Schubert cells in G/P for reductive G

All literature on the Schubert cells of the generalized flag varieties $$G/P$$ ("generalized" here means that $$P$$ is an arbitrary parabolic) assumes that $$G$$ is a semisimple complex group. I am interested in whether the same results also apply to arbitrary reductive complex $$G$$? In particular, that one can make decomposition of $$G/P$$ into Schubert cells and that these cells are isomorphic to affine spaces.

Maybe it is true that the flag variety $$G/P$$ for an arbitrary reductive $$G$$ and parabolic $$P$$ is isomorphic to the flag variety $$G'/P'$$ for a semisimple $$G'$$ and a parabolic $$P'\leq G,$$ thus one can use the knowledge on the Schubert cells on $$G'/P'$$ in order to get the same for $$G/P$$?

• In general, you have to work pretty hard to find an interesting structure-theoretic result about semisimple groups that isn't true for reductive ones. (With obvious exceptions, like "the centre is finite" and "the group is semisimple".) Feb 4, 2021 at 2:48

You already answered your question: the center of any reductive group lies in any parabolic, so if $$G$$ is reductive, and $$G_{\operatorname{ad}}$$ its adjoint quotient (which is, of course, semi-simple), then $$G/P\cong G_{\operatorname{ad}}/P'$$ (where $$P'$$ is the image of $$P$$ in $$G_{\operatorname{ad}}$$).
• Just to have it written for @Filip92—of course you already know. There are two common ways to pass from a reductive group to a semisimple one, namely, by passing to the adjoint quotient and by passing to the derived group. Since $G = \operatorname Z(G)\cdot\operatorname{Der}(G)$, the latter is just a finite cover of the former: $G_{\text{ad}} = G/\operatorname Z(G) \cong \operatorname{Der}(G)/(\operatorname Z(G) \cap \operatorname{Der}(G))$. Sometimes one or the other is better, but, since isogenous groups have the same gen'd flag varieties, in this case it doesn't matter. Feb 4, 2021 at 2:48
• @LSpice There is a good reason to use the adjoint quotient in this case, since that's always adjoint (as the name suggests) but the derived group can be any finite cover of the adjoint group. So, you get for free that G/P only depends on the Lie algebra, not the group (in fact, you can define it as the variety of Lie subalgebras conjugate to $\mathfrak{p}$ in the Grassmannian of $\mathfrak{g}$ of the right dimension). Feb 4, 2021 at 2:59