Lecture 9

Problem 50. Verify that (cπρ,Hom⁡(V,W))\big{(}c_{\pi}^{\rho},\operatorname{Hom}(V,W)\big{)} is a representation of GG.

Solution: Let T∈Hom⁡(V,W)T\in\operatorname{Hom}(V,W). Then for some g∈Gg\in G we have that π⁢(g−1)\pi(g^{-1}) and ρ⁢(g)\rho(g) are in GL⁡(V)\operatorname{GL}(V) and GL⁡(W)\operatorname{GL}(W) respectively, so ρ⁢(g)⁢T⁢π⁢(g−1)\rho(g)T\pi(g^{-1}) is in Hom⁡(V,W)\operatorname{Hom}(V,W). Let g∈Gg\in G be arbitrary, then

cπρ⁢(g)⁢(S+λ⁢T)=ρ⁢(g)⁢(S+λ⁢T)⁢π⁢(g−1)=ρ⁢(g)⁢S⁢π⁢(g−1)+λ⁢ρ⁢(g)⁢T⁢π⁢(g−1)=cπρ⁢(g)⁢(S)+λ⁢cπρ⁢(g)⁢(T),c_{\pi}^{\rho}(g)(S+\lambda T)=\rho(g)(S+\lambda T)\pi(g^{-1})=\rho(g)S\pi(g^{% -1})+\lambda\rho(g)T\pi(g^{-1})=c_{\pi}^{\rho}(g)(S)+\lambda c_{\pi}^{\rho}(g)% (T),

for all S,T∈Hom⁡(V,W)S,T\in\operatorname{Hom}(V,W) and λ∈ℂ\lambda\in{\mathbb{C}}. Thus cπρ⁢(g)c_{\pi}^{\rho}(g) is linear for each g∈Gg\in G. Also

cπρ⁢(g)⁢cπρ⁢(g−1)⁢(T)=ρ⁢(g)⁢ρ⁢(g−1)⁢T⁢π⁢(g)⁢π⁢(g−1)=ρ⁢(g⁢g−1)⁢T⁢π⁢(g⁢g−1)=Tc_{\pi}^{\rho}(g)c_{\pi}^{\rho}(g^{-1})(T)=\rho(g)\rho(g^{-1})T\pi(g)\pi(g^{-1% })=\rho(gg^{-1})T\pi(gg^{-1})=T

for all T∈Hom⁡(V,W)T\in\operatorname{Hom}(V,W) so cπρ⁢(g)c_{\pi}^{\rho}(g) is invertible for each g∈Gg\in G. Hence

cπρ:G⟶GL⁡(Hom⁡(V,W)).c_{\pi}^{\rho}:G\longrightarrow\operatorname{GL}(\operatorname{Hom}(V,W)).

Let gg and hh be arbitrary in GG. Then

cπρ⁢(g⁢h)⁢(T)=ρ⁢(g⁢h)⁢T⁢π⁢(h−1⁢g−1)\displaystyle c_{\pi}^{\rho}(gh)(T)=\rho(gh)T\pi(h^{-1}g^{-1}) =ρ⁢(g)⁢ρ⁢(h)⁢T⁢π⁢(h−1)⁢π⁢(g−1)\displaystyle=\rho(g)\rho(h)T\pi(h^{-1})\pi(g^{-1})
=ρ⁢(g)⁢(cπρ⁢(h)⁢(T))⁢π⁢(g−1)=cπρ⁢(g)⁢cπρ⁢(h)⁢(T),\displaystyle=\rho(g)\left(c_{\pi}^{\rho}(h)(T)\right)\pi(g^{-1})=c_{\pi}^{% \rho}(g)c_{\pi}^{\rho}(h)(T),

for all T∈Hom⁡(V,w)T\in\operatorname{Hom}(V,w). Thus cπρc_{\pi}^{\rho} is a group homomorphism and (cπρ,Hom⁡(V,W))(c_{\pi}^{\rho},\operatorname{Hom}(V,W)) is a representation of GG.

The next few problems give an alternative proof of the orthogonality statement in Theorem 8.7 (actually a slightly stronger result is proved).

Problem 51. Let (π,V)(\pi,V) be a finite-dimensional representation of a group GG. Suppose that VV is equipped with a GG-invariant inner product ⟨⋅,⋅⟩\langle\cdot,\cdot\rangle. Show that

χπ⁢(g)=∑i=1dim⁡V⟨π⁢(g)⁢vi,vi⟩,\chi_{\pi}(g)=\sum_{i=1}^{\operatorname{dim}V}\langle\pi(g)v_{i},v_{i}\rangle,

where {vi}i=1,…,dim⁡(V)\{v_{i}\}_{i=1,\ldots,\operatorname{dim}(V)} is any orthonormal basis for VV with respect to the invariant inner product.

Remark: A function Φ:G→ℂ\Phi:G\rightarrow{\mathbb{C}} of the form Φ⁢(g)=⟨π⁢(g)⁢v1,v2⟩\Phi(g)=\langle\pi(g)v_{1},v_{2}\rangle, where v1,v2∈Vv_{1},v_{2}\in V for some unitary representation (π,V)(\pi,V) of GG is called a matrix coefficient.

Solution:

The result holds more generally. Let A:V→VA:V\rightarrow V be any linear operator. Then for any orthonormal basis (w.r.t. ⟨⋅,⋅⟩\langle\cdot,\cdot\rangle) {vi}\{v_{i}\} of VV, the matrix of AA w.r.t. the basis is given by

(⟨A⁢vi,vj⟩)1≤i,j≤dim⁡V.\big{(}\langle Av_{i},v_{j}\rangle\big{)}_{1\leq i,j\leq\operatorname{dim}V}.

In particular, each diagonal element is of the form ⟨A⁢vi,vi⟩\langle Av_{i},v_{i}\rangle, hence

tr⁡A=∑i=1dim⁡V⟨A⁢vi,vi⟩.\operatorname{tr}A=\sum_{i=1}^{\operatorname{dim}V}\langle Av_{i},v_{i}\rangle.

