# Unipotent linear algebraic groups

Let $U_1$ be a unipotent group inside some Chevalley group $G$. For now, think of $G$ as being $SL_n(K)$ where $K$ is a field; then we can take $U_1$ to be a bunch of strictly upper triangular matrics. Assume if you like that $K$ is algebraically closed.

Now suppose that $U_1$ is normalized by a non-trivial torus $T_1$. Are there any general statements that can be made about the structure of $U_1\ ?$

For instance: let us assume that $T_1$ is $1$-dimensional, as this is the limiting case. I suspect that the following is true: if $r(t)$ is not equal to $s(t)$ for all positive roots $r,s$, and all elements $t$ in $T_1$, then $U_1$ is a product of root subgroups.

I haven't written down a proof of this statement, but doodling suggests that it is true! Indeed I suspect it is true of $T_1$ contains ANY element $t$ satisfying the given condition. I would like a more general statement though: covering the case where $r(t)=s(t)$ for particular positive roots $r$ and $s$.

I should note that I tend to automatically ignore small characteristic cases! Funny things can happen in this situation...

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I expanded my answer. Hope it's clearer now, if not ask. – user175348 Dec 12 '09 at 23:44

Let U be a smooth connected unipotent group over an arbitrary field k, and let T be a k-split k-torus equipped with a left action on U such that the T-action on Lie(U) contains no occurrence of the trivial weight. In the "classical" case when weight spaces are 1-dimensional and no two distinct weights are positive rational multiples of each other then indeed U as a scheme is a product of "root groups" (under multiplication, in whatever order one wishes). More generally, U is a T-equivariant direct product of groups U_i which in turn admit a T-equivariant composition series {U_{ij}} whose successive quotients U_{ij}/U_{i,j+1} are vector groups admitting a unique linear structure relative to which T acts through a single character (increasing positive integral multiples of some fixed nontrivial character of T depending on i). In the "classical" case this gives that each U_i is G_a (as a k-group), equipped with a linear T-action. In the general case it follows that U is necessarily k-split as a unipotent k-group (i.e., this didn't have to be assumed at the outset, such as if k is imperfect).

For a proof, resting on some elegant ideas of Gabber, see Prop. 3.3.6, Lemma 3.3.8, and Theorem 3.3.11 in the book "pseudo-reductive groups". (In Theorem 3.3.11, take the \Psi_i there to be weights lumped according to being positive rational multiples of each other in X(T)_Q.)

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@Brian: You should probably register your account. Also, welcome MO. – Harry Gindi Feb 10 '10 at 9:17

Assume characteristic 0. I do not know how much of this extends to finite characteristic.

Let $\mathbf u$ be the Lie algebra of the unipotent subgroup $U$, and $\mathbf t$ that of the torus (1- dimensional or not, it doesn't matter).

Define $\Delta(\mathbf g,\mathbf t)$ as the sets of roots of $\mathbf g$ w.r.t. $\mathbf t$ (the usual definition is fine, even if $\mathbf t$ is not maximal, however the root spaces will in general not be 1-dimensional). Let $C$ denote the centralizer.

Then you have $\mathbf u=C_{\mathbf u}(\mathbf t)\oplus\sum \mathbf u_\alpha$ for $\alpha\in\Delta(\mathbf g,\mathbf t)$. Here $\mathbf u_\alpha=\mathbf u\cap\mathbf g_\alpha$ or equivalently the set {$X\in\mathbf u\mid [H,X]=\alpha(H)X \forall H\in\mathbf t$}.

Let now $\mathbf t_{max}$ be a maximal torus containing $\mathbf t$, and $\Delta(\mathbf g,\mathbf t_{max})$ the corresponding root system (this is the "usual" root system). An element $T$ of $\mathbf t_{max}$ is called regular if $\alpha(T)\neq\beta(T)$ and $\alpha(T)\neq 0$ for all roots $\alpha\neq\beta\in\Delta(\mathbf g,\mathbf t_{max})$.

If the torus $\mathbf t$ contains a regular element $T$, the roots w.r.t. $\mathbf t$ are in bijection with those w.r.t. $\mathbf t_{max}$, and in particular the root spaces are 1-dimensional. It follows that if $\mathbf u_\alpha\neq 0$ then $\mathbf u_\alpha=\mathbf g_\alpha$, and $\mathbf u$ is a sum of root spaces.

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I'm sorry, I don't understand Andrea's comment. Please explain some more! – Nick Gill Dec 1 '09 at 13:04

If $T_1$ is something like having first entry on the diagonal in $GL_n(k)$ allowed to be anything in the non-zero elements of the field $k$, i.e. $k^*$, with the remaining diagonal elements $1$, then the normaliser is $GL_{n-1}$. Now all the roots coming from this $GL_{n-1}$ all have $r(t)=s(t)=0$ for all $t\in T_1$. And so you can have any old unipotent group in $GL_{n-1}(k)$ you like. This is a hard (even wild?) problem, and is akin (worse if not assuming $U$ is connected) to classifying all $p$-groups, at least in characteristic $p$. In characteristic $0$, it surely can't be much better.

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