Let $i:H \to G$ be a homomorphism of compact Lie groups. The induced representation $\iota_*V := \mathrm{Map}^H(G,V)$ of an $H$-representation $V$ does not give an element of the representation ring $R(G)$ of $G$ (because $\iota_*V$ is usually infinite-dimensional).

But $\iota_*V$ is an element of a larger group $R.\!(G)$. Here $R.\!(G)$ is defined to be the group of maps from the set $\hat G$ of irreducible representations of $G$ to $\mathbb Z$. The abelian group $R.\!(G)$ is naturally isomorphic to the dual group $\mathrm{Hom}(R(G),\mathbb Z)$.

($\ast$) By Peter-Weyl theorem, $\iota_*V$ is uniquely described as the direct sum $\oplus W_i^{\oplus m_i}$ of irreducible $G$-representations. Here $m_i$ is the multiplicity of $W_i$ and $W_i$ is an irreducible $G$-representation. Then we define $\iota_*V:\hat G \to \mathbb Z$ by $$(\iota_*V)(P) = \begin{cases}m_i&P=W_i \text{ for some } i\\ 0&\text{else}\end{cases}\quad(P \in \hat G).$$ So we obtain $\iota_*:R(H) \to R.\!(G)$.

On the other hand, the pullback of representations via $i$ gives rise to a homomorphism $i^*:R(G) \to R(H)$. Taking its transpose, we obtain a homomorphism $i_*:R.\!(H) \to R.\!(G)$.

Now we can naturally embed $R(H)$ into $R.\!(H)$. (Take the irreducible decomposition of an $H$-representation, then assign to each irreducible $H$-representation its multiplicity.)

Question: Does $\iota_*:R(H) \to R.\!(G)$ agree with $i_*:R(H) \to R.\!(G)$?


  1. This question arose when I was reading the paper "The representation ring of a compact Lie group" by Segal. He does not seem to give an explicit definition of $\iota_*$. (It is unclear at least for me.) I suspect that the above paragraph ($\ast$) is wrong.

  2. I moved the answer which was here below.

  • $\begingroup$ It would be helpful to add the online link at Numdam to this 1968 IHES paper by Segal, who gets his analogue of induction from work of Bott (Annals of Math. 1957). $\endgroup$ Sep 8, 2013 at 16:48
  • $\begingroup$ Your question doesn't strike me as it being of particular importance that it handles about the representation theory of compact Lie groups, but about $\bigoplus\limits_{n \in \mathbb{N}} \mathbb{N}$ and $\prod\limits_{n \in \mathbb{N}} \mathbb{N}$, right? $\endgroup$
    – Marc Palm
    Sep 9, 2013 at 8:10
  • $\begingroup$ @MarcPalm Sorry, I can't get it. Do you mean that there is a wrong definition in my question? $\endgroup$
    – H. Shindoh
    Sep 9, 2013 at 14:14
  • $\begingroup$ No, probavly you should remove the representation theory from the question. I don't see why it matters that you are dealing with the representation theory of compact groups. $\endgroup$
    – Marc Palm
    Sep 10, 2013 at 6:06
  • $\begingroup$ The representation ring of non- compact groups is not semisimple. I recall that there is a paper by Atiyah et al., in which the discrete series of semisimple subgroups are constructed via K-theory $\endgroup$
    – Marc Palm
    Sep 10, 2013 at 6:12

1 Answer 1


I think that I have found an answer to my question. The definition ($\ast$) is not wrong, but not preferable. For a $G$-representation $M$, define $\mu^G_M \in R.\!(G)$ by $$\mu^G_M(N) := \dim \mathrm{Hom}^G(N,M)\qquad(N\ \text{representation of }G)$$ By the universal property, $\mu^G_M$ gives rise to a group homomorphism $R(G) \to \mathbb Z$. Then $\iota_*$ is given by $$(\iota_*V)(M) = \dim \mathrm{Hom}^G(M,\mathrm{Map}^H(G,V)).$$

On the other hand, for $i_*V$ is given by $$(i_*V)(M) = i_*(\mu^H_V)(M) = \mu^H_V(i^*M) = \dim\mathrm{Hom}^H(i^*M,V).$$

Since $\mathrm{Hom}^G(M,\mathrm{Map}^H(G,V))=\mathrm{Hom}^H(i^*M,V)$, $i_*=\iota_*$.


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