Let $(W,S)$ be an irreducible affine Coxeter group, $M=(m_{ij})$ be its Coxeter matrix, and $\{\alpha_s\}_{s\in S}\in V$ be the system of simple roots in the standard geometric realization, so the $\{\alpha_s\}$ are linearly independent. The bilinear form $\bullet$ given by $$\alpha_s\bullet\alpha_s = \begin{cases}-\cos\frac{\pi}{m_{s,t}}, & m_{s,t}<\infty\\ -1, & m_{s,t}=\infty\end{cases}$$ has signature $(n-1, 0)$.
It's well-known that ${\rm Rad}(\bullet)$ is one-dimensional, and is spanned by $\delta=\sum_{s}c_s\alpha_s$ where all $c_s>0$. Let $C$ be the Tits cone in $V^\ast$, it's also well-know that the closure of $C$ is the half-space bounded by $\delta$: $\overline{C}=\{v\in V^\ast\mid \langle\delta,v\rangle\geq0\}$.
My question is: is $C$ the open half-space plus the origin, i.e. $C=\{0\}\cup\{\delta>0\}$, or is $C$ equal to the closed half-space, i.e. $C=\{\delta\geq0\}$ ?
The reason that I think $C$ does not contain points $x\ne0$ on the hyperplane $\delta=0$ is as follows: if such $x\in C$, then since $C=\cup w\overline{D}$, where $D$ is the fundamental chamber, there are $w\in W$ and $y\in\overline{D}$ that $x=wy$. Then since $W$ fixes $\delta$, we have $$0 = \langle\delta,x\rangle =\langle\delta,wy\rangle=\langle w^{-1}\delta,y\rangle=\langle\delta,y\rangle.$$ But $\delta$ is a positive linear combination of the simple roots and $y$ belongs to $\overline{D}$, and the simple roots form a basis of $V$, this forces $y=0$, hence $x=0$.
