A partial linear space of order $(s,t)$ is a geometry of points and lines such that every pair of distinct points lie on at most one line, there are $s+1$ points on every line, and there are $t+1$ lines through any point. We also require a non-degeneracy condition such as we don’t have all the points incident with just one line, and $s,t>1$. Now an antiflag is a point $P$ and line $\ell$ that are not incident, and the antiflag numbers for a partial linear space are the possible values for the number of lines on $P$ that are concurrent with $\ell$. For example, a projective plane of order $q$ is a partial linear space of order $(q,q)$ with single antiflag number $q+1$. A generalised quadrangle has just one antiflag number, and it is $1$, whereas a polar space of rank at least $3$ has antiflag numbers $1$ and $s+1$.

Now if there is a single antiflag number $\alpha$, then the partial linear space is said to be a partial geometry. The collinearity graph of a partial geometry is the undirected graph whose vertices are the points, and they are adjacent if they are collinear. It is not difficult to prove that the collinearity graph of any partial geometry is strongly regular; that is, it is a regular graph and there are two constants $\lambda$ and $\mu$ such that for any pair of adjacent (resp. non-adjacent) vertices, there are $\lambda$ (resp. $\mu$) common neighbours.

The antiflag condition “1 or $s+1$” is the defining axiom of a polar space, and this is a highly nontrivial result of Buekenhout and Shult (1974). Moreover, we know that a polar space of rank at least 3 is “classical” by a theorem of F. D. Veldkamp and J. Tits, and so the collinearity graph of such a partial linear space must have a rank 3 automorphism group. It then follows that the collinearity graph of a polar space is strongly regular; the rank 2 case is simpler as then we have generalised quadrangles and they are partial geometries.

So we know that the collinearity graphs of polar spaces are strongly regular, but is there a simpler proof of this fact?

Well, if we have something like line-transitivity then it can be done easily, but I don’t know how to prove that the number of planes on a line is constant, and this is what is needed. First we know that the collinearity graph is regular of degree $s(t+1)$ by considering the points lying on the $t+1$ lines on any point. We show now that there is a $\mu$ value. Consider two points $X$ and $Y$ that are not collinear. Then given a line $\ell$ on $Y$, there are $1$ or $s+1$ points on $\ell$ collinear with $X$. But since one of the points on $\ell$, namely $Y$, is not collinear with $X$, we see that there is a unique point on $\ell$ collinear with $X$. Since this is true of all $t+1$ lines on $Y$, it follows that for every pair of non-adjacent vertices, there are $\mu:=t+1$ vertices adjacent to both.

Now suppose we have two distinct points $X$ and $Y$ lying on a common line $\ell$. There are two types of lines on $X$ other than $XY$: (i) those whose only point collinear with $Y$ is $X$, and (ii) those for which every point is collinear with $Y$. It suffices to count the number $\beta$ of the latter type of line, since then we would arrive at a constant number of common neighbours to two adjacent vertices, and this constant would be $s-1 + s\beta$. Another way to describe these lines of type (ii) would be that they span a plane with $XY$. That is, if we join up all of the points on these two lines and then consider the points lying on these lines, we end up with a singular subspace (in the language of point-line geometries), and in particular, a projective plane of order $s$. We then take all of the singular subspaces that contain $XY$. All that remains is to show that the number of planes on a line is constant …

## 6 thoughts on “The collinearity graph of a polar space”

1. Ferdinand Ihringer says:

Nice! A counting problem!

I solved it if I understand the problem correctly. Sadly I already thought that I solved it at least three times since I read your post yesterday, but now I am confident that my solution is correct. Do you still need a prove? Can I use some kind of LaTeX in comments?

2. Thanks Ferdinand. As promised by email, here is your solution to my problem! It’s so simple and elegant, I can’t believe I over-looked it.

Fix a point $P$. For a line $\ell$, define
$\lambda_\ell:=\{X\in\mathcal{P}: X\text{ is collinear with }P\text{ and }X\ne P\}$.
Now double count the ordered pairs $(Q,R)$ where
$Q\in \ell$ but $Q\ne P$,
$R$ is not collinear with $P$
$Q$ and $R$ are collinear.

There are $s$ possibilities for $Q$, and once we have chosen $Q$, there are $k-\lambda_\ell-1$ possibilities for $R$.
On the other hand, there are $n-k-1$ possibilities for $R$, and once we’ve chose $R$, there is a unique point $Q$ on $\ell$ collinear with $R$ (it can’t be “all” since $P$ is not collinear with $R$). Therefore,
$s(k-\lambda_\ell-1)=n-k-1$
and we see that $\lambda_\ell$ is independent of $\ell$. Therefore,
$\lambda=k-1-\frac{n-k-1}{s}$.

3. A natural question arising from the fact that the parameter $\lambda$ of the collinearity graph depends on $n$ is as follows: Are there two polar spaces with the same order $(s, t)$ but different number of vertices?

Based on the classification of polar spaces, it looks like the answer is no, but I don’t know how to show it axiomatically/combinatorially.

1. As we discussed this before privately, so I will put the one comment, which I still had, here … It is easy to see that for generalized quadrangles (GQ) or finite classical polar spaces the above is true. A slightly more elaborate calculation shows that there is also no GQ with the same parameters $(s, t)$ as a finite classical polar space (if I did this correctly). But in this case at least one interesting thing happens: A GQ with parameters $(2, 2(2^{2d-2}-1)/3$ and a parabolic quadric of rank d over $\mathbb{F}_2$ have the same number of points. Of course the Krein condition $t \leq s^2$ implies that there is no GQ of such order for $d > 2$, but this is not completely trivial.

The above example works for general $q$ and other polar spaces.

I wanted to emphasize that (1) one can maybe use Krein conditions in proving your conjecture and (2) any combinatorial argument which does not use the Krein conditions, might be particularly interesting.

1. Nice observation. The bound t 2 that has an order (s, t), satisfies t > s^2.

2. The previous comment got messed up for some reason. Here is what I was saying:
“The bound $t \leq s^2$ can be proved combinatorially. Perhaps that proof can help us here. Also, can we combinatorially show that for any abstract polar space of rank $> 2$, if it has an order $(s, t)$ then either $t > s^2$ or $s > t^2$?