Degree of induced representations

Posted: February 23, 2011 in Representations of Finite Groups
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Note. We will keep all the notation and hypothesis given in this Definition.

Lemma 1. If W is a left \mathbb{C}[H]-module and g \in G, then g \mathbb{C}[H] \otimes_{\mathbb{C}[H]} W \cong W as \mathbb{C}-vector spaces.

Proof.  Clearly both g\mathbb{C}[H] \otimes_{\mathbb{C}[H]} W and W are \mathbb{C}-vector spaces. Define \varphi_1 : g \mathbb{C}[H] \times W \longrightarrow W and \psi_1 : W \longrightarrow g \mathbb{C}[H] \otimes_{\mathbb{C}[H]} W by \varphi_1(gr,w)=rw and \psi(w)=g \otimes_{\mathbb{C}[H]}w, for all r \in \mathbb{C}[H] and w \in W. Since \varphi_1 is \mathbb{C}[H]-balanced, it induces an abelian group homomorphism \varphi : g \mathbb{C}[H] \otimes_{\mathbb{C}[H]} W \longrightarrow W defined by \varphi(gr \otimes_{\mathbb{C}[H]}w)=rw, for all r \in \mathbb{C}[H] and w \in W. Obviously \varphi is also a \mathbb{C}-linear and \varphi \psi and \psi \varphi are identity maps. So \varphi is a \mathbb{C}-linear isomorphism. \Box

Lemma 2. Let [G:H]=m and suppose that \{g_1H, g_2H, \cdots , g_mH \} are the set of left cosets of H in G. Then \mathbb{C}[G] = g_1 \mathbb{C}[H] \oplus g_2 \mathbb{C}[H] \oplus \cdots \oplus g_m \mathbb{C}[H], as right \mathbb{C}[H]-modules.

Proof. By definition, every element of \mathbb{C}[G] is uniquely written as a finite \mathbb{C}-linear combination of elements of G and every element of G is in g_i H \subset g_i \mathbb{C}[H], for some 1 \leq i \leq m. So \mathbb{C}[G]=\sum_{i=1}^m g_i \mathbb{C}[H]. To show that this sum is direct, suppose that

\sum_{i=1}^m g_i u_i = 0, \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

for some u_i \in \mathbb{C}[H]. We need to prove that u_i=0, for all i. To do that, let H=\{h_1, \cdots , h_k \} and u_i=\sum_{j=1}^k c_{ij} h_j, where 1 \leq i \leq m and c_{ij} \in \mathbb{C}. Then

0 = \sum_{i=1}^m g_iu_i = \sum_{i,j}c_{ij}g_ih_j. \ \ \ \ \ \ \ \ \ \ \ (2)

Now if in (2), g_ih_j = g_rh_s, for some i,j,r,s, then g_i \in g_rH and hence i=r, because the cosets are disjoint. So h_j=h_s, i.e. j=s. Thus, in (2), the elements g_ih_j are pairwise distinct and hence, since every element of \mathbb{C}[G] is uniquely written as a \mathbb{C}-linear combination of elements of G, we get c_{ij}=0, for all i,j. Therefore u_i=0 for all i. \ \Box

Theorem. If \rho: H \longrightarrow \text{GL}(W) is a representation of H, then \deg \text{Ind}_H^G \rho = [G:H] \deg \rho.

Proof.  Let [G:H]=m and suppose that \{g_1H, g_2H, \cdots , g_mH \} are the set of left cosets of H in G. Recall that tensor product distributes over direct sum. Thus by Lemma 2

\mathbb{C}[G] \otimes_{\mathbb{C}[H]} W \cong \bigoplus_{i=1}^m (g_i \mathbb{C}[H] \otimes_{\mathbb{C}[H]} W)

and hence by Lemma 1

\deg \text{Ind}_H ^G \rho = \dim_{\mathbb{C}} (\mathbb{C}[G] \otimes_{\mathbb{C}[H]} W) = \sum_{i=1}^m \dim_{\mathbb{C}} W = \sum_{i=1}^m \deg \rho = m \deg \rho. \ \Box

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