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?


1 Answer 1


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 $(g\cdot f(g'))=f(g^{-1}g').$

Even in this setting, the statement of Frobenius reciprocity holds.

  • $\begingroup$ Can you please provide a reference for that / proof for that? $\endgroup$
    – SimB4
    Commented Apr 19, 2023 at 5:05
  • $\begingroup$ I think you can find it in "Compact Lie groups" by Sepanski. $\endgroup$ Commented Apr 19, 2023 at 8:25
  • $\begingroup$ thanks for the reply, corresponding chapter in the book is very well written for me. $\endgroup$
    – SimB4
    Commented Apr 20, 2023 at 6:02

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