# Why are coroots needed for the classification of reductive groups?

As we know reductive groups up to isomorphism corresponds to root data up to isomorphism. My question is why in the definition of root data do we need the coroots?

Let's break it down to two questions:

1. Can you give an example of two non-isomorphic reductive groups $$G_1$$ and $$G_2$$ for which one gets the same roots? (I.e., if $$\Phi_1$$ are the roots of the first root datum whose character group is $$X_1$$, and $$\Phi_2$$ is the roots of the second root datum whose character group is $$\Phi_2$$, then there exists and isomorphism $$X_1\rightarrow X_2$$ which reduces to a bijection of $$\Phi_1$$ with $$\Phi_2$$.) If I understand correctly, I don't think that's ever possible if $$G_1$$ and $$G_2$$ are centrally isogenous... And of course a minimal requirement for such an example is that $$G_1$$ and $$G_2$$ have the same rank.
2. Heuristically, what information do the coroots provide?
• The pairs $(X,R)$ attached to $\mathbb{G}_{m}^{r-1}\times PGL_{2}$ and $\mathbb{G}_{m}^{r-2}\times GL_{2}$ are isomorphic. – anon Oct 29 '20 at 0:03

$$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Zent{Z}\newcommand\Q{\mathbb Q}\newcommand\Z{\mathbb Z}$$The collections of roots and the coroots, as abstract root systems, provide the same information (each being recoverable as the dual of the other), which may be why it seems non-obvious that the co-roots matter. The point is that we are given not just $$(R, R^\vee)$$ but $$(X, R, X^\vee, R^\vee)$$, which is to say the way that the root and co-root systems sit inside dual integral lattices; or, to say it differently, we have $$R^\vee$$ not just as an abstract root system, but as a collection of elements of $$\Hom(X, \Z)$$.

For semisimple groups, this, too, is unnecessary: since $$X \otimes_\Z \Q$$ is the $$\Q$$-span of $$R$$, one can recover $$R^\vee$$ inside $$X^\vee \otimes_\Z \Q = \Hom_\Z(X, \Q)$$ in the sense of abstract root systems. (To say it less fancily, we know the pairing of $$R^\vee$$ with $$R$$, hence with $$\Q R = X \otimes_\Z \Q$$.) It is part of the structure theory that the resulting subset of $$X^\vee \otimes_\Z \Q$$ actually lies in $$X^\vee = \Hom_\Z(X, \Z)$$.

I had a hunch that $$(X, R)$$ was really telling us the ingredients of $$(G/\Zent(G), \Zent(G))$$, and that the rôle of additionally specifying $$R^\vee$$ as a subset of $$X^\vee$$ was to specify the particular extension $$1 \to \Zent(G) \to G \to G/\Zent(G) \to 1$$; but, thanks to comments of @JayTaylor and @DavidESpeyer, I realise that I was not quite right (although only finitely wrong, which I guess is a good amount to be wrong). In fact we recover $$(G/\Zent(G)^\circ, \Zent(G)^\circ)$$, and need to specify the extension there.

To be specific, there is a (maybe weakly?) terminal group with a given root system $$(X, R)$$, namely, the group $$G(X \cap \Q R, R) \times D(X/X \cap \Q R)$$, where $$G(X \cap \Q R, R)$$ is the semisimple group with the indicated root system, and $$D(X/X \cap \Q R)$$ is the torus with the indicated character lattice. Given any other group $$G$$ with the same root system, we simply map $$G \to G/\Zent(G)^\circ \times G/[G, G]$$. Now the natural map $$\Zent(G)^\circ \to G/[G, G]$$ is not an isomorphism, but it is an isogeny, and isogenous tori over an algebraically closed field are isomorphic—they have the same rank, and that's all that there is to say about a torus over an algebraically closed field. This is the precise sense in which I say that $$(X, R)$$ ‘exactly knows’ $$(G/\Zent(G)^\circ, \Zent(G)^\circ)$$.

Now to the additional information carried in the co-roots, viewed as elements of $$X^\vee$$. Since there is an almost-direct product decomposition, in the form of a canonical isogeny $$[G, G] \times \Zent(G)^\circ \to G$$, we have that there is also an isogeny $$[G, G] \to G/\Zent(G)^\circ$$. This isogeny is determined by the corresponding map on root data (not systems); and, whereas the root datum of $$G/\Zent(G)^\circ$$ is $$(X \cap \Q R, R, X^\vee/R^\perp, R^\vee)$$, that of $$[G, G]$$ is $$(X/R^{\vee\,\perp}, R, X^\vee \cap \Q R^\vee, R^\vee)$$, where $$R^\perp = \{\lambda \in X^\vee \mathrel: \text{\langle\alpha, \lambda\rangle = 0 for all \alpha \in R}\}$$ and $$R^{\vee\,\perp} = \{\chi \in X \mathrel: \text{\langle\chi, \alpha^\vee\rangle = 0 for all \alpha^\vee \in R^\vee}\}$$. The morphism is the canonical one. We see how what's important is exactly the way that $$R^\vee$$ acts on $$X$$.

