6
$\begingroup$

Let $k$ be an algebraically closed field of positive characteristic and let $G$ be a connected split reductive group. We know $G$ is the product of its center $Z(G)$ and derived group $[G, G]$ and $[G, G]$ is semisimple so it's a product of simple algebraic groups $G_1, \ldots, G_n$ corresponding to the decomposition of it's root space. So $G = G_1\cdots G_nZ(G)$.

My questions are about the Lie algebras of these groups.

  1. Is it true that $\mathrm{Lie}([G, G]) = \mathrm{Lie}(G_1) \times \cdots \times \mathrm{Lie}(G_n)$?

  2. Is it true that $\mathrm{Lie}(G) = \mathrm{Lie}([G, G]) \times \mathrm{Lie}(Z(G))$?

I think the first is true. The dimension is right and the $G_i$ commute with each other so their Lie algebras should commute as well. I'd like the second to be true as well for basically the same reason.

I'm a little worried that maybe something funny can happen due to the products not being direct products (there can be a finite intersection which if taken scheme theoretically might have a Lie algebra?) and also maybe there should be a condition on the characteristic of the field. All the references that discuss these structure theorems (I have Malle and Testerman and all the LAG's: Humphreys, Borel, Springer) never talk about what this means for the Lie algebras. I'd be happy with a straight up answer or a reference if anyone knows of one.

$\endgroup$
  • $\begingroup$ What if you take the algebraic simply connected cover of the neutral component of $G$? $\endgroup$ – Orest Bucicovschi Feb 6 '15 at 20:24
  • $\begingroup$ @orangeskid: By neutral component, do you mean the connected component containing the identity? I meant to assume that $G$ is connected, I'll edit that into the question. $\endgroup$ – Jim Feb 6 '15 at 21:48
  • $\begingroup$ Oh, OK, so $G$ connected, right. So my guess is that in the case of simply connected groups the decomposition is direct. That may not be true in positive characteristic, I haven't thought about it. $\endgroup$ – Orest Bucicovschi Feb 6 '15 at 23:48
  • $\begingroup$ Well, a direct product of simply connected groups is simply connected, so by the classification will be the unique simply connected group of that type. And if the product is direct then it's definitely true that the Lie algebra is the product of the Lie algebras. So I think you're right about the simply connected groups and I think it doesn't depend on the characteristic. $\endgroup$ – Jim Feb 7 '15 at 3:11
  • $\begingroup$ This is a good question. I felt like this was something that had to be mentioned in Jantzen, but I could not find anything quite like it, even when we replace the Lie algebra by the algebra of distributions. That it does not seem to be mentioned there makes me believe that it probably fails, but it might just be that it was not a useful thing for the purposes of representations. $\endgroup$ – Tobias Kildetoft Feb 10 '15 at 8:39
0
$\begingroup$

Edit: There are some issues with the ideas in this answer discussed in the comments.

As was already discussed in the comments, 1. is clear if $[G,G]$ is a direct product of simple algebraic groups, in particular, if $[G,G]$ is simply connected.

Note that for any semisimple group $G$, we have an isogeny $\pi : G_{sc} \rightarrow G$ inducing an isomorphism on the Lie algebras (this is Proposition 9.15 in Malle-Testerman), so $\text{Lie}(G) \cong \text{Lie}(G_{sc})$.

We have $[G,G] = G_1\cdots G_n$ a product decomposition into simple algebraic groups $G_i$ and consequently $[G,G]_{sc} = (G_1)_{sc} \times \cdots \times (G_n)_{sc}$. Using the isomorphisms between Lie algebras induced by $[G,G]_{sc} \rightarrow [G,G]$ and $(G_i)_{sc} \rightarrow G_i$ we get $$ \text{Lie}([G,G]) \cong \text{Lie}([G,G]_{sc}) \cong \prod_{i = 1}^n \text{Lie}((G_i)_{sc}) \cong \prod_{i = 1}^n \text{Lie}(G_i)$$ and this gives 1.

For any connected reductive $G$ we have a surjective morphism $[G,G] \times Z(G) \rightarrow G$ of algebraic groups given by multiplication. The kernel of this morphism is finite, hence its Lie algebra is trivial (e.g. Thm. 7.4 a in Malle-Testerman) giving an isomorphism of Lie algebras (by Thm. 7.9 in Malle-Testerman) $$\text{Lie}(G) \cong \text{Lie}([G,G] \times Z(G)) \cong \text{Lie}([G,G]) \times \text{Lie}(Z(G)).$$

$\endgroup$
  • $\begingroup$ Proposition 9.15 is actually missing an assumption, you need that $p$ doesn't divide the order of the fundamental group of $G$. For example the isogeny $\mathrm{SL}_2(k) \to \mathrm{PGL}_2(k)$ does not induce an isomorphism of Lie algebras when the characteristic of $p$ is $2$. $\endgroup$ – Jim Feb 9 '15 at 18:50
  • $\begingroup$ Also I'm not sure about your argument for $Z(G) \times [G, G]$. Take for example $G = \{(a, b, c) \ | \ a^p = b + c\}$ and assume $k$ is algebraically closed; then there is an isogeny $\mathbb G_a \times \mathbb G_a \to G$ given by $(x, y) \mapsto (x + y, x^p, y^p)$. The kernel of this isogeny is finite (it's $0$ actually) but the map on Lie algebras is $(v, w) \mapsto (v + w, 0, 0)$ so the map on Lie algebras has kernel $(v, -v)$. $\endgroup$ – Jim Feb 9 '15 at 18:54
  • $\begingroup$ I see your first point. For the second one, I believe the main difference in you example and the situation above is that the induced quotient map is not an isomorphism of algebraic groups. But if I recall correctly, then quotient maps induced by multiplication always induce isomorphisms (you need some kind of separability argument there, but I do not know enough geometry to make this precise). $\endgroup$ – Matthias Klupsch Feb 10 '15 at 7:03
  • $\begingroup$ What do you mean by "quotient maps induced by multiplication"? I think my counterexample can be a multiplication map: I could define closed subgroups $H = \{(x, x^p, 0)\}$ and $K = \{(x, 0, x^p)\}$ and then look at the multiplication map $H \times K \to G$. It's again a bijection but the map on Lie algebras still has a kernel. $\endgroup$ – Jim Feb 10 '15 at 7:09
  • $\begingroup$ Yes, you are right. I really thought I knew what was going on here but now it seems I did not at all. Sorry for that. $\endgroup$ – Matthias Klupsch Feb 10 '15 at 7:54

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.