4.3 Tensor Products of Unitary Representations
4.3.1 Tensor Products of Hilbert Spaces
Definition 4.3.1. LetHandKbe Hilbert spaces. Then K (x, y),(x0, y0)
:=hx0, xihy0, yi
defines on the product setH × Ka positive definite kernel because it is a prod-uct of two positive definite kernels (Proposition 3.2.1(d)). The corresponding Hilbert space in CH×K is called thetensor productof HandK and is denoted byHb⊗K.
Then the realization map
γ: H × K → H⊗K,b γ(x, y) :=K(x,y) is bilinear and we write
x⊗y:=K(x,y)
for the image of (x, y) under this map. These elements span a dense subspace and their scalar products are given by
hx⊗y, x0⊗y0i=hK(x,y), K(x0,y0)i=K((x0, y0),(x, y)) =hx, x0ihy, y0i.
Remark 4.3.2. (a) One can obtain a more concrete picture of the tensor prod-uct by choosing orthonormal bases (ej)j∈J in H and (fk)k∈K in K. Then the family (ej⊗fk)(j,k)∈J×Kis orthonormal in the tensor product and spans a dense subspace, so that it is an orthonormal basis. That it spans a dense subspace follows directly from the continuity of the bilinear map γ (Proposition 3.3.5), which implies that
x⊗y=γ(x, y) = X
j∈J,k∈K
hx, ejihy, fki ·ej⊗fk.
(b) Similarly, we find that the subspaces ej⊗ K of Hb⊗K are pairwise or-thogonal and span a dense subspace, so that
H⊗K ∼b =⊕bj∈J(ej⊗ K) (cf. Exercise 1.3.5). In addition, we have
hej⊗v, ej⊗wi=hv, wi,
so that the inclusion maps
K → H⊗K,b v7→ej⊗v are isometric embeddings. This implies that
H⊗K ∼b =`2(J,K) (cf. Example 1.3.8).
(c) With a slight generalization, we can form tensor products of finitely many Hilbert spacesH1, . . . ,Hn by using the kernel
K(x, x0) :=
n
Y
j=1
hx0j, xji for x= (x1, . . . , xn)∈
n
Y
j=1
Hj,
which leads to a reproducing kernel space
⊗bnj=1Hj:=HK⊆C
Qn j=1Hj. One easily verifies that
⊗bnj=1Hj ∼= ⊗bn−1j=1Hj
⊗Hb n,
so that an alternative construction is to apply the construction of twofold tensor products several times.
To form tensor products of two representations, we first verify that pairs of linear operators define operators on the tensor product space:
Lemma 4.3.3. Let A ∈ B(H) and B ∈ B(K). Then there exists a unique bounded linear operatorA⊗B onH⊗Kb with
(A⊗B)(v⊗w) := (Av)⊗(Bw) for v∈ H, w∈ K. (4.2) It satisfies
kA⊗Bk ≤ kAkkBk and (A⊗B)∗=A∗⊗B∗.
Proof. Since the elementsv⊗wspan a dense subspace, the operatorA⊗B is uniquely determined by (4.2). It therefore remains to show its existence. To this end, we fist consider the caseA=1.
Identifying Hb⊗K with `2(J,K) (Remark 4.3.2(b)), we see that B defines an operator Be on `2(J,K) ∼= ⊕bj∈JK by B(xe j) := (Bxj), and kBek = kBk (Exercise 4.2.3). This proves the existence of 1⊗B. We likewise obtain an operatorA⊗1withkA⊗1k=kAk, and we now put
A⊗B:= (A⊗1)(1⊗B).
It satisfies
(A⊗B)(v⊗w) = (A⊗1)(1⊗B)(v⊗w) = (A⊗1)(v⊗Bw) =Av⊗Bw
and
kA⊗Bk=k(A⊗1)(1⊗B)k ≤ kA⊗1kk1⊗Bk ≤ kAk · kBk.
From
h(A⊗B)(v⊗w), v0⊗w0i=hAv, v0ihBw, w0i=hv, A∗v0ihw, B∗w0i
=hv⊗w,(A∗⊗B∗)(v0⊗w0)i we derive that (A⊗B)∗=A∗⊗B∗.
Lemma 4.3.4. Let (πj,Hj) be a continuous unitary representation of Gj for j= 1,2. Then
(π1⊗π2)(g1, g2) :=π1(g1)⊗π2(g2)
defines a continuous unitary representation of the product group G1×G2 in H1⊗Hb 2.
Proof. From the uniqueness part of Lemma 4.3.3 we derive that (π1(g1)⊗π2(g2))(π1(h1)⊗π2(h2)) =π1(g1h1)⊗π2(g2h2) and
(π1(g1)⊗π2(g2))∗=π1(g−11 )⊗π2(g2−1).
Thereforeπ1⊗π2 is a unitary representation ofG1×G2. Forvj, wj ∈ Hj, we further have
h(π1⊗π2)(g1, g2)(v1⊗w1), v2⊗w2i=hπ1(g1)v1, w1ihπ2(g2)v2, w2i, which is a continuous function onG1×G2. Now the continuity ofπ1⊗π2follows from the fact that the elements v1⊗v2, v1 ∈ H1, v2 ∈ H2, form a total subset (Lemma 1.2.6).
Definition 4.3.5. If (π1,H1) and (π2,H2) are unitary representations of the same group G, then we define their tensor product as the representation on H1⊗Hb 2, given by
(π1⊗π2)(g) :=π1(g)⊗π2(g).
