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The following appears in my notes:


Suppose $V$ and $V'$ are vector spaces over a field $F$. Let $G$ be a group, and let $\rho : G \to GL(V)$ and $\rho' : G \to GL(V')$ be representations of $G$. Let $\phi : V \to V'$ be a linear map. We say $\phi$ is a $G$-homomorphism if $ \rho'(g)\circ\phi = \phi\circ\rho(g)$ for all $g$ in $G$. We also say that $\phi$ intertwines $\rho$ and $\rho'$.

$\mathrm{Hom}_G(V,V')$ is the $F$-space of all of these.


It seems strange to me that $\mathrm{Hom}_G(V,V') $ should be independent of $\rho, \rho'$ as the notation suggests. If $\rho$ and $\rho'$ are both the trivial representation, then $\mathrm{Hom}_G(V,V') $ contains all linear maps $V \to V'$. So, does "$\mathrm{Hom}_G(V,V')$" only make sense in the context of specified $\rho, \rho'$, or am I missing something?

Thanks

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    It is only independent in the sense that $V$ and $V^{\prime}$ are $FG$-modules, rather than just vector spaces. The $FG$-module structure implicitly carries an action by $G$ as linear transformations on the underlying $F$-vector spaces. This is equivalent to specifying $\rho$ and $\rho^{\prime}.$2012-01-16
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    Writing $\hom_G((V,\rho),(V',\rho'))$, or anything like that, very rapidly becomes annoying and useless.2012-01-16
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    @GeoffRobinson I think my definition of an $FG$-module is a bad one (in fact, it's not really a definition). I'd appreciate a link to a decent one online2012-01-16

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