# The Unapologetic Mathematician

## Dual Frobenius Reciprocity

Our proof of Frobenius reciprocity shows that induction is a left-adjoint to restriction. In fact, we could use this to define induction in the first place; show that restriction functor must have a left adjoint and let that be induction. The downside is that we wouldn’t get an explicit construction for free like we have.

One interesting thing about this approach, though, is that we can also show that restriction must have a right adjoint, which we might call “coinduction”. But it turns out that induction and coinduction are naturally isomorphic! That is, we can show that

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

Indeed, we can use the duality on hom spaces and apply it to yesterday’s Frobenius adjunction:

\displaystyle\begin{aligned}\hom_H(W\!\!\downarrow^G_H,V)&\cong\hom_H(V,W\!\!\downarrow^G_H)^*\\&\cong\hom_G(V\!\!\uparrow_H^G,W)^*\\&\cong\hom_G(W,V\!\!\uparrow_H^G)\end{aligned}

Sometimes when two functors are both left and right adjoints of each other, we say that they are a “Frobenius pair”.

Now let’s take this relation and apply our “decategorifying” correspondence that passes from representations down to characters. If the representation $V$ has character $\chi$ and $W$ has character $\psi$, then hom-spaces become inner products, and (natural) isomorphisms become equalities. We find:

$\displaystyle\langle\psi\!\!\downarrow^G_H,\chi\rangle_H=\langle\psi,\chi\!\!\uparrow_H^G\rangle_G$

which is our “fake” Frobenius reciprocity relation.

December 3, 2010

## (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.

December 3, 2010