Let $V=\mathbb{F}_p^n$ be a vector space over a finite prime field. One knows then which subsets $A\subset V$ satisfy $A+A=A$, the sets that are stable under addition: these are the kernels of linear maps, that is, each such set $A$ is the zero set of a bunch of linear forms.

Now we consider a two-coordinates analogous problem, that is, we consider a set $A\subset V\times V$ and define the sets

$\displaystyle{P\stackrel{V}{+}P= \{(x,y_1+ y_2)\mid (x,y_1),(x,y_2)\in P\}}$

and $P\stackrel{H}{+}P$, V and H meaning vertical and horizontal. What are the sets that satisfy both $P\stackrel{V}{+}P=P$ and $P\stackrel{H}{+}P=P$? Call such a set a transverse set. Natural exemples are cartesian products of vector spaces (which we shall call rectangles) as well as zero sets of bilinear forms. More generally, a set of the form $\displaystyle{\{(x,y)\in W_1\times W_2\mid Q_1(x,y)=\cdots=Q_r(x,y)=0\}}$ for some subspaces $W_2,W_2\leq V$ and some bilinear forms $Q_i$ is horizontally and vertically closed. Call such a set a bilinear set. Is it the only example?

For a transverse set $P\subset V\times V$ and $x\in V$, let $P_x$ be the fiber above x. Thus $P_x$ is a vector space. Actually $P_x$ depends only the projective class $[x]\in P(V)$. Moreover, the stability under horizontal operations is equivalent to $P_{x+y}\supset P_x\cap P_y$; more generally if $[z]$ is on the projective line spanned by $x$ and $y$, we have $P_z\supset P_x\cap P_y$.

The codimension 1 case

Suppose $P_0=V$ while all other subspaces $P_x$ are hyperplanes. Write $P_x=\xi(x)^\perp$ with $\xi(x)\in P(V)$ some vector uniquely defined up to homothety. In other words, we have a map $\xi : P(V)\rightarrow P(V)$ satisfying the property that if $z$ is on the projective line spanned by $x$ and $y$, then $\xi_{z}$ lies on the projective line spanned by $\xi_x$ and $\xi_y$. Thus $\xi$ maps projective lines onto lines.

If we suppose it is injective, too, then by the fundamental theorem of projective geometry, we conclude that it is a projective map. Thus it is induced by a linear map $\xi : V\rightarrow V$. Hence writing $Q(x,y)=\xi(x)\cdot y$, we have shown that $P$ is the zero set of the bilinear form $Q$.

In general $\xi$ is far from being injective. It can for instance be constant on $P(V)$ (equal to $H^\perp$ for some hyperplane $H$). In this case $P=\{0\}\times V\cup V\times H$. This is neither a rectangle, nor the zero form of a quadratic form, because if it was, $\xi$ could be taken linear, and because $\xi_x=0\leftrightarrow x=0$, it should be injective. However $P$ obviously contains a very large rectangle.

I can at least find a large bilinear set in the transverse set $P$. If $\xi$ is not injective on $P(V)$, there exist two linearly independent vectors $x,x'$ where $\xi$, takes the same value, which it then takes at least on $\text{span}(x,x')\setminus\{0\}$. Take a complement $V_1$ of this plane. If there $\xi$ is injective, we’re done. Otherwise, go on. After $k$ steps, we have $D_1,\cdots,D_k$ planes whose sum $D$ is of dimension 2k and for which there is a subspace $H$ of codimension at most $k$ such that $D\times H$ is in $P$. Either we never find injectivity before $k=n/3$ steps, in which case we basically end up with a rectangle $W_1\times W_2$ made up from two subspaces of dimension $2n/3$, or we find injectivity on a space $W$ of dimension at least $n/3$. Then we have a bilinear set $\{(x,y)\in W\times V\mid Q(x,y)=0\}$ for some bilinear form $Q$.

Arbitrary codimension

A more general potential counterexample, communicated to me by Ben Green, is a set of the form $\bigcup_i V_i\times W_i$ where $V_i, W_i$ are sequences of increasing (resp. decreasing) subspaces.  Suppose for instance that $V_0=W_r=\{0\}$ while $W_0=V_r=V$. Thus for $x\in V$, we have $P_x=W_{i(x)}$ where $i(x)=\min\{i\mid x\in V_i\}$. I can’t quite prove it’s not a bilinear set though.

Anyway it contains a large rectangle. Indeed, one of the $V_i\times W_i$ must have size at least $\lvert P\rvert /n$.

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