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. 



relate parabolic subalgebras to gradings?
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.


