Nilpotent symplectic alternating algebras II

A symplectic alternating algebra (SAA) is a symplectic vector space L, whose associated alternating form is non-degenerate, that is furthermore equipped with a binary alternating product $\cdot$ $:L\times L\mapsto L$ with the extra requirement that

for all $x,y,z\in L$. This condition can be expressed equivalently by saying that $(u\cdot x,v)=(u,v\cdot x)$ for all $u,v,x\in L$ or in other words that multiplication from the right is self-adjoint with respect to the alternating form.

Symplectic alternating algebras originate from a study of powerful $2$-Engel groups [1, 5] and there is in a 1-1 correspondence between a certain rich class of powerful $2$-Engel $3$-groups of exponent $27$ and SAAs over the field $\mbox{GF}(3)$. We will consider this in more details later in the introduction.

Let $2n$ be a given even integer and $\mathbb{F}$ a fixed field. Let $V$ be the symplectic vector space over the field $\mathbb{F}$ with a non-degenerate alternating form. Fix some basis $u_{1},u_{2},\ldots,u_{2n}$ for $V$. An alternating product $\cdot$ that turns $V$ into a symplectic alternating algebra is uniquely determined by the values

Let $L$ be the resulting symplectic alternating algebra. We refer to the data above as a presentation for $L$ with respect to the basis $u_{1},\ldots,u_{2n}$.

If $m(n)$ is the number of symplectic alternating algebras over a finite field $\mathbb{F}$ then $m(n)=|\mathbb{F}|^{\frac{4n^{3}}{3}+O(n^{2})}$ [$2$]. Because of the sheer growth, a general classification does not seem to be within reach although this has been done for small values of $n$. Thus it is not difficult to see that $m(0)=m(1)=1$ $\mbox{and }m(2)=2$. For higher dimensions the classification is already difficult. It is though known that when $\mathbb{F}=\mbox{GF}(3)$ we have $m(3)=31$ [2]. Some general structure theory is developed in  [6, 4, 2, 3]. In particular there is dichotomy result that is an analog to a corresponding theorem for Lie algebras, namely that $L$ either contains a non-trivial abelian ideal or is a direct sum of simple symplectic alternating algebras [6]. We also have that any symplectic algebra that is abelian-by-nilpotent must be nilpotent while this is not the case in general for solvable algebras [4].

Here we focus on the subclass of nilpotent symplectic alternating algebras (NSAAs) and this paper is a sequel to [2] where a rich general structure theory was developed for NSAAs with a number of beautiful properties. Before discussing these we need to introduce some notation. Firstly we can always pick a basis $x_{1},y_{1},\ldots,x_{n},y_{n}$ with the property that $(x_{i},x_{j})=(y_{i},y_{j})=0$ and $(x_{i},y_{j})=\delta_{ij}$ for $1\leq i\leq j\leq k\leq n$. We refer to a basis of this type as a standard basis. It turns out that for any nilpotent symplectic alternating algebra one can always choose a suitable standard basis such that the chain of subspaces

with $I_{k}=\mathbb{F}x_{n}+\cdots+\mathbb{F}x_{n-k+1}$ for $k>0$, is a central chain of ideals. One can furthermore see from this that $x_{i}y_{j}=0$ if $j\leq i$ and that $I^{\perp}_{n-1}$ is abelian. It follows that a number of the triple values $(uv,w)$ are trivial. Listing only the values that are possibly non-zero it suffices to consider

