The Unapologetic Mathematician

Mathematics for the interested outsider

(Real) Frobenius Reciprocity

Now we come to the real version of Frobenius reciprocity. It takes the form of an adjunction between the functors of induction and restriction:

\displaystyle\hom_H(V,W\!\!\downarrow^G_H)\cong\hom_G(V\!\!\uparrow_H^G,W)

where V is an H-module and W is a G-module.

This is one of those items that everybody (for suitable values of “everybody”) knows to be true, but that nobody seems to have written down. I’ve been beating my head against it for days and finally figured out a way to make it work. Looking back, I’m not entirely certain I’ve ever actually proven it before.

So let’s start on the left with a linear map f:V\to W that intertwines the action of each subgroup element h\in H\subseteq G. We want to extend this to a linear map from V\!\!\uparrow_H^G to W that intertwines the actions of all the elements of G.

Okay, so we’ve defined V\!\!\uparrow_H^G=\mathbb{C}[G]\otimes_HV. But if we choose a transversal \{t_i\} for H — like we did when we set up the induced matrices — then we can break down \mathbb{C}[G] as the direct sum of a bunch of copies of \mathbb{C}[H]:

\displaystyle\mathbb{C}[G]=\bigoplus\limits_{i=1}^nt_i\mathbb{C}[H]

So then when we take the tensor product we find

\displaystyle\mathbb{C}[G]\otimes_HV=\left(\bigoplus\limits_{i=1}^nt_i\mathbb{C}[H]\right)\otimes_HV\cong\bigoplus\limits_{i=1}^nt_iV

So we need to define a map from each of these summands t_iV to W. But a vector in t_iV looks like t_iv for some v\in V. And thus a G-intertwinor \hat{f} extending f must be defined by \hat{f}(t_iv)=t_i\hat{f}(v)=t_if(v).

So, is this really a G-intertwinor? After all, we’ve really only used the fact that it commutes with the actions of the transversal elements t_i. Any element of the induced representation can be written uniquely as

\displaystyle v=\sum\limits_{i=1}^nt_iv_i

for some collection of v_i\in V. We need to check that \hat{f}(gv)=g\hat{f}(v).

Now, we know that left-multiplication by g permutes the cosets of H. That is, gt_i=t_{\sigma(i)}h_i for some h_i\in H. Thus we calculate

\displaystyle gv=\sum\limits_{i=1}^ngt_iv_i=\sum\limits_{i=1}^nt_{\sigma(i)}h_iv_i

and so, since \hat{f} commutes with h and with each transversal element

\displaystyle\begin{aligned}\hat{f}(gv)&=\sum\limits_{i=1}^n\hat{f}(t_{\sigma(i)}h_iv_i)\\&=\sum\limits_{i=1}^nt_{\sigma(i)}h_i\hat{f}(v_i)\\&=\sum\limits_{i=1}^ngt_i\hat{f}(v_i)\\&=g\hat{f}\left(\sum\limits_{i=1}^nt_iv_i\right)\\&=g\hat{f}(v)\end{aligned}

Okay, so we’ve got a map f\mapsto\hat{f} that takes H-module morphisms in \hom_H(V,W\!\!\downarrow^G_H) to G-module homomorphisms in \hom_G(V\!\!\uparrow_H^G,W). But is it an isomorphism? Well we can get go from \hat{f} back to f by just looking at what \hat{f} does on the component

\displaystyle V=1V\subseteq\bigoplus\limits_{i=1}^nt_iV

If we only consider the actions elements h\in H, they send this component back into itself, and by definition they commute with \hat{f}. That is, the restriction of \hat{f} to this component is an H-intertwinor, and in fact it’s the same as the f we started with.

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December 3, 2010 - Posted by | Algebra, Group theory, Representation Theory

2 Comments »

  1. […] proof of Frobenius reciprocity shows that induction is a left-adjoint to restriction. In fact, we could use this to define […]

    Pingback by Dual Frobenius Reciprocity « The Unapologetic Mathematician | December 3, 2010 | Reply

  2. […] branching rule down, proving the other one is fairly straightforward: it’s a consequence of Frobenius reciprocity. Indeed, the branching rule tells us […]

    Pingback by The Branching Rule, Part 3 « The Unapologetic Mathematician | January 31, 2011 | Reply


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