Let G be a connected affine algebraic group scheme over a field of positive characteristic. H be any closed subgroup of G. In which cases is the quasi projective variety G/H known to have a smooth compactification ?

  • 1
    $\begingroup$ You at least need to assume that $G/H$ is smooth! Perhaps it is best to consider the case when $G$ and $H$ are both smooth (which implies that $G/H$ is also smooth). $\endgroup$ Jun 23, 2014 at 19:03
  • $\begingroup$ Thanks for pointing. Should have mentioned G/H is given to be smooth. $\endgroup$
    – Amit H
    Jun 23, 2014 at 19:15
  • 3
    $\begingroup$ A 1-dimensional smooth connected $G$ with $H=1$ can fail to satisfy this condition when the unique regular compactification is not smooth at infinity. Consider Rosenlicht's example $y^p=x-ax^p$ for $a\in k-k^p$ (with $p={\rm{char}}(k)>0$). This has closure $y^p=xz^{p-1}-ax^p$ in $\mathbf{P}^2_k$ with one point $[1,a^{1/p},0]$ on the line at infinity, it is regular there (local ring $k[y,z]_{(y^p-a)}/(z^{p-1}=y^p-a)$ is a discrete valuation ring with uniformizer $z$) but is not smooth there if $p > 2$ (as $z^{p-1}=(y-a^{1/p})^p$ is not $\overline{k}$-smooth at $(a^{1/p},0)$ if $p>2$). $\endgroup$
    – user27920
    Jun 23, 2014 at 20:08
  • $\begingroup$ @AmitHogadi: Maybe you should clarify what precisely you are looking for. Do you want counterexamples over non-perfect fields, as user52824 already gave you? Do you want positive examples? Are you interested in particular types of group schemes, e.g., reductive or semisimple group schemes? Do you want your compactifications to be equivariant for the natural action of $G$? If you do not say more, then the question is a bit too general. $\endgroup$ Jun 24, 2014 at 13:47


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.