Let $G$ be a semisimple algebraic group over an algebraically closed field $k$, so that $G$ is an almost-direct product of its minimal closed connected normal subgroups of positive dimension, $G_1,\ldots,G_m$. Let $\varphi:G_1\times\ldots\times G_m\to G$ be the product map, so that $\varphi$ is an isogeny (surjective with finite kernel).

The $G_i$ are simple, so there is an isogeny $(G_{i})_{sc}\to G_i$, where $(G_i)_{sc}$ is the algebraic group with the same root system as $G_i$, whose character group is the full weight lattice. In other words, $(G_i)_{sc}$ is the simply-connected form of $G_i$.

If we take a product of all of these maps, and compose with $\varphi$, we obtain an isogeny $(G_{1})_{sc}\times\ldots\times (G_{m})_{sc}\to G$. Is it conventional to call the group $(G_{1})_{sc}\times\ldots\times (G_{m})_{sc}$ the simply-connected form of $G$? If so, then it seems it is possible to have two nonisomorphic semisimple groups with isomorphic simply-connected forms (as long as they have the same simple factors). Or do we only speak of simply-connected forms of simple groups, not semisimple groups?

  • $\begingroup$ Is it a problem to have nonisomorphic groups with isomorphic semisimple forms? This happens even for simple groups. I mean, if $G_{sc}$ is the semisimple form of the non-simply connected simple group $G$ then aren't $G$ and $G_{sc}$ nonisomorphic groups with simply connected form $G_{sc}$? $\endgroup$ – Jim Dec 23 '14 at 8:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.