Reductive group with simply connected derived group has all root groups $\mathrm{SL}_2$

Question

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PGL{PGL}$Motivation: I am trying to understand why the Deligne-Langlands conjectures are only stated for $p$-adic reductive groups with connected center. I know that in such a case, the Langlands dual has simply connected derived group. I am trying to understand why this is a desirable condition.

1. Suppose $G$ is a complex reductive group (playing the role of the Langlands dual in the statement of the Deligne-Langlands conjecture) with datum $(X,Y, R, R^{\vee}, \Pi)$. Suppose that the derived group of $G$ is simply connected. Is it the case that all root groups in $G$ are in fact $\SL_2$? More precisely, is it true that all root homomorphisms $h_{\alpha}: \SL_2\to G$ has image isomorphic to $\SL_2$? What is the precise obstruction for a root homomorphism taking image $\PGL_2$?

2. Assuming that the answer to the answer to question 1 is affirmative, does this necessarily imply that for all $\alpha\in \Pi$, we have $\alpha^{\vee}\notin 2Y$? Probably this one is easy but I cannot see it.

It seems we need both of the above questions to have positive answer; otherwise, I am rather lost.

Answer 1

$\newcommand\Z{\mathbb Z}$Point 1 has already been answered by @spin, and by @nfdc23 at Centralizers of subtori in reductive groups, derived subgroups. It is also Proposition 2.1(i) of Kac and Weisfeiler - Coadjoint action of a semi-simple algebraic group and the center of the enveloping algebra in characteristic $p$.

In light of point 1, point 2 is the same as asking whether there are a group $G$ with derived group $G' = \operatorname{SL}_2$, a maximal torus $T$ in $G$, and a root $\alpha$ of $T$ in $\operatorname{Lie}(G)$ such that $\lambda = \frac1 2\alpha^\vee$ lies in the cocharacter lattice of $T' = G' \cap T$. This cannot happen.

Indeed, suppose it did. Note that the adjoint quotient $G_\text{ad} = G/\operatorname Z(G)$ is $\operatorname{PGL}_2$. Since the composition $\operatorname{GL}_1 \xrightarrow\lambda T' \to T_\text{ad} \mathrel{:=} T/\operatorname Z(G) \xrightarrow\alpha \operatorname{GL}_1$ is the identity, we find that $\operatorname{im} \lambda$ is a subtorus of $T'$ that projects isomorphically to $T_\text{ad}$. Since $T'$ is $1$-dimensional, actually it equals $\operatorname{im} \lambda$; that is, the adjoint quotient restricts to an isomorphism $T' \to T_\text{ad}$. However, the adjoint-quotient map $G' \cong \operatorname{SL}_2 \to G_\text{ad} \cong \operatorname{PGL}_2$ has a non-trivial kernel on every maximal torus.

Answer 2

I think looking at Steinberg's lecture notes might be instructive for this question.

Following the notation $h_{\alpha}(t)$ defined in Steinberg's lecture notes (Lemma 19), the homomorphism $\operatorname{SL}_2 \rightarrow G$ corresponding to a root $\alpha$ has image $\operatorname{PGL}_2$ if and only if $h_{\alpha}(-1) = 1$.

By Lemma 28 (c) in Steinberg, you have $h_{\alpha}(-1) = 1$ if and only if $\langle \mu, \alpha \rangle \in 2 \mathbb{Z}$ for all weights $\mu$ in the weight lattice corresponding to $G$.

If $G$ is simply connected, you have the full weight lattice, so for any root $\alpha$ there exists a weight $\mu$ such that $\langle \mu, \alpha \rangle = 1$. Therefore for all roots $\alpha$, the corresponding homomorphism $\operatorname{SL}_2 \rightarrow G$ has image $\operatorname{SL}_2$.

EDIT: Suppose that $G$ is simple. Using the result above and examining the root systems case-by-case, we conclude the following:

The homomorphism $\operatorname{SL}_2 \rightarrow G$ corresponding to a root $\alpha$ has image $\operatorname{PGL}_2$ if and only if $G$ is of adjoint type $B_n$ and $\alpha$ is a short root.

proof: In the simply connected case there always exists a weight $\mu$ such that $\langle \mu, \alpha \rangle = 1$. In the simply laced case (except adjoint $A_1 = B_1 = C_1$), for every root $\alpha$ there exists another root $\beta$ such that $\langle \beta, \alpha \rangle = 1$. This is also true for type $C_n$ if $n \geq 3$. (It suffices to check this for simple $\alpha$, and then it is clear from the Dynkin diagram.)

The remaining case is $G$ of adjoint type $B_n$. In type $B_n$ for $n \geq 2$, for each long root $\alpha$ there exists a root $\beta$ such that $\langle \beta, \alpha \rangle = 1$.

For a short root $\alpha$ in type $B_n$ ($n \geq 1$), we have $\langle \beta, \alpha \rangle \in 2 \mathbb{Z}$ for all roots $\beta$. In the adjoint case the weight lattice is spanned by the roots, so we conclude that for adjoint type $B_n$, a map $\operatorname{SL}_2 \rightarrow G$ corresponding to a short root has image $\operatorname{PGL}_2$.

(Another edit: Just a minor note, the simple case immediately answers the question for reductive $G$ as well.)

Answer 3

This restriction that the dual group have simply-connected derived subgroup is not actually needed for the Kazhdan-Lusztig parametrization to hold. This is worked out in Reeder's paper Isogenies of Hecke algebras and a Langlands correspondence for ramified principal series representations, available here.

In general this condition is convenient in this setting because some very useful things are nicer in $G^\vee$-equivariant $K$-theory when $G^\vee$ is simply connected.