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In "Linear algebraic groups, 2nd ed. T.A.Spinger, Birkhauser" 8.4.5, one finds a characterization of parabolic subgroups via co-characters, as follows:

for simplicity, assume that $k$ is a base field of characteristic zero which is algebraically closed, with linear groups standing for affine algebraic groups over $k$. Let $G$ be a connected semi-simple group, and $$\mu:\mathbb{G}_m\rightarrow G$$

a co-character. Then the elements (rather, local sections as a description in term of functor of points) $g\in G$ such that the limit $\lim_{t\rightarrow 0}\mu(t)g\mu(t)^{-1}$ exists form a closed subgroup $P(\mu)$ of $G$. $P(\mu)$ is parabolic in $G$, and every parabolic subgroup of $G$ arise in this way.

(for $k$-non-algebraically closed, see B.Conrad's comment below)

At the level of Lie algebras, a parabolic Lie subalgebra $\mathfrak{p}$ of a semi-simple Lie algebra $\mathfrak{g}$ comes as follows: there is a homomorphism of Lie algebra $\mu:k\rightarrow \mathfrak{g}$ of semi-simple image, which gives a grading of $\mathfrak{g}$ under the adjoint representation $\mathfrak{g}=\oplus_n\mathfrak{g}(n)$, such that $\mathfrak{p}=\oplus_{n\geq 0}\mathfrak{g}(n)$. This follows from the above proposition by taking differentials at the origin.

My question: what kind of $\mathbb{Z}$-grading $\mathfrak{g}=\oplus_n\mathfrak{g}(n)$ can leads to a parabolic subalgebra of the form $\mathfrak{p}=\oplus_{n\geq0}\mathfrak{g}(n)$?

At first sight, one notices that if $\mathfrak{g}=\oplus \mathfrak{g}(n)$ is given by a co-character $\mu:k\rightarrow \mathfrak{g}$, with $k$ acting on $\mathfrak{g}$ through the adjoint representation. Thus $[\mathfrak{g}(m),\mathfrak{g}(n)]\subset \mathfrak{g}(m+n)$. Moreover the image of $\mu$ is a one0dimensional semi-simple Lie subalgebra in $\mathfrak{g}(0)$, with $\mathfrak{g}(0)$ equal to its centralizer.

Conversely, if $\mathfrak{g}=\oplus \mathfrak{g}(n)$ is a grading, such that $[\mathfrak{g}(m),\mathfrak{g}(n)]\subset \mathfrak{g}(m+n)$, can one find a co-character $\mu$ such that $\mu$ gives the same grading via the previous procedure? or can one find an action of a one-dimensional torus on $\mathfrak{g}$ that preserves the Lie algebra structure?

This seems problematic. In fact given a $\mathbb{Z}$-grading for $\mathfrak{g}$ a semi-simple Lie algebra, assuming that $\mathfrak{g}(0)$ is also reductive, I don't see how this should really come from a co-character.

Thanks for attention.

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Strictly speaking, Springer assumes $k$ alg. closed (of any char., but that's a side issue). Over any field $k$ the limit method yields all parabolic $k$-subgps in a conn'd reductive $k$-gp, but that requires more work. In char. 0 the aut. scheme of $\mathfrak{g}$ has identity component $G/Z_G$, $\mathcal{D}(\mathfrak{h})$ is Lie alg. of a conn'd ss $k$-subgp of $G$ for any Lie subalg. $\mathfrak{h}$ in $\mathfrak{g}$, and $P(\mu)$ "works" for any action $\mu:\mathbf{G}_m \times G \rightarrow G$ (consider the resulting semidirect product). Those facts should help (along with Levi decomp). – BCnrd Dec 28 2010 at 15:25
[small correction above: $\mathcal{D}(\mathfrak{h})$ does "exponentiate" to a connected closed $k$-subgroup of $G$, but is semisimple iff $\mathcal{D}(\mathfrak{h})$ is its own derived algebra.] To flesh out the above "hints" a bit, the specified weights define an action of $\mathbf{G}_m$ on $\mathfrak{g}$ as a Lie algebra, and hence a $k$-homomorphism of $\mathbf{G}_m$ into the Aut-scheme of $\mathfrak{g}$, so into its identity component $G/Z_G$ (using that $G$ is ss). This action $\mu$ define a $k$-parabolic $P_{G/Z_G}(\mu)$ in $G/Z_G$. Then take its preimage in $G$ to conclude. – BCnrd Dec 30 2010 at 16:38

2 Answers

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Recall that when $\mathfrak{g}$ is a Lie algebra over a field $k$, a derivation of $\mathfrak{g}$ is a $k$-linear map $D: \mathfrak{g} \rightarrow \mathfrak{g}$ such that $$D( [X,Y] ) = [DX, Y] + [X, DY],$$ for every $X,Y \in \mathfrak{g}$.

