Representation ring and induced representation

Question

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 $i_*:R(H) \to R.\!(G)$?

Notes:

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.

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