Problem 52. Let (π,V)(\pi,V) and (ρ,W)(\rho,W) be two representations of a group GG. For T∈Hom⁡(V,W)T\in\operatorname{Hom}(V,W), define TGT^{G} by

TG=1|G|⁢∑g∈Gπ2⁢(g)⁢T⁢π1⁢(g−1).T^{G}=\frac{1}{|G|}\sum_{g\in G}\pi_{2}(g)T\pi_{1}(g^{-1}).

Show that TG∈HomG⁡(V,W)T^{G}\in\operatorname{Hom}_{G}(V,W).

Solution: Since TGT^{G} is the linear combination of compositions of linear operators, TGT^{G} is also a linear operator. It remains to show that it intertwines with the GG-actions on VV and WW. We compute: for any g∈Gg\in G,

TG⁢π1⁢(g)=1|G|⁢∑h∈Gπ2⁢(h)⁢T⁢π1⁢(h−1)⁢π1⁢(g)=1|G|⁢∑h∈Gπ2⁢(h)⁢T⁢π1⁢(h−1⁢g).\displaystyle T^{G}\pi_{1}(g)=\frac{1}{|G|}\sum_{h\in G}\pi_{2}(h)T\pi_{1}(h^{% -1})\pi_{1}(g)=\frac{1}{|G|}\sum_{h\in G}\pi_{2}(h)T\pi_{1}(h^{-1}g).

We now do the change of variables h−1~=h−1⁢g\widetilde{h^{-1}}=h^{-1}g, hence h=g⁢h~h=g\widetilde{h}, giving

=1|G|⁢∑h~∈Gπ2⁢(g⁢h~)⁢T⁢π1⁢(h~)=π2⁢(g)⁢TG.=\frac{1}{|G|}\sum_{\widetilde{h}\in G}\pi_{2}(g\widetilde{h})T\pi_{1}(% \widetilde{h})=\pi_{2}(g)T^{G}.

Problem 53. Show that if (π,V)(\pi,V) and (ρ,W)(\rho,W) are two non-isomorphic irreducible unitary representations of GG, then their matrix coefficients are orthogonal with respect to the standard L2L^{2}-inner product on ℂ{\mathbb{C}}-valued functions on GG, i.e.

1|G|⁢∑g∈G⟨π⁢(g)⁢v1,v2⟩V⁢⟨ρ⁢(g)⁢w1,w2⟩W¯=0∀v1,v2∈V,w1,w2∈W.\frac{1}{|G|}\sum_{g\in G}\big{\langle}\pi(g)v_{1},v_{2}\big{\rangle}_{V}% \overline{\big{\langle}\rho(g)w_{1},w_{2}\big{\rangle}_{W}}=0\qquad\forall\,v_% {1},v_{2}\in V,\,w_{1},w_{2}\in W.

Hint: Define the operator T∈Hom⁡(V,W)T\in\operatorname{Hom}(V,W) by

T⁢(v)=⟨v,v1⟩V⁢w1∀v∈V.T(v)=\langle v,v_{1}\rangle_{V}w_{1}\qquad\forall v\in V.

Then use Problem 43 and Schur’s lemma to study ⟨w2,TG⁢v2⟩W\langle w_{2},T^{G}v_{2}\rangle_{W}.

Solution: Let v1,v2,w1,w2v_{1},v_{2},w_{1},w_{2} be as in the statement of the problem, and define T∈Hom⁡(V,W)T\in\operatorname{Hom}(V,W) by T⁢(v)=⟨v,v1⟩V⁢w1T(v)=\langle v,v_{1}\rangle_{V}w_{1}. By Problem 43, TG∈HomG⁡(V,W)T^{G}\in\operatorname{Hom}_{G}(V,W), and since (π,V(\pi,V and (ρ,W)(\rho,W) are non-isomorphic, Schur’s lemma gives TG=0T^{G}=0. In particular, TG⁢v2=0T^{G}v_{2}=0, hence

0=⟨w2,TG⁢v2⟩W=1|G|⁢∑g∈G⟨w2,ρ⁢(g)⁢T⁢(π⁢(g−1)⁢v2)⟩W.0=\langle w_{2},T^{G}v_{2}\rangle_{W}=\frac{1}{|G|}\sum_{g\in G}\langle w_{2},% \rho(g)T(\pi(g^{-1})v_{2})\rangle_{W}. (0.0.1)

Now, by definition and the fact that π1\pi_{1} is unitary,

T(π(g−1)v2=⟨π(g−1)v2,v1⟩Vw1=⟨v2,π(g)v1⟩Vw1=⟨π⁢(g)⁢v1,v2⟩V¯w1.T(\pi(g^{-1})v_{2}=\langle\pi(g^{-1})v_{2},v_{1}\rangle_{V}w_{1}=\langle v_{2}% ,\pi(g)v_{1}\rangle_{V}w_{1}=\overline{\langle\pi(g)v_{1},v_{2}\rangle_{V}}w_{% 1}.

This gives

ρ⁢(g)⁢T⁢(π⁢(g−1)⁢v2)=⟨π⁢(g)⁢v1,v2⟩V¯⁢ρ⁢(g)⁢w1,\rho(g)T(\pi(g^{-1})v_{2})=\overline{\langle\pi(g)v_{1},v_{2}\rangle_{V}}\rho(% g)w_{1},

which, when substituted into (0.0.1), gives

0=1|G|⁢∑g∈G⟨w2,⟨π⁢(g)⁢v1,v2⟩V¯⁢ρ⁢(g)⁢w1⟩W=\displaystyle 0=\frac{1}{|G|}\sum_{g\in G}\langle w_{2},\overline{\langle\pi(g% )v_{1},v_{2}\rangle_{V}}\rho(g)w_{1}\rangle_{W}= 1|G|⁢∑g∈G⟨π⁢(g)⁢v1,v2⟩V⁢⟨w2,ρ⁢(g)⁢w1⟩W\displaystyle\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)v_{1},v_{2}\rangle_{V}% \langle w_{2},\rho(g)w_{1}\rangle_{W}
=1|G|⁢∑g∈G⟨π⁢(g)⁢v1,v2⟩V⁢⟨ρ⁢(g)⁢w1,w2⟩W¯,\displaystyle=\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)v_{1},v_{2}\rangle_{V}% \overline{\langle\rho(g)w_{1},w_{2}\rangle_{W}},

which was what was to be shown.

