Let be a vector space over a finite prime field. One knows then which subsets satisfy , the sets that are stable under addition: these are the kernels of linear maps, that is, each such set is the zero set of a bunch of linear forms.

Now we consider a two-coordinates analogous problem, that is, we consider a set and define the sets

and , V and H meaning vertical and horizontal. What are the sets that satisfy both and ? 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 for some subspaces and some bilinear forms is horizontally and vertically closed. Call such a set a *bilinear* set. Is it the only example?

For a transverse set and , let be the fiber above x. Thus and is a vector space. Actually depends only the projective class . Moreover, the stability under horizontal operations is equivalent to the property if is on the projective line spanned by and , we have .

**The codimension 1 case**

Suppose and each subspace has codimension at most 1. It is easy to see that the set of such that is a vector subspace which we call . Furthermore, if , we have , so that descends to an injective map on . Thus it is enough to study the case where has codimension exactly one unless . Write with some vector uniquely defined up to homothety. In other words, we have a map satisfying the property that if is on the projective line spanned by and , then lies on the projective line spanned by and . Thus maps projective lines into lines. Actually it can be quickly checked that such a map is either bijective or constant on each single projective line.

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 . Hence writing , we have shown that is the zero set of the bilinear form .

In general is far from being injective. It can for instance be constant on (equal to for some hyperplane ). In this case . This is neither a rectangle, nor the zero form of a quadratic form, because if it was, could be taken linear, and because , it should be injective. In fact, if it is the intersection of zero sets of a family of bilinear forms, there needs to be bilinear forms in this family, such as the forms if (wlog) is the hyperplane . However obviously contains a very large rectangle.

These are actually the only cases: such a map is either injective or constant. Indeed, if is just a line, it’s obvious, so suppose . Suppose that maps projective lines into lines (in the sense above) but is neither injective nor constant. That is, there exist two distinct points such that , and a third point satisfying . This implies that span a projective plane. Take a point on the line spanned by . Because is a bijection on both lines , you can find on such that . Now consider the intersection . Then you have , so that on the line , the map is neither a line nor injective.

**Arbitrary codimension**

A more general potential counterexample, communicated to me by Ben Green, is a set of the form where are sequences of increasing (resp. decreasing) subspaces. Thus for , we have where . Again you can make it a bilinear set, but you need atrociously many bilinear forms.

However it contains a large rectangle. In fact, while so that the density of is at most . Thus . Finally by the pigeonhole principle, one of the cartesian products have density at least .

In general, we have a vector space . Suppose for instance each has codimension 2. If we manage to find maps such that $latex \text{span}(\xi_1(x),\xi_2(x))$ for each , we are in good shape. Though it is not clear how to show that such a consistent choice of bases can be made…

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