Coset Complexes in Finite Groups

Poset and Homotopy

Let $(P,⩽)$ be a poset. The order complex $\Delta P$ is defined as the set of all finite chains in $P$.

Example. For a set $P:=\{a,b,c,d\}$ with $a>b>c$ and $a>d$, we have that

$$\Delta P=\left\{\{a,b,c\},\{a,d\},\{a,b\},\{a,c\},\{b,c\},\{a\},\{b\},\{c\},\{d\},\varnothing \right\}.$$

where $\{a\}$ is called a $0$-chain and $\{a,b,c\}$ is called a $2$-chain.

Homotopy

Abstract Complex

We say $K$ is a complex over a set $V$, if the following holds:

• $K\subseteq 2^V$
• $\{v\}\in K$ for all $v\in V$
• $A\in K$ and $B\subseteq A$ yields that $B\in K$.

Simplicial Complex

See 5 Simplicial complexes.

So there is a natural map $\Delta P\to |\Delta P|$ from $n$-chain to $n$-simplex.

For example, see here.

Homotopy Equivalence

Let $f,g\in C(X,Y)$. If there exists $H:X\times I\to Y$ such that for any $x\in X$, $H(x,0)=f(x)$ and $H(x,1)=g(x)$, then $f$ and $g$ are homotopy, which is denoted by $f\simeq g$.

Let $\phi :X\to Y$ be a continuous map. We say that another continuous map $\psi :Y\to X$ is a homotopy inverse for $\phi$ if $\psi \circ \phi \simeq {\mathrm{id}}_X$ and $\phi \circ \psi \simeq {\mathrm{id}}_Y$. If there exists a homotopy inverse for $\phi$, then $\phi$ is called a homotopy equivalence and $X$ is homotopy equivalent to $Y$.

Poset Map

Definition

We say $f:\mathcal{P}\to \mathcal{Q}$ is a poset map if $x⩽y$ in $\mathcal{P}$ implies $f(x)⩽f(y)$, where $\mathcal{P}$ and $\mathcal{Q}$ are posets.

Then $f$ induces a map $\Delta (f):\Delta \mathcal{P}\to \Delta \mathcal{Q}$ by $\{a_1,\cdots ,a_n\}\mapsto \{f(a_1),\cdots ,f(a_n)\}$. Furthermore, it induces $|\Delta (f)|:|\Delta \mathcal{P}|\to |\Delta \mathcal{Q}|$ naturally.

Proposition

Let $\mathcal{P}$ and $\mathcal{Q}$ be two posets. Suppose that $f,g:\mathcal{P}\to \mathcal{Q}$ are two poset maps. Then:

• $f$ induces a continuous map $|\Delta f|:|\Delta \mathcal{P}|\to |\Delta \mathcal{Q}|$.
• If $f(x)⩽g(x)$ for all $x\in \mathcal{P}$, then $|\Delta f|$ and $|\Delta g|$ are homotopic, simply denoted by $f\simeq g$.

Quillen's Fiber Theorem

For $z\in \mathcal{Q}$, define ${\mathcal{Q}}_{⩽z}:=\{y\in \mathcal{Q}:y⩽z\}$.

Let $\mathcal{P}$ and $\mathcal{Q}$ be two posets. Let $f:\mathcal{P}\to \mathcal{Q}$ be a poset map such that $f^{-1}({\mathcal{Q}}_{⩽z})$ is contractible for all $z\in \mathcal{Q}$. Then $f$ induces a homotopy equivalence of $\mathcal{P}$ with $\mathcal{Q}$.

Brown-Quillen Complexes

Define a set $S_p(G)=\{1\ne H⩽G:H\text{mbox}isap\text{mbox}-group\}$. Then we get a poset $(S_p(G),\subseteq )$.

• (Brown) $\tilde{\chi }(\Delta S_p(G))\equiv 0\mathrm{mod}|G{|}_p$, where $\tilde{\chi }=\chi -1$ and $\chi$ is the Euler characteristics.
  • example: for some simple groups, there is $\tilde{\chi }(\Delta S_p(G))=\pm |G{|}_p$. (DS. Smith, Subgroup complexes)

Sylow

For any $P,Q\in {\mathrm{Syl}}_p(G)$ with $P\ne Q$, suppose that $P\cap Q=1$. Define $f:S_p(G)\to {\mathrm{Syl}}_p(G)$. Then by ^b9d401, $\Delta S_p(G)$ and $\Delta {\mathrm{Syl}}_p(G)$ are homotopy equivalent.

Hence $\tilde{\chi }(\Delta S_p(G))=\tilde{\chi }({\mathrm{Syl}}_p(G))\equiv 0\mathrm{mod}|G{|}_p$.

Question

When does $\tilde{\chi }(S_p(G))=0$?

Quillen(?)‘s conjecture: $G$ is a finite group. Then $|\Delta S_p(G)|\simeq \ast$ iff $O_p(G)\ne 1$.

The direction ← can be proved by ^50889b.

For example, consider $S_2(\mathrm{SL}(2,q))$ and $S_2(\mathrm{SL}(2,q)/Z(\mathrm{SL}(2,q)))$.