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.