Due on 06th March - ETH Z

REPRESENTATION THEORY EXERCISES AND SOLUTIONS

1. Due on 06th March

Definition 1.1. A (continuous) quotient representation of a (continuous) representation(ρ, V ) of a group G is a representation π of G on a quotient space V/W , where W is Ginvariant, such that

π(g).(v +W ) = ρ(g).v +W, g ∈ G, v ∈ V.

Definition 1.2. A coinvariant representation of a representation (ρ, V ) of a group G is aquotient representation (ρG, VG) as above, where VG = V/W , and W is the subspace generatedby {g.v − v | g ∈ G, v ∈ V }.

1.1. Exercise.

(1) Let π be the representation of R on R3 given by the map

t 7→

1 t1 t

1

.

Compute πG.(2) Let V be a Hilbert space and W ⊆ V . Then V/W ∼= W⊥.(3) Let π be a unitary representation then πG ∼= πG.(4) Produce a counter-example of (3) in non-unitary case.

1.2. Solution.

(1) Let v = (v1, v2, v3)t. If π(t)v − v = t(v2, v3, 0)t. Clearly the subspace generated bysuch vectors is R2. Hence πG = V/R2 ∼= R which can be realized as R(0, 0, 1)t.

(2) As V = W ⊕W⊥ there exists a projection map

Pr : V → W.

This map is linear and surjective as W ⊆ V is mapped identically. Kernel of thismap is W⊥. Hence the claim follows from the first isomorphism theorem and the fact(in a Hilbert space) that W⊥⊥ = W .

(3) First we note that if W be the subspace generated by {gv − v | g ∈ G, v ∈ V } thenW⊥ = V G. To see that, if w ∈ W⊥ then

〈w, gv − v〉 = 0 ⇐⇒ 〈w − g−1w, v〉 = 0,

for all g ∈ G and v ∈ V , using unitarity. Using non-degeneracy of the inner productwe conclude that gw = w for g ∈ G, so w ∈ V G. So using (2) we have a vector spaceisomorphism induced by the projection map Pr : V → V G. It is enough to prove

1

2 REPRESENTATION THEORY EXERCISES AND SOLUTIONS

that this map commutes with corresponding representations. To see that, note thatV = V G ⊕W as a Hilbert space direct sum. Then

πG(g) ◦ Pr(v +W ) = πG(g)(v) = v = π(g)v = Pr(π(g)v +W ) = Pr ◦πG(g)(v +W ).

This completes the proof.

1.3. Exercise. Let Sn be the symmetric group of n letters for n ≥ 1. Suppose that ω(σ) isthe number of disjoint cycles in the decomposition of a permutation σ ∈ Sn. Prove that forz ∈ C

1

n!

∑σ∈Sn

zω(σ) =n∏j=1

(1 +

z − 1

j

).

1.4. Solution. As both sides of the equation are polynomials it is enought o prove theformula for z = m ∈ N. Let V = Cm and W = V ⊗n. Let Sn acts on W as follows: Fix abasis {v1 . . . , vm} of V . An element σ ∈ Sn acts on a basis vector of W by

σ(vi1 ⊗ . . .⊗ vin) := viσ(1) ⊗ . . .⊗ viσ(n) .

for 1 ≤ ij ≤ m, 1 ≤ j ≤ n. We call this representation π. Therefore each π(σ) is apermutation matrix. So the character of π

χπ(σ) := Tr(π(σ))

is the number of fixed basis vectors by Sn, i.e. the number of n-tuple (i1, . . . , in) which areinvariant by the action σ. Note that for any fixed choice of such tuple the function j 7→ ijwill be fixed by σ along the cycles of it. But for the indices we can arbitrarily choose ij in{1, . . . ,m}. So total number of fixed vector is mω(σ). Now the left hand side of our desiredexpression

1

n!

∑σ∈Sn

mω(σ) =1

|Sn|∑g∈Sn

χπ(g) = dim(W Sn),

from character theory of the finite groups. Also, as any representation of a finite group isunatry (unitarizable), using previous question we have

W Sn ∼= WSn = V ⊗n/〈σ(vi1 ⊗ . . .⊗ vin) = (vi1 ⊗ . . .⊗ vin)〉 = SymnV.

It is a well-known easy to check fact that

dim(SymnV ) =

(n+m− 1

m− 1

)=

n∏j=1

(1 +

m

j− 1

j

).

This concludes the proof.

Definition 1.3. A continuous linear operator T on a Hilbert space H to itself is calledcompact if there exists a sequence {Tn} of continuous linear operators with dim Im(Tn) <∞such that Tn → T in the operator norm topology.

REPRESENTATION THEORY EXERCISES AND SOLUTIONS 3

1.5. Exercise. Let T be a compact normal operator on a Hilbert space H. Using theSpectral Theorem stated in the class prove that there is an at most countable set S ⊆ Csuch that the eigenspace ker(T − λ) is non-zero iff λ ∈ S and is finite dimensional if λ 6= 0.Also prove that the Hilbert space orthogonal direct sum decomposition

H = ⊕λ∈Sker(T − λ).

