Can any pair of associate parabolics be related by opposite parabolics?

Question

Let $G$ be a reductive group, say over an algebraically closed field of characteristic zero.

We have the following definitions for a pair of parabolic subgroups $P_1$ and $P_2$ with Levi quotients $M_1$ and $M_2$:

• Recall that $P_1$ and $P_2$ are said to be associate if the projections $P_1 \cap P_2 \to M_1$ and $P_1 \cap P_2 \to M_2$ are surjective.
• Furthermore, they are called opposite if the intersection $P_1 \cap P_2$ is a Levi subgroup.
• I will say that $P_1$ and $P_2$ are somewhat opposite if there exists a parabolic $P$ containing both, with the property that the images of $P_1$ and $P_2$ in the Levi quotient of $P$ are opposite parabolics.

Is it true that any pair of associate parabolics are connected by a sequence of pairwise somewhat opposite parabolics?

Answer 1

I have found a proof that any two associate parabolics can be connected by a sequence somewhat opposite parabolics, but a reference would still be appreciated.

First let me remark two parabolic subgroups $P_1$ and $P_2$ are associate if and only if there exists $M \hookrightarrow P_1 \cap P_2$ which is a Levi subgroup of both $P_1$ and $P_2$.

Next let us do some warm-ups about pairs $(P_1, P_2)$ of associate parabolics. We fix a common Levi subgroup $M \hookrightarrow P_1 \cap P_2$ for convenience.

1. If $P_1$ and $P_2$ both contain a Borel subgroup $B$, then $P_1 = P_2$.
2. If $P_1$ is submaximal (this means that $P_1$ is a maximal element in the poset of proper parabolics) then $P_2$ is either equal or opposite to $P_1$.
3. Suppose that $Q$ is any parabolic subgroup containing $M$, and write $L$ for its Levi quotient. Also write $P_1^L$ and $P_2^L$ for the images of $P_1 \cap Q$ and $P_2 \cap Q$ in $L$, respectively. Then $P_1^L$ and $P_2^L$ are associate parabolics in $L$ containing the image of $M$ as a common Levi.

Proof of (1). This is clear because $P_1 \cap P_2$ is a parabolic subgroup.

Proof of (2). Choose a maximal torus $T \subset M$ and a Borel $T \subset B \subset P_1$. The assumption that $P_1$ is submaximal means that $P_1$ contains every positive simple root except for, say, $\alpha$. If $P_2$ contains $\alpha$ then $P_2 = P_1$; otherwise $P_2$ contains $-\alpha$ and is the opposite of $P_1$ with respect to our pinning.

Proof of (3). It suffices to show that the image of $M$ in $L$ is a Levi subgroup of $P_1^L$. Since the composition $P_1 \cap Q \hookrightarrow Q \to L$ has unipotent kernel, this is equivalent to the assertion that $M \hookrightarrow P_1 \cap Q$ is a Levi subgroup. But the composition $P_1 \cap Q \hookrightarrow P_1 \to M_1 = M$ has unipotent kernel by the same token and the inclusion $M \hookrightarrow P_1 \cap Q$ splits this map.

Now let me begin the proof of the original assertion with the same notation as above. That is, $P_1$ and $P_2$ are associate parabolics and we choose a common Levi $M$. For a parabolic $Q$ containing $P_1$ submaximally (by this I mean that the image of $P_1$ in the Levi quotient $Q \to L$ is a submaximal parabolic), I will describe a procedure that either does nothing to $P_1$ or creates a parabolic subgroup $P_1'$ with the following properties:

1. It is somewhat opposite to $P_1$.
2. It is associate to $P_2$.
3. Its intersection with $P_2$ is strictly larger than $P_1 \cap P_2$.

Consider the parabolic subgroups $P_1^L$ and $P_2^L$ of $L$, which are either equal or opposite by (2) and (3) of the warm-up. In the first case we do nothing. Otherwise, define $P_1'$ to be the preimage of $P_2^L$ along $Q \to L$. It has the requisite properties: (1) holds by construction and (2) holds because $P_1'$ contains $M$ as a Levi subgroup. Finally (3) is verified by choosing a maximal torus in $M$ and comparing roots.

By virtue of this procedure, it remains to verify the following claim: if $P_1^L = P_2^L$ for every parabolic $Q$ containing $P_1$ submaximally, then $P_1 = P_2$.

Proof of the claim. Let's choose a pinning as in the proof of (1) and check that $P_2$ contains every positive simple root $\alpha$. This would imply that $P_2$ contains $B$ and hence equals $P_1$. Write $I$ for the subset of the Dynkin diagram corresponding to $P_1$.

• If $\alpha \in I$ then $\alpha$ belongs to $M$ and hence $P_2$ also.
• If $\alpha \notin I$ then we consider the parabolic $Q_\alpha$ corresponding to $I \cup \{\alpha\}$. Note that $Q_\alpha$ contains $P_1$ submaximally and that $\alpha$ belongs to the Levi of $Q_\alpha$.