LECTURE 21: SCHUR ORTHONONALITY
1. Schur’s Lemma
Lemma 1.1 (Schur’s Lemma). Let V , W be irreducible representations of G.
(1) If f : V → W is a G-morphism, then either f ≡ 0, or f is invertible.
(2) If f1 , f2 : V → W are two G-morphisms and f2 6= 0 , then there exists λ ∈ C
such that f1 = λf2 .
Proof. (1) Suppose f is not identically zero. Since ker(f ) is a G-invariant subset in V ,
it must be {0}. So f is injective. In particular, f (V ) is a nonzero subspace of W . On
the other hand, it is easy to check that f (V ) is a G-invariant subspace of W . It follows
that f (V ) = W , and thus f is invertible.
(2) Since f2 6= 0, it is invertible. So f = f2−1 ◦ f1 is a G-morpism from V to V itself.
Let λ be one of the eigenvalues of the linear map f . Then f − λ·Id is a G-morphism
from V to V which is not invertible. It follows that f −λ·Id ≡ 0, and thus f1 = λf2 . Note that in the proof we showed in particular
Corollary 1.2. Let V be an irreducible representation of G, then HomG (V, V ) = C·Id.
Conversely, we have
Lemma 1.3. If (π, V ) is a unitary representation of G, and HomG (V, V ) = C·Id, then
(π, V ) is an irreducible representation of G.
Proof. Let 0 6= W ⊂ V be a G-invariant subspace. We need to show that W = V .
Let P : V → W be the orthogonal projection (with respect to the given G-invariant
inner product). Since both W and W ⊥ are G-invariant, we have for any g ∈ G and
any v = w + w⊥ ∈ V ,
P (g · v) = P (g · w + g · w⊥ ) = g · w = g · P (v),
i.e. P : V → W ⊂ V is a G-morphism. It follows that P = λ·Id for some λ ∈ C. Now
P 2 = P implies λ = 1, and thus W = V .
Recall that the center Z(G) of a Lie group G is
Z(G) = {h ∈ G : gh = hg, ∀g ∈ G}.
Corollary 1.4. If (π, V ) is an irreducible representation of G, then for any h ∈ Z(G),
π(h) = λ·Id for some λ ∈ C.
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2
LECTURE 21: SCHUR ORTHONONALITY
Proof. Suppose h ∈ Z(G), then for any g ∈ G,
π(h)π(g) = π(hg) = π(gh) = π(g)π(h).
In other words, π(h) : V → V is a G-morphism, and the conclusion follows.
Corollary 1.5. Any irreducible representation of an abelian Lie group is one dimensional.
Proof. Since G is abelian, Z(G) = G. By the previous corollary, for any g ∈ G, π(g) is
a multiple of the identity map on V . It follows that any subspace of V is G-invariant.
So V has no nontrivial subspace, which is equivalent to dim V = 1.
2. Schur Orthogonality for Matrix Coefficients
Let (V, π) be a representation of a Lie group G. If we choose a basis e1 , · · · , en of
V , we can identify V with Cn , and represent any g ∈ G by a matrix:

 
π11 (g) · · · π1n (g)
v1
.
.
.
.. 



.
.
.
π(g)v =
.
.
.
.
πn1 (g) · · · πnn (g)
vn
P
for v =
vi ei . So if we take Lj : V → C be the function
X
Lj (
vi ei ) = vj ,
i
then πij is the function on G given by
πij (g) = Li (π(g)ej ).
Definition 2.1. For any v ∈ V, L ∈ V ∗ , the map
φ : G → C,
φ(g) = L(π(g)v)
is called a matrix coefficient of G.
Obviously any matrix coefficients of G is a continuous function on G. In fact, they
form a subring of C(G):
Proposition 2.2. If φ1 , φ2 are matrix coefficients for G, so are φ1 + φ2 and φ1 · φ2 .
Proof. Let (πi , Vi ) be representations of G, vi ∈ Vi , Li ∈ Vi∗ such that φi (g) =
Li (πi (g)vi ). Then (π1 ⊕ π2 , V1 ⊕ V2 ) is a representation of G, L1 ⊕ L2 ∈ V1∗ ⊕ V2∗ =
(V1 ⊕ V2 )∗ and
(L1 ⊕ L2 )((π1 ⊕ π2 )(g)(v1 , v2 )) = φ1 (g) + φ2 (g).