Problem 54. Let (π,V)(\pi,V) be an irreducible unitary representation of GG. Show that

1|G|⁢∑g∈G⟨π⁢(g)⁢v1,v2⟩V⁢⟨π⁢(g)⁢u1,u2⟩V¯=⟨v1,u1⟩V⁢⟨v2,u2⟩V¯dim⁡V∀v1,v2,u1,u2∈V.\frac{1}{|G|}\sum_{g\in G}\big{\langle}\pi(g)v_{1},v_{2}\big{\rangle}_{V}% \overline{\big{\langle}\pi(g)u_{1},u_{2}\big{\rangle}_{V}}=\frac{\langle v_{1}% ,u_{1}\rangle_{V}\overline{\langle v_{2},u_{2}\rangle_{V}}}{\operatorname{dim}% V}\qquad\forall\,v_{1},v_{2},u_{1},u_{2}\in V.

Hint: Define the operator T∈Hom⁡(V)T\in\operatorname{Hom}(V) by

T⁢(v)=⟨v,v1⟩V⁢u1∀v∈V.T(v)=\langle v,v_{1}\rangle_{V}u_{1}\qquad\forall v\in V.

Then compute ⟨u2,TG⁢v2⟩V\langle u_{2},T^{G}v_{2}\rangle_{V} using Problem 43 and Schur’s lemma (compare with the proof of Proposition 25; show that tr⁡TG=tr⁡T\operatorname{tr}T^{G}=\operatorname{tr}T).

Solution: Letting u1,u2,v1,v2u_{1},u_{2},v_{1},v_{2} be as stated in the problem, we define

T⁢(v)=⟨v,v1⟩V⁢u1.T(v)=\langle v,v_{1}\rangle_{V}u_{1}.

By Problem 43 and Schur’s lemma, TG=λ⁢IdT^{G}=\lambda{\,\mathrm{Id}}, and so taking the trace of both sides gives tr⁡TG=λ⋅dim⁡V\operatorname{tr}T^{G}=\lambda\cdot\operatorname{dim}V, i.e. λ=tr⁡TGdim⁡V\lambda=\frac{\operatorname{tr}T^{G}}{\operatorname{dim}V}. We now compute tr⁡TG\operatorname{tr}T^{G}:

tr⁡TG=tr⁡(1|G|⁢∑g∈Gπ⁢(g)⁢T⁢π⁢(g−1))=1|G|⁢∑g∈Gtr⁡(π⁢(g)⁢T⁢π⁢(g−1))=1|G|⁢∑g∈Gtr⁡T=tr⁡T.\displaystyle\operatorname{tr}T^{G}=\operatorname{tr}\left(\frac{1}{|G|}\sum_{% g\in G}\pi(g)T\pi(g^{-1})\right)=\frac{1}{|G|}\sum_{g\in G}\operatorname{tr}% \big{(}\pi(g)T\pi(g^{-1})\big{)}=\frac{1}{|G|}\sum_{g\in G}\operatorname{tr}T=% \operatorname{tr}T.

Since T=0T=0 on v1⟂v_{1}^{\perp}, we have that

tr⁡T=⟨T⁢(v1‖v1|),v1‖v1|⟩V=⟨u1,v1⟩V;\operatorname{tr}T=\langle T\big{(}{\textstyle\frac{v_{1}}{\|v_{1}|}}\big{)},{% \textstyle\frac{v_{1}}{\|v_{1}|}}\rangle_{V}=\langle u_{1},v_{1}\rangle_{V};

thus

TG=⟨u1,v1⟩Vdim⁡V⁢Id.T^{G}=\frac{\langle u_{1},v_{1}\rangle_{V}}{\operatorname{dim}V}{\,\mathrm{Id}}.

This gives

⟨u2,TG⁢v2⟩V=⟨u2,⟨u1,v1⟩Vdim⁡V⁢v2⟩V=⟨v1,u1⟩V⋅⟨u2,v2⟩Vdim⁡V=⟨v1,u1⟩V⋅⟨v2,u2⟩V¯dim⁡V.\langle u_{2},T^{G}v_{2}\rangle_{V}=\left\langle u_{2},{\textstyle\frac{% \langle u_{1},v_{1}\rangle_{V}}{\operatorname{dim}V}}v_{2}\right\rangle_{V}=% \frac{\langle v_{1},u_{1}\rangle_{V}\cdot\langle u_{2},v_{2}\rangle_{V}}{% \operatorname{dim}V}=\frac{\langle v_{1},u_{1}\rangle_{V}\cdot\overline{% \langle v_{2},u_{2}\rangle_{V}}}{\operatorname{dim}V}. (0.0.2)

On the other hand, from the definition of TGT^{G} we obtain

⟨u2,TG⁢v2⟩V=\displaystyle\langle u_{2},T^{G}v_{2}\rangle_{V}= 1|G|⁢∑g∈G⟨u2,π⁢(g)⁢T⁢π⁢(g−1)⁢v2⟩V=1|G|⁢∑g∈G⟨u2,π⁢(g)⁢⟨π⁢(g−1)⁢v2,v1⟩V⁢u1⟩V\displaystyle\frac{1}{|G|}\sum_{g\in G}\langle u_{2},\pi(g)T\pi(g^{-1})v_{2}% \rangle_{V}=\frac{1}{|G|}\sum_{g\in G}\langle u_{2},\pi(g)\langle\pi(g^{-1})v_% {2},v_{1}\rangle_{V}u_{1}\rangle_{V} (0.0.3)
=1|G|⁢∑g∈G⟨u2,π⁢(g)⁢⟨π⁢(g−1)⁢v2,v1⟩V⁢u1⟩V=1|G|⁢∑g∈G⟨v1,π⁢(g−1)⁢v2⟩V⁢⟨u2,π⁢(g)⁢u1⟩V\displaystyle=\frac{1}{|G|}\sum_{g\in G}\langle u_{2},\pi(g)\langle\pi(g^{-1})% v_{2},v_{1}\rangle_{V}u_{1}\rangle_{V}=\frac{1}{|G|}\sum_{g\in G}\langle v_{1}% ,\pi(g^{-1})v_{2}\rangle_{V}\langle u_{2},\pi(g)u_{1}\rangle_{V}
=1|G|⁢∑g∈G⟨π⁢(g)⁢v1,v2⟩V⁢⟨π⁢(g)⁢u1,u2⟩V¯.\displaystyle=\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)v_{1},v_{2}\rangle_{V}% \overline{\langle\pi(g)u_{1},u_{2}\rangle_{V}}.

