Let $G$ be a connected reductive group with $G_{der}$ simply connected and $T$ a maximal torus over an algebraically field $k$.

We consider a extension $\tilde{T}$ of the maximal torus $T$ by a torus $Z$.

Is it possible to lift this extension to a central extension of $\tilde{G}$ of $G$ by $Z$ such that $\tilde{G}$ has still its derived group simply connected?


There is always the uninteresting example $\widetilde{G} = G \times Z$ relative to a choice of splitting of $\widetilde{T}$ as a central extension of $T$ by $Z$ (as may be chosen since you assumed the ground field to be algebraically closed), and we claim that it is the only one (up to isomorphism of central extensions), ultimately due to the Isomorphism Theorem. Moreover, we claim that the hypothesis that $\widetilde{G}$ has simply connected derived group is redundant.

Let's pose and answer the analogous question over any ground field $k$ (specializing to algebraically closed fields will then yield the above claims). Let $G$ be a connected reductive $k$-group such that $\mathscr{D}(G)$ is simply connected, let $T$ be a maximal $k$-torus of $G$, and let $\widetilde{T} \rightarrow T$ be a surjective homomorphism of $k$-tori with kernel $Z$ (a $k$-group scheme of multiplicative type which for clarity we do not assume to also be a $k$-torus). We seek to determine necessary and sufficient conditions under which there exists a central extension $$1 \rightarrow Z \rightarrow \widetilde{G} \stackrel{q}{\rightarrow} G \rightarrow 1$$ with a connected reductive $k$-group $\widetilde{G}$ such that $q^{-1}(T) \simeq \widetilde{T}$ (as extensions of $T$ by $Z$), and in such cases we wish to classify such central extensions $\widetilde{G}$ up to $k$-isomorphism. (We recall that all reasonable definitions of "simply connected" in the context of connected semisimple $k$-groups are insensitive to extension of the ground field $k$.)

The scheme-theoretic intersection $T' = T \cap \mathscr{D}(G)$ is a maximal $k$-torus of $\mathscr{D}(G)$, and we claim that a central extension exists as above if and only if the pullback central extension $\widetilde{T} \times_T T'$ of $T'$ by $Z$ is split as such, in which case we claim that $\mathscr{D}(\widetilde{G}) \rightarrow \mathscr{D}(G)$ is an isomorphism (so $\mathscr{D}(\widetilde{G})$ is simply connected) and that the possibilities for $\widetilde{G}$ up to isomorphism (as central extensions over $k$) are a torsor for the cokernel of ${\rm{Hom}}_k(T,Z) \rightarrow {\rm{Hom}}_k(T',Z)$. Note in particular that if $k = k_s$ and $Z$ is a $k$-torus then this cokernel is trivial, which would then give the uniqueness assertion at the start.

Assuming such a central extension over $k$ exists, the restriction $q':\mathscr{D}(\widetilde{G}) \rightarrow \mathscr{D}(G)$ is a central quotient map between connected semisimple $k$-groups, so it must be a central isogeny. The same then holds over $k_s$, yet by the Isomorphism Theorem over $k_s$ we see that a connected semisimple $k_s$-group which is simply connected has no nontrivial central extension by a connected semisimple $k_s$-group. Thus, $q'_{k_s}$ is an isomorphism and hence $q'$ is an isomorphism. The pullback of $q$ along the inclusion $\mathscr{D}(G) \hookrightarrow G$ therefore splits, so restricting over $T'$ gives the necessary condition that the central extension $q^{-1}(T') = \widetilde{T} \times_T T'$ of $T'$ by $Z$ splits as a central extension over $k$.