In particular, if T 6= 0, S is not singleton.

1.6. Solution. Let T 6= 0 be a compact operator (if T = 0 then whole statement is trivial).Then (equivalently),

{Tv | ‖v‖= 1} is compact.

First we prove that eigenvalues of T is upper and lower bounded and any nonzero eigenspaceis finite dimensional. First claim can be restated as, for and ε > 0 there are only finitelymany eigenvalues whose absolute values are larger than ε. If not, we may choose {vi} a setof orthonormal vectors such that vi is λi-eigenvector with λi|≥ ε. Then {Tvi} is a infiniteset which has no limit point, as,

‖λivi − λjvj‖2= |λi|2+|λj|≥2ε2,

contradicting compactness of T . A similar argument also shows that each 0 6= λ-eigenspaceis finite dimensional. Hence, number of eigenvalues is at most countable as set of eigenvaluesis,

∪∞n=1{λ is T eigenvalue | |λ|> 1/n}.So it is now enough to prove the spectral decomposition of H. First we need to show thatT has an eigenvalue which equals to

λ := sup{|〈Tv, v〉|| ‖v‖= 1}.(exists as T is compact). There exists vn ∈ H with ‖vn‖= 1 such that

|〈Tvn, vn〉|→ λ.

As T is compact we pass through a subsequence to get w := limTvn. Now as

‖Tvn‖≥ 〈Tvn, vn〉 → λ

we have that ‖w‖≥ λ > 0. Now using a well known fact of a bounded linear operator on aHilbert space that

sup{‖Tv‖| ‖v‖= 1} =: ‖T‖= λ := sup{|〈Tv, v〉|| ‖v‖= 1},and the fact that 〈Tvn, vn〉 ∈ R as T is self-adjoint, we conclude that

‖Tvn − λvn‖2= ‖Tvn‖2−λ2 − 2λ〈Tvn, vn〉 ≤ 2λ2 − 2λ〈Tvn, vn〉 → 0.

But, as Tvn → w from above we conclude that λvn → w. Because T is bounded linear wealso have that λTvn → Tw. In other words, Tw = λw. Now a usual argument using Zorn’slemma after finding maximal sum of eigensubspace completes the proof.


1.7. Exercise. Prove that any sub-representation of T := R/Z on L2(T) is of the form

EX = ⊕n∈XCen,for some subset X ⊆ Z. en(z) = exp(2πinz).

1.8. Solution. Let us callFY := ⊕n∈YCen

for some Y ⊂ Z. For a sub-representation E of π (as right regular representation) on L2(T)let us define

X := {n ∈ Z | f(n) 6= 0 for some f ∈ E}.Now E ⊂ FX is trivial to show; because if 0 6= f ∈ E then from the Fourier expansion of ffor all n such that f(n) 6= 0 the character en ∈ FX . Now we show that FX ⊂ E. It is enoughto show that if n ∈ X then en ∈ E. Let n ∈ X so by the definition of X there exists f ∈ Esuch that f(n) 6= 0. Let us consider the L2(T)-valued function

φ : x 7→∫Tπ(t)f(x)en(−t)dt.

By a change of variable in the integral we can see that φ(x) = en(x)f(n). As f(n) 6= 0 byassumption it is enough to show that φ ∈ E to show that en ∈ E. But φ(x) can be writtenas a limit, from the theory of Riemann integral, as

φ(x) :=

∫ 1

0

π(t)f(x)en(−t)dt = limm→∞

1

m

m∑i=0

ρ(i/m)f(x)en(i/m).

But we note that each summand in the RHS is a translate of f ∈ E hence belongs to E.Therefore, their sum and the limit of the sum also belong to E as E is a closed subspace,completing the proof.



Definition 2.1. Given finite dimensional representations (πi, Vi) of Gi for i = 1, 2 we definethe external tensor product representation π1 � π2 of the group G1 ×G2 on V1 ⊗ V2 by

π1 � π2(g1, g2)(v1 ⊗ v2) = π1(g1)v1 ⊗ π2(g2)v2.

2.1. Exercise. Check that π1 � π2 is indeed a representation and is unitary if πi’s are so.

2.2. Solution. Checking that the tensor product a representation is easy. To check unitarityone notes that,

〈π1 � π2(g1, g2)v1 ⊗ v2, π1 � π2(g1, g2)w1 ⊗ w2〉= 〈π1(g1)v1 ⊗ π2(g2)v2, π1(g1)w1 ⊗ π2(g2)w2〉= 〈π1(g1)v1, π1(g1)w1〉〈π2(g2)v2, π2(g2)w2〉= 〈v1 ⊗ v2, w1 ⊗ w2〉.

2.3. Exercise. Let π be a finite dimensional representation of a finite group G with thecharacter χπ. Prove that

π is irreducible ⇐⇒ 1

|G|∑g∈G

|χπ(g)|2= 1.

