(Pk, dω) = 1, there should be a basis vk of Vn so that vk ↔ Pk in the sense that for each x ∈ g, the set {ihx, ξi : ξ ∈ Pk} (typically something like an interval) is roughly the set of eigenvalues that show up if tx we write vk as a sum of dπn(x)-eigenvectors. (Here dπ(x) := ∂t=0π(e ).) In particular, this says that the full distribution of eigenvalues of dπn(x) should be approximately {ihx, ξi : ξ ∈ O}; in the special case x = J3, these two sets may be readily veriﬁed to be respectively

{i(−n), i(−n + 2), . . . , in} and [−i(n + 1), i(n + 1)], which look similar in the n → ∞ limit (cf. problem 2 on Homework 1). For another example, if each Pk is concentrated near some ξk ∈ O, then the vk given x a basis of approximate eigenvectors for πn(e ) when x ∈ g is small enough, with eigenvalues eihx,ξki (cf. problem 3 on Homework 1), hence Z x X ihx,ξki ihx,ξi χn(e ) ≈ e ≈ e dω, k O recovering an approximate form of the (exact) Kirillov formula (ignoring the p j(x) factor, which is ≈ 1 for small x). We want to explain to what extent this naive idea is true and how it can be useful for estimating certain quantities that arise in representation theory and the study of periods of automorphic forms. We also want to explain how it ﬁts into the broader picture of the orbit method and how the latter addresses the main problems in representation theory (e.g., determining the unitary dual and its topological structure, explicit description of restriction/induction functors, character formulas, computing Plancherel measure, etc.).

6 Kirillov formula: the general statement

Last time, we proved that the Fourier transform of the characters of irreducible representations of SU(2) (pulled back to the Lie algebra and twisted by the square root of the Jacobian of the exopnential map) have a particularly nice shape. However, our discussion was completely ad hoc. In particular, it wasn’t clear at all how we might have “predicted” the normalization of the volume forms on the spheres without just doing the calculation. We now explain how that normalization arises naturally, putting the formula proved last time in some context.

12 Let G be a Lie group. We denote by g its Lie algebra and by g∗ := ∗ HomR(g, R) the linear dual of its Lie algebra. For x ∈ g and ξ ∈ g , we denote by hx, ξi ∈ R their natural pairing. Let Ad : G → GL(g) be the adjoint representation. By duality, we obtain the coadjoint representation Ad∗ : G → GL(g∗), characterized by hx, ξi = hAd(g)x, Ad∗(g)ξi. We may diﬀerentiate these to obtain Lie algebra representations

ad : g → End(g), ad∗ : g → End(g∗).

A coadjoint orbit is an Ad∗(G)-orbit O ⊆ g∗. An important observation (of Kirillov–Kostant–Souriau–Lie–Poisson) is that the Lie bracket on g gives rise to a canonical symplectic form (cf. §4.2.2) σ on each coadjoint orbit O, characterized by the identity

σ(x∗, y∗) = [x, y], where • for x ∈ g, x∗ denotes the vector ﬁeld on O given by x∗(ξ) := ∗ tx ∂t=0 Ad (e )ξ;

• [x, y] ∈ g =∼ g∗∗ deﬁnes the function O 3 ξ 7→ h[x, y], ξi More explicitly, σ should smoothly assign to each ξ ∈ O an alternating bilinear form σξ on the tangent space TξO. We may describe σξ as follows. Let ξ ∈ O. Then O is the Ad∗(G)-orbit of ξ, thus1 we may identify

∗ ∗ TξO = ad (g)ξ = {[x, ξ]: x ∈ g}, [x, ξ] := ad (x)ξ.

The 2-form σ is then deﬁned by

σξ([x, ξ], [y, ξ]) = h[x, y], ξi for all x, y ∈ g.

A few things need to be checked here:

• that σξ is well-deﬁned, i.e., that h[x, y], ξi = 0 if either x or y belongs to gξ, as follows from the identity

h[x, y], ξi = hx, [y, ξ]i = −hy, [x, ξ]i; (18)

• that σξ is non-degenerate, i.e., that if y ∈ g has the property that σξ(x, y) = 0 for all x ∈ g, then y =∈ gξ, as follows again from (18); • that σ is closed, which boils down to the Jacobi identity [[x, y], z] + [[y, z], x] + [[z, x], y] = 0.

