# Hyperbolic Coxeter groups, Humphreys' book

Let $$(W,S)$$ be an irreducible Coxeter system with non-degenerate bilinear form $$B$$ on the Euclidean vector space $$V$$. The simple root attached to $$s\in S$$ is denoted by $$\alpha_s\in V$$. Let $$\{\omega_s:s\in S\}$$ be the dual basis of $$\{\alpha_s:s\in S\}$$. In Section 6.8 of Humphreys' book “Reflection groups and Coxeter groups” he defines $$C=\{v\in V:\forall s\in S:B(v,\alpha_s)>0\}$$ and $$D=\overline{C}$$ (closure with respect to the Euclidean metric). He claims that $$D$$ is the convex hull of the vectors $$\omega_s$$. I can't believe this, since for instance, $$2\omega_s$$ belongs to $$C\subseteq D$$. Perhaps he meant conical hull?

What bothers me more is the proof of Proposition 6.8. In the last paragraph, I understand that $$N$$ has two connected components and both are convex. Each $$\omega_s$$ is contained in one of the components. Now he wants to argue with convexity, but why lie all $$\omega_s$$ in the same component? Here is the excerpt:

Humphreys' (very accurate) errata on his homepage does not cover this part of the book. Sadly, he died last year from Covid.

• Sorry I don't have a complete answer for you, I spent some time thinking about it but couldn't close out the argument in Proposition 6.8. What I can say is that when Humphreys says "convex hull", he means "the cone over the convex hull", or more precisely the $\mathbb{R}^{\ge0}$ span of the simplex spanned by those vectors. For the second part of your question, you might like to compare to Exercise 13 in Chapter V of Lie Groups and Lie Algebras, by Bourbaki, where they walk you through the proof of this Proposition. Aug 22, 2021 at 17:57
• Thank you for the Bourbaki reference! I will get into that later today. Aug 23, 2021 at 6:00
• I was flicking through my copy of M Davis' book and found a discussion of some relevant ideas on page 98 under section 6.8. If you are still thinking about this, that might give you a fresh approach Sep 30, 2021 at 22:17
• Thanks again for your suggestion! Bourbaki and Davis use some unfamiliar vocabulary (to me), but I just spent some time translating to problem into linear algebra. I'm not quite there, but I will write some sort of answer. Perhaps you can take a look once it is online. Oct 2, 2021 at 11:38

Please check page 25 of these notes: https://mathweb.ucsd.edu/~ssam/old/21S-264C/notes-264C.pdf

Below is another proof I figured out myself, it's a bit more tedious than the note given above, but it proved a stronger result: For hyperbolic groups, the closure of the Tits cone is equal to exactly one connected component of $$H=\{v\in V\mid (v,v)\leq 0\}$$, say $$H^+$$, and has 0 as the only intersection with the other component $$H^-$$.

A reference is Howlett's paper: https://link.springer.com/article/10.1007/BF02677488. I will directly cite the results proved in this paper. I'll also keep the notations consistent with Howlett.

The proof requires some knowledge of the dual cone: the dual cone of the Tits cone $$U$$ is defined as

$$U^\ast = \{v\in V\mid (v,x)\geq 0 \text{ for all x\in U}\}.$$

It's well-known that $$(x,x)\leq0$$ for any $$x\in U^\ast$$, regardless of the signature of the inner product $$(\cdot,\cdot)$$. Hence $$U^\ast$$ is contained in $$(v,v)\leq0$$. Howlett's paper proved that $$U^\ast$$ is completely contained in one component (prop3.7), say $$H^-$$. Taking dual this gives $$\overline{U}=(U^\ast)^\ast\supset H^+$$, thus $$\overline{U}$$ contains $$H^+$$.

note: here we used the fact that the dual of dual cone is the closure of the original cone: $$(C^\ast)^\ast = \overline{C}$$ for a cone $$C$$.

On the other hand, since all the $$\{\omega_s\}$$ are non-space-like vectors, and $$W$$ preserves the inner product, we see that the Tits cone $$U$$ lies in $$H$$, hence also $$\overline{U}$$.

To prove $$\overline{U}\cap H^-=\{0\}$$, note that the dual cone $$U^\ast$$ must contain at least one time-like vector: $$z\in U^\ast$$ and $$(z,z)<0$$. Otherwise, if $$U^\ast$$ contains only light-like vectors, it's spanned by a single light-like vector $$(\delta,\delta)=0$$, taking dual gives $$\overline{U} = \{\delta\geq0\}$$. This is a half-space certainly contains space-like vectors, contradicts to $$\overline{U}\subset H$$.

Fix a $$z\in U^\ast$$ and $$(z,z)<0$$, so $$z$$ must be in $$H^-$$. Any vector $$x\ne0\in H^-\cap U$$ satisfies $$(z,x)<0$$, contradicts the definition of the dual cone.

• In the future, you should not vandalize your own post by deleting it, or by replacing it with just a hyperlink. Nov 5, 2023 at 15:38
• Thank you for your answer! At the moment, I don't have time to think about, but I try to come back as soon as possible. Nov 9, 2023 at 19:45
• Now I have read the proof in the notes you have linked. There is one part, which I do not understand: "If $v\in N$, then either $v^+ \in N$ or $v^-\in N$, and in either case, its negative is also in $N$. We conclude that the two connected components of $N$ correspond to when all the coefficients (in the $\alpha$ basis) are all positive or all negative." Can you elaborate on this? Dec 16, 2023 at 19:40
• This is because any vector $(v,v)<0$ when expressed as a linear combination of the simple roots $v=\sum c_s\alpha_s$, all the coefficients $c_s$ are nonzero and have the same sign. Dec 17, 2023 at 1:30

I tend to avoid all geometric interpretations. Let $$M=(\alpha_s,\alpha_t)_{s,t}$$ be the Gram matrix of the Coxeter system. We know that $$M$$ has signature $$(n-1,1)$$. The dual basis can be expressed as $$\omega_s=M^{-1}e_s$$ where $$e_s$$ is the $$s$$-th standard basis vector. Using that $$\det(M)<0$$ and every principal minor $$\det(M_{ss})>0$$, we see that $$(\omega_s,\omega_s)=e_sM^{-1}e_s=\det(M_{ss})/\det(M)<0$$ (this is basically the lemma before the proposition). We need to show that $$vM^{-1}v<0$$ for every non-negative integer vector $$v\ne 0$$. Why is that?