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I am trying to study (finite) spherical buildings from a very combinatorial point of view : Every rank 3 spherical building is a finite simplicial complex of dimension 3, so one can define its density as the ratio #triangles/#vertices, expressed as a function of the number $n$ of vertices. Maybe it is the wrong name for a very classical notion but I did not find it anywhere. One of the first distinctions in buildings is the concept of thick buildings. In a way, this density measures the thickness of a given building, but I never saw this notion.

I am wondering what is the density of the spherical buildings, and which one are the densest ?

Starting from the classification of spherical buildings, it is enough to compute the density of the buildings of type $A_3$ and $C_3$. I know how to do it for $A_3$, since the building in this case is the flag complex of a projective space on $\mathbb{F}_q^4$, and it seems to give $\frac{[4]_q!}{\binom{4}{2}_q}=O(q^2)=O(\sqrt{n})$ where $n$ is the number of vertices, with the usual q-analog notations.

But $C_3$ looks scary to me, as there are multiple possible geometries, and I don't know of a systematic way to extract the (simple) combinatorial information out of the groups. Is this something well known, or is there a simple way to do this ?

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You write $\frac{[4]_q!}{\binom{4}{2}_q}\in O(n^{3/2})$, where I guess the $n$ should be a $q$. But when I compute this quotient, I get $q^2+2q+1 \in O(q^2)$... Huh? –  Max Horn Jun 25 '13 at 11:27
    
You are right of course, I got messed up between several edits. It should be correct now. –  Arnaud Jun 25 '13 at 13:01
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1 Answer

up vote 3 down vote accepted

While I have never seen this notion of "density" of a finite building before, let's see what we can do...

A thick finite $C_3$ building corresponds to a thick finite rank 3 polar space. These are well-known, they are $W(5,q)$, $Q(6,q)$, $Q^-(7,q)$, $H(5,q^2)$ and $H(6,q^2)$, corresponding to the groups $Sp(6,q)$, $O(7,q)$, $O^-(8,q)$, $U(6,q)$ and $U(7,q)$.

To compute the densities, I took a look at a table indicating the number of points contained in these spaces as well as the number of points collinear to a single fixed point (in my case, I looked at the book Diagram Geometry by Buekenhout and Cohen, p. 485; there are other, older sources, but I already had that book sitting next to me... Google just showed me another source here).

From this and the knowledge how many points are contained in a line and a plane (namely $q+1$ and $q^2+q+1$), it is a matter of simple combinatorics to count the total number of points, lines and planes, as well as chambers (= maximal flags = triples of a point, line and plane containing each other). Each point, line and plane corresponds to a vertex in the simplicial complex, while the chambers correspond to triangles. Thus the "density" as you defined it can be computed via the formula $$ \frac{\#chambers}{\#points + \#lines + \#planes}.$$

Plugging in the numbers of points and perp sizes for all of the $C_3$ buildings listed above, the "density" turns out to be in $O(q^3)$ in all cases (modulo mistakes on my side resp. in the small GAP program I wrote to compute all of this). So it is a big larger than for projective spaces, where I confirmed it to be in $O(q^2)$.

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Thank you very much, I do not know much about polar spaces and missed a good reference. Would you happen to have an asymptotic bound on the density in terms of the number of vertices, i.e. n=#points+#lines+#planes ($q$ is not a very informative measure) ? I will play with Mathematica on my side.. –  Arnaud Jun 25 '13 at 13:15
    
The number of vertices varies between $O(q^7)$ and $O(q^{30})$ (it can also be $O(q^{10})$ and $O(q^{27})$). So the density in terms of $n$ varies between $O(n^{3/7})$ and $O(n^{3/30})=O(n^{1/10})$. –  Max Horn Jun 25 '13 at 13:42
    
Sweet. So it looks like according to this measure, the projective space is the densest one. Many thanks for this great answer. –  Arnaud Jun 25 '13 at 13:47
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