Induction of representation commutes with dual

Question

Suppose we have finite groups $H \leq G$ and a $\mathbb{K}[H]$ module $V$. I would like to say that $\operatorname{Ind}_H^G(V^*) \cong (\operatorname{Ind}_H^G(V))^*$.

If I have good hypothesis on $\mathbb{K}$ (a.e char does not divide the cardinality of $G$ )I know to do it with Frobenius reciprocity,but otherwise I have not really idea of how to do it.

Answer 1

You don't need any assumptions about the characteristic.

Let $g_1,\ldots,g_n$ be a set of left coset reps for $H$ in $G$. Here's a map $(kG\otimes_{kH} V)^* \to kG\otimes_{kH}V^*$: $$ \Phi(f) = \sum_i g_i \otimes_{kH}(v \mapsto f(g_i\otimes v)).$$ You can check that $\Phi$ is a $G$-map as follows: fix $x \in G$ and note that three are elements $h_i \in H$ and a permutation $\sigma \in S_n$ such that $$ x^{-1}g_i = g_{\sigma(i)} h_i.$$ Now $$ \begin{align} \Phi(xf) & = \sum_i g_i \otimes (v \mapsto (xf)(g_i\otimes v)) \\ &= \sum_i g_i \otimes (v \mapsto f(x^{-1}g_i \otimes v)) \\ &= \sum_i g_i \otimes (v \mapsto f(g_{\sigma(i)} \otimes h_iv)) \\ &= \sum_i g_i \otimes h_i^{-1} (v \mapsto f(g_{\sigma(i)} \otimes v)) \\ &= \sum_i g_i h_i^{-1} \otimes v \mapsto f(g_{\sigma(i)} \otimes v) \\ &= \sum_i x g_{\sigma(i)} \otimes v \mapsto f(g_{\sigma(i)} \otimes v) \\ &= x \sum_i g_{\sigma(i)} \otimes v \mapsto f(g_{\sigma(i)} \otimes v) \\ &= x \Phi(f). \end{align}$$

For a map in the other direction, for $g \in G$ define $\delta_g \in kG^*$ by $\delta_g(x) = 1$ if $x=g$ and 0 otherwise (a bimodule isomorphism between $kG$ and $kG^*$), then define $$\Psi(g \otimes \alpha) = x \otimes v\mapsto \sum_{h \in H}\delta_g(xh)\alpha(h^{-1}v).$$

You need to check that $\Psi$ is well-defined and that it is inverse to $\Phi$ (then it is automatically an $G$-homomorphism, so you don't need to check that). I'm not going to write out the well-definedness checks here, but for inverses: $$\begin{align} \Psi(\Phi(f)) &= \Psi(\sum_i g_i \otimes (v \mapsto f(g_i\otimes v)) \\ &= x\otimes w \mapsto \sum_i \sum_h \delta_{g_i}(xh)f(g_i\otimes h^{-1}w) \end{align}. $$ The delta function is nonzero iff $g_i=xh$, so $x^{-1}g_i=h$. Those are the only values of $h$ and $i$ making a nonzero contribution to the sum, and you get $$ f(g_i \otimes h^{-1}w) = f(xh \otimes h^{-1}w) = f(x\otimes w).$$ Therefore $\Psi \Phi = 1$.

Answer 2

Here's a sketch of a proof by universal properties, not using any assumption on the characteristic :

$\hom (W, (\mathrm{Ind} V)^*)\cong \hom (W\otimes \mathrm{Ind}V, k) \cong\hom (\mathrm{Ind}(\mathrm{Res}W\otimes V), k) \cong \hom (\mathrm{Res}W\otimes V, \mathrm{Res}k) \cong \hom(\mathrm{Res}W, V^*)\cong \hom(W, \mathrm{CoInd}(V^*)) $

So we get (since they have the same universal property) $(\mathrm{Ind}V)^* \cong \mathrm{CoInd}(V^*)$, but for $[G:H]<\infty$, $\mathrm{CoInd}\cong \mathrm{Ind}$ so we do get $\mathrm{Ind}(V^*)\cong (\mathrm{Ind}V)^*$ (you can see that m_t_ also used the hypothesis $[G:H]<\infty$ in their answer by choosing $g_1,...,g_n$ representatives)

I'll leave it to you to fill in the gaps: in particular what does $\hom$ mean at each stage, what structure does $k$ have, which tensor product is it but these should be straightforward. The only nonstraightforward part is the isomorphism $W\otimes \mathrm{Ind}(V) \cong \mathrm{Ind}(\mathrm{Res}(W)\otimes V)$ but it can be done using universal properties as well.