Equating (0.0.2) and (0.0.3) gives

1|G|⁢∑g∈G⟨π⁢(g)⁢v1,v2⟩V⁢⟨π⁢(g)⁢u1,u2⟩V¯=⟨v1,u1⟩V⋅⟨v2,u2⟩V¯dim⁡V,\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)v_{1},v_{2}\rangle_{V}\overline{\langle% \pi(g)u_{1},u_{2}\rangle_{V}}=\frac{\langle v_{1},u_{1}\rangle_{V}\cdot% \overline{\langle v_{2},u_{2}\rangle_{V}}}{\operatorname{dim}V},

as desired.

Problem 55. Use Problems 51 to 54 to give an alternative proof of the fact that the characters of irreducible representations form an orthonormal set in C⁢F⁢(G)CF(G).

Solution: Let (π,V)≇(ρ,U)(\pi,V)\not\cong(\rho,U) be two irreducible representations of GG. By Problems 51 and 52, we may assume that these are both unitary representations. Let {vi}\{v_{i}\} be an orthonormal basis of VV and {uj}\{u_{j}\} an orthonormal basis of UU. By Problem 51,

χπ⁢(g)=∑i⟨π⁢(g)⁢vi,vi⟩V,χρ⁢(g)=∑j⟨ρ⁢(g)⁢uj,uj⟩U.\chi_{\pi}(g)=\sum_{i}\langle\pi(g)v_{i},v_{i}\rangle_{V},\qquad\chi_{\rho}(g)% =\sum_{j}\langle\rho(g)u_{j},u_{j}\rangle_{U}.

Then

⟨χπ,χρ⟩G=∑i,j⟨⟨π⁢(⋅)⁢vi,vi⟩V,⟨ρ⁢(⋅)⁢uj,uj⟩U⟩G.\langle\chi_{\pi},\chi_{\rho}\rangle_{G}=\sum_{i,j}\bigg{\langle}\langle\pi(% \cdot)v_{i},v_{i}\rangle_{V},\langle\rho(\cdot)u_{j},u_{j}\rangle_{U}\bigg{% \rangle}_{G}.

By Problem 53,

⟨⟨π⁢(⋅)⁢vi,vi⟩V,⟨ρ⁢(⋅)⁢uj,uj⟩U⟩G=1|G|⁢∑g∈G⟨π⁢(g)⁢vi,vi⟩V⁢⟨ρ⁢(g)⁢uj,uj⟩U¯=0,\bigg{\langle}\langle\pi(\cdot)v_{i},v_{i}\rangle_{V},\langle\rho(\cdot)u_{j},% u_{j}\rangle_{U}\bigg{\rangle}_{G}=\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)v_{i% },v_{i}\rangle_{V}\overline{\langle\rho(g)u_{j},u_{j}\rangle_{U}}=0,

hence ⟨χπ,χρ⟩G=0\langle\chi_{\pi},\chi_{\rho}\rangle_{G}=0.

On the other hand, by Problems 51 and 54, we have

⟨χπ,χπ⟩G=∑i,j⟨⟨π⁢(⋅)⁢vi,vi⟩V,⟨π⁢(⋅)⁢vj,vj⟩V⟩G=∑i,j1|G|⁢∑g∈G⟨π⁢(g)⁢vi,vi⟩V⁢⟨π⁢(g)⁢vj,vj⟩V¯\displaystyle\langle\chi_{\pi},\chi_{\pi}\rangle_{G}=\sum_{i,j}\bigg{\langle}% \langle\pi(\cdot)v_{i},v_{i}\rangle_{V},\langle\pi(\cdot)v_{j},v_{j}\rangle_{V% }\bigg{\rangle}_{G}=\sum_{i,j}\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)v_{i},v_{% i}\rangle_{V}\overline{\langle\pi(g)v_{j},v_{j}\rangle_{V}}
=∑i,j⟨vi,vj⟩V⋅⟨vi,vj⟩V¯dim⁡V.\displaystyle=\sum_{i,j}\frac{\langle v_{i},v_{j}\rangle_{V}\cdot\overline{% \langle v_{i},v_{j}\rangle_{V}}}{\operatorname{dim}V}.

Since {vi}\{v_{i}\} is an orthonormal basis of VV, ⟨vi,vj⟩V⋅⟨vi,vj⟩V¯=1\langle v_{i},v_{j}\rangle_{V}\cdot\overline{\langle v_{i},v_{j}\rangle_{V}}=1 if j=ij=i and zero otherwise, hence

⟨χπ,χπ⟩G=∑i∑j⟨vi,vj⟩V⋅⟨vi,vj⟩V¯dim⁡V=∑i1dim⁡V=1.\langle\chi_{\pi},\chi_{\pi}\rangle_{G}=\sum_{i}\frac{\sum_{j}\langle v_{i},v_% {j}\rangle_{V}\cdot\overline{\langle v_{i},v_{j}\rangle_{V}}}{\operatorname{% dim}V}=\sum_{i}\frac{1}{\operatorname{dim}V}=1.

