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Let $G$ be an affine algebraic group defined over a field $k$. The notion of Borel subgroup and Parabolic subgroup exists. For classical groups $GL(n), SO(2n), SO(2n+1), Sp(2n)$, we know for a fact that any Parabolic subgroup is stabilizer of a flag. Note that here every group has a standard representation.

Every affine algebraic group has a faithful representation defined over $k$. Thus it makes sense to ask

1.Do all parabolic subgroup arise as stabilizer of a flag in a given faithful representation?

2.Maybe a weaker question would be for every parabolic subgroup, does there exists a (faithful) representation and a flag whose stablizer is the parabolic subgroup we began with?

Apriori question 2 seems weaker than question 1. Is Q2 strictly weaker than Q1? i.e. Is it possible that question 2 has an affirmative answer but question 1 cannot.

I think I have the the answer to the question 1. in affirmative sense when $G$ is reductive using the dynamic description of parabolic subgroups. In characteristic zero we have levi decomposition of a group and it seems we have the solution in this case also. But the method of using Levi decomposition fails in positive characteristic.

Any hints/solution are welcome.

random123
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1 Answers1

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In some sense it's a definition. If G is a semisimple algebraic group and P is a parabolic subgroup, the quotient space G/P is defined to be a generalized flag variety. So P stabilizes the flags, i.e., the elements of G/P.

Not a grad student
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  • Thank you for your answer. I have heard such a definition but never came across the proof that it is equivalent to the definition which says that a parabolic is a subgroup which contains the Borel. Assuming this definition, i hope that my question makes sense. – random123 Feb 13 '18 at 05:41
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    I meant that as a definition of flag. A parabolic subgroup of an algebraic group over an algebraically closed field is one that contains a Borel subgroup, but this would not be true over arbitrary fields since algebraic groups over arbitrary fields need not contain a Borel subgroup defined over the ground field. A parabolic subgroup should be a subgroup such that G/P is complete. Over algebraically closed fields this is equivalent to containing a Borel subgroup. – Not a grad student Feb 14 '18 at 04:05
  • Yes. That is correct. I was a bit mistaken when i wrote it contains a borel. Lets assume we are over algebraically closed fields and then could we argue that we can find a representation of $G$ in $GL(V)$ such that there exists a flag $0 \subset V_1 \subset V_2 \subset \dots \subset V_k = V$ of vector spaces whose stabilizer is the given parabolic? – random123 Feb 14 '18 at 04:54
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    All parabolic subgroups of $GL_n$ are stabilizers of flags. This is because all parabolic subgroups of $GL_n$ are conjugate to block upper triangular matrices. I think the parabolic subgroups of G are the inverse images of the parabolic subgroups of $GL_n$, so they would be the elements in $G$ that stabilize a flag. – Not a grad student Feb 14 '18 at 13:35