4
$\begingroup$

Let $k$ be an algebraically closed field. Consider a class $C$ of linear algebraic groups over $k$ such that

  • every reductive group is in $C$.
  • If $H_1 \hookrightarrow G$, $H_2 \hookrightarrow G$ such that $H_i, G \in C$, then $H_1 \cap H_2 \in C$.

Can we describe the smallest class satisfying the property? Do any/every solvable group lie in it ?

Motivation: the intersection of two reductive subgroups inside a reductive group may not be reductive.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

There are solvable groups inside your class C. Let $G=SL_3$ and $H=SL_2$ passing through the highest root and its negative root (the one which fixes the second basis vector $e_2$, and leaves stable the span of $e_1,e_3$). The intersection of all the conjugates of $H$ under the Borel subgroup of upper triangular matrices in $G$ (being a finite intersection since $G$ is Noetherean) lies in the class C. But is is easy to see that this intersection is just the group of unipotent upper triangular matrices in the subgroup $H$ and is therefore not reductive.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you! I know similar example before, like $H_1=GL_2$ in $G=GL_3$ $H_2=gH_1g^{-1}$ with $g=\begin{pmatrix} 1 & 0 & 1 \\ 0 &1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$. I shall mean that whether every solvable group lie in the class... Btw, one can prove that although $H_1 \cap H_2$ may not be reductive, $H_1 \cap gH_2g^{-1}$ is indeed reductive for some $g$. $\endgroup$
    – Zhiyu
    Commented May 10, 2019 at 2:36
  • $\begingroup$ well, one of your questions was whether solvable groups lie in your class. $\endgroup$
    – Venkataramana
    Commented May 10, 2019 at 2:38
  • $\begingroup$ I am a little careless.. thank you anyway. $\endgroup$
    – Zhiyu
    Commented May 10, 2019 at 3:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .