$(X_0,R_0)$ is a root system

Question

I asked this question on stackexchange (https://math.stackexchange.com/questions/1964415/maximal-k-split-torus-and-the-weyl-group-why-is-this-a-root-system), but did not get an answer.

$G$ is a semisimple algebraic group defined over a field $F$, maximally split maximal torus $T$ defined over $F$ and maximal $F$-split subtorus $A_0 \subseteq T$. $R$ is the set of roots of $T$ in $G$, $X$ is the group of characters of $T$, and $X_0 = A_0^{\perp}$. The notes I'm reading claim that $R_0 := R \cap X_0$ is such that $(\mathbb{Q} \otimes_{\mathbb{Z}} X_0,R_0)$ is a root system. Moreover, it is claimed that the $\mathbb{Q}$-span of $R_0$, intersected with $X$, is equal to $X_0$.

I haven't been able to verify these claims. In fact, I believe that $R_0$ may even be the empty set even when $X_0 \neq 0$. Here is the example I used:

Example: $F = \mathbb{R}$, $G$ is the derived group of $U(2,1)$, where $U(2,1)(\mathbb{C}) = \textrm{GL}_3(\mathbb{C})$, and $U(2,1)(F) = \{ x \in \textrm{GL}_3 : w \space ^t\overline{x}^{-1}w = x \}$. Here $w = \space ^t\textrm{Diag}(1,-1,1)$, and bar denotes element wise complex conjugation. Then

$$ T = \{ \begin{pmatrix} a & & \\ & b & \\ & & \frac{1}{ab} \end{pmatrix} \}$$ is a maximal torus of $G$, defined over $F$, and I believe

$$A_0 = \{ \begin{pmatrix}x & & \\ & 1 & \\ & & \frac{1}{x} \end{pmatrix} \}$$ is a maximally split $F$-split subtorus of $G$, contained in $T$. We can take $\chi_1, \chi_2$ as a basis of $X$ by restricting the standard characters on the usual maximal torus of $\textrm{GL}_3$. Then $$X_0 = \{ n \chi_2 : n \in \mathbb{Z} \}$$

$$R = \{ \pm (\chi_1 - \chi_2), \pm(2 \chi_1 + \chi_2), \pm (\chi_1 + 2\chi_2) \}$$ and we see that $R \cap X_0$ is empty. $\blacksquare$

Is my example incorrect? This fact about $(X_0, R_0)$ being a root system is used in the notes I'm reading to prove facts about a "relative root system" inside $X/X_0$. But it seems to be like $(X_0,R_0)$ need not be a root system.

Answer 1

I think the writer of the notes is mostly right. The sets $X_0$, $R_0$ consists of all weights, roots which restrict to $0$ on $A_0$. So $(X_0,R_0)$ is indeed a root system, namely that of the anisotropic kernel, say $L$, of $G$ but in the more general sense that $R_0$ is not required to rationally span $X_0$. That would be the case if and only if $L$ is semisimple, as well. In your example $L$ is a $1$-dimensional anisotropic torus so $R_0$ is empty while $X_0\cong\mathbb Z$. In general, the center of $L$ comes from conjugate non-compact simple roots in the Satake-Tits diagram of $G$. So $L$ is semisimple if e.g. $G$ is of inner type. Maybe that's what the author had in mind.

Answer 2

EDIT : After the answer of Friedrich Knop, I see that his interpretation is surely the right one (the notation $R_0$ for "roots restricting to $0$" is an indication). What I was describing was in fact the relative root system, and I've now edited my answer to describe both root systems (the one coming from the anisotropic kernel denoted $(X_0,R_0)$ by the OP, and the relative one).

Let me give a try at clearing confusion and notation (I make guesses from what it should be and from what the notation hints at). First $A_0$ is a maximal $F$-split torus and $T$ is a maximal torus containing it (I don't know what "maximally split maximal torus" means). Since $A_0\hookrightarrow T$, you have dually $\pi \colon X(T)\twoheadrightarrow X(A_0)$ (where $X(\cdot)$ denotes the group of characters over a splitting field for $T$). Your $X_0$ should then really be $\text{ker }\pi$, and $R_0$ should be $\text{ker }\pi \cap R$ (well, given the definition of $A_0^{\perp}$ in the comments, that is exactly the definitions the OP give in the question, sorry for the redundancy).

Note also that $A_0^{\perp}$ does not mean much (to me, but see the comments), but here is a guess for what this notation hints at. Let $V = X(T)\otimes \mathbf{Q}$ and $V_0 = X(A_0)\otimes \mathbf{Q}$. Using the map $\pi$, $V_0$ is naturally seen as a quotient of $V$, say $p\colon V\twoheadrightarrow V_0$. With these notations, your $\mathbf{Q}\otimes X_0$ is just $\text{ker }p$, and $R_0$ becomes $R\cap \text{ker }p$. Strangely, we like to denaturalize this situation, and identify $V_0$ as a subspace of $V$. To do this, a standard procedure is to choose an inner product on $V$, and then identify $V_0$ with $(\text{ker }p)^{\perp}$. Which inner product ? Well, just take a (essentially unique up to scaling, see the comments) Weyl invariant inner product. In the end, we indeed recover $\mathbf{Q}\otimes X_0 = V_0^{\perp} $

Finally, let me clear notations for your example (as asked by commenters) and check what the above discussion gives: to define $SU(2,1)$, it is usual to proceed as you do, but taking $w$ to be the anti-diagonal matrix having coefficients $1$ on the anti-diagonal ($w$ has determinant $-1$, so conjugation by $w$ preserves the determinant). From what you describe, it seems clear that you define $\chi_i (\text{Diag}(a_1,a_2,\frac{1}{a_1a_2})) = a_i$ (for $i=1,2$). Fortunately, taking $\lbrace \chi_1,\chi_2\rbrace$ to be an orthonormal basis is a Weyl invariant inner product ! Finally, the set of roots $R$ is indeed as you describe it, and $R_0$ is indeed empty (as pointed out by Friedrich Knop, that's as it should be).

Going further, we can also describe the relative root system $Q$ to be the projection of $R$ on $ V_0 = (\text{ker }p)^{\perp} $. We get the root system $$Q=\lbrace \pm\chi_1,\pm2\chi_1\rbrace $$ This is the non-reduced root system $BC_1$, as it should be !