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Let $G$ be a finite group, and $H\subseteq G$ a subgroup. Let $F$ be a field. Let $W$ be a finite-dimensional $F[H]$-module. Let $T$ be a left transversal of $H$ in $G$. Then we can define:

$W^G=\sum_{t\in T}t\otimes W$

which is the $F[G]$-module induced from $W$. I'm following the notation used in I.M. Isaacs' "Character Theory of Finite Groups" (see Chapter 6).

This induced module "mixes" the $F$-representation afforded by $W$, and the permutation representation given by the action by $G$ on the left cosets of $H$ via left multiplication. I say this because $G$ acts on the subspaces $\{t\otimes W\}_{t\in T}$ by permuting them (one might call this the "coarse" action, since we are considering the action at the level of subspaces and not at the level of individual elements of the module).

Is there a more general operation we can perform on modules, which "mixes" any given pair of $F$-representations in some analogous way?

Clarification:

Is there an operation $f$ that can "mix" any pair $(W,V)$, where $W$ is a $F[H]$-module and $V$ is a $F[G]$-module, and yield an $F[G]$-module $f(W,V)$, such that $f(W,V)=W^G$ in the special case that $V$ is the permutation module given by the action by $G$ on the cosets of $H$?

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  • $\begingroup$ I've looked it up but it isn't quite what I'm looking for. I have appended a clarification to the main post. $\endgroup$ Commented Jun 13, 2023 at 10:42
  • $\begingroup$ Certainly you'd like a functor; next decide if you want a left or right exact operation. Depending on your choice, you will want tensor or hom with a bimodule. (I also recommend keeping track of left vs right modules). How about W is a right module, V is a left module, and we tensor $W \otimes_H B \otimes_G V$. Then plug in bimodules $H$ and $G$ for $W$ and $V$ to get a formula for $B$. You get that $B$ should be whatever $f(H, G)$ is where the bimodule structure comes from functoriality of $f$ applied to multiplication on the side opposite the action. Probably you just want $B={}_HG_G$! $\endgroup$ Commented Jun 14, 2023 at 11:55

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I'm not sure this answers your question, but here is a general procedure:

Fix any representation $W$ of $G\times H$. Then for any representation $V$ of $H$, the hom space $\hom_H(W,V)$ carries a $G$-action, and hence produces a functor $H\mathrm{-Rep}\to G\mathrm{-Rep}$ (here, note that I do not require $H\subset G$). Some examples:

  • Suppose $H\subset G$, and let $W=\mathbb C[G]$, which carries a $G\times H$-action, by making $G$ act by left multiplication and $H$ act by right multiplication. Then $\hom_H(\mathbb C[G],V)$ is exactly the induced representation $V^G$.
  • Suppose you have a homomorphism $\varphi\colon G\to H$, and let $W=\mathbb C[H]$, which carries a $G\times H$-action, as above (explicitly, $(g,h)\cdot x:=\varphi(g)xh^{-1}$). Then the $G$-action on $\hom_H(\mathbb C[H],V)=V$ is the composition $G\xrightarrow{\varphi}H\xrightarrow\pi\mathrm{GL}(V)$, where $\pi$ is the representation of $H$. In particular: if $G\subseteq H$, then this is the restriction, and if $G\to H=G/N$ is a quotient, then this is the inflation.

  • A bit different in flavor, but Deligne-Lusztig theory provides a similar construction. Letting $X_n$ be the variety $(\det(t_i^{q^{j-1}})_{i,j})^{q-1}=(-1)^{n-1}$, the étale cohomology $H_c^i(X_n,\overline{\mathbb Q}_\ell)$ carries an action of $\mathrm{GL}_n(\mathbb F_q)\times\mathbb F_{q^n}^\times$, which allows the construction of a representation of $\mathrm{GL}_n(\mathbb F_q)$ from a character of $\mathbb F_{q^n}^\times$.
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  • $\begingroup$ Sorry, in the first example, could you explicitly describe the action by $G\times H$ on $\mathbb{C}[G]$? Did you perhaps mean to say: for $x\in\mathbb{C}[G]$, the action is given by $x\cdot (g,h) = (x\cdot g)\cdot h$? But this doesn't define an action... $\endgroup$ Commented Jun 14, 2023 at 2:12
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    $\begingroup$ Sorry, there was a typo: $G\times H$ acts on $G$ by $(g,h)\cdot x:=gxh^{-1}$. This naturally extends to a $G\times H$-action on $\mathbb C[G]$. $\endgroup$ Commented Jun 14, 2023 at 2:42
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    $\begingroup$ I'm sure you know this, the third example isn't that much different, as changing the variety to GLn/B and the multiplicative group to $\mathbb{F}^\times _q {}^n$ gives a case of the first example $\endgroup$ Commented Jun 14, 2023 at 8:37

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