1

I am (still) reading Evasiveness of Graph Properties and Topological Fixed Point Theorems. On page 52 it states:

Suppose that $H$ is a group of order $p^m$. With $m \geq 1$, which acts on $\Delta$ in such a way that $\Delta^h$ is a subcomplex for any $h \in H$. Then $\Delta^H$ is $\mathbb{F}_p$ acylic.

I am assume that $h : \Delta \to \Delta$ where $\Delta$ is a simplicial complex. Thus $\Delta^h$ I believe is just the image of $h$ on $\Delta$. But what is $\Delta^H$? Stating that $\Delta^H$ is acyclic leads me to believe it is a simplicial complex. However, as every $h \in H$ is also a simplicial complex I am lead to believe $\Delta^H$ is a collection of simplicial complexes. What is $\Delta^H$?

Dair
  • 3,064
  • 1
    Is $\Delta^H$ the set of points of $\Delta$ fixed under all $h\in H$? If so it will be the intersection of all $\Delta^h$, and so a subcomplex as all of them are. – Angina Seng Apr 30 '17 at 02:46
  • @LordSharktheUnknown: I'm begining to feel like it might be the union as of page 25 here: https://arxiv.org/pdf/cs/0205031.pdf However, these texts are extremely dense. I could be misreading something haha. – Dair May 01 '17 at 00:56

1 Answers1

1

Here the superscripts denote fixed points, as is common when talking about group actions. Namely, $\Delta^h=\{x\in\Delta:hx=x\}$ denotes the fixed points of the action of $h$ on $\Delta$ and $\Delta^H=\{x\in\Delta:hx=x\text{ for all }h\in H\}$ denotes the fixed points of the action of the entire group $H$ on $\Delta$.

(Note that it definitely would not make sense for $\Delta^h$ to mean the image of $h$, since $h$ must be a bijection since it is part of a group action.)

Eric Wofsey
  • 330,363