Here is the next instalment in our Phan systems study group which was held yesterday.  I forgot to bring my laptop  so I took notes by hand. Blogging the study group has already had one benefit as Gordon is now attending. Having been initially put off by the title he has realised that we are actually doing lots of stuff that he is interested in learning.

John began the discussion by working through the geometry of the symplectic polar space of rank 2.  Let $J=\left(\begin{array}{cc} 0_2 &I_2\\ -I_2& 0_2\end{array}\right)$ and let $\beta$ be the bilinear form on $V=K^4$, for some field K, given by $\beta(v,w)=vJw^T$. Note we are using row vectors, and if $v=(v_1,v_2,v_3,v_4)$, $w=(w_1,w_2,w_3,w_4)$ we have $\beta(v,w)=v_1w_3-v_3w_1+v_2w_4-v_4w_2$. Then $\beta$ is an alternating form, that is $\beta(v,w)=-\beta(w,v)$ and $\beta(v,v)=0$ for all $v,w\in V$. The totally isotropic subspaces of V, are those subspaces W for which the restriction $\beta_{\mid W}$ is zero, that is, $\beta(v,w)=0$ for all $v,w\in W$. We can then form a point-line geometry whose points are the totally isotropic 1-spaces of V and the lines are the totally isotropic 2-spaces of V.  Note that all 1-spaces of V are totally isotropic, but V does not contain any totally isotropic 3-spaces. A hyperbolic pair is a pair of vectors $(e,f)$ such that $\beta(e,f)=1$.

Let $e_1=(1,0,0,0), e_2=(0,1,0,0), f_1=(0,0,1,0)$ and $f_2=(0,0,0,1)$. Then $(e_1,f_1)$ and $(e_2,f_2)$ are hyperbolic pairs, while $\langle e_1,e_2\rangle,\langle f_1,f_2\rangle,\langle e_1,f_2\rangle$ and $\langle e_2,f_1\rangle$ are totally isotropic. We have $V=\langle e_1,f_1\rangle\perp\langle e_2,f_2\rangle$, that is, V can be written as an orthogonal direct sum of two hyperbolic lines. One of the chambers in this geometry is $\langle e_1\rangle\subset \langle e_1,e_2\rangle$.

The point-line geometry we have just defined is an example of a generalised quadrangle. That is, it satisfies the following two axioms:

• Any two points lie on at most one line.
• Given  a line $\ell$ and a point p not incident with $\ell$, there is a unique point q on $\ell$ which is incident with p.

Such a geometry is often referred to as a geometry of type $B_2$  and is denoted by the diagram

In general, we can define an alternating bilinear form on $K^{2n}$ by using the matrix $\left(\begin{array}{cc} 0_n& I_n\\-I_n&0_n\end{array}\right)$. The set of all totally isotropic subspaces of $K^{2n}$ forms the symplectic polar space of rank n.  The maximal totally isotropic subspaces have dimension n and $K^{2n}$ can be written as an orthogonal direct sum of n hyperbolic lines. This geometry has the diagram $B_n$:where the nodes of the diagram represent from left to right the totally isotropic 1-spaces, totally isotropic 2-spaces through to the totally isotropic n-spaces. Recall that the edge between nodes i and j represents the geometry formed by the residue of a flag containing elements of all types except i and j. So for example, if we take a flag $F=U_3\subset U_4\subset \cdots\subset U_n$ where each $U_i$ is a totally isotropic i-space then the residue of F is the set of all totally isotropic 1-spaces and 2-spaces incident with F. Since $U_3$ is totally isotropic, this is all the 1-spaces and 2-spaces contained in $U_3$ and hence is a projective plane. This is why a single edge is placed between the nodes representing 1-spaces and 2-spaces. At the other end of the diagram, if we take a flag $F'=U_1\subset U_2\subset \cdots \subset U_{n-2}$ with each $U_i$ totally isotropic of dimension i, then the residue of $F'$ is the set of all totally isotropic n-spaces and $(n-1)$-spaces containing $U_{n-2}$. Since $U_{n-2}^\perp=\{v\in V\mid \beta(v,u)=0 \text{ for all } u\in U_{n-2}\}$ has dimension $2n-(n-2)=n+2$ and $U_{n-2}^\perp/U_{n-2}$ has dimension 4 and is isomorphic to our original symplectic polar space of rank 2, we have a double bond between the last two nodes in the diagram.

John mentioned that whereas a geometry with the same diagram as a projective space is indeed a projective space, a geometry with the diagram $B_n$ is either a symplectic polar space or the Neumaier geometry. Hopefully we learn more about this is subsequent weeks.

[added19/10/09: John has pointed out that I have misunderstood him here. The $B_2$ diagram is used to denote any generalised quadrangle, and a geometry with diagram $B_n$ is either a polar space or the Neumaier geometry. There are polar spaces other than the symplectic one with diagram $B_n$. This is something we should be covering later.]

Now onto the related groups.  The symplectic group $\mathrm{Sp}(4,K)$ is the group of all linear transformations of $V=K^4$ which preserves $\beta$. This is all matrices $A\in \mathrm{GL}(4,K)$ such that $AJA^T=J$. Writing A as $\displaystyle{ \left(\begin{array}{cc} A_1&A_2\\A_3&A_4\end{array}\right)}$ with each $A_i$ a $2\times 2$ matrix we see that we must have $A_3^TA_1-A_1^TA_3=0_2= A_4^TA_2-A_2^TA_4$ and $A_4A_1^T-A_3A_2^T=I_2$.