1 We’re taking for granted here basic properties of Lie group actions, typically proved in a ﬁrst course.

13 The nondegeneracy of σξ implies in particular that O is has even dimension, say 2n, and also that the n-fold wedge product σn is a volume form on O. Exercise 6. Determine the adjoint and coadjoint orbits for   1 RR G = Heis :=  1 R . 1 Verify in particular that the coadjoint orbits are all even-dimensional, while some of the adjoint orbits are odd-dimensional. We will refer to 1 σ ω := ( )n n! 2π as the canonical symplectic volume form on O. (Kirillov would write this as eσ/2π, and omit the 2π, which may be considered an artefact of our normalization of Fourier transforms.) We’ll eventually prove some chunk of the following: Theorem 8. Let G be a connected Lie group that is either nilpotent or compact. For each irreducible unitary representation (π, V ) of G there is a coadjoint orbit O = Oπ so that the character χ = χπ of π is given by Z χ(ex)pj(x) = eihx,ξi dω(ξ), ξ∈O where ω is the canonical symplectic volume form on O. Exercise 7. Take G = SU(2). Choose coordinates g =∼ R3 and g∗ =∼ R3 as in the ﬁrst lecture, by writing ix3 ix2 + x1 x = , hx, ξi = x1ξ1 + x2ξ2 + x3ξ3. ix2 − x1 −ix3

Verify that the coadjoint orbits are the spheres O = {ξ ∈ R3 : |ξ| = λ}, λ ≥ 0. ∗ ∼ 3 Assume now that λ > 0. Set ξ = (0, 0, λ) ∈ O ⊆ g = R , and let J1,J2,J3 be the basis of g considered above. Verify that [J1, ξ] = (0, −2λ, 0), [J2, ξ] = (2λ, 0, 0), and h[J1,J2], ξi = 2λ, hence that σξ((1, 0, 0), (0, 1, 0)) = 1/2λ. Deduce that the canonical symplectic measure ω is given by µ/4πλ, where µ is the standard surface measure. [Hint: µ identifies with the rotation invariant 2-form for which µξ((1, 0, 0), (0, 1, 0)) = 1.] Conclude that Theorem 8 specializes to Theorem 7. Some refinements: 1. If G is nilpotent and simply-connected, then the correspondence between representations π and coadjoint orbits O turns out to be bijective. In particular, every coadjoint orbit arises. By contrast, for SU(2), only the spheres with positive integral radius were relevant. 2. The formula holds when G is reductive (generalizing the compact case) under the additional assumption that π is tempered, and with the caveat that in “rare” cases, one must take for O a finite union of orbits.

14 7 The Heisenberg group 7.1 We now consider the Lie group   1 RR G :=  1 R . 1

As generators we can take the elements

1 x 0 1 0 0 1 0 z x y z g1 :=  1 0 , g2 :=  1 y , g3 :=  1 0 1 1 1

x y z for x, y, z ∈ R. The maps x 7→ g1 , y 7→ g2 , z 7→ g3 are homomorphisms, the z elements g3 are all central, and we have the relation

x y xy y x g1 g2 = g3 g2 g1 . (19)

These remarks give a presentation for G.

7.2 We aim to work out the irreducible unitary representations of G. Motivated by Theorem 8, we might ﬁrst try working out the coadjoint orbits. The Lie algebra of G is given by       0 RR 0 x z   3 g =  0 R = [x, y, z] :=  0 y : x, y, z ∈ R =∼ R . 0  0 

The exponential map (given by the exponential series) is a diﬀeomorphism, given explicitly by

1 x z + xy/2 [x,y,z] z+xy/2 y x e =  1 y  = g3 g2 g1 . (20) 1

One checks readily that Haar measures both on G and on g may be given by dx dy dz, and that the Jacobian of the exponential map is trivial:

j(x) = 1 for all x ∈ g.

The adjoint action is given in the above optic by conjugation:

Ad(g)[x, y, z] = g[x, y, z]g−1.

15 The dual of the Lie algebra may be identiﬁed with the quotient space of matrices modulo upper-triangular matrices, with the pairing given by taking the trace of the product:  ∗ ∗ ∗ ∗   3 g = [α, β, γ] := α ∗ ∗ =∼ R ,  γ β ∗ 

h[x, y, z], [α, β, γ]i := trace([x, y, z][α, β, γ]) 0 x z ∗ ∗ ∗ = trace( 0 y α ∗ ∗) 0 γ β ∗ xα + zγ ∗ ∗ = trace( ∗ yβ ∗) ∗ ∗ 0 = xα + yβ + zγ.

Since trace(g[x, y, z]g−1g[α, β, γ]g−1) = trace([x, y, z][α, β, γ]), we see that the coadjoint action is also given in the above optic by conjugation: Ad∗(g)[α, β, γ] = g[α, β, γ]g−1.

∗ z Using this we compute readily that Ad (g3 ) is trivial, while ∗ x Ad (g1 )[α, β, γ] = [α, β − xγ, γ], (21) ∗ y Ad (g2 )[α, β, γ] = [α + xγ, β, γ]. (22) It follows that the coadjoint orbits for G are described as follows: • For each (α, β) ∈ R2, we have a zero-dimensional orbit

Oα,β := {[α, β, 0]}.

• For each γ ∈ R×, we have a two-dimensional orbit 2 ∼ 2 Oγ := [α, β, γ]:(α, β) ∈ R = R . 7.3

It’s easy to ﬁnd representations (πα,β,Vα,β) corresponding to the zero-dimensional orbits. The symplectic volumes of these orbits are 1 (by integrating the 0-form 0 ω = (σ/2π) /0! = 1 over a one-element set...), so we expect dim(Vα,β) = 1. The Kirillov formula tells us what the trace of πα,β should be: Z ih[x,y,z],·i ih[x,y,z],[α,β,0]i i(xα+yβ) trace(πα,β(exp([x, y, z]))) = e dω = e = e . Oα,β ∼ × Thus πα,β : G → GL(Vα,β) = GL1(C) = C is the one-dimensional representation given by i(xα+yβ) πα,β = e .

