Welcome to MathOverflow

Question

2

Let $G$ be a group and $X$ a set equipped with a transitive right $G$-action. Further, let $c: X\to X$ be a $G$-equivariant map. Is it true that $\text{Stab}(x) = \text{Stab}(c(x))$ for all $x\in X$?

This doesn't seem to be an interesting mathoverflow question on its own, but the reason I ask is the following: In Hovey's book on Model Categories Hovey proves some sort of five lemma for pointed model categories (Thm.6.5.3). In the end of the proof, he seems to conclude from $\alpha\text{Stab}(x)\alpha^{-1}\subset\text{Stab}(x)$ that equality holds; I don't understand how this can be done. The above statement comes from a try to fix this gap.

flag	asked Jan 16 2010 at 10:26 Hanno Becker 1,886●7●13

1

have you read the proof in the version which is available on google books? there he seems to eleborate the argument. – Martin Brandenburg Jan 16 2010 at 13:34

When hovey states his claim, does he say it's true for any $\alpha$? Because then it's a very easy exercise to verify that it makes sense, since if we have a containment relation between those two, we can also see that the inverse conjugate contains Stab(x), but since it's true for $\alpha$ and $\alpha^{-1}$, we have equality. – Harry Gindi Jan 16 2010 at 14:13

@Martin: The point is that I do not understand why $\alpha^{-1}$ should stabilize $\text{Stab}(x)$. Can you explain it to me? @Harry: Sorry, I should've said that $\alpha$ is some fixed element. Of course, if the relation would hold for all $\alpha$, then it'd be easy. – Hanno Becker Jan 16 2010 at 14:41

t3suji · Accepted Answer · 2010-01-16 13:49:00Z UTC

7

That's essentially the same as this question. So the answer is negative: if $X=H\backslash G$, the stabilizer of $Ha\in X$ is $a^{-1}Ha$; if $b$ is any element such that $bHb^{-1}\subset H$, the map $$Ha\mapsto Hba$$ is $G$-equivariant; if $bHb^{-1}\ne H$, we get a counterexample.

edited Jan 16 2010 at 13:49

answered Jan 16 2010 at 13:43

t3suji
2,717●1●8●13

Thank you very much, that's a nice counterexample! But this also means that Hovey's proof is still unclear to me. – Hanno Becker Jan 16 2010 at 14:44

Equivariant map preserves stabilizer

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.