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](http://www.numdam.org/item?id=PMIHES_1968__34__113_0)" 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.



Representation ring and induced representation
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 $\mathrm{Hom}^G(M,\mathrm{Map}^H(G,V))=\mathrm{Hom}^H(i^\*M,V)$, $i\_\*=\iota\_\*$.