2.4. Solution. Being a finite dimensional representation π is semisimple so that,

χπ =∑σ∈G

nπ(σ)χσ,

where nπ(σ) = dim HomG(π, σ). So

1

|G|∑g∈G

|χπ(g)|2=∑σ,τ∈G

nπ(σ)nπ(τ)1

|G|∑g∈G

χσ(g)χτ (g) =∑σ∈G

nπ(σ)2.

As each nπ(σ) ∈ N the LHS will be 1 iff nπ(σ) = 1 for some σ ∈ G and nπ(τ) = 0 if σ 6= τ ∈ Gwhich is equivalent of saying that π is isomorphic to the irreducible representation σ.

2.5. Exercise. Let G be the isomorphism class of unitary irreducible representation of agroup G. Show that the map

(π1, π2) 7→ π1 � π2

induces a bijection from G1 × G2 to G1 ×G2.


2.6. Solution. First we show that the map is well-defined. It amounts to show that when-ever πi’s are irreducible then π1 � π2 is also irreducible. Using the previous exercise it isenough to show that

1

|G|∑

g∈G1×G2

|χπ1�π2(g)|2= 1.

But the above sum can be written as

1

|G1||G2|∑g1∈G1

∑g2∈G2

|χπ1(g1)|2|χπ2(g2)|2=1

|G1|∑g1∈G1

|χπ1(g1)|2 1

|G2|∑g2∈G2

|χπ2(g2)|2,

which is 1 as πi’s are irreducible. Now we will show that the map is injective. It is enoughto prove that

χπ1�π2 = χπ′1�π

′2

=⇒ χπi = χπ′i, i = 1, 2.

The left side of the above implies that

χπ1(g1)χπ2(g2) = χπ′1(g1)χπ′

2(g2), ∀(g1, g2) ∈ G1 ×G2.

Plugging gi = ei ∈ Gi and using the fact that χπ(e) = dim(V ) we conclude that χπi = ciχπ′i

for some nonzero constants ci for i = 1, 2. From the irreducibility criterion of πi and of π′ione proves that ci = 1, which proves injectivity. Surjectivity follows from the fact that

|G|= |CG| and CG×H = CG × CH ,

where CG is the conjugacy classes of G.

Definition 2.2. Given a Banach space E, the dual Banach space E ′ of E is the space of allcontinuous linear maps E → C with the norm

‖λ‖E′ := sup‖v‖=1

|λ(v)|.

Given a representation π of a topological group G on E, i.e. a homomorphism π : G →BGL(E) we define the (topological) contragredient representation π of G on E ′ by the formula

〈π(g)λ, v〉 = 〈λ, π(g−1)v〉.

2.7. Exercise. Prove that if π is a continuous representation of a topological group G on aBanach space E with

supg∈G‖π(g)‖L(E)<∞,

then the contragredient of π is a continuous representation.


2.8. Solution. As π(g) is bounded for all g ∈ G and for any g, g′ ∈ G and λ, λ′ ∈ E ′ onehas

π(g)λ− π(g′)λ′ = π(λ− π(g−1g′)λ+ π(g−1g′)(λ− λ′).

it is enough to prove the continuity at the pair (1, 0) ∈ G× E ′. But we have

‖π(g)λ‖= sup‖v‖=1

|〈λ, π(g1−v〉|≤ supg∈G‖π(g)‖‖λ‖.

Letting λ enough close to 0 for arbitrary g we get the required continuity.

2.9. Exercise. Let π be a unitary representation of G on a Hilbert space H. Show that thefunctions

g 7→ 〈π(g)v, w〉,

for v, w ∈ H, are matrix coefficients of π.

2.10. Solution. First we note that for any Hilbert space H there is a canonical isomorphismbetween the complex conjugate H and the dual H ′ through the map

v 7→ {λv : w 7→ 〈w, v〉},

by Riesz representation theorem. Also by definition

π(g)λ(v) = λ(π(g−1)v) =⇒ π(g)λv(w) = 〈w, π(g)v〉 = λπ(g)v(w),

using unitarity of π. Now from the fact that

〈λv, λw〉H′ = 〈v, w〉H = 〈v, w〉Hwe get that all matrix coefficient of π are of the form

〈π(g)λv, λw〉 = 〈π(g)v, w〉,

for v, w ∈ H and g ∈ G.

2.11. Exercise. Let G be a group and H a subgroup. Let π be a finite-dimensional repre-sentation of G on a vector space V . Assume that the restriction of π to H is a direct sumof pairwise non- isomorphic irreducible representations Vi, 1 ≤ i ≤ k, of H. Show that anysubrepresentation of π is of the form

W =⊕i∈I

Vi

for some subset I of {1, . . . , k}.


2.12. Solution. Let W be a G-stable subspace of V . As as a H-representation

π |H= V =k⊕i=1

Vi,

with Vi pairwise non-isomorphic irreducible representation of H,

W ⊆⊕i∈I

Vi.

But Vi’s are, being pairwise non isomorphic, isotypic components. So any irreducible sub-space of W is an irreducible subspace of V , hence is equal to some Vi. So

Vi ⊆ W, ∀i ∈ I.Thus

W =⊕i∈I

Vi.