Given a $Z$-grading on a Lie algebra $\mathfrak{g}$ (i.e., a grading on the underlying $k$-vector space such that $[\mathfrak{g}(m), \mathfrak{g}(n)] \subset \mathfrak{g}(m+n)$, there is a unique derivation $D$ on $\mathfrak{g}$ satisfying $$\forall X \in \mathfrak{g}(n), D(X) = n \cdot X.$$

It's a theorem of Zassenhaus (I think in paper from 1939 that I can't get online at the moment) that if $\mathfrak{g}$ is finite-dimensional, with nondegenerate Killing form, then every derivation of $\mathfrak{g}$ is inner. Hence, given a $Z$-graded Lie algebra $\mathfrak{g}$, with nondegenerate Killing form, there exists $H \in \mathfrak{g}$ satisfying $$\forall X \in \mathfrak{g}(n), [H,X] = D(X) = n \cdot X.$$

This demonstrates that all $Z$-gradings on a semisimple Lie algebra in characteristic zero arise from an "$ad(H)$-eigenspace" cosntruction. In this setting, one can define a parabolic subalgebra as the direct sum of non-negatively graded pieces.

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thanks a lot. And by the way, how should one deduce the semi-simplicity or unipotency of $H$? As $\mathfrak{g}(0)$ should be the Levi subalgebra of a parabolic, is there other evidence that implies that $\mathfrak{g}(0)$ is reductive? – genshin Dec 30 2010 at 15:31
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 Note that if $deg(X) = m$, $deg(Y) = n$, then $ad(X) ad(Y)$ shifts degree by $m+n$ (hence upper/lower triang with zeros on diag., unless $m+n=0$). It follows that the Killing form places $\mathfrak{g}(n)$ in perfect duality with $\mathfrak{g}(-n)$ for all integers $n$. Use some linear algebra to relate the Killing form on $\mathfrak{g}(0)$ to the restriction of the Killing form on $\mathfrak{g}$. You'll eventually see nondegeneracy of the Killing form on the derived subalgebra of $\mathfrak{g}(0)$ (I think). – Marty Dec 30 2010 at 21:23
I'm confused. Say $\mathfrak{g}$ is the Lie algebra of a compact Lie group $G$, and $\mathfrak{t}$ is a 1-dimensional semi-simple Lie subalgebra, which is the Lie algebra of a 1-dimensional compact torus $T$ in $G$. Via the adjoint representation one also gets a $\mathbb{Z}$-grading on $\mathfrak{g}$, because the irreducible representations of $T$ is also parameterized by $\mathbb{Z}$. The grading is also compatible with the Lie bracket in the above sense, but the non-negative part does not corresponds to a parabolic subgroup of $G$, as $G$ is compact. Something missed in the statement? – turtle Jan 5 2011 at 13:40
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 Representations of 1-dimensional compact tori, on real vector spaces, don't decompose into pieces graded by integers. – Marty Jan 5 2011 at 16:31
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(I'm only addressing the characteristic 0, algebraically closed, adjoint case here).

A grading comes from a cocharacter if and only if $\mathfrak g(0)$ contains a Cartan sublagebra. The image of a cocharacter has to lie inside a maximal torus in the group, so the Lie algebra of that torus (a Cartan) lies in $\mathfrak g(0)$.

On the other hand, given a grading with a Cartan in $\mathfrak g(0)$, the graded pieces $\mathfrak g(n)$ are weight spaces for this Cartan, and in particular each weight space is homogeneous. This assigns a number to each root space, which is compatible with addition of roots; that is the same thing as a cocharacter of the adjoint group $G$ attached to $\mathfrak g$.

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thanks. is there any reference related? – genshin Dec 30 2010 at 14:09

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