1
$\begingroup$

Suppose we have two finite vector spaces A and B. Consider a finite group G that acts on A and B via linear transformations. Assume when G acts on A and B there exists only the trivial subspaces that are invariant with respect to the action.

We want to show that if $f: A \rightarrow B$ is equivariant under the group action then either $f$ is an isomorphism or f(x) = 0, $ \forall x \in A $

Solution Attempt: They tell us only the trivial subspace can be invariant with respect to group action. Now from linear algebra we know ker($f$) is a subspace of A, and to show its invariant subspace of A we need to show ker($f$) gets mapped to an element in A. But what I am wondering is where do group actions come into play here? Because after we have show ker($f$) is invariant then we know there will only be two possibilities for it and can conclude things about f.

$\endgroup$
3
  • 2
    $\begingroup$ $f(x)=cx$ doesn't make sense unless $A=B$. $\endgroup$
    – anon
    Aug 6 '13 at 3:10
  • $\begingroup$ @anon I agree, I have edited the question to only include the first condition. $\endgroup$
    – user77404
    Aug 6 '13 at 3:44
  • 1
    $\begingroup$ But the first condition by itself is incorrect; an equivariant map need not be the zero map. The content of Schur's lemma is that the vector space $\hom_G(A,B)$ of equivariant maps $A\to B$ is one-dimensional (if the scalar field is algebraically closed, which allows eigenvalues). In particular if $A=B$ then every equivariant map is multiplication by some scalar. All of this is still under the assumptions that $A$ and $B$ are irreducible, that is have no proper nontrivial invariant subspace. $\endgroup$
    – anon
    Aug 6 '13 at 4:23
2
$\begingroup$

The reason you can conclude things about $f$ is from studying the group action. The very statement "the kernel of $f$ is invariant" is a statement about the group action. Furthermore, to show this statement is true, one must use the fact that $f$ is $G$-equivariant. Without it, all hope is lost.

As anon mentions, $f(x) = cx$ doesn't (usually) make sense. I believe you meant that $f$ is an isomorphism, and possibly thinking about the case $A = B = \mathbb{R}$.

Edit:

To show the claim, that $\ker f$ is invariant under the $G$-action, suppose that $\alpha \in \ker f$. Then we want to show that $g \cdot \alpha$ is also in the $\ker f$. This follows immediately from $f$ being $G$-equivariant by writing $f(g \cdot \alpha) = g \cdot f(\alpha)$.

$\endgroup$
3
  • $\begingroup$ I have edited the question and took out f(x) = cx. Could you start me of on how we would show the kernel of f is invariant? We are assuming in the question f is G-equivariant. $\endgroup$
    – user77404
    Aug 6 '13 at 3:46
  • $\begingroup$ Your edit makes the question wrong. It is also possible (as Chris explicitly points out) that $f$ is an isomorphism, i.e. that $\ker f = 0$. $\endgroup$
    – RghtHndSd
    Aug 6 '13 at 4:10
  • $\begingroup$ There must be a mistake in the question then. But at least I learned more about Group actions and how to show ker(f) is equivariant. Thanks. $\endgroup$
    – user77404
    Aug 6 '13 at 4:14
1
$\begingroup$

Here's a counterexample. Let $A=B=\mathbb C$ and $G=\{1,i,-1,-i\}$, so that $G$ acts on $A$ and $B$ by multiplication. Define $f:A\to B$ by $f(z)=iz$. Then $f$ is equivariant under the actions of $G$ on $A$ and $B$, but $f$ is not trivial. Moreover, if we think of $\mathbb C$ as a 2-dimensional real vector space, $f$ is not multiplication by a scalar.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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