8
$\begingroup$

I am looking for a reference where the relative root system, the relative system of simple roots, and parabolic $\Bbb R$-subgroups for the real algebraic group ${\rm SU}(p,q)$ are explicitly computed.

More generally, let $F$ be a field of characteristic 0, $L/F$ a quadratic extension, and $H$ be an $L/F$-Hermitian form on $L^n$. Write $G={\rm SU}(L^n,H)$, which is a semisimple $F$-group. I am looking for a reference where the relative system of simple roots and the conjugacy classes of $F$-parabolics in $G$ are explicitly computed. Here "relative" means with respect to a maximal split $F$-torus.

The case of a special orthogonal group is treated in Borel's 1966 paper "Linear algebraic groups" in the Boulder proceedings "Algebraic Groups and Discontinuous Subgroups", Proc. Sympos Pure Math. vol. 9. Borel writes that for an orthogonal group, the $F$-paraboics are the stabilizers of the $F$-rational isotropic flags.

Question. Is it true for $G={\rm SU}(L^n,H)$ that the $F$-parabolics in $G$ are the stabilizers of the $F$-rational isotropic flags in $(L^n, H)$ ?

$\endgroup$
17
  • 1
    $\begingroup$ You start off with $\mathbb R$-groups, but then ask for a field of characteristic $0$. If the non-Archimedean case is of interest, then you may already know that Section 3.10 of Tits - Reductive groups over local fields discusses a special case. $\endgroup$
    – LSpice
    Mar 8 at 15:13
  • 1
    $\begingroup$ @LSpice: No, I did not know that. Thank you for the reference. Anyway, at the moment I am not really interested in the non-Archimedean case. $\endgroup$ Mar 8 at 16:39
  • 1
    $\begingroup$ I do not know of a good reference, but surely with an Inertia Theorem and a Witt Extension theorem removing ambiguities, your question has the expected positive answer...? $\endgroup$ Mar 8 at 17:56
  • 1
    $\begingroup$ The Atlas software computes the conjugacy classes of real parabolics (in any real group), as well as the Weyl groups. I can check some examples of that is of interest. $\endgroup$ Mar 8 at 20:13
  • 1
    $\begingroup$ I think that my answer (which does use Witt's theorem, as @paulgarrett predicted) now handles the case of a non-trivial anisotropic kernel. (Actually I think that the first version already did, but I faked myself out.) $\endgroup$
    – LSpice
    Mar 8 at 22:24
4
$\begingroup$

I have edited in some remarks from the comments (1 2 3). To start, I am used to thinking of the stabiliser of a self-dual flag, but of course every self-dual flag has an isotropic flag as its “bottom half”, and every isotropic flag can be completed to a self-dual flag by tossing the duals “on top”. (Yu - An introduction to explicit Bruhat–Tits theory is where I first encountered the description of Bruhat–Tits buildings of symplectic groups in this ‘self-dual’ language.)

$\DeclareMathOperator\tr{tr}$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 split torus $S$ in $G$, consisting of the transformations that preserve the $F$-line through each basis vector. Then $S$ is a maximal split torus in $G$, since its centraliser 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 basis of $V^*$ dual 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 of $L^n$ that annihilate all $v_{i'}$ with $i' \ne i, i^*$, and that carry $W$ into $L v_i$.

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 of $L^n$ that annihilate all $v_{j'}$ with $j' \ne j, j^*$, and that carry $L v_j$ into $L v_i$. (Note that $a_{i i^*} = 2a_i$, so, to get the full root algebra for $a_i$—if it is a relative root—we need to take the sum of the $a_i$ and $a_{i i^*}$ root spaces.)

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. The relative root system is of type $\mathsf C_{\dim(V)}$ if $W = 0$ and $\mathsf{BC}_{\dim(V)}$ if $W \ne 0$. If we identify $I$ with $\{1, \dotsc, m\}$, then one choice of simple roots is the union of $\{a_{i(i + 1)} \mathrel: 1 \le i < m\}$ with $\{a_{m m^*}\}$ if $W = 0$, or with $\{a_m\} if $W \ne 0$.

I will now appeal to the parameterisation of (rational) parabolics by (rational) cocharacters, according to which we attach to a (rational) cocharacter $\lambda$ of $G$ the parabolic $P_G(\lambda) = \{g \in G \mathrel: \text{$\lim_{t \to 0} \lambda(t)g\lambda(t)^{-1}$ exists}\}$. (This parameterisation has the advantage that it also singles out a Levi component $M_G(\lambda) = \operatorname{Cent}_G(\lambda)$ of $P_G(\lambda)$.)

Fix a rational cocharacter $\lambda$ of $G$. After replacing it by a $G$-conjugate, we may assume that it takes values in $S$; and, after replacing it by a further Weyl conjugate, we may assume that $\langle a_{i j^*}, \lambda\rangle \ge 0$ for all $i, j \in I$. Then $P_G(\lambda)$ is the stabiliser of the flag $$ \Bigl\{\bigoplus_{\substack{j \in I \\ \langle a_{i j}, \lambda\rangle \ge 0}} L v_j \mathrel: i \in I\Bigr\} $$ (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.