2.13. Exercise. Let Vm be the subspace of degree m homogeneous polynomials in C[x, y].We consider the representation πm of SL2(C) on Vm by

πm(g)f : (x, y) 7→ f((x, y)g).

Prove that for each m ≥ 0 the representation πm is irreducible.Hint: consider the restriction of πm to the subgroup T of diagonal matrices, and its actionon a natural basis of Vm to decompose it into irreducible representations.

2.14. Solution. We fix a basis of Vm to be

{ei}mi=0, ei := xiym−i.

We note that the diagonal subgroup A of SL2(C), namely

A :=

{a(λ) :=

(λ

λ−1

)| λ ∈ C×

}acts on ei by

a(λ)ei = λ2i−mei.

Hence A stabilizes each one dimensional subspace Cei. Therefore, as an A-representation

Vm =m⊕i=0

Cei.

Because each Cei is one dimensional they are trivially irreducible. Hence, applying previousquestion we can conclude that if 0 6= V ⊆ Vm is any SL2(C)-stable irreducible subspace then

V =⊕i∈I

Cei, I ⊆ {0, . . . ,m}.


As V nonzero there exists ei ∈ V for some 0 ≤ i ≤ m. Also as V is SL2(C)-stable(1 1

1

)ei,

(11 1

)ei ∈ V.

But this implies thatxi(x+ y)m−i, (x+ y)iym−i ∈ V.

Now expanding by the binomial theorem we see that

xi(x+ y)m−i =m∑j=i

(m− ij − i

)ej ∈ V.

From the irreducible decomposition of V we conclude that j ∈ I,∀j ≥ i. Similarly, expandingthe other monomial we can conclude j ∈ I,∀j ≤ i. Hence I = {0, . . . ,m} or V = Vm. Fromirreducibility assumption of V we conclude that Vm is irreducible.



G is a compact group unless mentioned otherwise.

3.1. Exercise. The regular representation of G on L2(G) decomposes as a Hilbert spacedirect sum

L2(G) =⊕π

M(π),

with notations from the lecture. Let

A :=⊕π

M(π),

be the algebraic direct sum of M(π)’s. Prove that TFAE

(1) f ∈ L2(G) is in A.(2) span(λG(g)f) is finite dimensional.(3) span(ρG(g)f) is finite dimensional.(4) span(νG(g)f) is finite dimensional.

3.2. Solution. We will show that (1) ⇐⇒ (2). Rests will follow similarly. First of all, Ais set of finite linear combinations of matrix coefficients. If f ∈ A then f can be written as

f =n∑i=1

cifi, fi ∈M(πi).

Clearly, any λ(g)f will be in⊕n

i=1M(πi), which is finite dimensional as each M(πi) is. Hence(2). Conversely, if span(λ(g)f) is finite dimensional, hence closed, being G-stable, shoulddecompose into finitely many irreducible representations. So f = λ(e)f is contained in afinite sum of M(π)’s. Hence (1).

3.3. Exercise. Let G be the circle group S1 = R/Z. Identify all the irreducible representa-tions of G. Also compute λG(f) for f ∈ L1(R/Z).

3.4. Solution. We have already discussed all sub-representations of regular representationof the circle group. As f ∈ L1(R/Z) it has a Fourier expansion, namely,

f(x) =∑n∈Z

f(n)e(nx), f(n) =

∫R/Z

f(x)e(−nx)dx.

Then by definition, for φ ∈ L2(R/Z)

λ(f)φ(g) =

∫R/Z

f(x)λ(x)φ(g)dx =

∫R/Z

f(x)φ(g − x)dx

=

∫R/Z

∑n∈Z

f(n)e(n(g − x))φ(x)dx =∑n∈Z

f(n)φ(n)e(ng).


Definition 3.1. Let ρ be a unitary representation of G and π ∈ G. The π-isotypic componentof ρ is the closure of the sum of all sub-representations which are isomorphic to π.

3.5. Exercise. Prove that the projector on the π-isotypic component of ρ is ρ(dim(π)χπ).

3.6. Solution. We know that any unitary representation of G is Hilbert direct sum of irre-ducible representations. We first claim that

φπ = ρ(dim(π)χπ) := dim(π)

∫G

χπ(g)ρ(g)dµ(g)

is an intertwiner. To see this, we first note that χπ being a character of a representation isa class function i.e. conjugacy invariant. Now we note that for any unitary representation ρthe operator ρ(f) is an intertwiner with where f is an L1 class function:

ρ(f) ◦ ρ(g)v =

∫G

f(h)ρ(h)ρ(g)vdµ(h)

=

∫G

f(hg−1)ρ(h)vdµ(h)

=

∫G

f(g−1h)ρ(h)vdµ(h)

=

∫G

ρ(g)f(h)ρ(h)vdµ(h) = ρ(g) ◦ ρ(f)v.

