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I am quite interested in any partical answer to the following general (maybe a little bit vague) question: Is there some criterion about the connectedness of the intersection of two connected algebraic subgroups of a linear algebraic subgroup defined over a perfect field?

Indeed, the general question is motivated by the following special case of it: Let $G$ be a semisimple group defined over $k$, which is the algebraic closure of a finite field. Let $U\subset G$ be a connected unipotent subgroup and $H\subset G$ be a parabolic (or any connected) subgroup, then is $U\cap H$ is still connected?

Any comments are very welcome! Thanks in advance!

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  • $\begingroup$ In general, I think you're out of luck. (However, perhaps one could handle the case where $U$ is the unipotent radical of a parabolic, and hence filtered by $\mathfrak{gl}_1$'s, and that such groups have no smooth, finite subgroups?) It may be of interest to know Proposition 14.22(a) of Borel, which states that the intersection of two parabolics is connected; that seems close to your situation. $\endgroup$
    – LSpice
    Commented Jan 19, 2019 at 21:09
  • $\begingroup$ Also intersection with Levi subgroups of reductive groups often behaves well, as one shows using the fact they are centralisers of tori. $\endgroup$
    – LSpice
    Commented Jan 19, 2019 at 21:10

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Let $p=\mathrm{char}\,k$ and $G=GL(3,k)$. Then $$ \{\begin{pmatrix}1&0&0\\t-t^p&1&t\\0&0&1\end{pmatrix} \mid t\in k\} $$ is a group isomorphic to $\mathbf{G}_a$. Its intersection with the Borel subgroup of upper triangular matrices is $\mathbb{F}_p$.

Usually, the connectedness of $U\cap P$ is shown by exhibiting a torus which normalizes both $U$ and $P$ and whose action on $U$ is contracting.

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  • $\begingroup$ Just to be clear, that's the discrete group $\mathbb F_p$, not the group $\mathbf G_a$, even if it happens that $k = \mathbb F_p$. $\endgroup$
    – LSpice
    Commented Jan 21, 2019 at 12:24
  • $\begingroup$ (Err, I mean "the constant group scheme $\mathbb F_p$".) $\endgroup$
    – LSpice
    Commented Jan 21, 2019 at 22:51

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