Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
$g \phi$ 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 $x \mapsto g \phi(g^{-1} x)$. – Qiaochu Yuan Feb 18 '12 at 18:33
@QiaochuYuan Great, thanks (I'm missing the previous page, which I now assume introduces this action). – Matt Feb 18 '12 at 18:41
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.) – Ted Feb 18 '12 at 18:49
@QiaochuYuan Can you also convert this question to an answer? Again posting to the chat? – Julian Kuelshammer Jun 9 '13 at 19:34

This CW answer intends to remove the question from the unanswered queue.

As Qiaochu Yuan remarked in the comments the action of $g$ on homomorphisms is as follows $$(g\phi)(x):=g\phi(g^{-1}x)$$ Now if you have $g\phi=\phi$ then you have $(g\phi)(x)=g\phi(g^{-1}x)=\phi(x)$ for all $x$. Multiplying this by $g^{-1}$ from the left you get $\phi(g^{-1}x)=g^{-1}\phi(x)$. This is obviously the same as $\phi$ being in $\operatorname{Hom}_G(V,W)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.