In particular, φπ acts on the irreducible components of ρ by a constant, by Schur’s lemma.If τ is an irreducible component the constant by which φπ acts on it is

dim(τ)−1Tr(φπ) =dim(π)

dim(τ)

∫G

χπ(g)χτ (g)dµ(g).

By orthonormality of characters the above integral vanishes for all τ � π and is identity ifτ ∼= π. So φπ is identity on π-isotypic component and zero on other isotypic components,hence φπ is the projector.

3.7. Exercise. Let ρ : G ↪→ Un(C) be a faithful representation. Prove that for all π ∈ Gthere exist a, b ≥ 0 such that

π ↪→ ρ⊗a ⊗ ρ⊗b.Hint: Use Stone-Weierstrass theorem.

3.8. Solution. First we claim that given a set S ⊆ G, S = G if and only if the linear spanM of matrix coefficients of ρ ∈ S is dense in L2(G). This is easy to see. As if M is notdense then M⊥, being nonzero sub-representation of the regular representation, will containan irreducible representation which will be outside of S. On the other hand, if an irreduciblerepresentation π is not in S then M will automatically be not dense. Now let S be the set of


irreducible sub-representations of ρ⊗a ⊗ ρ⊗b for a, b ≥ 0. Thus, using the claim, it is enoughto prove that

M := { matrix coefficients of ρ⊗a ⊗ ρ⊗b for a, b ≥ 0 }is dense in C(G) in the supnorm topology. But this is clearly true from the Stone-Weierstrasstheorem.

3.9. Exercise. Let ρ : SUn(C) ↪→ Un(C) be a representation of SUn(C). Prove that

1 ↪→ ρ⊗m ⇐⇒ n | m,where 1 is the trivial representation.

3.10. Solution. First we assume that 1 ↪→ ρ⊗m. This means that there exists a nonzeroinvariant vector v, i.e. ρ⊗m(g)v = v. We write v has linear combination of standard basiselements:

v =∑

ci1,...,imei1 ⊗ . . .⊗ eim .As v is fixed by SU(n) it is fixed by its center which is isomorphic to Zn. In fact, the centeris

{λI | λn = 1}.From the fact that λI.v = v we claim that

λI.ei1 ⊗ . . .⊗ eim = λmei1 ⊗ . . .⊗ eim = ei1 ⊗ . . .⊗ eim .Thus λm = 1. Choosing λ to be primitive we conclude that n | m. Conversely, if n | m weneed to show that ρm contains an invariant vector. It is enough to show that 1 ↪→ ρ⊗n, asthen

1 ↪→ (ρ⊗n)⊗m/n ∼= ρ⊗m.

First we note that,ρm ∼= ρm ∼= Hom((Cn)⊗n,C).

Considering the determinant a multilinear map, i.e. det ∈ Hom((Cn)⊗n,C), we choose theimage of it in v ∈ ρ⊗n. We show that v is a nonzero invariant vector. v is nonzero as det isnonzero map. To show v is invariant we show that det is invariant. In fact,

g. det(v1⊗. . .⊗vn) = det(g−1v1⊗. . .⊗g−1vn) = det(g−1) det(v1⊗. . .⊗vn) = det(v1⊗. . .⊗vn),

concluding the proof.


4. Due on 10th April

Definition 4.1. A topological group G is called a complex Lie group if if G is complexanalytic manifold with a group structure such that inverse and multiplication operations ofG are holomorphic.

4.1. Exercise.

(1) Show that the groups Cn, GLn(C), SLn(C), Sp2n(C) are complex Lie groups.(2) Show that SU2(C) is not a complex Lie group. Give other examples of Lie groups

which are not complex Lie groups.(3) Let G be a complex Lie group. Then show that Lie(G) has a natural structure of a

complex Lie algebra.

(4) If G1f−→ G2 is a homomorphism of complex Lie groups then show that df is C-linear

morphism of respective Lie algebras.(5) Show that sl2(C), as a real Lie algebra, has irreducible representations sl2(C) →

gln(R) which are not isomorphic to any of ρn (notation as in lecture).

4.2. Solution.

(1) This is direct. Each of the groups is defined by a zero set of some holomorphicfunctions on CN for some N hence a complex analytic manifold. Showing that mul-tiplication and inverse operations are holomorphic is easy.

(2) This is clear from the real dimension of SU2(C). If it is a complex Lie group thenthe real dimension would be even. But dimension of SU2(C) is 3.

(3) Standard.(4) As f is a group morphism df is C-linear. We will show that df respects the Lie

bracket. Note that for any X, Y ∈ g we have

df([X, Y ]1) = df(∂t=0∂s=0(exp(tX), exp(sY )))

= ∂t=0df(∂s=0(exp(tX), exp(sY )))

= ∂t=0∂s=0f(exp(tX), exp(sY ))

= ∂t=0∂s=0(f(exp(tX)), f(exp(sY )))

= ∂t=0∂s=0(exp(tdf(X)), exp(sdf(Y )))

= [df(X), df(Y )]2

This confirms that df is a Lie algebra morphism.(5) Consider the vector space Vm,n of polynomials in variable (z1, z2, z1, z2) which are

homogeneous of degree m in z = (z1, z2) and homogeneous of degree n in z = (z1, z2),with SL2(C) action

πm,n(g)P (z, z) = P (zg, zg).