$\endgroup$
10
  • 1
    $\begingroup$ Shouldn’t the displayed formula have $v_j$? Also there is an “$i\in i$”. $\endgroup$ Mar 8 at 23:23
  • 1
    $\begingroup$ @MikhailBorovoi, standard parabolics (containing the Borel used to define 'simple') can be parameterised as you say, but all parabolics can be parameterised by cocharacters: the parabolic associated to $\lambda$ is the variety of all $g \in G$ such that $\lim_{t \to 0} \lambda(t)g\lambda(t)^{-1}$ exists; it is generated by the torus and those root subgroups corresponding to roots $\alpha$ for which $\langle\alpha, \lambda\rangle \ge 0$. In terms of simple roots, if $\lambda$ lies in the dominant chamber, then this corresponds to the set of simple roots satisfying the same condition. $\endgroup$
    – LSpice
    Mar 9 at 14:15
  • 1
    $\begingroup$ This approach also picks out a Levi (the centraliser of $\lambda$) and identifies the unipotent radical (those $g$ for which $\lim_{t \to 0} \lambda(t)g\lambda(t)^{-1} = 1$). It is classical, but it was Conrad, Gabber, and Prasad - Pseudo-reductive groups that drove home its importance for me. $\endgroup$
    – LSpice
    Mar 9 at 14:17
  • 1
    $\begingroup$ If you identify $I$ with $\{1, \dotsc, m\}$, then the root system is $\mathsf C_m$ if $W = 0$ and $\mathsf{BC}_m$ if $W \ne 0$. One set of simple roots is $\{a_{i(i + 1)} \mathrel: 1 \le i < m\} \cup \{a_{m m^*}\}$ if $W = 0$ and $\{a_{i(i + 1)} \mathrel: 1 \le i < m\} \cup \{a_m\}$ if $W \ne 0$. $\endgroup$
    – LSpice
    Mar 9 at 14:20
  • 1
    $\begingroup$ In the quasi-split case ($\dim W \le 1$), the root system we get is the one that comes from folding $\mathsf A_n$ using its order-2 diagram automorphism. See the lovely Stembridge - Folding by automorphisms. (That paper doesn't apply literally to $\dim W = 1$ since the "middle roots" are not orthogonal but belong to an orbit; but I think it is still reasonable to say that $\mathsf{BC}_m$ arises by folding in that case.) $\endgroup$
    – LSpice
    Mar 9 at 14:21
2
$\begingroup$

A slightly off-target answer, but possibly of some use. First, indeed, there is a large population who deliberately take the viewpoint, attributed to Harish-Chandra, that everything should be done for arbitrary reductive/semi-simple groups, not just "examples". Thus, as @LSpice's comment, "there's just one reductive group, $G$". :)

But, with some substance for the theory of integral representations of automorphic $L$-functions, the obvious/natural relations of classical groups prove non-trivial theorems... that do not have "intrinsic" analogues, in any useful sense, apparently.

So! "The classical groups". Over $\mathbb C$, there are really just $3$: general or special linear, orthogonal (nevermind the parity of dimension), and symplectic. Over $\mathbb R$ we have the subdivision...

Over $\mathbb R$, we can have Sylvester's Inertia theorems for the real forms that are described by "signatures", the simplest ones being $O(p,q)$, $U(p,q)$, $Sp^*(p,q)$. Yes, modeled by reals, complex, and quaternions. In light of Weil's "Classical groups/algebras with involutions", this is not a coincidence...

So, operationally, there are two types of classical groups: general-linear, and isometry (or similitude). In general-linear, parabolics are stabilizers of flags. In isometry... groups, parabolics are stabilizers of totally isotropic flags.

Choice of Levi component is equivalent to choice of complementary isotropic flag.

One can continue, as desired, to be able to re-specify "roots", etc.

Witt's theorem(s) show that there is only one isomorphism class of choices...

The "physicality" of Inertia Theorems as a part of "classification over $\mathbb R$" is (to my mind) use of inequalities, the intermediate value theorem, and such. Yes, this is all subsumed by the less-"physical" Witt-theorem business, but some of it is (to my mind) much easier to believe.

Yes, for example, Witt's Thm assures us that the leftover, after removing $W\oplus W'$ for maximal totally isotropic $W$ and $W'$ complementing each other, is independent of choices.

The latter bit is perhaps the easiest thing to talk about from a classical-groups viewpoint, but a bit clumsy (so far as I can understand) instrinsically. Anisotropic groups in the Levi component(s) of minimal ($\mathbb R$-) parabolics.

(I must confess that the times I've tried to give courses on "Lie theory" or "algebraic groups" emphasizing "classical groups", I've met considerable resistance from a considerable fraction of the audience, who could not find corroboration for the viewpoint "in the wild".)

After quite a few years of looking at reductive groups from both ends... I do finally think that even the basic facts are fairly well determined by the smallish number of data points we have from the split and quasi-split classical groups. Perhaps more usefully, the phenomena beyond that are easy to exemplify among classical groups, but somewhat clumsy to get-under-the-umbrella of intrinsic description, it seems to me.

(A funny business...)

(To be clearer: A. Weil's "Algebras with involutions and the classical groups"...)

$\endgroup$
3
  • $\begingroup$ When you say that $D = (W \oplus W')^\perp$ is independent of choices, you really mean that its $G$-orbit is well determined, right? From the 'intrinsic' point of view, I think of this as the conjugacy of maximal split tori $S$, with the isometry/similitude group of $D$ being the centraliser of $S$. But maybe I missed your point about what you were seeking to describe intrinsically (or about what counts as intrinsic). $\endgroup$
    – LSpice
    Mar 11 at 0:10
  • $\begingroup$ May I also make a suggestion that looking at reductive groups from both ends must indicate that you have been undertaking a careful study of endoscopy? :-) $\endgroup$
    – LSpice
    Mar 11 at 0:20
  • $\begingroup$ @LSpice, well, both $G$-orbit and isomorphism class without necessarily mentioning $G$-orbits... Many possible logical orderings... (And, yes, haha, just like Johny Carson's mythical "Dr. Al Bendova") $\endgroup$ Mar 11 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.