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I am studying representations of Lie groups and I still cannot find the intuition behind Frobenius Reciprocity theorem.

Why would group homomorphisms from representation of $G$ to the $\mathrm{Ind}^G_H(V)$ be the same as group homomorphisms from the same representation restricted on $H$to $V$?

I think my problem is that I cannot visualize the $\mathrm{Ind}^G_H(V)$ (definition below) and hence don´t understand what role it plays in the relation.

Source of the image: Sepanski - Compact Lie Groups

The definition of $Ind^G_H(V)$: enter image description here

Source of the images: Sepanski - Compact Lie Groups

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    $\begingroup$ Do you know what a (categorical) adjunction is? $\endgroup$ Commented Sep 6, 2021 at 8:24
  • $\begingroup$ @AlexWertheim Yes, if you mean pairs of adjoint functors... $\endgroup$ Commented Sep 6, 2021 at 8:24
  • $\begingroup$ @AlexWertheim but is there any evidence as to why we should expect $-\vert_H$ to have an adjoint? Just out of curiosity, I don’t know about representation theory… $\endgroup$ Commented Sep 6, 2021 at 9:12
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    $\begingroup$ @PrudiiArca $\operatorname{Res}^G_H$ is an exact functor $\operatorname{Rep} G\to\operatorname{Rep}H$, so it should have both left and right adjoints. $\endgroup$ Commented Sep 6, 2021 at 10:30
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    $\begingroup$ @TerezaTizkova - Note that Frobenius reciprocity is "usually" formulated as a natural isomorphism $Hom_H(W, U_H) \cong Hom_G(Ind^G_H(W),U)$. $\endgroup$
    – hm2020
    Commented Sep 6, 2021 at 11:10

1 Answer 1

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Question: "I am studying representations of Lie groups and I still cannot find the intuition behind Frobenius Reciprocity theorem."

Answer: Let $i:H \rightarrow G$ be an inclusion of arbitrary groups and let $W$ be a left $H$-module and $U$ a left $G$ module (let $W,U$ be $k$-vector spaces with $k$ a field). If you consider the group algebras $k[H], k[G]$ and view $W,V$ as modules over the group algebras, the above mentioned FR says there is a natural isomorphism

$$F:Hom_{k[H]}(W, U_{k[H]}) \cong Hom_{k[G]}(k[G]\otimes_{k[H]} W, U).$$

with inverse $F'$.

Given any map of $k[H]$-modules $\phi: W \rightarrow U_{k[H]}$ you get an induced map

$$F(\phi):k[G]\otimes W \rightarrow U$$

defined by $F(\phi)(x\otimes w)=x\phi(w) \in U$. Conversely given a map of $k[G]$-modules

$$\psi: k[G]\otimes W \rightarrow U$$

there is an induced map

$$F'(\psi): W \rightarrow U$$

defined by

$$F'(\psi)(w):=\psi(1\otimes w).$$

You may check that $F,F'$ are inverses of each other and that the above is an isomorphism for any $H,G,W,U$. With this formulation, the FR theorem becomes a statement about Hom and tensor products of modules over associative rings. The proof is straight forward: You must verify that the maps $F,F'$ defined above satisfy $F \circ F' = F' \circ F = Identity$ - their compositions are the "identity map".

Note: You can formulate a similar statement in terms of the universal enveloping algebras $U(Lie(H)), U(Lie(G))$ of the Lie groups $H,G$: There is a "natural isomorphism"

$$F:Hom_{U(Lie(H))}(W, U_{U(Lie(H))}) \cong Hom_{U(Lie(G))}(U(Lie(G)\otimes_{U(Lie(H))} W, U).$$

In fact for any map of associative $k$-algebras $\rho: A \rightarrow B$ with $W\in Mod(A), V\in Mod(B)$, there is a "natural isomorphism"

$$F(A;B): \text{ } Hom_A(W,V_A) \cong Hom_B(B\otimes_A W, V).$$

Frobenius reciprocity for $H \subseteq G$ is a special case of this isomorphism $F(A,B)$ - it is a statement about a relation between $Hom$ and $\otimes$ for modules over associative rings. When you take "derived functors" you get similar relations.

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