Buildings as Chamber Complexes
4.2 Examples
Exercises
4.12.Show that every thin building is a Coxeter complex.
4.13.Let ∆ be a building. For any simplex A ∈ ∆, show that the residue C(∆)≥A is a convex subset ofC(∆) in the sense of Definition 3.92.
4.14. (a) Given a simplex A in a building ∆, one could try to define the support of A by choosing an apartment Σ containing A and declaring suppA to be suppΣA, where the latter is the support of A in Σ (Defi- nition 3.98). Show that this does not work. In other words, if Σand Σ are two apartments containing A, suppΣAneed not equal suppΣA.
(b) Show, on the other hand, that the relation “suppA= suppB” is a well- defined relation on the simplices of ∆, i.e., if A and B have the same support in one apartment containing them, then they have the same support in every apartment containing them.
178 4 Buildings as Chamber Complexes
rank 1. In particular,∆ must have at least two vertices. Conversely, a rank-1 simplicial complex with at least two vertices is a building (with every 2-vertex subcomplex as an apartment). Thus the rank-1 buildings are precisely the flag complexes of the rank-1 incidence geometries with at least 2 points. [It is, of course, reasonable to demand that a 1-dimensional space, or “line,” have at least 2 points. In fact, one often even demands that there be at least 3 points, which is equivalent to requiring the flag complex∆ to be thick.]
Example 4.16.Suppose ∆ is a building of rank 2 (dimension 1). Then an apartmentΣ must be a 2m-gon for somem(2≤m≤ ∞). We will draw the Coxeter diagram as
m which should be interpreted as
ifm= 2 and as ifm= 3.
Let’s begin with the casem= 2. Then every apartment is a quadrilateral:
(As usual, the two colors, black and white, represent the two types of ver- tices.) It follows easily from the building axioms that every vertex of type • is connected by an edge to every vertex of type ◦, i.e., ∆ is a complete bi- partite graph. In the language of incidence geometry,∆is the flag complex of a “plane” in which every point is incident to every line. Conversely, the flag complex of such a plane is always a rank-2 building (with m= 2), provided that there are at least two points and at least two lines.
Note that we can also describe∆as the join of two rank-1 buildings. This suggests a general fact:
Exercise 4.17.If ∆ is a building whose Coxeter diagram is disconnected, show that ∆ is canonically the join of lower-dimensional buildings, one for each connected component of the diagram.
Returning now to Example 4.16, suppose next that m = 3. Then every apartment is a hexagon, which we may draw as the barycentric subdivision of a triangle:
This picture suggests a configuration of three lines in a plane (one line for each◦), whose pairwise intersections yield three points of the plane (one for each•). The existence of many such apartments, as guaranteed by (B1), makes it plausible that∆ is the flag complex of a projective plane. Recall the defin- ition of the latter:
Definition 4.18.Aprojective planeis a rank-2 incidence geometry satisfying the following three axioms:
(1) Any two points are incident to a unique line.
(2) Any two lines are incident to a unique point.
(3) There exist three noncollinear points.
With this definition, it is indeed the case that our building∆ is the flag complex of a projective plane. You may find it instructive to try to prove this as an exercise. [The exercise is not entirely routine; if you get stuck, you will see it again in Exercise 4.46.] Conversely, the flag complex of a projective plane is a building, with one apartment for every triangle in the projective plane. This converseis a routine exercise.
The most familiar example of a projective plane is the projective plane over a fieldk. By definition, the setP0of “points” is the set of 1-dimensional subspaces of the 3-dimensional vector spacek3; the setP1 of “lines” is the set of 2-dimensional subspaces ofk3; and “incidence” is given by inclusion, i.e., a pointx∈P0is incident to a lineL∈P1 ifx < Las subspaces ofk3.
It is now easy to construct concrete examples of buildings. LetP be the projective plane overF2, for instance, whereF2is the field with two elements.
This plane is also called the Fano projective plane. It has 7 points (each on exactly 3 lines) and 7 lines (each containing exactly 3 points). The resulting flag complex∆ is a thick building with 14 vertices and 21 edges. We will see in Exercise 4.23 below that the points ofP can be put in 1–1 correspondence with the 7th roots of unity ζj (ζ=e2πi/7, j = 0, . . . ,6) in such a way that the lines ofP are the triples
ζj, ζj+1, ζj+3
,j = 0, . . . ,6. This leads to the picture of ∆ shown in Figure 4.1. The interested reader can locate some of the apartments (there are 28 of them) and verify some cases of the building axioms.
Remark 4.19.This picture is misleading in one respect; namely, it fails to reveal how much symmetry∆has. One can see from the picture that∆admits an action of the dihedral group D14, but in fact Aut∆ is of order 336. The subgroup Aut0∆of type-preserving automorphisms is GL3(F2), which is the simple group of order 168.
Continuing with Example 4.16, one could analyze in a similar way the buildings corresponding tom= 4,5,6, . . .. Each value ofm corresponds to a particular type of plane geometry.
Definition 4.20.An incidence plane P is called a generalized m-gon if its flag complex is a building of type I2(m).
180 4 Buildings as Chamber Complexes
Fig. 4.1.The incidence graph of the Fano plane.
The terminology comes from the fact thatP has formal properties analo- gous to those of the geometry consisting of the vertices and edges of anm-gon.
See Van Maldeghem [264] or Tits–Weiss [262] for more information. See also Proposition 4.44 below, where we spell out precisely what it means for the flag complex to be a building of type I2(m).