for some $\alpha_{ijk},\beta_{ijk}\in\mathbb{F}$ where $1\leq i<j<k\leq n$. Such a presentation is called a nilpotent presentation. Conversely any such presentation describes a nilpotent SAA. The algebras that are of maximal class turn out to have a rigid ideal structure. In particular when $2n\geq 10$ we can choose our chain of ideals above such that they are all characteristic and it turns out that $I_{0},I_{2},I_{3},\ldots,I_{n-1},I^{\perp}_{n-1},I^{\perp}_{n-2},\ldots,I^{\perp}_{0}$ are unique and equal to both the terms of the lower and upper central series (see [2] Theorem 3.1 and 3.2). The algebras of maximal class can be identified easily from their nilpotent presentations. In fact, if $\mathcal{P}$ is any nilpotent presentation of L with respect to a standard basis $\{x_{1},y_{1},\ldots,x_{n},y_{n}\}$, and $2n\geq 8$, we have that L is of maximal class if and only if $x_{i}y_{i+1}\neq 0$ for all $i=2,\ldots,n-2$, and $x_{1}y_{2},y_{1}y_{2}$ are linearly independent (see [2] Theorem 3.4).

From the general theory of nilpotent SAAs one can also determine their growth. Thus if $k(n)$ is the number of nilpotent SAAs of dimension $2n$ over a finite field ${\mathbb{F}}$ then $k(n)=|{\mathbb{F}}|^{n^{3}/3+O(n^{2})}$ [2] Again the growth is too large for a general classification to be feasable. The algebras of dimension $2n$ for $n\leq 4$ are classified in [2] over any field. In this paper and its sequel we deal with the challenging classification of algebras of dimension $10$ over any field. As we will see the classification depends very much on the underlying field. It turns out that the classification of nilpotent symplectic alternating algebras of dimension 10 with a centre that is not isotropic can be easily reduced to the known classification of algebras of dimension $8$. The main bulk of the work is thus about algebras with isotropic centre that must lie between $2$ and $5$. In this paper we consider the situation when the isotropic centre is of dimension $3$ or $5$ leaving the remaining cases to the sequel. As we said above, the classification depends on the underlying field. We can read in particular from the classification that over a field that is algebraically closed there are $22$ NSAAs of dimension $10$.

In this section we will be assuming that $Z(L)$ is isotropic of dimension $3$. First we derive some properties that hold for these algebras. Here throughout

Proof  Otherwise $Z_{2}(L)=(L^{3})^{\perp}\nleq Z(L)^{\perp}=L^{2}$. Without loss of generality we can suppose that $y_{3}\in Z_{2}(L)\setminus L^{2}$. As $Z_{2}(L)\cdot L^{2}=\{0\}$, we then have $y_{3}\cdot L^{2}=\{0\}$. Now also $x_{2}\cdot L^{2}=\{0\}$. Let $\alpha=(x_{2}y_{4},y_{5})$ and $\beta=(y_{3}y_{4},y_{5})$. Notice that $\alpha,\beta\not=0$ as $x_{2},y_{3}\not\in Z(L)$. But then

that implies that $\beta x_{2}-\alpha y_{3}\in Z(L)$. This is absurd. $\Box$

Proof  Otherwise $\mbox{dim\,}L^{3}\leq 4$ and as $Z(L)\leq L^{3}\leq L^{2}=Z(L)^{\perp}$ we can choose our standard basis such that $Z(L)={\mathbb{F}}x_{5}+{\mathbb{F}}x_{4}+{\mathbb{F}}x_{3}$ and

This implies that ${\mathbb{F}}x_{5}+{\mathbb{F}}x_{4}+{\mathbb{F}}x_{3}+{\mathbb{F}}x_{2}+{\mathbb{F}}x_{1}+{\mathbb{F}}y_{1}\leq(L^{3})^{\perp}=Z_{2}(L)$ and (notice that $Z_{2}(L)\leq L^{2}$ as $Z(L)\leq L^{3}$) $L^{2}=Z_{2}(L)+Fy_{2}$ that implies that $L^{2}$ is abelian. Then for any $x\in L^{2}$ and $a,b,c\in L$, we have

and $L^{3}\leq(L^{2})^{\perp}=Z(L)$. Hence $L^{3}=Z(L)$ and $Z_{2}(L)=L^{2}$. Suppose $L=Z_{2}(L)+{\mathbb{F}}u_{1}+{\mathbb{F}}u_{2}+{\mathbb{F}}u_{3}$. Then $L^{2}=Z(L)+{\mathbb{F}}u_{1}u_{2}+{\mathbb{F}}u_{1}u_{3}+{\mathbb{F}}u_{2}u_{3}$ and we get the contradiction that $4=\mbox{dim\,}L^{2}-\mbox{dim\,}Z(L)\leq 3$. $\Box$

