4
$\begingroup$

Suppose $V$ and $W$ are both representations of a group $G$, where $V$ and $W$ are $k$-vector spaces. Define $\mathrm{Hom}(V,W)$ to be the space of $k$-linear maps $V \to W$. My notes say that:

$\mathrm{Hom}_G(V,W) = \{ \phi \in \mathrm{Hom}(V,W): g \phi = \phi \}$, and we have a linear projection $\mathrm{Hom}(V,W) \to \mathrm{Hom}_G(V,W)$ given by $ \displaystyle \phi \mapsto \frac{1}{|G|} \sum_{g \in G} g \phi$

I'm confused by this. $g \phi = \phi$ isn't enough for $\phi$ to be a $G$-linear map, is it? But yet the projection map fixes those maps that satisfy $g \phi = \phi$, which suggests that this isn't a mistake.

Thanks

  • 6
    $g \phi$ needs to be understood as the natural action of $G$ on $\text{Hom}(V, W)$, which isn't completely obvious; it sends the map $x \mapsto \phi(x)$ to the map $x \mapsto g \phi(g^{-1} x)$.2012-02-18
  • 0
    @QiaochuYuan Great, thanks (I'm missing the previous page, which I now assume introduces this action).2012-02-18
  • 2
    Another way of writing the natural action of G on Hom(V,W) is $(g \phi)(g x) = g\,\phi(x)$, which maybe looks "more natural". Or, if instead of writing $\phi(x)$ we write $\langle \phi, x \rangle$ then it becomes $\langle g \phi, g x \rangle = g \langle \phi, x \rangle$ (i.e., "g preserves $\langle \,, \rangle$", although we don't actually have a form here.)2012-02-18
  • 0
    @QiaochuYuan Can you also convert this question to an answer? Again posting to [the chat](http://chat.stackexchange.com/rooms/9141/the-crusade-of-answers)?2013-06-09

1 Answers 1