It seems that Ben is nevertheless right that the answer to the first question is "NO". Let G=SL(2,BbbC), and B be the subgroup of lower triangular matrices. Then the inclusion BtoG is an epi, since every algebraic representation of B that extends to G does so uniquely (on the nose, not just up to an isomorphism!). This follows from the fact that in any finite dimensional representation V of the Lie algebra sl(2), the operator e is determined by f and h. Indeed, the kernel K of e is spanned by vectors v satisfying hv=mv and fm+1v=0 for some integer mge0, and since V=BbbC[f]K, the operator e on V is uniquely determined.
So one might guess that a morphism of complex affine algebraic groups phi:HtoG is an epi if and only if G/phi(H) is connected and proper (but I did not check this).
No comments:
Post a Comment