# Tori acting on vector spaces

Let $T$ be a torus defined over a field $K$ of characteristic $p>0$. Suppose that $T$ acts (algebraically) on some vector space $V$ (over the same field $K$). Let $W$ be a subspace of $V$. Now consider $N_T(W)$, the normalizer in $T$ of $W$. I would like to be sure that this is a closed connected subgroup of $T$; i.e. I'd like it to be a torus. Is this true in general?

Here's a more specific question (supposing that the first question is too hard, or is not true in general). Consider $G=UT$ a solvable linear algebraic subgroup of $GL_r$. Suppose that $U_1$ is some subgroup of $U$. I'd like to be sure that $N_T(U_1)$ is a subtorus of $T$. I'm most interested in the situation when $p$ is large (I can imagine that this statement may only be true for $p>r$, say, but this is fine for the application I have in mind.)

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What do you mean by normalizer? The subgroup of elements in $T$ which map $W$ into itself? – Mariano Suárez-Alvarez Jul 26 '10 at 7:12
If so, then if $\{w_1,\dots,w_n\}$ is a basis of $W$, the condition that an element $t\in T$ preserve $W$ is equivalent to having the rank of the matrices $\{w_1,\dots,w_n,t\cdot w_i\}$ equal to $n$, for each $i\in\{1,dots,n\}$: in terms of $t$, this is a polynomial condition. – Mariano Suárez-Alvarez Jul 26 '10 at 7:16
No. Use scheme-theoretic normalizer: functor whose value on $K$-alg. $A$ is set of $t \in T(A)$ carrying $W \otimes_K A$ into itself. By Def. A.1.9ff in "Pseudo-reductive groups", rep'td by closed $K$-subgp scheme $N_T(W)$ in $T$ which acts on $W$ via auts; question is whether it's smooth & conn'd. Can assume $K$ alg. closed and $W \ne 0$. Passing to dim($W$)-th exterior powers, can assume $W$ is a line in $V$. Express basis of $W$ in terms of $T$-eigenbasis of $V$, so $N_T(W)$ is defined by equating some $T$-weights in $V$. Thus, $N_T(W)$ is often disconn'd and non-smooth, for any $p$. – BCnrd Jul 26 '10 at 7:57
Moreover, $N_T(W)$ can be disconnected in characteristic zero. Take $\mathbb{G}_m$ acting on $\mathbb{A}^2$ by $t : (x,y) \to (tx, t^3 y)$. Take $W$ to be the line spanned by some vector $(x,y)$, with $x$ and $y$ nonzero. Then $N_T(W) = \{ t : t^3=t \} = \{ \pm 1 \}$ which is disconnected in characteristic not $2$. – David Speyer Jul 26 '10 at 12:18
Speyer's example illustrates the potential for non-smoothness that BCnrd mentioned: $N_T(W)$ is not smooth in char. 2 (it is isomorphic to the finite diagonalizable group scheme $\mu_2$). – George McNinch Jul 26 '10 at 15:04

Your more specific question is not actually more specific. $\mathbb{G}_m^n \ltimes \mathbb{G}_a^n$ embeds in $GL_{2n}$ as $\left( \begin{smallmatrix} T & U \\ 0 & 1 \end{smallmatrix} \right)$, where $U$ is embedded along the diagonal. If $U$ is any $T$-rep, we can embed $T \ltimes U$ into $\mathbb{G}_m^n \ltimes \mathbb{G}_a^n$ for $n$ large enough (after a possible extension of the base field, see discussion below). And, of course, every vector subspace of $U$ is a subgroup. So all of the examples BConrad and I bring up will occur in examples of the form you discuss.
You can ensure that $N_T(U)$ is smooth for $p$ large: the equations defining $N_T(U)$ will all be of the form $\chi_i(t)=1$ for various characters $\chi_i$ of $T$. Once $p$ gets up above the torsion in $\mathrm{Char}(T)/\langle \chi_1, \chi_2, \ldots, \chi_N \rangle$, this will be smooth. But, as my example points out, you can't force this to be connected.
Can you really embed any torus $T$ into a split torus? The point counts don't seem to work out over a finite field. – S. Carnahan Jul 26 '10 at 16:37
Scott, since the normalizer scheme exists over the ground field and its formation commutes with any ext'n of the ground field, to verify if it is a torus it's harmless to extend scalars to an alg. closed ground field. Thus, one loses nothing by considering only split $T$. (Irrespective of finite fields, since the category of tori over any field is anti-equivalent to the category of Galois lattices via "geometric char. group", we see that any subtorus or quotient torus of a split torus is split, and likewise any extension of a split torus by a split torus is split, over any field.) – BCnrd Jul 26 '10 at 17:11
What is $r$? In any case, I suspect that a counter-example to your guess is to consider the sequence of two-dimensional vector spaces where $\mathbb{G}_m$ acts by $t: (x,y) \to (t x, t^N x)$, and the subspace $U_1$ spanned by $(1,1)$. Then, over $\mathbb{C}$, there are $N-1$ connected components of $N_T(U_1)$. So we can get as many components as we want, while keeping the dimensions of $U$, $U_1$ and $T$ fixed. – David Speyer Jul 27 '10 at 2:37
Can't you do $GL_4$, with $\left( \begin{matrix} t & 0 & x & 0 \\ 0 & t^N & 0 & y \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right)$? – David Speyer Jul 27 '10 at 17:50