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I am learning something about the induced representation of finite groups and trying to prove the Frobenius Reciprocity. One of the versions is:

Suppose $G$ is a finite group and $H < G$ be a subgroup. Let $\left ( \kappa, V \right ) $ be a repn. of $H$ on $V$, and $\left ( \lambda, W \right ) $ be a repn. of $G$ on $W$. Then $$\text{Hom}_{H} \left ( V, \text{Res}_{H}^{G} W\right ) \cong \text{Hom}_{G} \left (\text{Ind}_{H}^{G} V, W\right ) $$ which means two linear spaces are natrually isomorphic to each other.

Of course, as a vector space, $\text{Res}W$ is same as $W$.

I follow the algebraic construction of induced repn. throughout, i.e.

Let $H<G$, $\left ( \kappa, V \right ) $ be as the aforementioned. Let $[G:H] = n$.Define the induced representation of $G$ to be $\left ( \text{Ind}_{H}^{G}\kappa, \text{Ind}_{H}^{G}V \right ) $, s.t. $$\text{Ind}_{H}^{G}V= \bigoplus_{i=1}^n \sigma_i V, \sigma_i = g_i H \text{ is the coset of }H\text{ in }G. $$ and $\text{Ind}_{H}^{G}\kappa (g)$ acts on $\text{Ind}_{H}^{G}V$ by permuting $\sigma_i V$ to $\sigma_{j(i)} V$, and apply $\kappa (h_i)$ on $\sigma_{j(i)} V$, given that $gg_i = g_{j(i)}h_i$, for some $h\in H$.

To show the Reciprocity, we need a (linear) isomorphism from $\text{Hom}_{H} \left ( V, \text{Res} W\right )$ to $ \text{Hom}_{G} \left (\text{Ind}V, W\right )$. Given a map $\phi$ in $\text{Hom}_{H} \left ( V, \text{Res} W\right )$, we have $$ \text{Res}\lambda(h) \circ \phi = \phi \circ \kappa(h), \forall h\in H $$ and I was trying to construct some $\tilde{\phi}$ s.t. $$\lambda (g) \circ \tilde{\phi} = \tilde{\phi} \circ \text{Ind} \kappa (g), \forall g \in G.$$

Is everything OK up to this point? I was trying to restrict $\tilde{\phi}$ to a certain copy of $\sigma V$ as they will determine the map automatically.

Thank you for your reading and look forward to your kind help!

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Yes, that's the right idea. For a given $\phi \in {\rm Hom}_H(V, {\rm Res}_H^GW)$, the correct definition of the corresponding $\tilde \phi \in {\rm Hom}_G({\rm Ind}_H^GV,W)$ is $$ \tilde \phi(\sum_i\sigma_i v_i) = \sum_i \lambda(g_i)(\phi(v_i)).$$

You need to check that this $\tilde \phi$ obeys the condition $\lambda(g) \circ \tilde \phi = \tilde \phi \circ {\rm Ind}\ \kappa(g)$ that you mentioned in your post.

But you also need to check that every $\psi \in {\rm Hom}_G({\rm Ind}_H^GV,W)$ is constructed from some $\phi \in {\rm Hom}_H(V, {\rm Res}_H^GW)$ in this way. [Of course, the only $\phi$ that can possibly work is the restriction of $\psi$ to $\sigma_1V \cong V$, where $\sigma_1$ is the identity coset in $G/H$ represented by the group element $g_1 = e$. You need to show that it does work.]

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  • $\begingroup$ Thank you for your answer. Do you actually mean $\kappa (h_i) (v_i)$ for some $h_i \in H$ s.t. $gg_i = g_{j(i)} h_i$? $\endgroup$
    – S. D. Z
    Commented Mar 8, 2017 at 11:12
  • $\begingroup$ Oh yeah, I will edit it. $\endgroup$
    – Kenny Wong
    Commented Mar 8, 2017 at 11:13
  • $\begingroup$ Would you please specify $\lambda (\sigma_i)$? $\endgroup$
    – S. D. Z
    Commented Mar 8, 2017 at 11:19
  • $\begingroup$ Ah, you're using $\sigma_i$ for the coset. Gotcha. $\endgroup$
    – Kenny Wong
    Commented Mar 8, 2017 at 11:21
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    $\begingroup$ Actually it is. You chose your preferred representative of $\sigma_i$ in the beginning, do you agree? And $\lambda(g)$ is uniquely defined for any $g \in G$. $\endgroup$
    – Kenny Wong
    Commented Mar 8, 2017 at 11:32

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