Weyl chambers are connected components

Question

Let L be a semisimple finite dimensional complex Lie algebra and $\Phi$ its root system, $E=\text{span}_{\mathbb{R}} \{\alpha\in \Phi\}$. Denote for $\alpha\in \Phi$ the hyperplane orthogonal to $\alpha$ by $P_{\alpha}=\{x\in E|(\alpha,x)=0\}$. Then the set of Weyl chambers is defined to be the set of connected components of $E\backslash \bigcup_{\alpha\in \Phi}P_{\alpha}$. The elements of $E\backslash\bigcup_{\alpha\in \Phi}P_{\alpha}$ are called regular elements. The claim now is that for a regular element $\gamma$, $$C(\gamma):=\{x\in E|\forall \alpha \in \Phi , (\gamma,\alpha)>0(resp.<) \Rightarrow (x,\alpha)>0 (resp.<)\} $$ is the Weyl chamber containing $\gamma$. It is clear to me that this should be true, but i cant formulate a good argument. Maybe someone can show why this is a connected component?

Answer 1

Basically all $C(\gamma)$ is saying is the elements on one side of each hyperplane. Clearly they are disconnected from one another so it only remains to show that each $C(\gamma)$ is connected.

In fact we can do even better. They are convex (and thus path-connected). Without loss of generality we can take $C(\gamma)$ to be the fundamental Weyl chamber, i.e. $(\gamma,\alpha) > 0$ for each positive root. Take any two elements $\lambda, \mu \in C(\gamma)$ and consider the straight line path $w(t) = (1-t)\lambda + t\mu$ which has $w(0) = \lambda,w(1)=\mu$. Then $(w(t),\alpha) = (1-t)(\lambda,\alpha) + t(\mu,\alpha)$. If $\alpha$ is a positive root $(\lambda,\alpha), (\mu,\alpha)>0$ and for $t \in [0,1]$ we have $t,(1-t)>0$ so then we have $(w(t),\alpha)>0$ and we have a path in $C(\gamma)$ from $\lambda$ to $\mu$.