Welcome to MathOverflow

Question

1

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.)

edited Jul 27 2010 at 0:13

Scott Morrison♦
4,460●4●26●44

asked Jul 26 2010 at 7:08

Nick Gill
2,881●7●18

What do you mean by normalizer? The subgroup of elements in $T$ which map $W$ into itself? – Mariano Suárez-Alvarez Jul 26 2010 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 2010 at 7:16

4

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 2010 at 7:57

1

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 2010 at 12:18

1

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 2010 at 15:04

David Speyer · Accepted Answer · 2010-07-26 17:22:25Z UTC

3

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.

edited Jul 26 2010 at 17:22

answered Jul 26 2010 at 15:54

David Speyer
55k●3●102●236

1

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 2010 at 16:37

Oh, you are right. All the counter-examples above are split, but the general argument doesn't work. – David Speyer Jul 26 2010 at 16:43

3

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 2010 at 17:11

1

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 2010 at 2:37

1

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 2010 at 17:50

show 3 more comments

Tori acting on vector spaces

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.