Skip to main content
3 of 9
We only need j \ne i, not j \ne i^*; and, oh hey, we handle the mixed-anisotropic case too; no need to order I
LSpice
  • 12.9k
  • 4
  • 45
  • 69

$\DeclareMathOperator\tr{tr}$Write $L^n = V^* \oplus \overline V \oplus W$ for some $L$-vector space $V$, where $(W, H)$ is anisotropic and $H(v^* \oplus \overline v) = \tr_{L/F} \langle v^*, v\rangle$. A choice of $L$-basis $(v_i)_{i \in I}$ for $V$ gives rise to a maximal split torus $S$ in $G$, consisting of the transformations that preserve the $F$-line through each basis vector. (Proof: the centraliser of $S$ in $G$ is the product of the anisotropic group $\operatorname{SU}(W, H)$ with the torus consisting of the transformations that preserve the $L$-line through each basis vector.)

Write $(v_{i^*})_{i \in i}$ for the dual basis to $(v_i)_{i \in I}$. Put $J = I \cup I^*$.

For each $i \in J$, the map $a_i : S \to \operatorname{GL}_1$ that sends $s \in S$ to $s v_i/v_i$, in the hopefully obvious notation, is a relative root if $W \ne 0$; and its root space consists of all those skew-adjoint endomorphisms $E$ of $L^n$ such that the restriction of $E$ to $W$ has image in $L v_i$, and such that $E$ annihilates all $v_{i'}$ with $i' \ne i, i^*$.

For each pair $i, j \in J$ such that $i \ne j$, the map $a_{i j} = a_i - a_j$ is a relative root; and its root space is the set of all skew-adjoint endomorphisms $E$ of $L^n$ such that the restriction of $E$ to $L v_j$ has image in $L v_i$, and such that $E$ annihilates $W$ and all $v_{j'}$ with $j' \ne j, j^*$. (Note that $a_{i i^*} = 2a_i$.)

Since these spaces, together with $\operatorname C_{\mathfrak{su}(L^n, H)}(S)$, span $\mathfrak{su}(L^n, H)$, we have found all relative roots.

For every cocharacter $\lambda$ of $S$, we can assume up to Weyl conjugacy that $\langle a_i, \lambda\rangle \ge 0$ for all $i \in I$, and $\langle a_{i j^*}, \lambda\rangle \ge 0$ for all $i, j \in I$. Now the parabolic corresponding to $\lambda$ is the stabiliser of the flag $$ \Bigl(\bigoplus_{\substack{j \in I \\ \langle a_{i j}, \lambda\rangle \le 0}} L v_i\Bigr)_{i \in I} $$ (where I've put $a_{ii} = 0$ for each $i \in J$).

Since every isotropic flag can be conjugated into $V$, we have shown that the parabolics are precisely the stabilisers of isotropic flags.

LSpice
  • 12.9k
  • 4
  • 45
  • 69