Similarly we have a linear functional L1 ⊗ L2 on V1 ⊗ V2 satisfying (L1 ⊗ L2 )(v1 ⊗ v2 ) =
L1 (v1 )L2 (v2 ), and thus
(L1 ⊗ L2 )((π1 ⊗ π2 )(g)(v1 ⊗ v2 )) = φ1 (g)φ2 (g).
LECTURE 21: SCHUR ORTHONONALITY
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Now suppose G is a compact Lie group, and dg the normalized Haar measure on G.
Recall that L2 (G), the space of square-integrable functions with respect to this Haar
measure, is the completion of the space of continuous functions on G with respect to
the inner product
Z
hf1 , f2 iL2 =
f1 (g)f2 (g)dg.
G
Theorem 2.3 (Schur’s Orthogonality I). Let (π1 , V1 ) and (π2 , V2 ) are two non-isomorphic
irreducible representations of a compact Lie group G. Then every matrix coefficient of
π1 is orthogonal in L2 (G) to every matrix coefficient of π2 .
Proof. Fix G-invariant inner products on V1 and V2 respectively. Suppose
φi (g) = hπi (g)vi , wi i, i = 1, 2
are matrix coefficients for πi , where vi , wi ∈ Vi . Fix a basis of V1 such that e1 = v1 .
Define a linear map f : V1 → V2 by f (e1 ) = v2 and f (ek ) = 0 for all k ≥ 2. Consider
the map
Z
F : V1 → V2 ,
π2 (g)f (π1 (g −1 )v)dg.
v 7→ F (v) =
G
F is linear since f is. It is also G-equivariant, since
Z
Z
−1
−1
−1
F (π1 (h )v) =
π2 (g)f (π1 (hg) v)dg = π2 (h ) π2 (hg)f (π1 (hg)−1 v)dg = π2 (h−1 )F (v).
G
G
By Schur’s lemma, F (v) = 0 for any v, and in particular, hF (v), w2 i = 0. On the other
hand, for any j,
X
π2 (g)f (π1 (g −1 )ej ) = π2 (g)f (
π1 (g −1 )kj ek ) = π1 (g −1 )1j π2 (g)(v2 ),
k
−1
−1
where π1 (g )kj = hπ1 (g )ej , ek i is the matrix coefficients of π1 with respect to the
basis {e1 , · · · , en }. It follows that
Z
hπ1 (g −1 )ej , e1 ihπ2 (g)v2 , w2 idg = 0
G
for any j. So by linearity,
Z
hπ1 (g −1 )w1 , v1 ihπ2 (g)v2 , w2 idg = 0.
G
Note that the inner product is G-invariant,
hπ1 (g −1 )w1 , v1 i = hw1 , π1 (g)v1 i = hπ1 (g)v1 , w1 i = φ1 (g).
So the theorem follows.
Theorem 2.4 (Schur’s Orthogonality II). Let (π, V ) be an irreducible representation
of a compact Lie group G, with G-invariant inner product h·, ·i. Then
Z
1
hπ(g)w1 , v1 ihπ(g)w2 , v2 idg =
hw1 , w2 ihv1 , v2 i.
dim V
G
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LECTURE 21: SCHUR ORTHONONALITY
Proof. Define the linear maps f, F : V → V as above. Then F is G-equivariant,
and thus F = λ · Id for some λ = λ(v1 , v2 ) ∈ C. On the other hand, when we take
π1 = π2 = π, the computation in the previous proof shows
Z
λ(v1 , v2 )hw1 , w2 i = hF (w1 ), w2 i =
hπ(g −1 )w1 , v1 ihπ(g)v2 , w2 idg.
G
Since the Haar measure is invariant under the inversion map g → g −1 , the right hand
side is invariant if we exchange w1 with v2 , and exchange w2 with v1 . It follows that
λ(v1 , v2 ) = Chv2 , v1 i = Chv1 , v2 i
for some constant C. Finally if we take v1 = v2 = e1 a unit vector, then
Z
Z
Z
Z
−1
−1
Tr(F ) = Tr π(g)◦f ◦π(g )dg =
Tr(π(g)◦f ◦π(g ))dg =
Tr(f )dg =
1dg = 1.
G
G
G
On the other hand, in this case F = C·Id. It follows from Tr(F ) = 1 that C =
G
1
.
dim V