# $(X_0,R_0)$ is a root system

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.

• Could you please give a link to the notes you are reading? – Mikhail Borovoi Nov 19 '16 at 14:46
• I think you should take for $X_0$ the character group of $A_0$ rather than $A_0^\bot$. Also it is not clear what you mean by $^t{\rm Diag}(1,-1,1)$, because transposing does not change a diagonal matrix. Also it is not quite clear what you mean by $\chi_1,\chi_2$: what do you mean by "the standard characters on the usual maximal torus of ${\rm GL}_3$"? – Mikhail Borovoi Nov 19 '16 at 14:55
• As Borovoi notes, the formulation needs more precision as well as references. For example, your notions of "algebraic group" and "root system" are fuzzy. The basic ideas here go back to the 1965 paper by Borel-Tits along with Chapter VI of Bourbaki's treatise (which they relied on), but further variants occur in later sources. Also, your tag 'ag' isn't useful here. – Jim Humphreys Nov 19 '16 at 16:21
• Note, incidentally, that the empty set is a perfectly good root system. – LSpice Nov 20 '16 at 14:14

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.

• Thanks for answering. Sorry, what is the anisotropic kernel? Is that related to the maximal anisotropic subtorus of $T$? – D_S Nov 20 '16 at 23:32
• The anisotropic kernel is the complement of $A_0$ in its centralizer in $G$. See section 2.2 of Tits, J. Classification of algebraic semisimple groups. 1966 Algebraic Groups and Discontinuous Subgroups (Proc. Sympos. Pure Math., Boulder, Colo., 1965) pp. 33-62 Amer. Math. Soc., Providence, R.I., 1966 – Friedrich Knop Nov 21 '16 at 7:24
• The (semisimple) anisotropic kernel is the derived group of the centraliser of $A_0$ in $G$. (Maybe this is what you mean by 'complement'?) – LSpice Nov 23 '16 at 16:55
• No, complement means $C_G(A_0)=A_0L$ with $A_0\cap L$ finite. So $L$ is is the derived subgroup of the centralizer times the anisotropic part of its center. This is sometimes called the reductive anisotropic kernel. – Friedrich Knop Nov 23 '16 at 19:05

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 !

• Thank you very much for your answer. I also need to understand the relative root system. – D_S Nov 20 '16 at 23:28
• With the $\perp$ notation, I meant the correspondence between subtori of $T$ and cotorsion free subgroups of $X = X(T)$. So $$X_0 = A_0^{\perp} = \{ \chi \in X : \chi(t) = 1 \textrm{ for all } t \in A_0 \}$$ and we can recover $A_0$ as $$X_0^{\perp} = \bigcap\limits_{\chi \in X_0} \textrm{Ker } \chi$$ – D_S Nov 20 '16 at 23:30
• Is the Weyl group invariant inner product unique up to scaling? I thought that was the case if and only if $G$ is semisimple and the root system is irreducible – D_S Nov 21 '16 at 22:08
• Indeed, sorry for the inaccuracy. I've edited accordingly. – thierry stulemeijer Nov 23 '16 at 10:14