In light of the answers given to this question, I would like to pose a more general one: Do all Grassmannian spaces have genus 0? If so, do there exist any flag manifolds with non-zero genus?


EDIT: My first answer was confusing and not quite accurate. Let me try again.

In arbitrary characteristic, the structure sheaf of any homogeneous space $G/P$ (for $G$ a semisimple group) has no higher cohomology. This is an instance of Kempf's vanishing theorem. The space of sections is 1-dimensional, so this implies that the arithmetic genus is 0.

Now using Serre duality, we conclude that only the top cohomology of the canonical sheaf is nonzero, which means that the geometric genus is also 0.

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  • $\begingroup$ ... and the arithmetic genus? $\endgroup$ – Aston Smythe Jun 11 '10 at 17:24
  • $\begingroup$ I'm not sure what your comment is warning me about. I'm taking the definition of geometric genus to be the dimension of the sections of the canonical sheaf. If all cohomology vanishes except in top degree, this number is 0... $\endgroup$ – Steven Sam Jun 11 '10 at 20:34
  • $\begingroup$ Sorry Sam, I think I am not fully awake at the moment. You are right, there is nothing to be worried about. So let me put it differently: geometric genus is zero and there are no globally defined holomorphic forms on the flag variety, simply because the variety is rational, and these objects are birational invariants. In characteristic zero, arithmetic genus is also a birational invariant, so we are done: Kepmf's theorem is only necessary in positive characteristic. $\endgroup$ – t3suji Jun 11 '10 at 20:59
  • $\begingroup$ Birational invariance is a much simpler reason if one is only interested in geometric genus. Thank you for the response. $\endgroup$ – Steven Sam Jun 11 '10 at 21:08
  • $\begingroup$ Steven, do you mean "$\mathit{complete}$ homogeneous space", i.e. $G/P$? That's certainly implied by the question, but the answer is a but unclear. $\endgroup$ – Victor Protsak Jun 12 '10 at 4:08

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