Now we assume that $\widetilde{T} \times_T T'$ splits as a central extension of $T'$ by $Z$ over $k$. Let ${\rm{Ex}}_k(H,Z)$ denote the abelian group of $k$-isomorphism classes of central extensions of $H$ by $Z$ for an affine $k$-group scheme $H$ of finite type. (Note that such extensions might not be smooth and connected, unless of course $H$ is smooth and connected and the multiplicative type $k$-group scheme $Z$ is a torus.) This is contravariant in $H$ via pullback. General nonsense shows that if $1 \rightarrow H' \rightarrow H \rightarrow H'' \rightarrow 1$ is a short exact sequence of $k$-group schemes of finite type and $H''$ is connected then there is an exact sequence of abelian groups $${\rm{Hom}}_k(H,Z) \rightarrow {\rm{Hom}}_k(H',Z) \rightarrow {\rm{Ex}}_k(H'',Z) \rightarrow {\rm{Ex}}_k(H,Z) \rightarrow {\rm{Ex}}_k(H',Z)$$ (the point being that any extension of a connected $k$-group scheme by $Z$ must be central, as the automorphism scheme of $Z$ is etale since $Z$ is of multiplicative type).

The preceding shows that the pullback map ${\rm{Ex}}_k(G,Z) \rightarrow {\rm{Ex}}_k(\mathscr{D}(G),Z)$ has trivial restriction to the classes of smooth connected central extension of $G$ by $Z$, as such extensions are necessarily connected reductive (since $G$ is connected reductive and $Z$ is of multiplicative type). Applying the above nonsense with $H = G$ and $H' = \mathscr{D}(G)$, since ${\rm{Hom}}_k(\mathscr{D}(G),Z) = 1$ (due to the commutativity of the $k$-group scheme $Z$ and the perfectness of the smooth connected $\mathscr{D}(G)$) we see that $\theta:{\rm{Ex}}_k(G/\mathscr{D}(G),Z) \rightarrow {\rm{Ex}}_k(G,Z)$ is injective and hits the classes of smooth connected central extensions.

Hence, the possibilities for $\widetilde{G}$ (as a central extension up to isomorphism) correspond precisely to the $k$-isomorphism classes of central extensions $E$ of the $k$-torus $G/\mathscr{D}(G)$ by the $k$-group $Z$ such that the pullback $E \times_{G/\mathscr{D}(G)} G$ is smooth and connected with $\widetilde{T}$ equal to the preimage of $T$ (as central extensions by $Z$). Note that $E$ must be a $k$-group scheme of multiplicative type (and if $Z$ is a $k$-torus then $E$ must be one too). But $T/T' \rightarrow G/\mathscr{D}(G)$ is an isomorphism, so we conclude that the possibilities for $\widetilde{G}$ correspond to the classes $[E]$ in ${\rm{Ex}}_k(T/T',Z)$ such that $E \times_{T/T'} G$ is smooth and connected with $E \times_{T/T'} T \simeq \widetilde{T}$ as extensions of $T$ by $Z$. But inspection of an open cell of $(G,T)_{k_s}$ shows that the "smooth connected" condition on $E \times_{T/T'} G$ is automatic since we are assuming $E \times_{T/T'} T \simeq \widetilde{T}$ (as $\widetilde{T}$ is smooth and connected).

Thus, the possibilities for $\widetilde{G}$ are classified by the preimage of $[\widetilde{T}]$ under the natural map ${\rm{Ex}}_k(T/T',Z) \rightarrow {\rm{Ex}}_k(T,Z)$. But for any $k$-torus $S$ we have ${\rm{Ex}}_k(S,Z) = {\rm{Ext}}^1_k(S,Z)$ (Ext in the abelian category of commutative $k$-group schemes of finite type), so the preimage of $[\widetilde{T}]$ is non-empty if and only if $[\widetilde{T}]$ is killed by the natural map ${\rm{Ex}}_k(T,Z) \rightarrow {\rm{Ex}}_k(T',Z)$, and this annihilation is exactly the hypothesis that $\widetilde{T} \times_T T'$ splits as an extension of $T'$ by $Z$. The link with Ext$^1$ also shows that the preimage of $[\widetilde{T}]$ is a torsor for the cokernel of ${\rm{Hom}}_k(T,Z) \rightarrow {\rm{Hom}}_k(T',Z)$.

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