$Z_{\alpha}$ tells me about the positive root and its reflection? Let $G$ be a reductive algebraic group with maximal torus $T$. Let $\Phi(G,T)$ be the roots of $G$ relative to $T$. Let $\alpha\in \Phi(G,T)$, then we define $T_\alpha = (\text{ker}(\alpha))^\circ$, where the $^\circ$ means to take the connected component of the identity.
Let $Z_{\alpha}=C_G(T_\alpha)$ be the centraliser of $T_\alpha$. Then this contains $T$, and also two Borel subgroups that contain $T$, right? Does that mean that given $\alpha\in \Phi$, I can get the two Borel subgroups that have unipotent radicals $U_{\alpha}$ and $U_{-\alpha}$ which descend to the $\mathfrak{n}_{\alpha}$ and $\mathfrak{n}_{-\alpha}$, for free.
What I mean is, $Z_{\alpha}$ actually tells me about both a positive root, and its reflection simultaneously?
I guess so, it seems that $\Phi(Z_\alpha, T)=\{\pm \alpha\}$ and we obtain $W(Z_\alpha,T)=\sigma_\alpha\in W$.

The reason why I am not sure this is true, is that if $\rho:G\to \text{GL}(V)$ is a rational representation. Let $\alpha\in \Phi$ be a root. Then $\rho(U_\alpha)$ maps $V_\lambda$ to $\sum_{k\in\Bbb N} V_{\lambda + k\alpha}$ .  I don't know why this $k$ is not simply $1$, unless I am not getting Lie algebra $\mathfrak{n}_{\alpha}\subset \mathfrak{s}_{\alpha}\cong \mathfrak{sl}_2$.
 A: I think this example might help a bit, and it's too large for a comment even if it isn't much of an answer. 
Let $G=\mathrm{GL}_n(\mathbb{C})$ and $\alpha=\epsilon_i-\epsilon_{i+1}$ be a root. Then your $Z_\alpha$ is $D_{i-1}\times\mathrm{GL}_2(\mathbb{C})\times D_{n-i-2}$. Then $Z_\alpha$ clearly contains $T$, but it cannot contain Borel subgroups of $G$; its dimension is too small (at least for almost all $n$). 
It's generally true that $Z_\alpha$ has semisimple rank 1, though, and contains two Borel subgroups $B_\alpha$ and $B_{-\alpha}$ of $Z_\alpha$ with $\mathrm{Lie}(B_\alpha)=\mathfrak{t}\oplus\mathfrak{g}_\alpha$. If $U_\alpha$ denotes the unipotent part (which is only the same as the unipotent radical because $B_\alpha$ is solvable) of $B_\alpha$, then $\mathrm{Lie}(U_\alpha)=\mathfrak{g}_\alpha$.
So yes, $Z_\alpha$ tells you about a reflection in $W(G,T)$ and a root and its negative. This is what people mean when they say reductive groups are "made from" copies $\mathrm{SL}_2$ (consider the derived subgroup of $Z_\alpha$) and the algebra-level analogue are the "$\mathfrak{sl}_2$ triples".
