Alice led the discussion again this week.  At the end of the session we realised that this will be the last study group for the year due to various people going away for holidays/conferences/meetings in the lead up to christmas and of course the school’s christmas party. We will resume sometime early in the new year.

We first began by discussing the comments on the post on last week’s study group.

Recall from last week how given a building with apartments isomorphic to the Coxeter complex $(W,S)$, we can define the W-distance on chambers. We look again at the chamber system we obtain from the Fano plane.

Let $s_1$ denote differ by a point and $s_2$ denote differ by a line. Let x be the chamber whose line is the orange line and point is the bottom left hand corner of the diagram and y be the chamber whose line is the red line and point is the top of the diagram. Then $\delta(x,y)=s_1s_2=w$. There are four chambers which are neighbours of y. These are $z_1$, which has line the red line and point the point in the middle of the line, $z_2$ which has line the pink line and point the top point,  $z_3$ which has line the orange line and point the top point, and $z_4$ which has line the red line and point the bottom right hand corner.  Then $\delta(y,z_1)=s_1$ and $\delta(x,z_1)=s_1s_2s_1=ws_1$. Also $\delta(y,z_2)=\delta(y,z_3)=s_2$ and $\delta(x,z_3)=s_1=ws_2$ while $\delta(x,z_2)=s_1s_2=w$.  Moroever, $\delta(x,z_4)=ws_1$. Thus we have the property that if $z$ is a chamber such that $\delta(y,z)=s\in S$ then $\delta(x,z)\in\{w,ws\}$ and if $\ell(ws)=\ell(w)+1$,  (where $\ell(g)$ is the length of a reduced expression for g in terms of elements of S), then $\delta(x,z)=ws$.

This motivates the following definition of a building: A building of type $(W,S)$ is a pair $(C,\delta)$ with C a set (whose elements are called chambers) and $\delta:C\times C\rightarrow W$, such that for $x,y\in C$ with $\delta(x,y)=w$ the following hold:

• $w=1$ if and only if $x=y$
• If $z$ is a chamber such that $\delta(y,z)=s\in S$ then $\delta(x,z)\in\{w,ws\}$ and if $\ell(ws)=\ell(w)+1$, then $\delta(x,z)=ws$.
• If $s\in S$ then there exists $z\in C$ such that $\delta(y,z)=s$ and $\delta(x,z)=ws$.

This alternative definition of a building yields a chamber system such that the relations are defined by $x\sim_s y$ if and only if $\delta(x,y)\in\langle s\rangle$.

Alice says that this definition is what she uses in her papers on buildings as this is usually easier to work with than the other two definitions.

Given a subset J of S, the residue of type J, denoted $R_J(c)$, is (using the chamber system definition) the connected component containing c when only taking colours in J. Using the distance W-distance definition, it is the set $\{x\in C\mid\delta(x,c)\in W_J\}$ where $W_J=\langle s\mid s\in J\rangle$.

The usual graph distance between x and y is equal to $\ell(\delta(x,y))$.

Using this definition we have the Gate property: Given $x\in C$ and R a residue, there exists a unique $c\in R$ such that $\ell(x,c)\leq \ell(x,y)$ for all $y\in R$.  Moreover, $\delta(x,y)=\delta(x,c)\delta(c,y)$ and $\ell(x,y)=\ell(x,c)+\ell(c,y)$.  That is there is a  shortest path from x to any $y\in R$ which goes through c. The element c is called the projection of x onto R.

Let $\mathcal{B}$ be a thick building with apartment set $\mathcal{A}$ and let $G\leqslant \mathrm{Aut}(\mathcal{B})$ act strongly transitively on $\mathcal{B}$, that is acts transitively on $\{(A,c)\mid c\in A, A\in \mathcal{A}\}$. Pick one of these pairs $(A_o,c_0)$. Let $B=G_{c_0}$,$N=G_{A_0}$ and $T=B\cap N=G_{A_0,c_0}$, which fixes $A_0$ chamberwise. Then $N/T=G_{A_0}^{A_0}\cong W$. Moreover, $s\in S$ is the reflection of $A_0$ mapping $c_0$ onto its s-neighbour in $A_0$.

