3
$\begingroup$

EDIT: As L Spice pointed out, there is an error in the observation. The question is void therefore

Let $p$ be a prime and $q=p^r$. Let $V$ be a $\mathbb F_q$-vector space of dimension four, with a symplectic form $\omega$. Choose a Darboux basis $x_1,x_2,y_1,y_2\in V$ with $\omega(x_i,y_j)=\delta_{ij}$ and $\omega(x_i,x_j)=\omega(y_i,y_j)=0$. A two-dimensional subspace $P$ of $V$ is called a Lagrangian plane, if $\omega(v,w)=0$ for any $v,w\in P$. Denote $\mathfrak L$ the set of all Lagrangian planes.

Using the group structure of $V$, we may define the group algebra $A:=\mathbb F_q\!\left<V\right>$, which as a $\mathbb F_q$-vector space is spanned by symbols $\{Z_i\,,\, i\in V\}$. Multiplication is defined by $Z_i\cdot Z_j :=Z_{i+j}$, so $1=Z_0$. Then $A$ is a local $\mathbb F_q$-algebra with maximal ideal $\mathfrak m$ generated by the $(Z_i-1)$.

For a plane $P\subset V$, we define $S_P := \sum_{i\in P} Z_i$. We consider the linear $L$ subspace of $A$, spanned by all $S_P$ for $P\in\mathfrak L$. I observed that $L$ is in fact an ideal for $q=2,3,4,5,7$ and I think that this is true for every $q$.

My question is: how to prove that $L$ is an ideal?

The curious thing is that the choice of the set of planes $\mathfrak L$ seems to be completely arbitrary, since the construction of $A$ does not depend on $V$ being symplectic. But in fact, if we replace $\mathfrak L$ by another set of planes, such as the non-isotropic ones, the resulting linear span will not be an ideal. So there are two questions: How to characterize sets of planes $\mathfrak S$ such that the span $\left< S_P\,,\,P \in \mathfrak S\right>$ is an ideal? How to generalize this to higher dimensions?

Remark 1: for $q=2,3,4,5,7$ we have $\dim_{\mathbb F_q} L=10,25,50,91,225$, respectively.

Remark 2: If $q=p$, then $A$ is isomorphic to $\mathbb F_q[Z_{x_1},Z_{x_2},Z_{y_1},Z_{y_2}]/J$, where $J$ is the ideal generated by the $(Z_i-1)^p$. If $P$ is spanned by vectors $v$ and $w$, then $S_P=(Z_v-1)^{p-1}(Z_w-1)^{p-1}$.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ It seems that this is not an ideal unless $q = 2$. Notice that $L$ is contained in $A' \mathrel{:=} \{\sum_{v \in V} c_v Z_v : \text{$c_v = c_{\lambda v}$ for all $v \in V$ and $\lambda \in \mathbb F_q^\times$}\}$. On the other hand, if $P \in \mathfrak L$ and $y \in V \setminus P$, then $Z_y S_P$ is not in $A'$ (unless $q = 2$). $\endgroup$
    – LSpice
    Commented Feb 24, 2016 at 7:18
  • $\begingroup$ Ough, you're right. Now I found the error in my calculation. $\endgroup$
    – Nicolas Malebranche
    Commented Feb 24, 2016 at 9:45
  • $\begingroup$ If you now regard the question as void, then it is probably appropriate for you to close it. $\endgroup$
    – LSpice
    Commented Feb 24, 2016 at 13:47
  • 5
    $\begingroup$ I'm voting to close this question because of the initial remark added by the OP $\endgroup$
    – Stefan Kohl ♦
    Commented Feb 24, 2016 at 17:23

1 Answer 1

Reset to default
0
$\begingroup$

As we discussed in comments, your space is not an ideal. In fact it is the same as the space spanned by the elements of the form $S_\ell \mathrel{:=} \sum_{v \in \ell} Z_v$ as $\ell$ runs over lines in $V$. Indeed, note that every (isotropic) plane contains $p + 1$ lines, and every line is contained in $p + 1$ isotropic planes, so that $S_P = \sum_{\ell : \ell \subseteq P} S_\ell$ and $S_\ell = \sum_{P : \ell \subseteq P} S_P$. In particular, if $\sum_{v \in V} c_v Z_v \in L$, then $c_v = c_{\lambda v}$ for all $v \in V$ and all $\lambda \in \mathbb F_q^\times$.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .