# Frobenius reciprocity for reductive groups

Let $$G$$ be finite group and $$H$$ its subgroup. We assume that everywhere the base field is $$\mathbb{C}$$. For any characters $$\phi$$ of $$G$$-representation and $$\psi$$ of $$H$$-representation Frobenius reciprocity states that: $$\langle Ind^G_H(\psi), \phi \rangle_G = \langle \psi, Res^G_H(\phi) \rangle_H \qquad (1)$$ In particular, this means that multiplicity of $$G$$-representation in induction of $$H$$ is equal to the multiplicity of $$H$$-representation in the restriction of $$G$$-representation (correct me please if I am wrong somewhere).

Now, let $$G$$ be reductive Lie group and $$H$$ its Lie subgroup. Since any $$G$$-representation is also an $$H$$-representation, we may define restriction of $$G$$-representation to subgroup $$H$$, which will decompose into direct sum of $$H$$-irreducibles with some multiplicities.

Question: Is there an induction process that builds $$G$$-representation by given $$H$$-representation (1)? Does this works?

Yes, there is. Given a representation $$\rho: H\to GL(V)$$ you can define $$Ind_H^G(\rho)$$ to be the space of continuous functions $$f:G\to V$$ such that $$f(gh) = \rho(h^{-1})f(g)$$ for $$h\in H$$ and $$g\in G$$.
$$G$$ acts on this space via $$(g\cdot f(g'))=f(g^{-1}g').$$