# Why is $(\textrm{Ker } \beta)^0 = (\textrm{Ker } \alpha)^0$

Let $$G$$ be a connected linear algebraic group, $$\beta$$ a root of a maximal torus $$T$$ of $$G$$, $$S = (\textrm{Ker } \beta)^0$$, and assume that $$Z_G(S)$$ is not solvable. Then $$T$$ is a maximal torus of $$Z_G(S)$$, and the Weyl group $$W(Z_G(S),T)$$ has order $$2$$. Hence $$Z_G(S)$$ has semisimple rank $$1$$ (Theorem 25.3, Humphreys). Hence $$H = Z_G(S)/R_u$$ is reductive of semisimple rank one, where $$R$$ is the radical of $$Z_G(S)$$.

The restriction of the map $$\pi: Z_G(S) \rightarrow H$$ to $$T$$ is an isomorphism of $$T$$ onto a maximal torus $$\pi(T)$$ of $$H$$, because $$T \cap R_u = 1$$. So we can think of $$\beta$$ as a character $$\beta'$$ of $$\pi(T)$$.

If we know something about reductive groups of semisimple rank one like $$H$$, we know that there are only two roots $$\alpha', -\alpha'$$ of $$H$$ whose weight spaces are each one dimensional. We can interpret these characters as characters $$\alpha, -\alpha$$ of $$T$$.

The claim in the book I'm reading (Springer, Linear Algebraic Groups) is that $$(\textrm{Ker } \alpha)^0 = (\textrm{Ker } \beta)^0$$ I don't see why, is this obvious?

My attempt at proving this was to note that the Lie algebra of $$Z_G(S)$$ contains the weight space $$\mathfrak g_{\beta}$$. So there exists a $$0 \neq X \in \mathscr L(Z_G(S))$$ such that $$\textrm{Ad }t(X) = \beta(t)X, t \in T$$

Going modulo $$\mathscr L(R_u)$$, and using the facts that $$\mathfrak t \cap \mathscr L(R_u) = 0$$ (since $$T \cap R_u = 1$$), and that $$\mathscr L (Z_G(S)/R_u) \cong \mathscr L(Z_G(S))/\mathscr L(R_u)$$

we get that the image of $$X$$ in $$\mathscr L(H)$$ is in the root space of $$\beta'$$. (However, the image of $$X$$ might be $$0$$, in fact we might have $$\mathscr L(\mathfrak g_{\beta}) \subseteq \mathscr L(R_u)$$ for all we know) That would imply that $$\beta'$$ is a root of $$\pi(T)$$ in $$H$$, which implies $$\beta = \pm \alpha$$. This is stronger than what Springer claims. But of course this might not happen, as I just mentioned in parentheses.

I guess after changing your $\beta$ to a constant multiple of it (which still lies in $P'$ automatically), you will be able to choose such an X in the lie algebra which is nonzero when passing to the quotient.

This is because, if this does not happen, centralizer of $ker(\alpha)$ would consist of your maximal torus plus something in the unipotent radical. And this would make it solvable which contradicts the fact that it is in $P'$

(It being 5 years later, you've certainly worked this out by now; but no answer was accepted, so I thought I'd write it out for anyone else who encountered the issue.)

It's often important to understand while reading Springer that all constructions are performed in the category of smooth groups. Thus Springer means by $$\ker(\beta)^\circ$$ what is really the smooth reduced scheme $$\ker(\beta)^\circ_\text{red}$$ underlying the identity component of the scheme-theoretic kernel of $$\beta$$ (so that Springer's $$\ker(\beta)^\circ$$ is a torus, rather than an infinitesimal extension of a torus). In terms of character lattices, the difference is that the character group of the scheme-theoretic $$\ker(\beta)^\circ$$ might have $$p$$-torsion, whereas the character group of Springer's $$\ker(\beta)^\circ$$ quotients out this $$p$$-torsion (hence is torsion-free, the other torsion having been removed by passage to the identity component).

$$\DeclareMathOperator\X{X}$$The character lattice of $$T/\ker(\beta)$$ is $$\mathbb Z\beta \subseteq \operatorname X^*(T)$$, more or less by definition of the kernel as the smallest normal subgroup such that $$\beta$$ factors through the quotient. The character lattice of $$T/\ker(\beta)^\circ$$ in the scheme-theoretic sense kills off any possible étale quotient of $$\ker(\beta)$$ (on the group side) by killing off all prime-to-$$p$$ torsion in $$\X^*(\ker(\beta)) \cong \X^*(T)/\X^*(T/\ker(\beta))$$. That is, the character lattice of $$T/\ker(\beta)^\circ$$ in the scheme-theoretic sense consists of all characters of $$T$$ some prime-to-$$p$$ multiple of which lie in $$\X^*(T/\ker(\beta)) = \mathbb Z\beta$$. In other words, it is $$\mathbb Z_{(p)}\beta \cap \X^*(T)$$. Finally, using Springer's sense, the character lattice of $$T/\ker(\beta)^\circ_\text{red}$$ consists of all characters of $$T$$ some multiple of which lie in $$\mathbb Z\beta$$, where now we allow multiples divisible by $$p$$. In other words, it is $$\mathbb Q\beta \cap \X^*(T)$$. This shows that two characters of $$T$$ have the same smooth connected kernel if and only if one is a non-$$0$$ rational multiples of the other.

I had to go hunting for the definition of $$P'$$, which is in §7.1.4. The characters $$\pm\alpha'$$—from which I'll drop the primes because, as Springer points out, they are really characters of $$T$$—are the non-$$0$$ weights of (the image of) $$T$$ on $$\operatorname{Lie}(G_\beta)$$. Since $$\ker(\beta)^\circ_\text{red}$$ acts trivially on $$\operatorname{Lie}(G_\beta) = \operatorname{Lie}(\operatorname C_G(\ker(\beta)^\circ_\text{red})) = \operatorname C_{\operatorname{Lie}(G)}(\ker(\beta)^\circ_\text{red})$$, we have that $$\alpha$$ lies in $$\X^*(T/\ker(\beta)^\circ_\text{red}) = \mathbb Q\beta \cap \X^*(T)$$. Since $$\alpha$$ is non-trivial by Lemma 7.2.3, we are done.

(Incidentally, after having worked with the smaller connected kernel, we can show that we would have got the same answer had we worked with the larger one; for, since $$\ker(\beta)^\circ$$ in the scheme-theoretic sense contains Springer's $$\ker(\beta)^\circ_\text{red}$$, the containment of centralisers is reversed, giving that $$\operatorname C_G(\ker(\beta)^\circ)$$ is contained in $$\operatorname C_G(\ker(\beta)^\circ_\text{red})$$. Thus we may replace $$G$$ by $$C_G(\ker(\beta)^\circ_\text{red})$$, and so suppose that $$\ker(\beta)^\circ_\text{red}$$ is central in the first place. I claim that $$\ker(\beta)^\circ$$, and even $$\ker(\beta)$$ is central in $$G$$. But then $$G$$ is of semisimple rank $$1$$, and the classification in §7.3.2 shows that there are only two non-$$0$$ weights of $$T$$ on $$\operatorname{Lie}(G)$$, which must therefore be $$\pm\beta$$, so that, as you suspected, $$\alpha = \pm\beta$$, and the centre of $$G$$ is precisely the kernel of $$\beta$$.)