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I'm currently studying Chapter 3 of the book "Introduction to compact transformation groups" by Bredon. The main matter that is discussed in this chapter are simplicial actions of a finite group $G$ over a simplicial complex $K$ (i.e., group actions of $G$ over the set of vertices of $K$, such that for all $g \in G$, the map $v \mapsto g \cdot v$ is a simplicial map.)

The author considers the following property about a simplicial action:

"If $g_0, g_1, \ldots, g_n$ are elements of $G$ and $\{ v_0, \ldots, v_n \}$ and $\{ g_0 \cdot v_0, \ldots, g_n \cdot v_n\}$ are both simplices of $K$, then there exists an element $g$ of $G$ such that $g \cdot v_i = g_i \cdot v_i$ for all $i$."

He then proves that for any simplicial action $G \curvearrowright K$, the induced action over the second barycentric subdivision of $K$ satisfies the previous property, and then goes on to define a regular simplicial action as one such that for all subgroups $H \leq G$, the induced (by restriction) action $H \curvearrowright K$ satisfies the property.

I have already reached a point where he actually uses that the definition involves all subgroups, in order to be able to make some inductive argument. Question is: is it clear that the fact that $G$ satisfies the property does not imply that all of its subgroups do? The book makes no comment about it, so at first you don't quite understand why the definition is made that way. Can someone find a counterexample that shows the necessity of the "all subgroups" condition?

Note. One can easily prove that the induced action on $K''$ is actually regular with this definition, so in practice it's not a big deal if this is really stronger or not, just curiosity...

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Maybe this will work.

Consider a 2-simplex $K = [v_1,v_2,v_3]$ and the action of the symmetric group $G = S_3$ on $K$ taking $\sigma \cdot v_i = v_{\sigma(i)}$. For $H$, take the alternating subgroup $A_3 \cong \mathbb{Z}/3$. Then although there are elements of $H$ taking $v_1 \mapsto v_1$, taking $v_2 \mapsto v_3$, and taking $v_3 \mapsto v_2$, there's no element of $H$ simultaneously doing all these things.

This action doesn't meet the criterion above, but as you point out, if we take the first barycentric subdivision $L$ and let $G$ act on $L$ before, then the action of $G$ does satisfy the criterion, though the action of $H$ does not, so the $G$ action is not a regular simplicial action.

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  • $\begingroup$ I think that example does not satisfy the property for $G$ either. In fact, the author proves that the property implies the following weaker condition (which is not satisfied in your example): "If $v \in K$ and $g \in G$ are such that $v$ and $g \cdot v$ belong to the same simplex, then $g \cdot v = v$." $\endgroup$ – Matías May 11 '14 at 1:20
  • $\begingroup$ It seems clear that the $S_3$ on one 2-simplex example doesn't satisfy this new condition you've stated, but it does seem to satisfy the original property you stated above. Perhaps I just don't understand the condition properly. $\endgroup$ – jdc May 12 '14 at 3:37
  • $\begingroup$ I think the problem is that $v_0, \ldots, v_n$ are not required to be distinct. In fact, to prove that this implies the weaker condition, one just takes $v_0=v_1=v$, $g_0=1$, $g_1=g$. $\endgroup$ – Matías May 13 '14 at 4:39
  • $\begingroup$ That's ... confusing to me; suppose $g_0 = 1$ and $g_1 \cdot v = v' \neq v$; then how can any $g$ take $[v,v] = [v]$ to $[v,v']$? $\endgroup$ – jdc May 13 '14 at 22:47
  • $\begingroup$ That's the point of it. There cannot be such a $g$, therefore, it has to be $g \cdot v = v$. This is why the condition I stated in my post implies the weaker condition that I stated in my first comment. $\endgroup$ – Matías May 14 '14 at 0:20

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