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Dear all,

Consider a flag $V=V_1\supset V_2\supset \cdots \supset V_k\supset V_{k+1}=\{0\}$ of a vector space $V$ over a field of $p$ elements. Let $I$ be a subset of the index set $\{1,2,...,k\}$.

(1) I am working at the subgroup of $GL(V)$ consisting of all linear automorphisms $h$'s of $V$ such that $h$ preserves the flag and its induced isomorphism $\overline {h}: V_i/V_{i+1}\rightarrow V_i/V_{i+1}$ is the identity for every $i\in I$. This subgroup is a subgroup of the parabolic subgroup associated to the flag. When $I=\{1,2...,k\}$ then this subgroup is exactly the unipotent radical of the parabolic subgroup. My question is:

What is a good name for this subgroup? Is uni-parabolic a good name?

(2) Let $\varepsilon_1,...\varepsilon_s$ be a basis of $V$ such that $\varepsilon_{s-s_i+1},...,\varepsilon_s$ is a basis for $V_i$. We consider the subgroup of $GL(V)$ consisting of all linear automorphisms $h$'s of $V$ such that $h$ preserves the flag and its induced isomorphism $\overline {h}: V_i/V_{i+1}\rightarrow V_i/V_{i+1}$ is a permutation on the basis $\{[\varepsilon_{s-s_i+1}],...,[\varepsilon_s]\}$ of $V_i/V_{i+1}$ for every $i\in I$. Another question is:

What is a good name for this subgroup? Is symmetry-parabolic a good name?

Thank you very much for your help.

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For (1) unipotent radical of the parabolic is the standard name. –  Amritanshu Prasad Mar 31 '13 at 15:33
    
Hi Amritanshu. The subgroup considered in (1) is the unipotent radical of a parabolic subgroup only when $I$ is the whole set {1,2...,k}. If $I$ is a proper subset of {1,2,...,k} then I think it is not the unipotent radical anymore. –  Hung Nguyen Mar 31 '13 at 19:13
    
Dear Hung Nguyen: Amritanshu is completely correct. Try to think about some examples. –  user30379 Mar 31 '13 at 21:26
    
I look at this example: $dimV=2$ and a flag $V=V_1\supset V_2\supset V_3=0$ where $dim V_2=1$ and $I={2}$. Then the subgroup defined in (1) consists of matrices \begin{pmatrix}\ast & 0\\ \ast& 1 \end{pmatrix} Unless I am missing something, this subgroup is not unipotent, so how can it be the unipotent radical of a parabolic subgroup of $GL(V)$? –  Hung Nguyen Apr 1 '13 at 1:48
    
@Hung Nguyen, pranavk: Hung Nguyen is right; I did not read the question carefully enough. It's something between the parabolic and its unipotent radical (hence the suggestion uni-parabolic). –  Amritanshu Prasad Apr 1 '13 at 4:08
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2 Answers 2

In Galois theory of algebraic number fields while discussing a prime lying above a prime of the base filed the inertial group is defined as one inducing identity at the residue field level. Your definition is analogous to that. SO inertial subgroup of the parabolic could be a candidate reasonable name.

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@pranavk: This question comes from a friend of mine working on algebraic topology. I am not an expert in this area but I will try to explain.

Let $V$ be an elementary $p$-group of rank $s$. Then $V$ can be considered as an $s$-dimensional vector space over $F_p$, the field of $p$ elements. Let $H^\ast(V)$ denote the mod $p$ cohomology of the group $V$. Now the linear group $GL(V)$ acts on $V$ and hence on $H^\ast(V)$. So if $G$ is a subgroup of $GL(V)$ then we can define the invariant $H^\ast(V)^{G}$ consisting of all elements in $H^\ast(V)$ invariant under the action of $G$. It turns out that $H^\ast(V)^{G}$ has a structure of a module over the Steenrod algebra.

My friend is looking at some homomorphisms between the invariants $H^\ast(V)^{G}$. He can handle ``easy'' cases where $G$ is the entire $GL(V)$ or the symmetric group of degree $s$ (which can be embedded naturally into $GL(V)$). Now he wants to extend the work to other natural subgroups of $GL(V)$ like the ones we are discussing. I hope this makes sense.

@P Vanchinathan: Thank you for your suggestion. In character theory of finite groups we also have inertial subgroup.

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Dear Hung Nguyen: Thanks for the explanation, though I'm a bit puzzled as to why your friend thinks that the class of subgroups as in your question is a "natural" one. Of course, the notion of a "natural" subgroup of ${\rm{GL}}(V)$ is a matter of taste, but parabolic subgroups and their unipotent radicals and Levi factors seem to be rather more "natural" than the things in your question. But maybe your friend has some good reason to regard the subgroups in your question as "natural"? –  user30379 Apr 14 '13 at 5:01
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