My question is why a particular element of the Weyl group of $O(8)$ seems to contradict a theorem about root systems. But to tell you the particular element I have to tell you specifically how I'm thinking of $O(8)$.
Consider $8 \times 8$ matrices and let $$K = \begin{bmatrix} &&&&&&& 1 \\ &&&&&& 1 \\ &&&&& 1 \\ &&&& 1 \\ &&& 1 \\ && 1 \\ & 1 \\ 1 \end{bmatrix}$$ I define the orthogonal group $O(8)$ to be $O(8) = \{A \in GL_8 \mid A^{\mathrm{tr}}KA = K\}$. These are the matrices such that taking the transpose across the other diagonal (the one that goes top-right to bottom-left) yields the inverse.
Diagonal matrices $$D(t_1, t_2, t_3, t_4) = \begin{bmatrix} t_1 \\ & t_2 \\ && t_3 \\ &&& t_4 \\ &&&& t_4^{-1} \\ &&&&& t_3^{-1} \\ &&&&&& t_2^{-1} \\ &&&&&&& t_1^{-1} \end{bmatrix}$$ give a maximal torus $T$. I let $e_i \in X(T)$ be the character $D(t_1, t_2, t_3, t_4) \mapsto t_i$. Then $O(8)$ is type $D_4$ with roots $\Phi = \{\pm e_i \pm e_j \mid i < j\}$. Choose simple roots $\alpha_i = e_i - e_{i + 1}$ for $i = 1,2,3$ and $\alpha_4 = e_3 + e_4$.
Now my question: Consider the matrix $$A = \begin{bmatrix} 1 \\ & 1 \\ && 1 \\ &&& 0 & 1 \\ &&& 1 & 0 \\ &&&&& 1 \\ &&&&&& 1 \\ &&&&&&& 1\end{bmatrix}$$ Clearly $A \in O(8)$ and $AD(t_1, t_2, t_3, t_4)A^{-1} = D(t_1, t_2, t_3, t_4^{-1})$ so $A$ represents a non-zero element of the Weyl group $W = N_{O(8)}(T)/C_{O(8)}(T)$. But the action of $A$ on $X(T)$ fixes $\alpha_1$, $\alpha_2$ and swaps $\alpha_3, \alpha_4$. This contradicts the theorem that Weyl groups act simply transitively on the Weyl chambers. So something must be wrong. The only thing I can think of is that $A$ lies in $O(8)$ but not in $SO(8)$; maybe something I'm doing here works only for connected reductive groups but I have no clue what that something might be.