# What is the Frobenius Reciprocity theorem saying?

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.

The definition of $$Ind^G_H(V)$$:

Source of the images: Sepanski - Compact Lie Groups

• Do you know what a (categorical) adjunction is? Sep 6, 2021 at 8:24
• @AlexWertheim Yes, if you mean pairs of adjoint functors... Sep 6, 2021 at 8:24
• @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… Sep 6, 2021 at 9:12
• @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. Sep 6, 2021 at 10:30
• @TerezaTizkova - Note that Frobenius reciprocity is "usually" formulated as a natural isomorphism $Hom_H(W, U_H) \cong Hom_G(Ind^G_H(W),U)$. Sep 6, 2021 at 11:10

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.