A Borel subgroup B is the stabiliser of a chamber. Using our chamber $\langle e_1\rangle\subset\langle e_1,e_2\rangle$ we have that B is all matrices in $\mathrm{Sp}(4,K)$ with $A_1$ lower triangular, $A_2=0$ and $A_4=A_1^{-T}$, the inverse-transpose of $A_1$.

A frame is $\mathcal{F}=\{\langle e_1\rangle,\langle e_2\rangle,\langle f_1\rangle,\langle f_2\rangle\}$. Then N is the setwise stabiliser of $\mathcal{F}$ and so is a group of monomial matrices. The group N must preserve the partition $\{\{\langle e_1\rangle,\langle f_2\rangle,\{\langle e_2\rangle,\langle f_2\rangle\}$ and so $N^{\mathcal{F}}=C_2\mathrm{ wr } C_2$. The base group of this wreath product is generated by the matrices

$\displaystyle{\left(\begin{array}{cccc} 0&0&1&0\\ 0&1&0&0\\1&0&0&0\\ 0&0&0&1\end{array}\right), \left(\begin{array}{cccc} 1&0&0&0\\ 0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\end{array}\right)}$.

These matrices interchange $\langle e_1\rangle$ and $\langle f_1\rangle$, and $\langle e_2\rangle$ and $\langle f_2\rangle$  respectively. The top group is given by all matrices

$\displaystyle{\left(\begin{array}{cc}P&0_2\\ 0_2&P^{-T}\end{array}\right)}$

where P is a $2\times 2$ permutation matrix. The kernel T of the action of N on $\mathcal{F}$ is all diagonal matrices of the form

$\displaystyle{\left(\begin{array}{cccc} \lambda_1&0&0&0\\ 0&\lambda_2&0&0\\ 0&0&\lambda_1^{-1}&0\\ 0&0&0&\lambda_2^{-1}\end{array}\right)}$.

So the Weyl group of $\mathrm{Sp}(4,K)$ is $N/T\cong C_2\mathrm{ wr }C_2$.

This can all be generalised to $\mathrm{Sp}(2n,K)$ where we have that the Weyl group is $N/T\cong C_2\mathrm{ wr }S_n$, the Coxeter group of type $B_n$. This is also thought of as the group of signed permutations of the set $\{ \pm1,\pm2,\ldots,\pm n\}$.

To finish off, Alice spoke briefly about the apartments in this setting. Whereas I had initially interpreted her as saying that in the $A_n$ case an apartment was just the same as a frame, it is actually more subtle than that. In fact an apartment is the set of all flags which you can get from the elements of a frame. Thus in the case of $\mathrm{Sp}(4,K)$, the apartment corresponding to $\mathcal{F}=\{\langle e_1\rangle,\langle e_2\rangle,\langle f_1\rangle,\langle f_2\rangle\}$ is the set of all flags whose elements are in $\mathcal{F}$.This can be represented in the following picture where the vertices are 1-spaces and the edges are 2-spaces. Each vertex-edge incident pair is a rank 2 flag.

The chamber system of this apartment is then given below. I have only labelled three of the nodes.The red edges represent that the 1-spaces of the chamber are different while the green edges represent that the 2-spaces of the chamber are different. You can define reflections $s_1,s_2$ of the apartment (the square picture above) such that $s_1$ is the reflection about the vertical axis of the square and $s_2$ is the reflection about the diagonal axis from the top left corner.  Then starting in the bottom left hand corner of the hexagon and working clockwise we get the sequence $1,s_2,s_2s_1,s_2s_1s_2,s_2s_1s_2s_1$ which maps the bottom left chamber around to the top right hand corner chamber. Working anticlockwise we get $1,s_1,s_1s_2,s_1s_2s_1,s_1s_2s_1s_2$ around to the top right hand corner. Since the diagram should commute we get $s_2s_1s_2s_1=s_1s_2s_1s_2$ and so $(s_1s_2)^4=1$, that is we get the Coxeter relation for groups of type $B_2$.

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1. Alice Devillers says:

One rectification:
the other polar spaces (quadratic, hermitian) are also of type $B_n$.

2. Neumaier’s $A_7$-geoemtry is an example of a geometry which has the $B_3$ (or more accurately, the $C_3$) diagram. Take the set {1,2,3,4,5,6,7} and consider the 30 Fano planes we can make out of this set (take an orbit of one under $S_7$). These projective planes break into two orbits of size 15 each under $A_7$). Now we define a geometry by…

POINTS: {1,2,3,4,5,6,7}
LINES: 3-element subsets of {1,2,3,4,5,6,7}
PLANES: One orbit of size 15 on Fano planes (under $A_7$).

Incidence is simply induced by set membership, and we obtain a rank 3 geometry with the residues giving us a $C_3$ geometry. To see this, if we fix one PLANE, that is, a Fano plane, the POINTS and LINES contained in it are the points and lines of the Fano plane. So this residue is clearly a projective plane. If we fix a POINT, we have 15 LINES containing it and every PLANE containing it. So we obtain a rank 2 residue with 15 elements of one type and 15 elements of the other type. We in fact obtain the smallest thick generalised quadrangle W(2).