In this post we will sketch a proof of the Bestvina–Brady Theorem. Let us first recall the definition of the Bestvina–Brady group, which was introduced in the last post. Consider a flag complex and its associated right–angled Artin group . Let be the homomorphism sending each generator of to . We define the Bestvina–Brady group, , to be the kernel of .
Our goal is to relate the finiteness properties of to the topological properties of . Before proceeding with the technical details, let us first give an informal overview of the argument. We will construct a space , and a map , so that is precisely the homomorphism induced by on the corresponding fundamental groups.
Lifting the map to the corresponding universal covers, we get a -equivariant Morse function . Note that acts nicely by deck transformations on . The key observation is that also acts nicely on the level sets of . We will analyze the topology of the level sets to get finiteness properties of . The topology of the level sets can be recovered from sets and via a Mayer–Vietoris sequence. These sets give information on ascending and descending links of vertices in , which in turn give finiteness properties of . We now begin the formal treatment.
For the rest of this post, we fix a finite flag complex , the associated right-angled Artin group , and the corresponding Bestvina–Brady group . We now explicitly construct a space . The details of the construction will be important in what follows. Consider the Euclidean space , where , with each standard basis vector corresponding to a vertex of . Let be the line segment joining the origin and the basis vector. Now, for every simplex , let be the smallest cube containing the edges . Finally, let be the image of the set
under the projection .
Intuitively, can be viewed as a CW–complex as follows. Its –skeleton consists of a single point. The –skeleton consists of a wedge of circles in one-to-one correspondence with the generators of . The -skeleton is obtained by attaching a –torus by for each edge . The -skeleton is obtained by attaching a –torus for each triangle in , and so on. From this point of view, it is clear that is a space.
Now, consider a linear map , defined by . It descends to a continuous map . Restricting to , gives a map .
Observation 1. The induced homomorphism on the fundamental groups, sends each generator of to . Consequently, the homomorphism coincides with .
Now consider the universal cover of and the universal cover of . The groups and act as deck transformations on and , respectively. Lifting to the universal covers yields a continuous -equivariant map . To simplify notation, throughout the post we will denote by .
Observation 2. is a Morse function on .
Proof: Since sends each edge of homeomorphically onto , we deduce that is non–constant on edges of , and hence is also non–constant on higher dimensional cells. It is not difficult to check that is affine on every cell of . Thus, is a Morse function on .
We omit the proof of the next observation, but the reasoning can be seen from the examples that follow. The proof can be found in Geoghegan’s Topologial Methods in Group Theory [Section 8.3].
Observation 3. The -links and -links of are homeomorphic to .
The Bestvina–Brady Theorem. Let be a finite flag complex, the associated right-angled Artin group, and the corresponding Bestvina–Brady group. Then
- if and only if is homologically -connected.
- if and only if is acyclic.
- is finitely presented if and only if is simply-connected.
We will first prove the implications for all three parts of the theorem, simultaneously. Then we will prove the implications for the first two parts of the theorem and provide a sketch for the third part. For convenience, we will use to denote .
Let us see how the topology of ascending and descending links gives information on finiteness properties of . We will need the following lemmas that appeared in one of the previous posts. Here .
Lemma 4. Consider two nonempty, connected subsets such that the set contains no vertices of . Then the inclusion is a homotopy equivalence.
Lemma 5 [Morse Lemma]. Let be a Morse function on . Consider two closed intervals such that and the set contains only one point of . Then is homotopy equivalent to with copies of coned off.
Corollary 6. Let be a Morse function and let be connected sets.
- If each -link and each -link is homologically -connected, then the inclusion induces an isomorphism on , for , and an epimorphism on .
- If each -link and each -link is simply-connected, then the inclusion induces an isomorphism on .
- If each -link and each -link is connected, then the inclusion induces an epimorphism on .
Proof: First, observe that the set is discrete. The proof now proceeds by induction using Lemma 4 and Lemma 5, as well as the Mayer-Vietoris and Seifert-Van Kampen Theorems, where the sets to be considered are and the cones on the descending/ascending links of the vertices .