Consequently, we get sl2(C) representation by differentiating above representation.Clearly this representation is not C-linear but R-linear unless either m or n is zero.


Hence it is not isomorphic to any of the ρn. This representations are irreduciblewhich can be proved similarly as the case of sl2(C).

4.3. Exercise. Let G = SL3(C) and g = sl3(C) Let T be the diagonal torus in G. Considerthe adjoint representation

ρ := AdG : G→ GL(g).

Construct the weight diagram of ρ with respect to T and find the weight spaces.

4.4. Solution. We will work over the Lie algebra. The Lie algebra t of the diagonal toruscan be parametrized as

t := {x = diag(x1, x2, x3) | x1 + x2 + x3 = 0, xi ∈ C}.Let ρ be the adjoint representation of g on gl(g). Eij denotes the standard elementarymatrices. First we note that

ad(x)Eij = (xi − xj)Eij.So weights are of the form Li − Lj where Li ∈ t∗ given by Li(x) = xi for x ∈ t. Hence theweight diagram form a Hexagon.

So weight spaces are gLi−Lj for i 6= j and i, j = 1, 2, 3. Here

gλ = {Y ∈ g | ad(x)Y = λ(x)E}.Consequently we have the weight space decomposition

g = t⊕⊕1≤i 6=j≤3gLi−Lj .


4.5. Exercise. Let

R := {±ei ± ej, i 6= j} ∪

{1

2

8∑i=1

±ei

}.

(in the first set, signs are chosen independently; in the second, even number of signs shouldbe pluses). Prove that R is a reduced root system with Dynkin diagram of E8.

4.6. Solution. Clearly R spans R8. Then we check every nαβ = n(α, β) is an integer.

We check case by case. Let p = 12

∑8i=1±ei with the prescribed property (with abuse of

notation).

n(±ei ± ej,±ek ± el) = ±δik ± δil ± δjk ± δjl ∈ {−2,−1, 0, 1, 2}is clearly an integer. Also

(p, p) = 1/4(∑±ei,

∑±ei) = 2

and(±ei,±ej,

∑±ei) ∈ {−2, 0, 2}

imply that n(±ei± ej, p) is an integer, in fact ∈ {−1, 0, 1}. Lastly we see that if p, q are twodifferent roots of p type then

n(p, q) = 1/4(number of positive signs− number of negative signs) ∈ {−2,−1, 0, 1, 2}which is also integer. Now we need to show that for any 2 roots α and β

sα(β) = β − n(β, α)α

is also a root. Again we check case by case. When α, β of the form ±ei,±ej result followsfrom n(α, β). When α = ±ei ± ej and β = p then description of p gives the result. Othercases follow similarly. Lastly we check that only multiple of a root which is also a root is±1.


5. Due on 10th May

Fix G = SUn(C).

5.1. Exercise. Let π be the representation of G on the space VN of polynomials of n complexvariables homogeneous of degree N , given by

π(g)P (z) = P (zg).

(1) Find all weights, the highest weight, and the dimension of the representation.(2) Use the Weyl dimension formula to prove that π is irreducible.

5.2. Solution. First we note that the complexified Lie algebra of G is g = sln(C). One canconstruct an explicit basis of VN :

B = {zk11 . . . zknn | k1 + . . . kn = N}.

Let ei ∈ h′ such that

ei(H) = Hi, H = diag(H1, . . . , Hn).

Clearly en = −∑n−1

i=1 ei. Let dπ be the representation of the Lie algebra. We will look atthe action on the basis vectors.

dπ(Hi) = ∂t=0π(exp(tHi))zk11 . . . zknn = kiz

k11 . . . zknn .

Thus weights are

{n∑i=1

kiei |n∑i=1

ki = N, ki ≥ 0}.

The highest weight is clearly Ne1 as ei > ej if i < j. As a complex vector space VN is

isomorphic to SymN(Cn), hence dimension is(N+n−1

N

).

By the Highest weight theorem we know that there exists an irreducible sub representationVNe1 of VN corresponding to the highest weight Ne1. So to prove that VN is irreducible it isenough to show that VNe1 has same dimension as VN . We will now calculate dimension ofVNe1 using the Weyl dimension formula.

We choose a positive system by

R+ = {ei − ej | i < j}.

Therefore the half sum of the positive roots is

ρ =n∑i=1

n+ 1− 2i

2ei.


By the Weyl dimension formula we get that

dim(VNe1) =∏α∈R+

〈Ne1 + ρ, α〉〈ρ, α〉

=∏i<j

〈Ne1, ei − ej〉+ 〈∑n

i=1n+1−2i

2ei, ei − ej〉

〈∑n

i=1n+1−2i

2ei, ei − ej〉

=n∏j=2

N + j − ij − i

=

(N + n− 1

n− 1

).

