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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.

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    $\begingroup$ $f(x)=cx$ doesn't make sense unless $A=B$. $\endgroup$
    – anon
    Aug 6, 2013 at 3:10
  • $\begingroup$ @anon I agree, I have edited the question to only include the first condition. $\endgroup$
    – user77404
    Aug 6, 2013 at 3:44
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    $\begingroup$ 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, 2013 at 4:23

2 Answers 2

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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 $f(g \cdot \alpha) = g \cdot f(\alpha)$.

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  • $\begingroup$ 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, 2013 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, 2013 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, 2013 at 4:14
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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.

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