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.

  • 1
    $\begingroup$ 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). $\endgroup$ – BCnrd Dec 28 '10 at 15:25
  • 1
    $\begingroup$ [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. $\endgroup$ – BCnrd Dec 30 '10 at 16:38

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.

  • $\begingroup$ 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? $\endgroup$ – genshin Dec 30 '10 at 15:31
  • 1
    $\begingroup$ 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). $\endgroup$ – Marty Dec 30 '10 at 21:23
  • 1
    $\begingroup$ 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? $\endgroup$ – turtle Jan 5 '11 at 13:40
  • 2
    $\begingroup$ Representations of 1-dimensional compact tori, on real vector spaces, don't decompose into pieces graded by integers. $\endgroup$ – Marty Jan 5 '11 at 16:31

(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 $\mathfrak g$.

  • $\begingroup$ thanks. is there any reference related? $\endgroup$ – genshin Dec 30 '10 at 14:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.