an element of the Weyl group fixing a vector in the fundamental Weyl chamber I am trying to solve exercise 10.12 from humphreys lie algebra book.
I need to prove that if an element $\sigma$ of the Weyl group is such that $\sigma v=v$ for $v$ a vector in the fundamental Weyl chamber then $\sigma$ must be the identity.
My attempt was trying to prove that $\sigma $ fixes all vectors in the weyl chamber which would imply that $\sigma$ fixes a basis of $E$ an therefore is the identity. Unfortunately I couldn't prove this so I don't know what to do.
 A: I will use the following result:

Corollary 10.2 - C (Humphreys Intro to Lie Algebras, pg. 50): If $\sigma = \sigma_1 \cdots \sigma_t$ is a expression for $\sigma \in \mathcal{W}$ in terms of reflections corresponding to simple roots, with $t$ as small as possible, then $\sigma(\alpha_t) \prec  0$. Here, $\sigma_i = \sigma_{\alpha_i}$.

Recall that, since $\Delta$ is a basis for $\Phi$, $\beta \in \Phi \Rightarrow \beta = \sum_{\alpha \in \Delta}k_\alpha \alpha$, for some $k_\alpha \in \mathbb{Z}$. If $\beta \prec 0$, then all $k_\alpha$ in the previous description are non positive.
Now to the problem: If $\gamma$ is in the fundamental Weyl chamber, then for every $\alpha \in \Delta$ it is true that $(\gamma, \alpha) > 0$. Since $\sigma(\gamma) = \gamma$, we have $$0 < (\gamma, \alpha) = (\sigma(\gamma), \sigma(\alpha)) = (\gamma, \sigma(\alpha)), \forall \alpha \in \Delta.$$ If $\sigma \neq 1$, one can write $\sigma$ in terms of reflections corresponding to simple roots, as follows: $$\sigma = \sigma_1 \cdots \sigma_t$$
with $t$ minimal. But, since $\alpha_t \in \Delta$, we have $0 < (\gamma, \sigma(\alpha_t))$, and by the corollary above, $\sigma(\alpha_t) = \sum_{\alpha \in \Delta}k_\alpha \alpha$ with every $k_\alpha \leq 0$. It follows that $$0 < \sum_{\alpha \in \Delta}k_\alpha (\gamma, \alpha).$$ By hypothesis, $(\gamma, \alpha) > 0$, so each term $k_\alpha (\gamma, \alpha)$ is non positive, a contradiction.
Therefore, $\sigma = 1$.