Proof  Otherwise we can choose our basis such that $L^{3}={\mathbb{F}}x_{5}+{\mathbb{F}}x_{4}+{\mathbb{F}}x_{3}+{\mathbb{F}}x_{1}+{\mathbb{F}}y_{1}$ and then $Z_{2}(L)=(L^{3})^{\perp}={\mathbb{F}}x_{5}+{\mathbb{F}}x_{4}+{\mathbb{F}}x_{3}+{\mathbb{F}}x_{2}+{\mathbb{F}}y_{2}$ and as $L^{2}\cdot Z_{2}(L)=\{0\}$, it follows that $x_{1}y_{2}=y_{1}y_{2}=0$. Then $L^{2}$ is abelian and thus we get the contradiction that $L^{3}\leq Z(L)$. $\Box$

Proof  We have seen that $\mbox{dim\,}L^{3}\geq 5$. So we can choose our standard nilpotent basis such that either

or

We consider the two cases in turn beginning with the first case. If $Z(L)\not\leq L^{4}$, then $\mbox{dim\,}Z(L)\cap L^{4}\leq 2$ and thus $\mbox{dim\,}L^{2}+Z_{3}(L)=\mbox{dim\,}(Z(L)\cap L^{4})^{\perp}\geq 8$. Suppose $L=L^{2}+Z_{3}(L)+{\mathbb{F}}u+{\mathbb{F}}v$. Then $L^{2}=L^{3}+Z_{2}(L)+{\mathbb{F}}uv=L^{3}+{\mathbb{F}}uv$ and we get the contradiction that $\mbox{dim\,}L^{2}\leq 5+1=6$. We now turn to the second case where $L^{3}={\mathbb{F}}x_{5}+{\mathbb{F}}x_{4}+{\mathbb{F}}x_{3}+{\mathbb{F}}x_{2}+{\mathbb{F}}x_{1}+{\mathbb{F}}y_{1}$. We argue by contradiction and suppose that $Z(L)\cap L^{4}<Z(L)$. Then we can choose our basis such that

and $Z(L)\cap L^{4}\leq{\mathbb{F}}x_{5}+{\mathbb{F}}x_{4}$. Now $y_{3}\in(L^{4})^{\perp}=Z_{3}(L)$ and as $Z_{3}(L)\cdot L^{3}=\{0\}$, it follows that

It follows from this that $x_{1}y_{2},y_{1}y_{2},y_{3}y_{2}\in{\mathbb{F}}x_{5}+{\mathbb{F}}x_{4}$. Thus in particular these three elements are linearly dependent and we have $(\alpha x_{1}+\beta y_{1}+\gamma y_{3})y_{2}=0$ where not all of $\alpha,\beta,\gamma$ are zero. Then $x_{2},\alpha x_{1}+\beta y_{1}+\gamma y_{3}$ commute with all the basis elements except possibly $y_{4}$ and $y_{5}$. Suppose

If $r=0$ then we get the contradiction that $x_{2}\in Z(L)$ and if $r\not=0$, we get the contradiction that $-sx_{2}+r\alpha x_{1}+r\beta y_{1}+r\gamma y_{3}\in Z(L)$. $\Box$

After these more general results we classify all the algebras where $Z(L)$ is isotropic of dimension $3$. We consider the two subcases $\mbox{dim\,}L^{3}=5$ and $\mbox{dim\,}L^{3}=6$ separately.