Advanced Algebra I
Induced representation
In order to introduce the notion of induced representation, we start by looking at some examples.
Example 0.1. Let H < G be a subgroup. Then there is a natural groups action G × G/H → G/H by translation. Hence one has a permutation representation G → C[G/H]. This representation can be described as following:
Let W = CeH where H denote the trivial coset in G/H. Let σW = Ceσ for σ ∈ G/H. Finally, we consider V = ⊕σ∈G/HσW . Then we obtain a representation of G on V = C[G/H].
One notices that this construction works for any W . So let ρ : G → GL(W ) be a representation of H. We consider vector spaces σW := {wσ|w ∈ W }. And we fix a representative sσ ∈ G for each coset σ ∈ G/H. Now let
V := ⊕σ∈G/HσW.
It’s easy to verify that there is a representation ˜ρ : G → V given as following:
For s ∈ G, ssσ = sτh for some h ∈ H and τ ∈ G/H. Then we consider
˜
ρs(wσ) = (ρhw)τ.
A careful reader might notices that ˜ρ depends on choices of represen- tatives sσ. However, different choices give isomorphic representations.
Thus it’s is unique up to isomorphism.
On the other hand, Let ρ : G → GL(V ) be a representation. It’s restriction gives ρH : H → GL(V ) a representation. If θ : H → GL(W ) is a subrepresentation of ρH, then we can find a G-invariant subspace P
σσW , where σW := ρsσW for any representative sσ of σ ∈ G/H.
Definition 0.2. We say that ρ : G → GL(V ) is a representation induced by the representation θ : H → GL(W ) if V = ⊕σ∈G/HσW .
We remark that on V there are now two representations ρ and ˜θ. We will prove that they are isomorphic by the following Lemma.
Lemma 0.3. Suppose that (V, ρ) is induced by (W, θ). Let ρ0 : G → GL(V0) be a representation of G and let f : W → V0 be a linear map such that f (θtw) = ρ0tf (w) for all t ∈ H and w ∈ W . Then there exists a unique linear map F : V → V0 which extends f and satisfying F ◦ ρs = ρ0s◦ F for all s ∈ G.
Proof. We first prove the uniqueness. If F ◦ ρs = ρ0s◦ F for all s ∈ G, then for w ∈ σW , we pick any s ∈ σ ∈ G/H, one has ρs−1w ∈ W and
F (w) = F (ρsρs−1w) = ρ0sF (ρs−1w) = ρ0sf (ρs−1w).
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Hence if F is determined by f . It implies that if F0is another extension, then F0 = F . This proves the uniqueness.
To prove the existence, for w ∈ σW , we define F (w) := ρ0sf (ρs−1w).
We first check that this is well-defined, i.e. independent of choice of s ∈ σ. Suppose we pick s0 = sh for some h ∈ H. Then
ρ0shf (ρ−1shw) = ρ0sρ0hf (ρ−1h ρ−1s w) = ρ0sf (θhρ−1h ρ−1s w) = ρ0sf (ρs−1w).
Since V = ⊕σW , F is thus defined on V . Moreover, it’s clear that
F |W = f . ¤
Corollary 0.4. Suppose that (V, ρ) and (V0, ρ0) are induced by (W, θ).
Then they are isomorphic.
Proof. Let f : W → V0 = ⊕σW be the inclusion. Since θt = ρ0t for t ∈ H, it’s easy to see that f (θtw) = ρ0tf (w) for all t ∈ H and w ∈ W . Then f extends to F : V → V0. Since F |W is an isomorphism, it follows that F |σW is an isomorphism by the construction of F . Therefore, (V, ρ)
and (V0, ρ0) are isomorphic. ¤
Corollary 0.5. Suppose that (V, ρ) is induced by (W, θ). Then ρ ∼= eθ.
Proof. Let ρ0 = ˜θ and V0 = ⊕σW . Then one sees that θ˜t(w) = θt(w) = ρtw
for t ∈ H and w ∈ W . We take f = idW then we are done. ¤ We now ready to compute the character.
Theorem 0.6. Let (V, ρ) be a representation induced by (W, θ). Let R = {sσ} be a system of representatives of G/H. For each u ∈ G, one has
χρ(u) = X
r∈R,r−1ur∈H
χθ(r−1ur) = 1
|H|
X
s∈G,s−1us∈H
χθ(s−1us).
Proof. Since ρ ∼= eθ, we can compute it by eθ and we may assume that ρ = eθ.
We first note that ρu maps σW to uσW , where uσ denote the coset of usσ. We have non-zero trace only at those subspace σW such that σW = uσW . This is equivalent to σ = uσ and hence equivalently, s−1σ usσ ∈ H.
Now, we compute trace on σW with s−1σ usσ = h ∈ H. By definition, θes(wσ) = (θhw)σ. Hence
tr(eθs|σW) = tr(θh) = χθ(s−1σ usσ).
Add them up, then we get the required equality. ¤ A convenience way to compute is
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Exercise 0.7. If C is the conjugacy class of u, and C ∩ H decompose into conjugacy classes D1, ..., Dr of H. Then
χρ(u) = |G|
|H|
Xr i=1
|Di|
|C|χθ(Di).
Example 0.8. Consider S3 < S4. There is a irreducible representation (W, θ) of degree 2 on S3. It induces a representation (V, eθ) of degree 8 on S4. It’s character has value 8, 0, −1, 0, 0 on 1, (12), (123), (1234), (12)(34) respectively. By orthogonal property, one has χeθ= χ3 + χ4+ χ5.
We now fix some notations. Let V be a representation of G and H <
G a subgroup. Then ResV denote the restriction of the representation of G on V to H.
On the other hand, if W is a representation of H, then IndW denotes the induced representation.
If V, V0 are representations of G, then HomG(V, V0) denotes the G- invariant linear transformation from V to V0.
Then the extension theorem can be rephrase as HomH(W, ResV0) ∼= HomG(IndW, V0).
Corollary 0.9 (Frobenius Reciprocity). If V is a representation of G and W is a representation of H < G, then
< χIndW, χV >G=< χW, χResV>H .
Proof. It suffices to prove this when W and V are irreducible. Note that < χIndW, χV >G is the number of copies of V appears in the decomposition of IndW, which equals to dim(HomG(IndW, V)). And similarly, < χW, χResV >H is the number of copies of W appears in the decomposition of ResV, which equals to dim(HomH(W, ResV)). ¤
