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Let $U$ be a smooth connected unipotent algebraic group over an algebraic closure $k$ of $\mathbf{F}_p$. Let $U' := [U,U]$ be the derived (= commutator) subgroup, and assume it is central in $U$. Let $V$ be a subgroup of the abstract group $U(k)/U'(k) = (U/U')(k)$.

Is there a subgroup $H$ of $U(k)$ such that

(a) $\pi(H) = V$,

(b) $H\cap U' =[H,H]$?

Note that $H\cap U'$ must contain $[H,H]$, which depends only on $V$ due to the centrality of $U'$ in $U$. So if we define the subgroup $\widetilde{V} = \pi^{-1}(V)$ in $U(k)$ then the question is really asking if $\widetilde{V}$ contains a subgroup $H$ that maps onto $V$ and meets $U'$ in precisely the commutator subgroup $[\widetilde{V}, \widetilde{V}]$.

The case of interest is $U = \mathcal{U}/[\mathcal{U},\mathcal{D}(\mathcal{U})]$, where $\mathcal{U}$ is the unipotent radical of a Borel subgroup of a connected semisimple $k$-group.

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No, V is just an arbitrary subgroup of U(K)/[U(K),U(K)]. H and V are arbitrary subgroups, not algebraic groups. When V is a "product of subgroup of roots", I know my conclusions; unfortunately, that situation is not sufficient for my purposes (i.e., I cannot assume it). It will not kill me if you assume that U is the unipotent radical of a Borel subgroup of a connected semisimple group. Just assume everything lives inside GL_n(F_q) in one way or the other. –  H A Helfgott Jul 6 '10 at 16:43
    
Sure. And yes, you have guessed completely right - the U we are discussing here really comes from U/U^2, where U is the unipotent radical of a Borel. –  H A Helfgott Jul 6 '10 at 16:56
    
Harald, here is an elementary test case. Consider a non-commutative central extension $U$ of $\mathbf{G}_ a$ by either $\mathbf{G}_ a$ (if any exist!) or by $\mathbf{G}_ a \times \mathbf{G}_ a$. (The commutator subgroup must be $(\mathbf{G}_ a)^2$ in the latter case is no examples of the former exist.) Taking $V$ to be a cyclic subgroup of order $p$, we get $[\widetilde{V}, \widetilde{V}] = 1$ and so you'd be asking if every nontrivial element of $U(k)/U'(k)$ lifts to a $p$-torsion element of $U(k)$. That sounds dubious in general, no? Could such $V$ be relevant to you? –  BCnrd Jul 6 '10 at 20:05

1 Answer 1

Here is an example of a "2-step nilpotent" unipotent group $E_1$ for which the quotient $V=E_1/[E_1,E_1]$ is a vector group (so every element of $V(k)$ is $p$-torsion) but for which not every element of $V(k)$ may be lifted to a $p$-torsion element of $E_1(k)$.

This confirms BCnrd's skepticism in one of the comments. I admit that this example may well not arise as a subgroup of $U/[U,[U,U]]$ for the unipotent radical $U$ of a Borel (though I don't see precisely how to argue that it doesn't).

To form $E_1$, I want to construct an extension of a vector group by the additive group $\mathbf{G}_a$ using the sum of two 2-cocycles. I'll first describe each of these separately.

First, recall that the additive group of length 2 Witt vectors $W_2$ is a self-extension of $\mathbf{G}_a$ $$0 \to \mathbf{G}_a \to W_2 \to \mathbf{G}_a \to 0$$ defined by a certain 2-cocycle $\sigma$. Addition is given by $(a_0,a_1) + (b_0,b_1) = (a_0 + b_0,\sigma(a_0,b_0) + a_1 + b_1)$ where $\sigma(X,Y) = \dfrac{1}{p}(X^p + Y^p - (X+Y)^p) \in \mathbf{Z}[X,Y]$.

Next, let $V$ be a finite dimensional $k$ vector space with $\dim V \ge 2$, viewed as a vector group over $k$. Let $\beta$ be a non-deg alternating form on $V$. Then $\beta$ defines a non-commutative central extension $$1 \to \mathbf{G}_a \to E \to V \to 1$$ which I'll write multiplicatively: the operation will be given by $(v,a)\cdot(w,b) = (v+w,\beta(v,w) + a + b)$. (I'm identifying the variety $E$ with $V \times \mathbf{G}_a$).

The extension I want is a hybrid. Fix a non-zero linear functional $\phi:V \to k$ and consider the extension $$1 \to \mathbf{G}_a \to E_1 \xrightarrow{\pi} V \to 1$$ with operation given by $(v,a)(w,b) = (v+w,\sigma(\phi(v),\phi(w)) + \beta(v,w) + a + b)$.

Since $\beta$ is non-degenerate we have $E_1/[E_1,E_1] \simeq V$. If $L \subset V$ is a line for which $\phi(L) \ne 0$, the subgroup $\pi^{-1}(L) \subset E_1$ is isomorphic to $W_2$. In particular, any element $(v,a) \in E_1(k)$ with $\phi(v) \ne 0$ has order $p^2$.

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Nice! Via gp Ex($V$, $W$) of central ex'tns, you pull back Witt class in Ex($\mathbf{G}_ a$, $\mathbf{G}_ a$) along $\phi:V \rightarrow \mathbf{G}_ a$ in 1st variable to get class in Ex($V$, $\mathbf{G}_ a$), which you add to class defined by $\beta$ to get the central ext'n $E_1$. To get non-commutativity (so $[E_1, E_1]$ as expected), need $\beta$ non-symmetric rather than alt'ing, so take $p > 2$ and $\beta \ne 0$. To get $\pi^{-1}(L) = W_2$ as central ext'n want $\beta$ alternating, but full force of non-degeneracy isn't needed. Anyway, let's see what Harald has to say. –  BCnrd Jul 8 '10 at 3:07

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