This corresponds to the restriction of the tensor product representation ofG×G to the diagonal ∆G={(g, g) :g∈G} ∼=G.
Definition 4.3.6. Let (ρ,H) be a continuous unitary representation ofGand (π,Hπ) be an irreducible one. Then the Banach space Mπ := BG(Hπ,H) is called themultiplicity space for πin ρ.
From Schur’s Lemma we know that BG(Hπ) = C1, so that we obtain a sesquilinear map
h·,·i: Mπ× Mπ→C, B∗A=hA, Bi1.
IfA6= 0 andv∈ Hπ is a unit vector, thenAv6= 0 because otherwiseAπ(G)v= π(G)Av= 0 leads toA= 0. We therefore obtain
hA, Ai=hA∗Av, vi=kAvk2>0,
showing that h·,·i is positive definite on Mπ, turning Mπ into a pre-Hilbert space. This argument also shows that the evaluation map
evv: Mπ → H, A7→Av is an isometric embedding.
Proposition 4.3.7. The multiplicity spaceMπ is a Hilbert space and the eval-uation map induces a unitary map
ev :Mπ⊗Hb π→ H[π], A⊗v7→Av
which is an equivalence of unitary representations if Mπ⊗Hb π is endowed with the representation1⊗πof G.
Proof. ReplacingHby the isotypic componentH[π] does not change the space Mπ(Remark 4.2.13), so that we may w.l.o.g. assume thatH=H[π]. In view of Proposition 4.2.11, we can writeHas ⊕bj∈JHπ ∼=`2(J,C)b⊗Hπ for some set J.
For eachx∈`2(J,C) we then obtain an elementAx∈ Mπ byAx(w) :=x⊗w for w ∈ Hπ. In particular, we obtain for the unit vector v from above the relation Ax(v) = x⊗v with kAxk2 =kv⊗xk2 =kxk2. Conversely, we have for each j ∈J an intertwining operatorPj:H → Hπ, x⊗w 7→ hx, ejiej⊗w, corresponding to the orthogonal projection ofHontoHj ∼=ej⊗ Hπ. For each A∈ Mπ we then havePjA∈BG(Hπ) =C1, so that
Av=X
j∈J
PjAv∈X
j∈J
Cej⊗v⊆`2(J,C)⊗v
implies that Av = x⊗v for some x ∈ `2(J,C). This proves that A = Ax. Collecting all this information, we see that
evv:Mπ →`2(J,C)⊗v∼=`2(J,C)
is unitary, and therefore thatMπ is a Hilbert space isomorphic to`2(J,C).
The preceding discussion also shows that the evaluation map Mπ⊗ Hπ→ H, A⊗v7→Av
induces an isomorphism of Hilbert spaces Mπ⊗Hb π→ H,
and even an equivalence of representations if Mπ⊗Hb π is endowed with the representation1⊗π.
The main advantage of the multiplicity space, as compared to the description ofH[π] as a tensor product`2(J,C)⊗Hb π, is thatMπand its scalar product are naturally defined in terms of the irreducible representation (π,Hπ) and the representation (ρ,H). We don’t have to refer to Zorn’s Lemma to define it.
Lemma 4.3.8. If (π,H) is an irreducible representation of Gand ρ:=1⊗π is the tensor product with the trivial representation1on M, then
ρ(G)0 =B(M)⊗1∼=B(M).
Proof. Let (ej)j∈J be an ONB inM, so that M=`2(J,C),M⊗H ∼b =L
j∈JH andρ∼=⊕j∈Jπ. We write
Pj:M⊗H → H,b x⊗v7→ hx, ejiw
for the projections which are in particular intertwining operators.
Clearly, B(M)⊗1 ⊆ ρ(G)0. If, conversely, A ∈ ρ(G)0, then, for i, j ∈ J, Aij:=PiAPj∗∈BG(H) =C1by Schur’s Lemma. Letaij ∈CwithAij =aij1 andv, w∈ Hbe unit vectors. Then
hA(ej⊗v), ei⊗wi=haijv, wi=aijhv, wi implies that the closed subspaces M ⊗v∼=MareA-invariant with
A(ej⊗v) = X
i∈I
aijei
⊗v.
Therefore the matrix (aij)i,j∈J defines a bounded operatorAeon MwithA= Ae⊗1. This proves thatA∈B(M)⊗1.
Proposition 4.3.9. If(πj,Hj),j = 1,2, are irreducible unitary representations of the groupsGj, then the tensor product representationπ1⊗π2 ofG1×G2 on H1⊗Hb 2 is irreducible.
Proof. Lemma 4.3.8 implies that
(π1⊗π2)(G1×G2)0= (π1(G1)⊗1)0∩(1⊗π2(G))0
= (1⊗B(H2))∩(B(H1)⊗1).
To see that this is not larger than C1, let A ∈ B(H1) and B ∈ B(H2) with C:=A⊗1=1⊗B. Let (ej)j∈J be an ONB inH1and (fk)k∈K be an ONB in H2. Then this operator preserves all subspacesH1⊗fk andej⊗ H2, so that
(H1⊗fk)∩(ej⊗ H2) =Cej⊗fk
(Remark 4.3.2(a)) implies that all elements ej ⊗fk are eigenvectors. Write C(ej⊗fk) =cjk(ej⊗fk). ThenC =A⊗1implies thatcjk does not depend onk, andC=1⊗B implies that it does not depend on j. ThereforeC=c1 forc=cjk.