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Let $G$ be a connected, reductive group over an algebraically closed field $k$. Let $B$ be a Borel subgroup with maximal torus $T$ and unipotent radical $U$. Let $\Phi^+ = \Phi(B,T)$ and $\Delta$ the base of $\Phi = \Phi(G,T)$ corresponding to $\Phi^+$.

A subset $\Psi$ of $\Phi^+$ is called closed if whenever $\alpha, \beta \in \Psi$, and $\alpha + \beta$ is a root, we have $\alpha + \beta \in \Psi$. If $\Psi$ is closed, then the root subgroups $U_{\alpha} : \alpha \in \Psi$ directly span a closed, connected subgroup $U_{\Psi}$ of $U$ which is normalized by $T$ and whose Lie algebra is $\bigoplus\limits_{\alpha \in \Psi} \mathfrak g_{\alpha}$.

Furthermore, if $\alpha \in \Phi^+$, and $\Psi \subseteq \Phi^+$ is closed, and all roots of the form $a \alpha + b \Psi$ for $a, b \in \mathbb{Z}^+$ lie in $\Psi$, then the root subgroup $U_{\alpha}$ normalizes $U_{\Psi}$. These facts are proved in Chapter 14 of Borel, Linear Algebraic Groups.

In particular, let $\Psi = \Phi^+ - \Delta$. It is easy to see that $\Psi$ is closed and normalized by every root subgroup $U_{\alpha} : \alpha \in \Phi^+$, and in particular, $U_{\Phi^+ - \Delta}$ is a normal subgroup of $U$. Moreover, it is a consequence of 8.32 in Springer's Linear Algebraic Groups that for any $x \in U_{\alpha}$ and $y \in U_{\beta}$, the commmutator $xyx^{-1}y^{-1}$ lies in $U_{\Phi^+ - \Delta}$. From here one can argue that $U_{\Phi^+ - \Delta}$ contains the derived group of $U$ by producing a homomorphism $U \rightarrow \prod\limits_{\alpha \in \Delta} \mathbf G_a$ with kernel $U_{\Phi^+ - \Delta}$.

My question is, is $U_{\Phi^+ - \Delta}$ exactly the derived group of $U$? For $G = \textrm{GL}_n$, this does seem to be the case.

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    $\begingroup$ There are exotic commutations among root groups in characteristics 2 and 3 for types B$_2$=C$_2$ and G$_2$, so probably one cannot expect a purely combinatorial recipe for computing the derived group of $U$ that works in all characteristics. In particular, you should check ${\rm{Sp}}_4$ and ${\rm{G}}_2$ directly (using the Chevalley commutation relations, which are given in Humphreys' book on linear algebraic groups, as well as in SGA3) with special attention to characteristics 2 and 3 before contemplating the general case. The rank-2 case is the essential case to understand. $\endgroup$ – nfdc23 Jun 20 '17 at 23:45
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    $\begingroup$ You gave no motivation, but I conjecture that what you seek (for which the above is unnecessary) is that $U$ has a canonical $B$-equivariant composition series (so independent of $T$!) with each successive quotient a vector group having a unique $B$-equivariant linear structure. This is true even for parabolic subgroup schemes of any reductive group scheme over any base scheme: see SGA3 XXVI 2.1, or Theorem 5.4.3 of the article Reductive Group Schemes in smf4.emath.fr/Publications/PanoramasSyntheses/2014/42-43/html/… (for a simpler proof via dynamic arguments). $\endgroup$ – nfdc23 Jun 21 '17 at 0:45
  • $\begingroup$ Thank you as always for your informative answers and references. Actually, my question was more motivated by representation theory: for $G$ quasisplit over a $p$-adic field $k$ and a choice of splitting $x_{\alpha}: \mathbf G_a \rightarrow U_{\alpha} : \alpha \in \Delta(B,T)$ such that $\sigma.x_{\alpha} = x_{\sigma.\alpha}$ for all $\sigma \in \textrm{Gal}(\overline{k}/k)$, the choice of an additive unitary character $\psi$ of $k$ allows you to define a unitary character $\chi: U(k) \rightarrow S^1$ via a composition of homomorphisms $\endgroup$ – D_S Jun 21 '17 at 1:17
  • $\begingroup$ $$U\rightarrow \prod\limits_{\alpha \in \Delta} U_{\alpha} \rightarrow \mathbf G_a$$ where the right most map is $(x_{\alpha}) \mapsto \sum\limits_{\alpha} x_{\alpha}$. In order to define the left map as a homomorphism of algebraic groups, one also needs a choice of root vectors $x_{\alpha} : \alpha \in \Phi^+ - \Delta$, and project. Although neither homomorphism is defined over $k$, the composition is.The existence of such a homomorphism with kernel $U_{\Phi^+ - \Delta}$ implies that $U_{\Phi^+ - \Delta}$ must contain the derived group of $U$, so this is what naturally led to my question. $\endgroup$ – D_S Jun 21 '17 at 1:20
  • $\begingroup$ But now I see what I'm doing is a special case of what you said, since the choice of a "$k$-splitting" gives the quotient group $U/U_{\Phi^+ - \Delta}$ the structure of a finite dimensional $k$-vector space. However, it's fine for me (at least for the near future) that this appears to depend on $T$. $\endgroup$ – D_S Jun 21 '17 at 1:26
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I think the answer is yes.