Problem 56. For each (σ,Wσ)∈Irr⁢(G)(\sigma,W_{\sigma})\in\mathrm{Irr}(G), let {wσ,i}i=1,…,dim⁡σ\{w_{\sigma,i}\}_{i=1,\ldots,\operatorname{dim}\sigma} be an orthonormal basis of WσW_{\sigma} with respect to an invariant inner product ⟨⋅,⋅⟩Wσ\langle\cdot,\cdot\rangle_{W_{\sigma}}. Show that {Φσ,i,j}σ∈Irr⁢(G),1≤i,j≤dim⁡σ\{\Phi_{\sigma,i,j}\}_{\sigma\in\mathrm{Irr}(G),1\leq i,j\leq\operatorname{dim% }\sigma} is an orthonormal basis of V={f:G→ℂ}V=\{f:G\rightarrow{\mathbb{C}}\} with respect to the inner product ⟨f1,f2⟩G=1|G|⁢∑g∈Gf1⁢(g)⁢f2⁢(g)¯\langle f_{1},f_{2}\rangle_{G}=\frac{1}{|G|}\sum_{g\in G}f_{1}(g)\overline{f_{% 2}(g)}, where

Φσ,i,j⁢(g):=dim⁡σ⁢⟨σ⁢(g)⁢wσ,i,wσ,j⟩Wσ.\Phi_{\sigma,i,j}(g):=\sqrt{\operatorname{dim}\sigma}\langle\sigma(g)w_{\sigma% ,i},w_{\sigma,j}\rangle_{W_{\sigma}}.

Solution: By Problem 53,

⟨Φσ,i,j,Φσ′,i′,j′⟩G=0\langle\Phi_{\sigma,i,j},\Phi_{\sigma^{\prime},i^{\prime},j^{\prime}}\rangle_{% G}=0

if σ≠σ′\sigma\neq\sigma^{\prime}. On the other hand, by Problem 54,

⟨Φσ,i,j,Φσ,i′,j′⟩G=σ⋅σ⋅⟨wσ,i,wσ,i′⟩Wσ⋅⟨wσ,j,wσ,j′⟩Wσ¯dim⁡σ.\langle\Phi_{\sigma,i,j},\Phi_{\sigma,i^{\prime},j^{\prime}}\rangle_{G}=\frac{% \sqrt{\sigma}\cdot\sqrt{\sigma}\cdot\langle w_{\sigma,i},w_{\sigma,i^{\prime}}% \rangle_{W_{\sigma}}\cdot\overline{\langle w_{\sigma,j},w_{\sigma,j^{\prime}}% \rangle_{W_{\sigma}}}}{\operatorname{dim}\sigma}.

Since {wσ,i}i\{w_{\sigma,i}\}_{i} is an orthonormal basis of WσW_{\sigma}, we then have

⟨Φσ,i,j,Φσ,i′,j′⟩G={1if⁢(i,j)=(i′,j′)0otherwise.\langle\Phi_{\sigma,i,j},\Phi_{\sigma,i^{\prime},j^{\prime}}\rangle_{G}=\begin% {cases}1\quad&\mathrm{if\;}(i,j)=(i^{\prime},j^{\prime})\\ 0\quad&\mathrm{otherwise}.\end{cases}

The set {Φσ,i,j}\{\Phi_{\sigma,i,j}\} is thus an orthonormal set in VV, and therefore forms a basis of span⁢{Φσ,i,j}\mathrm{span}\{\Phi_{\sigma,i,j}\}. Using Theorem 6.6, we get

dim⁡(span⁢{Φσ,i,j})=#⁢{Φσ,i,j}=∑σ∈Irr⁢(G)(dim⁡σ)2=|G|=dim⁡V,\operatorname{dim}(\mathrm{span}\{\Phi_{\sigma,i,j}\})=\#\{\Phi_{\sigma,i,j}\}% =\sum_{\sigma\in\mathrm{Irr}(G)}(\operatorname{dim}\sigma)^{2}=|G|=% \operatorname{dim}V,

giving V=span⁢{Φσ,i,j}V=\mathrm{span}\{\Phi_{\sigma,i,j}\}.

Problem 57. (Challenging!) For each (σ,Wσ)∈Irr⁢(G)(\sigma,W_{\sigma})\in\mathrm{Irr}(G) and f∈V={f:G→ℂ}f\in V=\{f:G\rightarrow{\mathbb{C}}\}, define f^⁢(σ)∈Hom⁡(Wσ)\widehat{f}(\sigma)\in\operatorname{Hom}(W_{\sigma}) by

f^⁢(σ)=1|G|⁢∑g∈Gf⁢(g)⁢σ⁢(g−1).\widehat{f}(\sigma)=\frac{1}{|G|}\sum_{g\in G}f(g)\sigma(g^{-1}).
  1. (a)
    ​

    Show that f↦f^⁢(σ)f\mapsto\widehat{f}(\sigma) is an intertwining operator between (ρ,V)(\rho,V), and (σ~,Hom⁡(Wσ))\big{(}\widetilde{\sigma},\operatorname{Hom}(W_{\sigma})\big{)}, where (ρ⁢(g)⁢f)⁢(h)=f⁢(h⁢g)(\rho(g)f)(h)=f(hg), ∀h∈G\forall h\in G, and σ⁢(g)~⁢(T)=σ⁢(g)⁢T\widetilde{\sigma(g)}(T)=\sigma(g)T for all T∈Hom⁡(Wσ)T\in\operatorname{Hom}(W_{\sigma}).

  2. (b)
    ​

    Show the Plancherel identity

    ⟨f1,f2⟩G=∑σ∈Irr⁢(G)dim⁡σ⁢tr⁡(f1^⁢(σ)⁢f2^⁢(σ)∗),\langle f_{1},f_{2}\rangle_{G}=\sum_{\sigma\in\mathrm{Irr}(G)}\operatorname{% dim}\sigma\,\operatorname{tr}\big{(}\widehat{f_{1}}(\sigma)\widehat{f_{2}}(% \sigma)^{*}\big{)},

    and use it to deduce the Fourier inversion formula

    f⁢(g)=∑σ∈Irr⁢(G)dim⁡σ⁢tr⁡(f^⁢(σ)⁢σ⁢(g)).f(g)=\sum_{\sigma\in\mathrm{Irr}(G)}\operatorname{dim}\sigma\,\operatorname{tr% }\big{(}\widehat{f}(\sigma)\sigma(g)\big{)}.