Generalized 3-gons are also calledgeneralized triangles. Thus a generalized triangle is the same thing as a projective plane. Generalized 4-gons are also called generalized quadrangles or polar planes. A generalized quadrangle has no triangles, but there do exist lots of quadrilaterals. Every quadrilateral yields an apartment in the flag complex, this apartment being an octagon (or barycentrically subdivided quadrilateral). See Exercise 4.24 below for a concrete example of a generalized quadrangle.
Finally, in case m = ∞, buildings of type I2(∞) are simply trees with no endpoints (where an endpoint of a tree is a vertex that is on only one edge). To see that such a tree is a building, simply take the apartments to be all possible subcomplexes that are lines (i.e.,∞-gons); the verification of the building axioms is a routine matter. The converse, that every building of this type is a tree, is more challenging. We will treat it in Proposition 4.44 below, along with our characterization of generalizedm-gons form <∞.
The two remaining examples are intended to provide a brief glimpse of some higher-dimensional buildings from the point of view of incidence geome- try. Details (including definitions of some of the terms), will be omitted; these can be found in Tits [247]. (See also Scharlau [207].) We will, however, give many details for the most important case in the next section. And we will show in Chapter 6 how to construct further examples of buildings via group theory rather than incidence geometry.
Examples 4.21.(a) IfP is an n-dimensional projective space, then its flag complex is a rank-nbuilding of type An, i.e., having Coxeter diagram
. . . (nvertices).
Every apartment is isomorphic to the barycentric subdivision of the bound- ary of an n-simplex, and there is one such apartment for every frame in the projective space (where a frame is a set of n+ 1 points in general position).
Conversely, every building of type Anis the flag complex of a projective space.
Whenn= 2, this example reduces to the casem= 3 of Example 4.16.
(b) If P is an n-dimensional polar space, then its flag complex is a rank-n building of type Cn, i.e., having Coxeter diagram
. . . 4 (nvertices).
Every apartment is isomorphic to the barycentric subdivision of the bound- ary of an n-cube (orn-dimensional hyperoctahedron), and there is one such apartment for every “polar frame” in the given polar space. Conversely, every building of type Cn is the flag complex of a polar space.
Whenn= 2, this example reduces to the casem= 4 of Example 4.16.
Remark 4.22.It is no accident that all of the examples in this section (except trees) have been defined as flag complexes. Indeed, we will see in Exercise 4.50 below thatevery building is a flag complex.
Exercises
4.23. (a) Let V be a 3-dimensional vector space overF2, and letP =P(V) be the projective plane in which the points are the nonzero vectors inV and the lines are the triples{u, v, w}withu+v+w= 0. Show thatP is isomorphic to the projective plane overF2.
(b) Let ∆(V) be the flag complex of P(V), with its canonical type func- tion. IfV∗ is the dual of V, show that the correspondence between sub- spaces of V and subspaces of V∗ induces a type-reversing isomorphism
∆(V) ∼= ∆(V∗). Consequently, any isomorphism V −→∼ V∗ induces a type-reversing automorphism of ∆(V). If the isomorphism V −→∼ V∗ comes from a nondegenerate symmetric bilinear form on V, show that the resulting automorphism of∆(V) is an involution.
(c) LetV be the fieldF8, viewed as a vector space overF2. Show that there is a 7th root of unity ζ ∈ F8 such that the lines in P =P(V) are the triplesLi=
ζi, ζi+1, ζi+3
, i∈Z/7Z.
(d) With V = F8 as in (c), recall that there is a nondegenerate symmetric bilinear form on V given by x, y = tr(xy), where tr :F8 → F2 is the trace. This induces a type-reversing involution σ of ∆ =∆(V) by (b).
Show that σ is given on vertices by ζi ↔ L6−i. Describeσ in terms of the picture of ∆in Figure 4.1.
4.24.In this exercise we will use some standard algebraic terminology con- cerning bilinear forms. Readers not familiar with this terminology can look
182 4 Buildings as Chamber Complexes
ahead at Section 6.6, where all the terms are defined. LetV be a 4-dimensional vector space over the fieldF2, with basise1, e2, f1, f2. There is a nondegenerate alternating bilinear form−,− onV such that
ei, fi=fi, ei= 1
for i= 1,2 and all other “inner products” u, v of basis vectors are 0. The purpose of this exercise is to construct a generalized quadrangle Q from V and −,−. Thepoints of Qare defined to be the nonzero vectors inV, and thelinesofQare the 2-dimensional totally isotropic subspaces ofV. A pointp isincident to a lineLifp∈L.
(a) Show that there are 15 points, each incident to 3 lines, and 15 lines, each incident to 3 points. Thus the flag complex∆ ofQ has 30 vertices and 45 edges.
(b) Show how to use the four given basis vectors to construct a quadrilateral inQand hence an octagon in∆. This octagon will be called thestandard apartment. More generally, any four vectors inV with inner products like those of the basis vectors give rise to an octagon in∆called anapartment.
[Four vectors of this form are said to form asymplectic basis ofV.]
(c) Show that∆is a building of type I2(4).
(d) Call two vertices of ∆ opposite if there is an apartment Σ containing them and they are opposite in Σ in the obvious sense (recall thatΣ is an octagon). Show that two vertices are opposite if and only if (i) they are noncollinear points ofQor (ii) they are nonintersecting lines ofQ.
(e) Show that∆contains 5 vertices (but not 6) that are pairwise opposite.