We can assume that $G$ is semisimple. As you have seen already, and which is is clear from the commutation relations, we have $[U, U] \leq \prod_{\alpha \in \Phi^+ - \Delta} U_\alpha$. I guess you could show equality by using the Chevalley commutator relations, although you need to be careful when $p$ is small (e.g. $G$ of type $B_n$ and $p = 2$).

Here is an idea for a different way to see this. The containment $[U, U] \leq \prod_{\alpha \in \Phi^+ - \Delta} U_\alpha$ shows that $$\dim U / [U, U] \geq \dim U / \prod_{\alpha \in \Phi^+ - \Delta} U_\alpha = \operatorname{rank} G$$

So it would be enough to show that $\dim U/[U,U] \leq \operatorname{rank} G$.

For this, a result of Steinberg states that there exists a regular unipotent element $u \in G$, which has the property that $\dim C_G(u) = \operatorname{rank} G$. One can show that $u$ is contained in a unique Borel subgroup $B$ of $G$, and that $C_G(u) = C_B(u) = C_U(u)Z(G)$, where $U$ is the unipotent radical of $B$. Proofs can be found in reference (*) below.

In any case, we have a unipotent element $u \in U$ such that $\dim C_U(u) = \operatorname{rank} G$. The conjugacy class $\operatorname{Cl}_U(u)$ of $u$ in $U$ is contained in the coset $u[U,U]$ (proof: $g^{-1}ug = u[u,g]$), so $$\dim \operatorname{Cl}_U(u) = \dim U - \dim C_U(u) \leq \dim [U,U]$$

giving $\dim U/[U,U] \leq \operatorname{rank} G$.

(*) See Chapter 4, sections 4.1 - 4.5 in

Humphreys, James E.: Conjugacy classes in semisimple algebraic groups. Mathematical Surveys and Monographs, 43. American Mathematical Society, Providence, RI, 1995.

or:

Theorem 3.1 in:

Steinberg, Robert: Regular elements of semisimple algebraic groups. Inst. Hautes Études Sci. Publ. Math. No. 25 1965 49–80.

and Lemma 4.3 in:

Springer, T. A.: Some arithmetical results on semi-simple Lie algebras. Inst. Hautes Études Sci. Publ. Math. No. 30 1966 115–141.


PS. For further terms of the derived series and the lower central series, you can take a look at the following paper, where they are described when $p$ is not too small for $G$.

Azad, H.; Barry, M.; Seitz, G.: On the structure of parabolic subgroups. Comm. Algebra 18 (1990), no. 2, 551–562.

