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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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To expand on what user Geoff Robinson has said, this is really only a notational suppression. It is completely analogous to saying given two real vector spaces $V$ and $W$ with scalar multiplication maps $\mu:\mathbb{R}\times V\to V$ and $\nu:\mathbb{R}\times W\to W$ a linear transformation $T:V\to W$ is an abelian group map such that $T$ intertwines (in the same sense) with $\mu$ and $\nu$. That said, you would probably not have said that the notation $\text{Hom}_\mathbb{R}(V,W)$ is ambiguous. The difference is that when we deal with vector spaces we suppress the multiplication maps in lieu of just writing a 'dot'. So, $\mu(r,v)$ gets replaced by $r\cdot v$ in which case a linear map is just an abelian group map such that $T(r\cdot v)=r\cdot T(v)$ for all $v$. If you decided given two representations $\rho:G\to\text{GL}(V)$ and $\psi:G\to\text{GL}(W)$ the operations $g\cdot v=\rho(g)(v)$ and $g\cdot w=\psi(g)(w)$ then an element of $\text{Hom}_G(V,W)$ is nothing more than a vector space map such that $T(g\cdot v)=g\cdot T(v)$ for all $g$.

The other way to think about representations are as elements of the functor category $\mathbf{Vect}_F^G$ in which case it actually seems more appropriate to write $\text{Hom}(\rho,\psi)$ but this is historically inconsistent.

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    I'd like to point out that the OP is already using such a notational suppression in asking the question: he talked about a group $G$ but suppressed its structure maps $m, ^{-1}$...2012-01-17