Solution:

  1. (a)
    ​

    Since f^\widehat{f} is a linear combination of elements of Hom⁡(Wσ)\operatorname{Hom}(W_{\sigma}), it must also be an element of Hom⁡(Wσ)\operatorname{Hom}(W_{\sigma}). It remains to show that f↦f^f\mapsto\widehat{f} intertwines with the GG-actions on VV and Hom⁡(Wσ)\operatorname{Hom}(W_{\sigma}). Given g∈Gg\in G, and f∈Vf\in V, we have

    ρ⁢(g)⁢f^=\displaystyle\widehat{\rho(g)f}= 1|G|⁢∑h∈G[ρ⁢(g)⁢f]⁢(h)⁢σ⁢(h−1)=1|G|⁢∑h∈Gf⁢(h⁢g)⁢σ⁢(h−1)⁢=(h′=h⁢g)⁢1|G|⁢∑h′∈Gf⁢(h′)⁢σ⁢((h′⁢g−1)−1)\displaystyle\frac{1}{|G|}\sum_{h\in G}[\rho(g)f](h)\sigma(h^{-1})=\frac{1}{|G% |}\sum_{h\in G}f(hg)\sigma(h^{-1})\underset{(h^{\prime}=hg)}{=}\frac{1}{|G|}% \sum_{h^{\prime}\in G}f(h^{\prime})\sigma\big{(}(h^{\prime}g^{-1})^{-1}\big{)}
    =1|G|⁢∑h∈Gf⁢(h)⁢σ⁢(g⁢h−1)=σ⁢(g)⁢(1|G|⁢∑h∈Gf⁢(h)⁢σ⁢(h−1))=σ~⁢(g)⁢f^⁢(σ),\displaystyle=\frac{1}{|G|}\sum_{h\in G}f(h)\sigma(gh^{-1})=\sigma(g)\left(% \frac{1}{|G|}\sum_{h\in G}f(h)\sigma(h^{-1})\right)=\widetilde{\sigma}(g)% \widehat{f}(\sigma),

    as desired.

  2. (b)
    ​

    For each σ∈Irr⁢(G)\sigma\in\mathrm{Irr}(G), let ⟨⋅,⋅⟩Wσ\langle\cdot,\cdot\rangle_{W_{\sigma}} be a GG-invariant inner product on WσW_{\sigma} as in Problems 42-46, and let {Φσ,i,j}σ,i,j\{\Phi_{\sigma,i,j}\}_{\sigma,i,j} be the orthonormal basis of VV as before. Then for each f∈Vf\in V,

    f=∑σ,i,j⟨f,Φσ,i,j⟩G.f=\sum_{\sigma,i,j}\langle f,\Phi_{\sigma,i,j}\rangle_{G}.

    Then

    ⟨f1,f2⟩G=∑σ,i,j⟨f1,Φσ,i,j⟩G⁢⟨f2,Φσ,i,j⟩G¯=∑σ∈Irr⁢(G)dim⁡(σ)⁢∑i,j⟨f1,Φσ,i,j⟩G⁢⟨f2,Φσ,i,j⟩G¯dim⁡(σ),\displaystyle\langle f_{1},f_{2}\rangle_{G}=\sum_{\sigma,i,j}\langle f_{1},% \Phi_{\sigma,i,j}\rangle_{G}\overline{\langle f_{2},\Phi_{\sigma,i,j}\rangle_{% G}}=\sum_{\sigma\in\mathrm{Irr}(G)}\operatorname{dim}(\sigma)\sum_{i,j}\frac{% \langle f_{1},\Phi_{\sigma,i,j}\rangle_{G}\overline{\langle f_{2},\Phi_{\sigma% ,i,j}\rangle_{G}}}{\operatorname{dim}(\sigma)},

    so it remains to show that

    tr⁡(f1^⁢(σ)⁢f2^⁢(σ)∗)=∑i,j⟨f1,Φσ,i,j⟩G⁢⟨f2,Φσ,i,j⟩G¯dim⁡(σ).\operatorname{tr}\big{(}\widehat{f_{1}}(\sigma)\widehat{f_{2}}(\sigma)^{*}\big% {)}=\sum_{i,j}\frac{\langle f_{1},\Phi_{\sigma,i,j}\rangle_{G}\overline{% \langle f_{2},\Phi_{\sigma,i,j}\rangle_{G}}}{\operatorname{dim}(\sigma)}.

    Recall that if A:Wσ→WσA:W_{\sigma}\rightarrow W_{\sigma} is a linear operator, A∗:Wσ→WσA^{*}:W_{\sigma}\rightarrow W_{\sigma} is defined through the formula ⟨A∗⁢w1,w2⟩Wσ=⟨w1,A⁢w2⟩Wσ.\langle A^{*}w_{1},w_{2}\rangle_{W_{\sigma}}=\langle w_{1},Aw_{2}\rangle_{W_{% \sigma}}. Using that tr⁢(A⁢B)=tr⁢(B⁢A)\mathrm{tr}(AB)=\mathrm{tr}(BA), we have

