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Over a base field $k$, linear $k$-groups stand for affine algebraic $k$-groups. For simplicity take $k$ to be a field of characteristic zero, as in this case one has the correspondence between connected linear $k$-groups and finite dimensional Lie $k$-algebras.

Let $\mathfrak{g}$ be the Lie algebra of a connected semi-simple linear $k$-group $G$. It is known that a (non-trivial) parabolic $k$-algebra $\mathfrak{p}$ of $\mathfrak{g}$ can be obtained as follows: there exists a (non-trivial) grading $$\mathfrak{g}=\oplus_{n\in\mathbb{Z}}\mathfrak{g}(n)$$ respecting the Lie bracket $$[\mathfrak{g}(m),\mathfrak{g}(n)]\subset\mathfrak{g}(m+n)$$ and that $\mathfrak{p}=\oplus_{n\geq 0}\mathfrak{g}(n)$. The grading on $\mathfrak{g}$ amounts to an action of $\mathbb{G}_m$ on $G$, which is given by conjugation through a co-character $\mu:\mathbb{G}_m\rightarrow G$, justified by the condition on Lie bracket. For convenience denote also by $\mu$ the grading on $\mathfrak{g}$, and $\mathfrak{p}=\mathfrak{p}^+(\mu)$ the parabolic Lie $k$-subalgebra associated as above, and $P^+(\mu)$ the corresponding parabolic $k$-subgroup. $\mu$ is only unique up to conjugation by the Levi $k$-subgroup $L(\mu)$ of $P^+(\mu)$, whose Lie algebra is $\mathfrak{g}(0)$.

My question is to extend these arguments to $k$-subgroups and Lie $k$-subalgebras. Keep the previous notations, and take $H\subset G$ a connected $k$-subgroup, with Lie $k$-subalgebra $\mathfrak{h}$. Then by restriction one gets a grading $$\mu_H:\mathfrak{h}=\oplus_n\mathfrak{h}(n)$$ with $\mathfrak{h}(n)=\mathfrak{h}\cap\mathfrak{g}(n)$. This grading satisfies $[\mathfrak{h}(m),\mathfrak{h}(n)]\subset\mathfrak{h}(m+n)$, hence is also given by some co-character $\mu_H:\mathbb{G}_m\rightarrow H$.

Assume that $\mu_H$ is non-trivial, and $H$ is reductive itself, how should one compare the two co-characters $\mu_H$ and $\mu$? From the arguments above one sees that $\mathfrak{p}^+(\mu_H)=\mathfrak{h}\cap\mathfrak{p}^+(\mu)$ and $P^+(\mu_H)=H\cap P^+(\mu)$, but if one extends $\mu_H$ to the cocharacter $\mathbb{G}_m\rightarrow H\rightarrow G$, what kind of difference might one find between $\mu_H$ and $\mu$?

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in short terms, let $H\subset G$ be a connected $k$-subgroup, with $Q\subsetneq H$ a parabolic $k$-subgroup, how to characterize the different parabolic $k$-subgroups $P$ of $G$ such that $P\cap H=Q$? –  genshin Dec 27 '10 at 13:32
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Many parts of the formulation are too vague and unlikely to be correct as stated: e.g., "the correspondence"; conditions on the given field such as being algebraically closed; source of your Lie algebra gradings and characterization of parabolic subalgebras; the restriction of a grading to a Lie subalgebra. It's essential to start with a known reference for the basics and then focus your question much more precisely. –  Jim Humphreys Dec 27 '10 at 14:34
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1 Answer 1

up vote 1 down vote accepted

(Edit)

In the given generality, I'm not sure the question has much hope for a tidy answer.

Consider $G=GL(V)$ and write $V = W \oplus W'$ where $W$ has dimension 2. So you get an embedding $H=GL(W) \to G$ in a natural way ($H$ acts trivially on $W'$).

The stabilizer $Q$ in $H$ of a line $L \subset W$ is a parabolic subgroup of $H$. And the parabolic subgroups of $G$ are the stabilizers of flags $F$ in $V$.

Consider a flag $$F = (0 \subset F_1 \subset F_2 \subset \cdots \subset F_r = V)$$ for which (i) $F_1 = L$, (ii) $F_2 = W$, and (iii) $F_i \cap W'$ is a complement in $F_i$ to $F_2=W$ for $i \ge 2$.

If $P$ is the stabilizer of $F$ in $G= GL(V)$, then $P \cap H = Q$. (Note that $H$ is contained in the stabilizer of the flag $F'=(0\subset F_2 \subset \cdots \subset F_r = V)$).

In general there are many such $P$.

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thanks! I was wondering if the intersection $P\cap H$ can really produce a parabolic subgroup for $H$ when $P$ is parabolic in $G$. But as is pointed out by Jim Humphreys in other pages, this is in general not true. So I think there's something wrong with my arguments characterizing parabolic subgroups via gradings and co-characters. –  genshin Dec 27 '10 at 18:19
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(My original statement wasn't quite right -- see edit for condition on $F_2$). Parabolics can indeed be characterized by co-characters as you wrote; see e.g. [Springer, Lin Alg Groups 2nd ed., 8.4.5]. If you start with a cocharacter of $H$ determining $Q$, of course you get a cocharacter of $G$, and hence a parabolic with $P \cap H = Q$. But there is no reason for all such $P$ to come about from cocharacters of $H$. –  George McNinch Dec 27 '10 at 18:31
    
thanks a great deal for providing a nice reference. as i said, one of the motivation for this MO item is to understand (the flaw behind) the argument that intersection of a parabolic subalgebra with a Lie subalgebra $h$ becomes a parabolic of $h$, which goes in the inverse sense contrary to the one you mention above. –  genshin Dec 27 '10 at 19:59
    
As for intersection of parabolic subgroups $P$ of $G$ with a subgroup $H$ of $G$, one is lead to consider the $H$-orbit of the class of $P$ in $G/P$. I'm wondering what happens if i trace the behavior of a maximal split torus $S$ of $H$. $S$ might be regular or singular in the larger group $G$. The orbit under $S$ or its normalizer might be more concretely described via root data, but it seems more difficult when one passes to a $H$-orbit. –  genshin Dec 27 '10 at 20:06
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