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I was looking at this proposition:

Let $(V,\rho)$ be a representation of a group $G$. If $W$ is a vector space isomorphic to $V$, then it can be turned into a representations of $G$ isomorphic to $V$.

It is interesting for me to see how linearity at level of vector spaces carries naturally over to $G$-linearity. I am actually surprised, so I was wondering whether all my steps are correct. Can anyone double check?

Consider the isomorphism $\varphi\colon V \rightarrow W$ given by hypothesis. Then the map \begin{equation*} \psi \colon \mathrm{GL}(V) \to \mathrm{GL}(W): A \mapsto \varphi \circ A \circ \varphi^{-1} \end{equation*} is an isomorphism. So we can construct a new representation $(W,\sigma)$ of $G$ as follows: \begin{equation*} \sigma \colon G \rightarrow \mathrm{GL}(W): g \mapsto \sigma(g) := \psi (\rho(g)). \end{equation*} This is in fact a representation of $G$: if $g,h \in G$, then \begin{align*} \sigma(gh) & = \psi(\rho(gh)) = \varphi \circ \rho(gh) \circ \varphi^{-1} = (\varphi \circ \rho(g) \circ \varphi^{-1}) \circ (\varphi \circ \rho(h) \circ \varphi^{-1}) \\ & = \psi(\rho(g)) \circ \psi(\rho(h)) = \sigma(g)\circ \sigma(h). \end{align*} Now we can prove that $(V,\rho)$ and $(W,\sigma)$ are isomorphic representations, i.e. that $\varphi$ is actually a $G$-linear map. If $v \in V,g \in G$, then \begin{align*} \sigma(g)(\varphi(v)) & = \psi(\rho(g))(\varphi(v)) = (\varphi \circ \rho(g) \circ \varphi^{-1})(\varphi (v)) \\ & = \varphi(\rho(g) v), \end{align*} which shows the claim.

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  • $\begingroup$ The statement in the current form is not really meaningful. $V$ is a representation of $G$, but $W$ is just a vector space, so it does not make sense to claim that $V$ and $W$ are isomorphic as representations of $G$ (what you show is simply that it is possible to turn $W$ into a representation which is isomorphic to $V$). $\endgroup$ Commented Jul 24, 2018 at 7:48
  • $\begingroup$ Yes I noticed this point. Thanks. $\endgroup$
    – Gibbs
    Commented Jul 24, 2018 at 8:25

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Yeah, that seems right. I mean, you just defined the action on $W$ in the unique way to make $\varphi$ equivariant, which is the natural thing to do.

Maybe think about it this way: If you have a group $G$ and a set $M$, and you have a (set-)bijection $f\colon G \to M$, then there is a unique groupstructure on $M$ such that $f$ becomes a group-isomorphism. You define for $m,n\in M$ $$ m \star n := f(f^{-1}(m) \cdot f^{-1}(n))$$ Indeed, now $(M,\star)$ is a group and you may verify that $f$ is an iso. Are you surprised by this?

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  • $\begingroup$ I agree with you. I was surprised at first, but after thinking for a while it looks fairly natural. Thank you for the feedback. $\endgroup$
    – Gibbs
    Commented Jul 24, 2018 at 7:53
  • $\begingroup$ @Gibbs If you agree with my answer, you could accept it, I guess? $\endgroup$
    – Babelfish
    Commented Jul 26, 2018 at 11:21

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