16 7.4

The representations (πγ ,Vγ ) corresponding to the two-dimensional orbits Oγ are more interesting. Let’s start by computing the canonical symplectic forms σ and volume forms ω attached to these orbits. Let J1 = [1, 0, 0],J2 = [0, 1, 0], J3 = [0, 0, 1] be the obvious basis of g. Then J3 is central, while [J1,J2] = J3. For ξ = [α, β, γ] ∈ Oγ , we see by diﬀerentiating (21) and (22) that [J1, ξ] = [0, −γ, 0] and [J2, ξ] = [γ, 0, 0], so that σξ([J1, ξ], [J2, ξ]) = h[J1,J2], ξi = hJ3, ξi = γ. It follows that σξ([1, 0, 0], [0, 1, 0]) = 1/γ, hence that in the standard coordinates 2 (α, β) ∈ R for Oγ , dα ∧ dβ dα ∧ dβ σ = , ω = . γ 2πγ

In particular, vol(Oγ , dω) = ∞, so we expect dim(Vγ ) = ∞. The Kirillov formula tells us what the character χγ should look like: Z [x,y,z] i(xα+yβ+zγ) dα dβ χγ (e ) = e . (23) 2π|γ| α,β∈R The integral doesn’t converge, but we can interpret it distributionally by integrating against a smooth compactly-supported measure on g, which we may ∞ ∼ ∞ 3 write as φ(x, y, z) dx dy dz for some φ ∈ Cc (g) = Cc (R ) with respect to Lebesgue measure dx dy dz. The precise way to interpret (23) is then that for each such φ, the integral operator Z [x,y,z] πγ (φ) := φ(x, y, z)πγ (e ) dx dy dz x,y,z∈R on Vγ is trace-class, with Z Z i(xα+yβ+zγ) dα dβ trace(πγ (φ)) = φ(x, y, z)e dx dy dz . (24) 2π|γ| α,β∈R x,y,z∈R By Fourier inversion, we may rewrite the desired trace identity (24) in the form Z 2π izγ trace(πγ (φ)) = φ(0, 0, z)e dz. (25) |γ| z∈R

How to construct the required representation πγ ? By noting that we may pull the factor eizγ out of the integral in (23), we might guess that the element [0,0,z] z izγ e = g3 should act as multiplication by the scalar e :

z izγ πγ (g3 ) := e (scalar multiplication).

x y We might then aim to ﬁnd a Hilbert space Vγ on which the actions of g1 and g2 respect the relation (19). It turns out, as we’ll explain later, that there is (up to isomorphism) only one way to do this. We’ll give the answer ﬁrst, and then explain how we could have arrived at it naturally.

17 7.5 We take 2 Vγ := L (R), x πγ (g1 )f(t) := f(t + x), y iγyt πγ (g2 )f(t) := e f(t). We compute readily that then

x y iγy(t+x) iγxy iγyt xy y x πγ (g1 g2 )f(t) = e f(t + x) = e e f(t + x) = πγ (g3 g2 g1 )f(t), so the relation (19) is respected, and we obtain a well-deﬁned unitary representation of G. ∞ Let’s compute the traces of the integral operators πγ (φ), φ ∈ Cc (g). Using (20), we obtain Z iγ(z+xy/2+yt) πγ (φ)f(t) = φ(x, y, z)e f(t + x) dx dy dz. x,y,z∈R

By the change of variables x 7→ x − t, we may rewrite the above as πγ (φ)f(t) = R f(x)Kφ(x, t) dx, where x∈R Z iγ(z+(x−t)y/2+yt) Kφ(x, t) = φ(x − t, y, z)e dy dz. y,z∈R Using that φ is smooth and compactly-supported and a bit of Fourier analysis, 2 we see that the kernel Kφ belongs to the Schwartz space on R . We’ll show eventually (postponing for now the functional-analytic details) that πγ (φ) is trace-class with Z Z iγ(z+ty) trace(πγ (φ)) = Kφ(t, t) dt = φ(0, y, z)e dt dy dz. t∈R t,y,z∈R Substituting t = γ−1β, dt = γ−1dβ and applying Fourier inversion to the y- integral, we readily derive the required formula (24). In summary, we’ve been able to produce, for each coadjoint orbit, a unitary representation whose trace is as predicted by the Kirillov formula. Some natural questions now arise: 1. Is there some more direct way we could have arrived at the representation Vγ from the coadjoint orbit Oγ , without just having to guess the answer and check that it works? (Yes.)

2. Are the representations Vγ irreducible? (We’ll see that the answer is yes.) 3. Do we get every irreducible unitary representation of G in this way? (We’ll see again that the answer is yes.) We take these up next.