    tr⁢(f1^⁢(σ)⁢f2^⁢(σ)∗)=tr⁢(f2^⁢(σ)∗⁢f1^⁢(σ))=∑i⟨f2^⁢(σ)∗⁢f1^⁢(σ)⁢wσ,i,wσ,i⟩Wσ\displaystyle\mathrm{tr}\big{(}\widehat{f_{1}}(\sigma)\widehat{f_{2}}(\sigma)^% {*}\big{)}=\mathrm{tr}\big{(}\widehat{f_{2}}(\sigma)^{*}\widehat{f_{1}}(\sigma% )\big{)}=\sum_{i}\langle\widehat{f_{2}}(\sigma)^{*}\widehat{f_{1}}(\sigma)w_{% \sigma,i},w_{\sigma,i}\rangle_{W_{\sigma}}
    =\displaystyle= ∑i⟨f1^⁢(σ)⁢wσ,i,f2^⁢(σ)⁢wσ,i⟩Wσ=∑i|G|−2⁢∑g,h∈Gf1⁢(g)⁢f2⁢(h)¯⁢⟨σ⁢(g−1)⁢wσ,i,σ⁢(h−1)⁢wσ,i⟩Wσ\displaystyle\sum_{i}\langle\widehat{f_{1}}(\sigma)w_{\sigma,i},\widehat{f_{2}% }(\sigma)w_{\sigma,i}\rangle_{W_{\sigma}}=\sum_{i}|G|^{-2}\sum_{g,h\in G}f_{1}% (g)\overline{f_{2}(h)}\langle\sigma(g^{-1})w_{\sigma,i},\sigma(h^{-1})w_{% \sigma,i}\rangle_{W_{\sigma}}
    =∑i|G|−2⁢∑g,h∈Gf1⁢(g)⁢f2⁢(h)¯⁢⟨σ⁢(h⁢g−1)⁢wσ,i,wσ,i⟩Wσ=∑i|G|−2⁢∑g,h∈Gf1⁢(g)⁢f2⁢(h)¯⁢Φσ,i,i⁢(h⁢g−1).\displaystyle=\sum_{i}|G|^{-2}\sum_{g,h\in G}f_{1}(g)\overline{f_{2}(h)}% \langle\sigma(hg^{-1})w_{\sigma,i},w_{\sigma,i}\rangle_{W_{\sigma}}=\sum_{i}|G% |^{-2}\sum_{g,h\in G}f_{1}(g)\overline{f_{2}(h)}\Phi_{\sigma,i,i}(hg^{-1}).

    We now expand f1,f2f_{1},f_{2} in terms of the matrix coefficients Φ\Phi:

    f1⁢(g)=∑σ′,j′,k′⟨f1,Φσ′,j′,k′⟩G⁢Φσ′,j′,k′⁢(g),f2⁢(h)¯=∑σ′′,j′′,k′′⟨f2,Φσ′′,j′′,k′′⟩G¯⋅Φσ′′,j′′,k′′⁢(h)¯.f_{1}(g)=\sum_{\sigma^{\prime},j^{\prime},k^{\prime}}\langle f_{1},\Phi_{% \sigma^{\prime},j^{\prime},k^{\prime}}\rangle_{G}\Phi_{\sigma^{\prime},j^{% \prime},k^{\prime}}(g),\qquad\overline{f_{2}(h)}=\sum_{\sigma^{\prime\prime},j% ^{\prime\prime},k^{\prime\prime}}\overline{\langle f_{2},\Phi_{\sigma^{\prime% \prime},j^{\prime\prime},k^{\prime\prime}}\rangle_{G}}\cdot\overline{\Phi_{% \sigma^{\prime\prime},j^{\prime\prime},k^{\prime\prime}}(h)}.

    This gives

    tr⁢(f1^⁢(σ)⁢f2^⁢(σ)∗)=\displaystyle\mathrm{tr}\big{(}\widehat{f_{1}}(\sigma)\widehat{f_{2}}(\sigma)^% {*}\big{)}= ∑σ′,σ′′,j′,j′′,k′,k′′⟨f1,Φσ′,j′,k′⟩G⁢⟨f2,Φσ′′,j′′,k′′⟩G¯\displaystyle\sum_{\sigma^{\prime},\sigma^{\prime\prime},j^{\prime},j^{\prime% \prime},k^{\prime},k^{\prime\prime}}\langle f_{1},\Phi_{\sigma^{\prime},j^{% \prime},k^{\prime}}\rangle_{G}\overline{\langle f_{2},\Phi_{\sigma^{\prime% \prime},j^{\prime\prime},k^{\prime\prime}}\rangle_{G}}
    ×∑i|G|−2∑g,hΦσ′,j′,k′(g)Φσ′′,j′′,k′′⁢(h)¯Φσ,i,i(hg−1).\displaystyle\qquad\qquad\times\sum_{i}|G|^{-2}\sum_{g,h}\Phi_{\sigma^{\prime}% ,j^{\prime},k^{\prime}}(g)\overline{\Phi_{\sigma^{\prime\prime},j^{\prime% \prime},k^{\prime\prime}}(h)}\Phi_{\sigma,i,i}(hg^{-1}).

    Starting with the inner sum, by Problems 44 and 45, we have

    1|G|⁢∑h∈GΦσ′′,j′′,k′′⁢(h)¯⁢Φσ,i,i⁢(h⁢g−1)=\displaystyle\frac{1}{|G|}\sum_{h\in G}\overline{\Phi_{\sigma^{\prime\prime},j% ^{\prime\prime},k^{\prime\prime}}(h)}\Phi_{\sigma,i,i}(hg^{-1})= 1|G|⁢∑h∈G⟨σ⁢(h)⁢σ⁢(g−1)⁢wσ,i,wσ,i⟩Wσ⁢⟨σ′′⁢(h)⁢wσ′′,j′′,wσ′′,k′′⟩Wσ′′¯\displaystyle\frac{1}{|G|}\sum_{h\in G}\langle\sigma(h)\sigma(g^{-1})w_{\sigma% ,i},w_{\sigma,i}\rangle_{W_{\sigma}}\overline{\langle\sigma^{\prime\prime}(h)w% _{\sigma^{\prime\prime},j^{\prime\prime}},w_{\sigma^{\prime\prime},k^{\prime% \prime}}\rangle_{W_{\sigma^{\prime\prime}}}}
    ={0if⁢σ′′≠σΦσ,i,j′′⁢(g−1)⁢⟨wσ,i,wσ,k′′⟩Wσ¯dim⁡σif⁢σ′′=σ.\displaystyle=\begin{cases}0\quad&\mathrm{if\;}\sigma^{\prime\prime}\neq\sigma% \\ \frac{\Phi_{\sigma,i,j^{\prime\prime}}(g^{-1})\overline{\langle w_{\sigma,i},w% _{\sigma,k^{\prime\prime}}\rangle_{W_{\sigma}}}}{\operatorname{dim}\sigma}&% \mathrm{if\;}\sigma^{\prime\prime}=\sigma.\end{cases}

