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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: enter image description here

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

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    $\begingroup$ 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. $\endgroup$ Aug 22, 2021 at 17:57
  • $\begingroup$ Thank you for the Bourbaki reference! I will get into that later today. $\endgroup$ Aug 23, 2021 at 6:00
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    $\begingroup$ 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 $\endgroup$ Sep 30, 2021 at 22:17
  • $\begingroup$ 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. $\endgroup$ Oct 2, 2021 at 11:38

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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.

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  • $\begingroup$ In the future, you should not vandalize your own post by deleting it, or by replacing it with just a hyperlink. $\endgroup$
    – cpiegore
    Nov 5, 2023 at 15:38
  • $\begingroup$ 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. $\endgroup$ Nov 9, 2023 at 19:45
  • $\begingroup$ 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? $\endgroup$ Dec 16, 2023 at 19:40
  • $\begingroup$ 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. $\endgroup$
    – Zhao_L
    Dec 17, 2023 at 1:30
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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?

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