Using Corollary 6, we can now prove the following result.
Theorem 7. Let be a -equivariant Morse function and let be the kernel of .
- If all -links and -links are homologically -connected, then
- If all -links and -links are acyclic, then .
- If all -links and -links are simply-connected, then is finitely presented.
- By Corollary 6, for any pair of real numbers , the inclusion induces an isomorphism on , for , and an epimorphism on . Recall that is acyclic since it is a universal cover of a space. Writing as , we deduce for each and all . A similar argument yields for each and all . Now using the decomposition , it follows from the Mayer–Vietoris sequence that is homologically -connected. We now claim that the group is of type . To see this, consider the cellular chain complex
Since acts on freely and cocompactly, we deduce that each group is finitely generated as a -module. Also observe that the chain complex is exact since is homologically –connected. Thus, the chain gives rise to a partial resolution of by finitely generated –modules.
- If all -links and -links are acyclic, we can proceed as above to show that is acyclic for any . Observe that is finite dimensional. Hence, in the cellular chain complex above, are non-zero for only finitely many 's. Furthermore, since is acyclic, we deduce that this cellular chain is exact and hence we obtain a finite resolution of by finitely generated -modules.
- If all -links and -links are simply-connected, then by Corollary 6, the inclusion induces an isomorphism on . So . Also, by part (1) of this theorem, is homologically -connected and so is connected. Thus, is simply-connected. Now observe that the orbit space of can be made into a space by adding cells of dimension and higher. Hence, .
We are now ready for —
Proof of the implication of the Bestvina–Brady Theorem:
- If is homologically -connected, then by Observation 3, -links and -links of are isomorphic to . Now, part (1) of Theorem 7 implies .
- If is acyclic, then by Observation 3, -links and -links of are acyclic. Now, part (2) of Theorem 7 implies .
- If is simply-connected, then by Observation 3, -links and -links of are simply-connected. Now, part (3) of Theorem 7 implies that is finitely presented.
We will now prove the implication for the first statement of the Bestvina–Brady Theorem. We will need the following theorem, which we state here without proof.
Theorem 8. For each , there is an isomorphism of -modules
Proof of the implication of part (1) of the Bestvina–Brady Theorem:
Let . Set . Let us assume that and arrive at a contradiction. By Theorem 8, . Since the Morse function, , was onto , we must have infinitely many vertices not in . So, is not finitely generated as a -module. Consider the cellular chain complex
Since acts cocompactly on , we deduce that each is a finitely generated free -module. Furthermore, from Theorem 8, we have for all , and hence the above chain is a partial projective resolution of finite type of length . By theorem 4.3 in Brown's Cohomology of groups, is finitely generated (to apply the theorem, we use the fact that ). But this implies that is a quotient of and hence must be finitely generated, giving us the desired contradiction. Thus, is homologically -connected.
We can now easily deduce the forward direction of (2) in the Bestvina–Brady Theorem.
Proof of the implication of part (2) of the Bestvina–Brady Theorem:
Suppose . Thus, there exists a finite resolution
by finitely generated projective -modules. Hence, for all , there exists a partial resolution of of length by finitely generated projective -modules. So, for all , we have and therefore by part (1) of the Bestvina–Brady theorem, is homologically -connected for all . Thus, is acyclic.
Proving the forward direction of the third part of the Bestvina–Brady Theorem requires much more work. Here is a sketch.
Proof of the implication of part (3) of the Bestvina–Brady Theorem:
If is not connected, then is not homologically -connected and by part (1) of the Bestvina–Brady Theorem we have , and so is not finitely presented. Now if is connected but not simply–connected, we can show that is generated by –translates of finitely many loops. Since is contractible, the loops are null-homotopic in , and so they must be null–homotopic in for some . Now since acts by horizontal translations, –orbits of loops are null-homotopic in . Therefore, the inclusion induces the trivial map on the fundamental groups. But it can be shown that this map must be an epimorphism on 's and therefore must be simply–connected, which cannot happen if is finitely presented.