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    $\begingroup$ Just to point out that this appears as Proposition 14.17 in Digne--Michel's book "Representations of Finite Groups of Lie Type". The argument they give is along the same lines as yours. $\endgroup$ – Jay Taylor Jun 21 '17 at 13:06
  • $\begingroup$ @JayTaylor: Thanks Jay! I was pretty sure this must be known, but I was never able to find a reference. $\endgroup$ – Mikko Korhonen Jun 21 '17 at 13:21
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Let me add a few comments in community-wiki format. There doesn't seem to be a convenient reference, apart from the one in Digne-Michel which Jay Taylor cites. But even here, the authors don't give a full proof of the existence of regular unipotent elements. Such regular elements and their properties are essential to the approach Mikko gives, though it's possible that there is a more elementary method yet to be found. What's clear is that a considerable amount of structure theory for semisimple groups is involved. (Of course, in characteristic 0 one can instead work more straightforwardly in the Lie algebra.)

Steinberg's treatment of regular elements (IHES, 1965) is available online through numdam.org, as is Springer's article (IHES, 1966). But note that Springer didn't succeed for bad primes in arriving at a proof of existence for regular unipotents. What he did was more direct than Steinberg's method, relying mainly on Chevalley's basis and commutation formula. Later on, students of Steinberg pushed this technique further for bad primes, but it's unclear how to extract a uniform theoretical approach. (Recently I wrote up some notes attempting to sort out the arguments used for both regular unipotents and regular nilpotents, posted here.)

As nsfc23 comments, it's tricky to work directly with Chevalley's commutator formula in some small characteristic cases. On the other hand, Steinberg's approach to regular unipotents requires fairly heavy machinery. As to finiteness of the number of unipotent clases (still conjectural in the mid-1960s), it remains an open problem to use modular representation theory of $G$ or its Lie algebra to get a more self-contained proof.

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  • $\begingroup$ Jim, this is just to say that Digne--Michel do give a full proof of the existence of regular unipotent elements, at least over $\bar{\mathbb{F}}_p$. This is Theorem 14.13. They present Lusztig's argument showing that there are finitely many unipotent classes and then deduce the existence of regular unipotent elements from this. In fact, as is mentioned in the introduction to Lusztig's paper, if one knows that there are finitely many unipotent classes over $\bar{\mathbb{F}}_p$ then one knows this over any algebraically closed field $K$ of characteristic $p$. Thus their proof works over $K$. $\endgroup$ – Jay Taylor Jun 22 '17 at 6:54
  • $\begingroup$ @Jay: What I wrote is probably too brief, but I agree that D-M have an existence proof for regular unipotents using the finiteness of the number of unipotent classes (for $p>0$). However, it's debatable whether they cover all the background for Lusztig's argument (foundations of etale cohomology, etc.). Anyway I'm still skeptical about using all this machinery to deal with the derived group of $U$, which should be more elementary. [P.S. Your reference to D-M 14.17 in a comment actually answers the question asked here apart from the method they use.] $\endgroup$ – Jim Humphreys Jun 22 '17 at 18:38
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    $\begingroup$ @JimHumphreys: I guess you know this, but regarding the last paragraph in your answer: I think that Steinberg's 1965 paper does not need finiteness of unipotent classes to prove the existence of a regular unipotent element. It takes some work, but does not need lots of machinery. $\endgroup$ – Mikko Korhonen Jun 22 '17 at 20:23
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    $\begingroup$ @Mikko: Yes, I've edited accordingly. I still expect to locate a more elementary proof in D_S's situation, preferably without case-by-case study of commutators or such. [It's amusing, by the way, that in his ICM Moscow 1966 talk, Steinberg himself regarded the finiteness of the number of unipotent clases to be a fairly easy problem. He had good instincts in most cases but underestimated the difficulty or was too optimistic about the correctness of some of his "problems" there.] $\endgroup$ – Jim Humphreys Jun 22 '17 at 21:01

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