Point-line geometries with a generating set that depends on the underlying field

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Problem: Reconstructed statement: for a classical polar space $\Delta(F)$ over a field $F$, with subgeometry $\Delta(F_0)$ over a subfield $F_0$, let $\Gamma$ be the $i$-Grassmann point-line geometry and $\Gamma_0$ its $F_0$-subgeometry. If $S$ generates $\Gamma_0$, then

$$\langle S\rangle_\Gamma=\{L\in\Gamma:\exists E\in\Delta_i(F_0)\text{ with }L\subseteq E^\perp,\ L\cap E\ne\varnothing\},$$

and $F_1=F_0$. The wording is ambiguous about the polar type; the standard context of the paper includes line-Grassmannians of symplectic polar spaces. The statement is false already for the rank $3$ symplectic line-Grassmannian.

Result: Let $F_0=\mathbb F_2$, $F=\mathbb F_4=\mathbb F_2(\omega)$, $\omega^2+\omega+1=0$. Let $V=F^6$ with symplectic form

$$B(x,y)=x_1y_4+x_2y_5+x_3y_6+x_4y_1+x_5y_2+x_6y_3.$$

Let $\Gamma$ be the line-Grassmannian of $W(5,F)$; its points are totally isotropic $2$-spaces, and its lines are pencils $[p,\Pi]$, where $p$ is a point and $\Pi$ a totally isotropic plane. Let $R$ denote the conjectured right-hand side.

A generated set $\langle S\rangle_\Gamma$ is always a subspace of $\Gamma$. Thus, if the conjecture were true, $R$ would have to be a subspace.

Set

$$u=(1,\omega+1,0,0,0,0),\quad v=(0,0,0,1,\omega,0),\quad z=(0,0,1,0,0,\omega+1),$$ $$p=u+v,\qquad \Pi=\langle u,v,z\rangle_F.$$

One checks $B(u,v)=B(u,z)=B(v,z)=0$, so $\Pi$ is totally isotropic. Define three lines through $p$ in $\Pi$:

$$L=\langle u,v\rangle_F,$$ $$A=\langle p,z+v\rangle_F,\qquad M=\langle p,z+\omega v\rangle_F.$$

Thus $A,M,L$ are points of one line of $\Gamma$.

Now $A,M\in R$. Indeed,

$$E_A=\langle(1,0,0,1,0,1),(0,1,1,1,1,1)\rangle_F$$

is an $F_0$-rational singular line, $A\cap E_A\ne0$, and direct calculation gives $A\subseteq E_A^\perp$. Similarly,

$$E_M=\langle(1,0,1,1,0,1),(0,1,1,1,1,0)\rangle_F$$

is $F_0$-rational singular, $M\cap E_M\ne0$, and $M\subseteq E_M^\perp$.

But $L\notin R$. The five points of $L$ are represented by $u+tv$, $t\in F$, and $v$. For each such point, the unique $F_0$-line containing it is either not singular, or is one of

$$\langle e_1,e_2\rangle,\quad \langle f_1,f_2\rangle,\quad \langle(1,0,0,0,1,0),(0,1,0,1,0,0)\rangle,$$

and in each of these three cases $L$ is not contained in the perp of that line. Hence no $F_0$-rational singular line $E$ satisfies $L\subseteq E^\perp$ and $L\cap E\ne0$.

Therefore $R$ contains two points $A,M$ of a line of $\Gamma$ but omits another point $L$ on that same line. Hence $R$ is not a subspace, while every generated closure $\langle S\rangle_\Gamma$ is a subspace. Taking $S$ to be the full point set of $\Gamma_0$, which certainly generates $\Gamma_0$, the conjectured equality fails.

Citation: No external resolution is used. Source of the reconstructed problem: R. J. Blok and A. Pasini, “Point-Line Geometries with a Generating Set that Depends on the Underlying Field,” in Finite Geometries, Developments in Mathematics 3, Kluwer/Springer, 2001, DOI: 10.1007/978-1-4613-0283-4_1.