$\newcommand{\mbf}{\mathbf}$ Hi all,

I've been thinking about the following question for a while now, and got a little stuck trying to solve it. Hopefully, someone here might be able to help.

For starters, let $K$ be some field, and led $G$ be the group of $K$-points of the group $\mbf{G}=\mathrm{Sp}_{2n}$, for some $n\in\mathbb N$. That is to say- $G$ is the group of isometries of a $2n$-dimensional vector space $V$ over $K$, endowed with a non-degenerate alternating form $(\cdot,\cdot):V\times V\to K$. It is well-known that in this situation there exists a Darboux basis for $V$ over $K$, i.e. a basis $\lbrace e_1,\ldots,e_n,f_1,\ldots,f_n\rbrace$ where $(e_i,f_j)=\delta_{i,j}$ and $(e_i,e_j)=(f_i,f_j)=0$ for all $i,j=1,\ldots,n$. This allows us to identify $G$ with the matrix group $$G=\lbrace\mbf{x}\in M_{2n}(K)\mid \mbf{x}^T\Omega\mbf{x}=\Omega\rbrace,$$ where $\Omega:=\left(\begin{matrix}\mbf{0}_n&\mbf{id}_n\\-\mbf{id}_n&\mbf{0}_n\end{matrix}\right)$.

A torus in $G$ is a connected abelian subgroup $T$ such that any $\mbf{y}\in T$ is diagonalizable in some field extension of $K$.

For example, if $K$ is algebraically closed, then any torus in $G$ is conjugate to the group $$\lbrace\left(\begin{smallmatrix}a_1\\&\ddots\\&&a_n\\&&&&a_1^{-1}\\&&&&&\ddots\\&&&&&&a_n^{-1}\end{smallmatrix}\right)\mid a_1,\ldots,a_n\in K^\times\rbrace.$$

My question is this-

Is it true that for any torus $T\subseteq G$, there exists a maximal isotropic subspace $W\subseteq V$ such that $TW\subseteq W$.

Recall that a subspace $W\subseteq V$ is isotropic if it satisfies $(w_1,w_2)=0$ for all $w_1,w_2\in W$.

In the case where $K$ is algebraically closed, the answer to this question is obviously true. However, my main concern is with the case of non-a.c. fields, and in particular- finite and local fields.

What I've managed to show- It is not very hard to show that the minimal polynomial $f(t)$ of any element of $G$ must satisfy the equality $$\tilde{f}(t)=f(t),$$ where $\tilde{f}(t)$ is defined by "reversing the coefficients of $f$", i.e. $$\tilde{f}(t):=\frac{t^{\deg(f)}}{f(0)}\cdot f(t^{-1}).$$ This implies that the minimal polynomial of any such elements can be written uniquely in the form \begin{equation}\tag{$*$}\prod_i f_i(t)^{r_i}\cdot\prod_j \left(g_j\cdot \tilde{g_j}(t)\right)^{q_i},\end{equation} where the $f_i$ are irreducible and satisfy $\tilde{f_i}=f_i$ and he $g_j$'s are irreducible with $g_j\ne\tilde{g_j}$.

In the case where a torus $T\subseteq G$ contains an element $y$ whose minimal polynomial is of the form $g(t)\cdot\tilde{g}(t)$, where the polynomials $g$ and $\tilde{g}$ have no common irreducible factors, the existence of a $T$-invariant is true, and the subspace can be realized as $W=\ker\left(g(y)\right).$

However, it is possible, to have tori which contain no such elements $y$ (for example, if $K=\mbf{F}_q$ is the field with $q$ elements and $G=\mathrm{Sp}_4(K)$, there is an emebedding of the group of elements of norm $1$ in the extension $\mbf{F}_{q^4}\mid \mbf{F}_{q^2}$ as a torus in $G$, and all elements of this group are either central, or have minimal polynomial whose irreducible factors satisfy $\tilde{f}=f$).

In any case, I would be very grateful if anyone could offer any insight for why this assertion is true or otherwise, point towards a counter example.

Thank you very much.

  • 2
    $\begingroup$ @PaulBroussous has mentioned that the answer is 'no' very generally, but, as a specific example, suppose that your field has odd (or zero) characteristic, and contains a non-square $\alpha$ such that the equation $x^2 - \alpha y^2 = 1$ has a solution over the field. Then $\begin{pmatrix} x & y \\ \alpha y & x \end{pmatrix}$ lies in a torus, and it preserves the 2-dimensional symplectic form but no rational line. $\endgroup$ – LSpice Feb 11 '16 at 19:29
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    $\begingroup$ Seconding Loren, questions about symplectic groups are best first tested on $\mathrm{Sp}_2=\mathrm{SL}_2$. $\endgroup$ – Lior Silberman Feb 12 '16 at 11:54

When $K$ is finite or $p$-adic the answer to your question is negative. Indeed there exist maximal tori which are anisotropic. Those tori are not included in any proper parabolic subgroup, so they cannnot stabilize a non trivial totally isotropic subspace.

  • 1
    $\begingroup$ Since you were (correctly) very careful to specify that it was a non-trivial totally isotropic subspace that was not being stabilised, I will chime in and point out that anisotropic tori are not contained in proper parabolic subgroups. $\endgroup$ – LSpice Feb 11 '16 at 20:14

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