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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.



Do tori in a symplectic group always have invariant maximal isotropic subspaces?
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. 