To put it more heuristically—at least for my value of ‘heuristic’—what we're really gaining is the ability to see the different ways that the connected centre intersects the derived subgroup. Notice the manifestation of this in @anon's example: for $$\operatorname{GL}_1 \times \operatorname{PGL}_2$$, the connected centre is the first factor and the derived subgroup is the second factor, and they intersect trivially; whereas, for $$\operatorname{GL}_2$$, the connected centre is the subgroup of scalar matrices, which intersects the derived subgroup $$\operatorname{SL}_2$$ in a subgroup of order 2. Specifically, the reason that this is interesting is that the kernel of $$[G, G] \to G/\Zent(G)^\circ$$ is $$\Zent(G)^\circ \cap [G, G]$$; and the character lattice of $$\Zent(G)^\circ \cap [G, G]$$ is $$X/((X \cap \Q R) + R^{\vee\,\perp})$$.

• I have a hunch, but I can't prove it yet, that $(X, R)$ is capturing exactly the pair $(G/\operatorname Z(G), \operatorname Z(G))$—certainly it's capturing *at least* that much, so the question is whether there's any more hidden information—and that specifying how $R^\vee$ sits inside $X^\vee$ is specifying which extension of $G/\operatorname Z(G)$ by $\operatorname Z(G)$ we have in mind. – LSpice Oct 29 '20 at 4:47
• Just to say, your viewpoint of the product map $G_{\mathrm{der}} \times Z^{\circ}(G) \to G$ is essentially the point of view taken in this paper: doi.org/10.1017/S0013091518000597 – Jay Taylor Oct 29 '20 at 10:58
• $X/\mathbb{Z}\Phi = X(Z(G))$ and $X/L = X(Z^{\circ}(G))$ (scheme theoretically). So your decomposition recovers $(G/Z^{\circ}(G),Z^{\circ}(G))$. Which makes sense as you should be able to recover $Z(G)/Z^{\circ}(G)$ from $X$ and $\Phi$. – Jay Taylor Oct 29 '20 at 11:09
• I was thinking about this last night, and I agree. Let $L$ be the lattice $X \cap \mathbb{Q} R$. It seems to me that $(X, L)$ is the root system of $G/Z^{\circ}(G)$ and that $X/L$ is the character lattice of $Z^{\circ}(G)$. As an abelian group, $X \cong L \oplus (X/L)$ so there isn't any more data than how $R$ sits in $L$ and the rank of $X/L$. – David E Speyer Oct 29 '20 at 11:09
• @LSpice Glad it worked, otherwise I would have had to shamefully admit that it's a link to one of my own papers. – Jay Taylor Oct 29 '20 at 14:44

(1) As anon says, an example is $$G_1 = \mathrm{GL}_2$$ and $$G_2 = \mathbb{G}_m \times \mathrm{PGL}_2$$. We can identify the root lattice and co-root lattice with $$\mathbb{Z}^2$$ (with the pairing being the standard dot product) so that the root and coroot systems are $$\Phi_1 = \{ \pm (1,-1) \},\ \Phi_1^{\vee} = \{ \pm (1,-1) \} \qquad \Phi_2 = \{ \pm (1,0) \},\ \Phi_2^{\vee} = \{ \pm (2,0) \}.$$

The automorphism $$(x,y) \mapsto (x,x+y)$$ of $$\mathbb{Z}^2$$ takes $$\Phi_1$$ to $$\Phi_2$$. However, no such automorphism can take $$\Phi_1^{\vee}$$ to $$\Phi_2^{\vee}$$, since the vectors in $$\Phi_2^{\vee}$$ are divisible by $$2$$ and those in $$\Phi_1^{\vee}$$ are not.

(2) Maybe this is too basic but: Fix a maximal torus $$T$$ in $$G$$. There are, up to conjugacy in the source, finitely many maps $$\mathrm{SL}_2 \to G$$ for which the maximal torus of $$\mathrm{SL}_2$$ lands in $$T$$. The coroots, thought of as one parameter subgroups of $$T$$, are the images of the torus of $$\mathrm{SL}_2$$ under those maps.

• Fixed, thank you! – David E Speyer Oct 29 '20 at 15:19
• I think that your last sentence might only identify the coroots up to $\pm$. (For example, there is only one non-trivial map $\operatorname{SL}_2 \to \operatorname{SL}_2$ up to conjugacy in the source, but there are two coroots.) – LSpice Oct 29 '20 at 18:28