Now given $s\in S$ we have $s=nT=Tn$. Moreover, if $nT=n'T$ then $nB=n'B$. Thus $sB$ makes sense and we can also refer to $sBs$, $BsB$ and $BwB$ for any $w\in W$.

Now $G, B, N, S$ have the following properties:

1. $G=\langle B,N\rangle$
2. $T=B\cap N\vartriangleleft N$ and $W=N/T$ admits a set of generators $S$ such that $(W,S)$ is a Coxeter system.
3. $sBs^{-1}\not\subseteq B$ for all $s\in S$, that is, the stabiliser of $c_o^s$ does not stabilise $s_0$. This follows from the thickness of the building.
4. $G=\bigsqcup_{w\in W} BwB$, this is called the Bruhat decomposition of G. Proof: Let $g\in G$. Then there exists an apartment A containing $c_0$ and $c_0^g$. By the strong-transitivity, there exists $b\in B$ mapping $(A,c_0)$ to $(A_0,c_0)$. Let $w=\delta(c_0,c_0^{gb})=\delta(c_0,c_0^g)$. Also $c_0^w=c_0^{gb}$ and so $c_0=c_0^{gbw^{-1}}$, that is $gbw^{-1}\in B$. Hence $gb\in Bw$ and so $g\in BwB$.
5. $P_J$,  the stabiliser of the J-residue, is equal to $\bigsqcup_{w\in W_J} BwB$. Such subgroups are called special subgroups.
6. $BwB BsB=BwsB$ if $\ell(ws)=\ell(w)+1$ and $BwBBsB=BwB\cup BwsB$ if $\ell(ws)=\ell(w)-1$.

Now any group G with subgroups B, N satisfying the above 6 properties (where $P_J$ in 5 is defined as the union instead of as the stabiliser of the J-residue), is called a group with a BN-pair. Equivalently we call $(G,B,N,S)$ a Tits system.

In fact, these 6 axioms are not independent and later it was realised that a group with a BN-pair could be defined by the following:

1. $G=\langle B,N\rangle$
2. $T=B\cap N\vartriangleleft N$ and $W=N/T$ admits a set of generators $S$ such that $sBs^{-1}\not\subseteq B$ for all $s\in S$.
3. $BwBBsB\subseteq BwB\cup BwsB$.

BN-pairs are not unique for a given group G. I suppose an example would be $G=PSp(4,3)=PSU(4,2)$, where the two different interpretations of G as a classical group would give two different buildings for G and so two different BN-pairs. Hopefully Alice can confirm this.

Now starting with a group G with a BN-pair, we can let C be the set of right cosets of B in G.  Then we can define $Bg\sim_s Bg'$ if and only if $P_sg=P_sg'$ where $P_s=B\cup BsB$. (It turns out that the axioms imply that the elements of S are involutions.) This gives a building where one apartment is $A_0=\{Bw\mid w\in W\}$ and the whole set of apartments is $\mathcal{A}=A_0^G$.

Axioms 2 and 3 above imply that $S$ consists only of involutions. Let $s\in S$. First substitute $w=s^{-1}$ in (3). Then $(Bs^{-1}B)(BsB)\subseteq (Bs^{-1}B)\cup B$. Now we can use (2) to see that $(Bs^{-1}B)(BsB)\ne B$ (see what happens otherwise!) and so
$(Bs^{-1}B)(BsB)=(Bs^{-1}B)\amalg B$ (the disjoint union). But we can take inverses of both sides, and we only change the right-hand side:
$(Bs^{-1}B)(BsB)=(BsB)\amalg B$
$Bs^{-1}B=BsB$ and $(BsB)(BsB)=BsB\amalg B$.
Now substitute $w=s$ in (3) and see what happens. Then you end up with
$(BsB)(BsB)=Bs^2B\amalg BsB$. and hence $Bs^2B=B$. Since $BsB\ne B$, it follows that $s$ has order 2 in $W$.