5.3. Exercise. Let 1 ≤ k ≤ n. Consider the representation V = Λk(Cn) of G.

(1) Describe all weights of V and describe corresponding weight subspaces.(2) Find the highest weight of V .(3) Prove that V is irreducible.

5.4. Solution. Consider the standard basis of V :

B = {vi1 ⊗ . . . vik | 1 ≤ i1 < · · · < ik ≤ n}.We will claim that

{ei1 + · · ·+ eik | 1 ≤ i1 < · · · < ik ≤ n}is the set of weight vectors. To see that we take Hi ∈ h in usual notation and note that

dπ(H)vi1 ⊗ . . . vik = (ei1 + · · ·+ eik)(H)vi1 ⊗ . . . vik .Clearly the highest weight is

ω := e1 + · · ·+ ek.

Dimension of V is(nk

).

Now we will compute the dimension of the irreducible representation Vω of the highestweight vector by the Weyl dimension formula. As similar to the previous exercise it is

dim(Vω) =∏α∈R+

〈ω + ρ, α〉〈ρ, α〉

=∏i<j

〈∑k

l=1 el, ei − ej〉+ 〈∑n

i=1n+1−2i

2ei, ei − ej〉

〈∑n

i=1n+1−2i

2ei, ei − ej〉

=k∏i=1

n∏j=k+1

1 + j − ij − i

=

(n

k

).

Using a same kind of argument as the previous exercise we conclude.

5.5. Exercise. Let g be the complex simple Lie algebra of type G2. Use the Weyl dimensionformula to find dimensions of the irreducible representations of g. Check that there is aunique one of dimension 7.


5.6. Solution. Let α1 be the short root and α2 be the long root. Then the fundamentalweights are

ω1 = 2α1 + α2, ω2 = 3α1 + 2α2.

Let Lλ be an irreducible representation with highest weights λ = aω1 + bω2 for a, b ∈ Z≥0.We recall the following.

(ω1, ω1) = 1, (ω1, ω2) = 3/2, and (ω2, ω2) = 3.

The positive roots are

ω1, ω2, 2ω1 − ω2, 3ω1 − ω2, ω2 − ω1, and 2ω2 − 3ω1.

Therefore the half sum of the positive root ρ = ω1 + ω2. After an extensive calculation onecan find that

dimLλ =(a+ 1)(a+ b+ 2)(2a+ 3b+ 5)(a+ 2b+ 3)(a+ bb+ 4)(b+ 1)

120.

For a = 1, b = 0 and a = 0, b = 1 we get the representations of dimensions 7 and 14respectively. For any other values of (a, b) it will clearly have larger dimension than 7. Sothere exists a unique 7 dimensional representation which is the standard representation C7.

5.7. Exercise. Let π : G→ GL(V ) be an irreducible representation of a compact connectedLie group G.

(1) What are the weights of the dual representations?(2) Prove that an irreducible representation is self dual if and only if the weights are

symmetric.(3) Prove that the adjoint representation is self-dual.(4) What are the duals of the representation associated to the fundamental weights of G

5.8. Solution. Weights of dual of a representations are negative of the original. To see this,let e be a weight of V . SO dπ(H)v = e(H)v. Therefore,

dπ(H)λ(v) = ∂t=0(λ(exp(−tH)v) = −e(H)λ(v),

which confirms the claim.Clearly if V ∼= V ∗ then ∀α weight of V it is a weight of V ∗, hence −α is also a weight of

V . So weights are symmetric. Conversely, let λ be the highest weight of V then −λ is alsoa weight of V because of symmetry. So λ is the highest weight of V ∗. So using the highestweight theorem V ∼= V ∗.

Using the previous, as the adjoint representation is irreducible it is enough to show thatroots are symmetric. Which is also the case.


6. Due on 29th May

6.1. Exercise. Prove that every matrix coefficient of the regular representation of a locallycompact group G vanishes at infinity. Prove that if G is non-compact then the regularrepresentation has no finite dimensional sub-representation.

6.2. Solution. Proof of this similar to the proof of the Riemann-Lebesgue lemma. First wesee that if f, g ∈ Cc(G) then the matrix coefficient is

φ(x) := 〈λG(t)f, g〉 =

∫G

f(t−1x)g(x)dx =

∫G

f(x)g(tx)dx.

Clearly support of φ is in supp(g)supp(f)−1 which is compact. So φ vanishes at infinity. Thegeneral case follows from the fact that Cc(G) is dense in L2(G). Let π be a finite dimensionalrepresentation of G inside λG. Let {vi}i be a orthonormal basis π. As π is unitary for everyg the matrix π(g) is unitary. Therefore, for every i the sum∑

j

〈π(g)vi, vj〉 = 1.

Hence, at least one of the matrix coefficients among 〈π(g)vi, vj〉i,j does not decay at infinity.This contradicts the presvious result.

6.3. Exercise. Let G be discrete group, and φ = δe be the Dirac function at the identity.Prove that φ is a function of positive type on G and the GNS-representation πφ associatedto φ is the regular representation of G.

