# Standard parabolic Lie subalgebras and conjugacy

Let $\mathfrak g$ be a given semisimple Lie algebra with corresponding adjoint Lie group $G$. A parabolic subalgebra is any subalgebra containing a Borel subalgebra.

We can pick a Borel $\mathfrak{b}= \mathfrak h \oplus \mathfrak n$ with associated root data $\Phi$ and pick a simple system $\Pi\subset \Phi$. Any parabolic subalgebra containing $\mathfrak{b}$ is called standard parabolic with respect to $\mathfrak b$, and they are in 1 to 1 correspondence with subsets $\Theta\subset \Pi$. This is easy to see; look at the root decomposition of the parabolics, and they must contain all the positive roots and some negative roots which correspond to some subset of simples. It also (more or less) follows from the conjugacy of Borel subalgebras under $G$ that any parabolic subalgebra is $G$-conjugate to a standard parabolic with respect to a fixed Borel.

To each standard parabolic, I can pick out the root spaces generated by $\Theta$ and combine them with the Cartan to get a reductive Lie algebra called a Levi subalgebra.

Question 1: Why is it the case that standard parabolics are non-conjugate? To be concrete, let's look at $\mathfrak{sl}_3$. Here, $G$-orbits are (I believe) just similarity classes and the nontrivial standard parabolics are of the form $$\begin{bmatrix}*&*&*\\*&*&*\\0&0&*\end{bmatrix}, \qquad \begin{bmatrix}*&*&*\\0&*&*\\0&*&*\end{bmatrix}$$ but it is not obvious to me how to show that these are not conjugate.

Question 2: Let $\Theta, \Theta'$ be subsets of $\Pi$. Why are the Levi subalgebras associated to these subsets $G$-conjugate if and only if the root systems generated by $\Theta$ and $\Theta'$ are conjugate under the Weyl group of $\Phi$?

I suspect both of these questions are easy but I lack Lie group intuition, having come at the subject via Lie algebras.

Edit: I should say that I encountered these claims in Collingwood and Mcgovern's "Nilpotent orbits in semisimple Lie algebras", and their only provided reference is an exercise in Humphreys' "Introduction to Lie algebras and representation theory" which does not address these questions. Any references for these facts would be appreciated.

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The parabolics in $sl_3$ are rather $\begin{bmatrix}*&*&*\\ *&*&*\\0&0&*\end{bmatrix}$, $\begin{bmatrix}*&*&*\\0&*&*\\0&*&*\end{bmatrix}$ – user8268 Aug 31 '12 at 7:23
Of course, I wrote the Levi subalgebras. Thanks for the correction. – user1306 Aug 31 '12 at 13:28
A1. The are not conjugate, since they corresponds to the set of positive roots $\{\epsilon_1-\epsilon_2\}$ and $\{\epsilon_2-\epsilon_3\}$, respectively. However, they are 'compatible', since the levi-factor of them are conjugate. – Qinghua Pi Jun 16 '14 at 11:43