    Since ⟨wσ,i,wσ,k′′⟩Wσ=0\langle w_{\sigma,i},w_{\sigma,k^{\prime\prime}}\rangle_{W_{\sigma}}=0 if k′′≠ik^{\prime\prime}\neq i and one if k′′=ik^{\prime\prime}=i, we now have

    tr⁢(f1^⁢(σ)⁢f2^⁢(σ)∗)=\displaystyle\mathrm{tr}\big{(}\widehat{f_{1}}(\sigma)\widehat{f_{2}}(\sigma)^% {*}\big{)}= 1dim⁡σ⁢∑σ′,j′,j′′,k′,i⟨f1,Φσ′,j′,k′⟩G⁢⟨f2,Φσ,j′′,i⟩G¯×|G|−1⁢∑g∈GΦσ′,j′,k′⁢(g)⁢Φσ,i,j′′⁢(g−1)\displaystyle\frac{1}{\operatorname{dim}\sigma}\sum_{\sigma^{\prime},j^{\prime% },j^{\prime\prime},k^{\prime},i}\langle f_{1},\Phi_{\sigma^{\prime},j^{\prime}% ,k^{\prime}}\rangle_{G}\overline{\langle f_{2},\Phi_{\sigma,j^{\prime\prime},i% }\rangle_{G}}\times|G|^{-1}\sum_{g\in G}\Phi_{\sigma^{\prime},j^{\prime},k^{% \prime}}(g)\Phi_{\sigma,i,j^{\prime\prime}}(g^{-1})

    Again using Problems 44 and 45, we have

    |G|−1⁢∑g∈GΦσ′,j′,k′⁢(g)⁢Φσ,i,j′′⁢(g−1)\displaystyle|G|^{-1}\sum_{g\in G}\Phi_{\sigma^{\prime},j^{\prime},k^{\prime}}% (g)\Phi_{\sigma,i,j^{\prime\prime}}(g^{-1}) =|G|−1⁢∑g∈GΦσ′,j′,k′⁢(g)⁢Φσ,j′′,i⁢(g)¯\displaystyle=|G|^{-1}\sum_{g\in G}\Phi_{\sigma^{\prime},j^{\prime},k^{\prime}% }(g)\overline{\Phi_{\sigma,j^{\prime\prime},i}(g)}
    =\displaystyle= ⟨Φσ′,j′,k′,Φσ,j′′,i⟩G={1if⁢(σ′,j′,k′)=(σ,j′′,i)0otherwise.\displaystyle\langle\Phi_{\sigma^{\prime},j^{\prime},k^{\prime}},\Phi_{\sigma,% j^{\prime\prime},i}\rangle_{G}=\begin{cases}1\quad&\mathrm{if\;}(\sigma^{% \prime},j^{\prime},k^{\prime})=(\sigma,j^{\prime\prime},i)\\ 0\quad&\mathrm{otherwise.}\end{cases}

    Entering this into the previous expression of tr⁢(f1^⁢(σ)⁢f2^⁢(σ)∗)\mathrm{tr}\big{(}\widehat{f_{1}}(\sigma)\widehat{f_{2}}(\sigma)^{*}\big{)} gives

    tr⁢(f1^⁢(σ)⁢f2^⁢(σ)∗)=1dim⁡σ⁢∑j′,i⟨f1,Φσ,j′,i⟩G⁢⟨f2,Φσ,j′,i⟩G¯,\mathrm{tr}\big{(}\widehat{f_{1}}(\sigma)\widehat{f_{2}}(\sigma)^{*}\big{)}=% \frac{1}{\operatorname{dim}\sigma}\sum_{j^{\prime},i}\langle f_{1},\Phi_{% \sigma,j^{\prime},i}\rangle_{G}\overline{\langle f_{2},\Phi_{\sigma,j^{\prime}% ,i}\rangle_{G}},

    as desired.

    We now show the inversion formula using the Plancherel formula: let 𝟏g:G→ℂ\mathbf{1}_{g}:G\rightarrow{\mathbb{C}} be the delta function at gg, i.e. 𝟏g⁢(h)=1\mathbf{1}_{g}(h)=1 if h=gh=g and zero otherwise. Then

    f⁢(g)=|G|⁢⟨f,𝟏g⟩G=|G|⁢∑σ∈Irr⁢(G)dim⁡σ⁢tr⁡(f^⁢(σ)⁢𝟏g^⁢(σ)∗)f(g)=|G|\langle f,\mathbf{1}_{g}\rangle_{G}=|G|\sum_{\sigma\in\mathrm{Irr}(G)}% \operatorname{dim}\sigma\,\operatorname{tr}\big{(}\widehat{f}(\sigma)\widehat{% \mathbf{1}_{g}}(\sigma)^{*}\big{)}

    From the definition of 𝟏g^⁢(σ)∗\widehat{\mathbf{1}_{g}}(\sigma)^{*}:

    𝟏g^⁢(σ)∗=1|G|⁢∑h∈G𝟏g⁢(h)⁢σ⁢(h)=1|G|⁢σ⁢(g),\widehat{\mathbf{1}_{g}}(\sigma)^{*}=\frac{1}{|G|}\sum_{h\in G}\mathbf{1}_{g}(% h)\sigma(h)=\frac{1}{|G|}\sigma(g),

    hence

    f⁢(g)=|G|⁢∑σ∈Irr⁢(G)dim⁡σ⁢tr⁡(f^⁢(σ)⁢1|G|⁢σ⁢(g))=∑σ∈Irr⁢(G)dim⁡σ⁢tr⁡(f^⁢(σ)⁢σ⁢(g)).f(g)=|G|\sum_{\sigma\in\mathrm{Irr}(G)}\operatorname{dim}\sigma\,\operatorname% {tr}\big{(}\widehat{f}(\sigma)\frac{1}{|G|}\sigma(g)\big{)}=\sum_{\sigma\in% \mathrm{Irr}(G)}\operatorname{dim}\sigma\,\operatorname{tr}\big{(}\widehat{f}(% \sigma)\sigma(g)\big{)}.