6.4. Solution. To see that φ is a function of positive type we take finitely many ci ∈ C andgi ∈ G, with say, hk are distinct. Then∑

i,j

cicjφ(g−1j gi) =

∑k

∑gki ,gkj=hk

cki ckj =∑k

|∑i

cki|2≥ 0.

Now as G is discrete we note that φ cyclically generate λG because L2(G) is closed linearspan of {δg}g∈G and δg = λG(g)δe. So from the GNS theorem it is enough to show that thematrix coefficient of δe is φ. We see that,

〈λG(x)δe, δe〉 =∑g∈G

δx(g)δe(g) = δx(e) = φ(x),

which completes the proof.

6.5. Exercise. Let π be a unitary representation of a locally compact group G and v ∈ π.Prove that for the positive type function φ(g) = 〈π(g)v, v〉π the GNS-representation πφ iscontained in π.


6.6. Solution. Consider the subrepresentation πξ of π which is cyclically generated from ξ.Clearly the function of positive type associated with πξ is the function

〈πξ(g)ξ, ξ〉 = 〈π(g)ξ, ξ〉 = φ(g).

So from the uniqueness of the GNS construction we get that

πφ ∼= πξ ↪→ π.

6.7. Exercise. Show that Raikov’s theorem is not true if P1(G) is replaced by P≤1(G).

6.8. Solution. Consider the group R and functions

χn : t 7→ eint

on R. These functions belong to P≤1(R): for a finite set of cj ∈ C and tj ∈ R we note that

∑j,k

cj ckχn(tj − tk) =∑j,k

cj ckeintjeintk =

∣∣∣∣∣∑i

cjeintj

∣∣∣∣∣2

≥ 0,

and χn(0) = 1. χn converges to the function 0 ∈ P≤1(R) which follows from the statndardRiemann-Lebesgue lemma. But χn does uniformly converge to the zero function. HenceRaikov’s theorem does not hold.

6.9. Exercise. Let φ be a continuous function of positive type on a topological group G.Show that

H := {x ∈ G | φ(x) = φ(e)}is a closed subgroup of G and that φ is constant on every double coset HxH, x ∈ G.

6.10. Solution. From the GNS-construction there exists (π, v) such that

φ(g) = 〈π(x)v, v).

Let x, y ∈ H. Then,

φ(x−1) = 〈π(x−1v, v〉 = 〈π(x)v, v〉 = φ(e).

Also

φ(xy) = 〈π(y)v, π(x−1)v〉 = 〈v, v〉−1〈π(y)v, v〉〈π(x)v, v〉 = 〈v, v〉 = φ(e).

So H is a group. H is closed because it is level set of a continuous function.Now we see that for h1, h2 ∈ H and x ∈ G

φ(h1xh2) = 〈π(x)π(h2)v, π(h−11 )v〉 = 〈π(x)v, v〉 = φ(x),

this proves the second assertion.


6.11. Exercise. Let H be an open subgroup of a topological group G and φ be a functionof positive type on H. Show that the trivial extension of φ to G, i.e.,

g 7→

{φ(g), if g ∈ H0, if g /∈ H

is of positive type. In fact, the characteristic function χH of H is a function of positive typeon G.

6.12. Solution. Let ci ∈ C and gi ∈ G for i = 1, . . . , N be a finite collection of points. Weorder {gi} in following way. Let n1 be such an positive integer such that g−1

1 gi ∈ H if i ≤ n,otherwise not. Among the rest of gi we choose an integer n2 such that g−1

n1+1gi ∈ H for alln1 + 1 ≤ i ≤ n1 + n2, otherwise not. We continue this process till it ends, say, k times, i.e.n1 + · · ·+ nk = N . Now we see that

g−1i gj ∈ H ⇐⇒ n1 + · · ·+ nl + 1 ≤ i, j ≤ n1 + · · ·+ nl + nl+1, for some 0 ≤ l ≤ k − 1.

SoN∑

i.j=1

cicjφ(g−1i gj) =

k−1∑l=0

n1+···+nl+1∑i,j=n1+···+nl+1

cicjφi(g−1i gj) ≥ 0.

This proves that φ is of positive type. Now χH is a function of positive type on H as it isidentically one there which proves the second part.

6.13. Exercise. Let G be locally compact abelian group with a finite non-negative measureµ on it. Define the Fourier transform

µ : G→ C χ 7→∫G

χ(g)dµ(g).

Prove that µ is a function of positive type on G Also show that for any φ of positive typeon G there exists a finite Borel measure µ on G such that µ = φ.

6.14. Solution. We note that for any finite set of ci ∈ C and χi ∈ G∑i,j

cicjµ(χ−1j χi) =

∑i,j

cicj

∫G

χjχi(g)dg =

∫G

∣∣∣∣∣∑i

ciχi(g)

∣∣∣∣∣2

dg ≥ 0

which proves the first assertion. For the second, which is known as the Bochner’s theoremcan be found in Theorem D.2.2 in Bekka-Harpe-Vallette.

Due on 06th March - ETH Z

Documents

Transcript of Due on 06th March - ETH Z