# 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.

## 2 Comments	»

1. The last chi is missing it’s /.

Very interesting posts and well-written stuff, btw. Thanks.

Comment by Greg Simon | December 4, 2010 | Reply

• Thanks, and thanks.

Comment by | December 4, 2010 | Reply

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