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 is a vector space. Actually depends only the projective class . Moreover, the stability under horizontal operations is equivalent to ; more generally if is on the projective line spanned by and , we have .

**The codimension 1 case**

Suppose while all other subspaces are hyperplanes. 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 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 . 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. However obviously contains a very large rectangle.

I can at least find a large bilinear set in the transverse set . If is not injective on , there exist two linearly independent vectors where , takes the same value, which it then takes at least on . Take a complement of this plane. If there is injective, we’re done. Otherwise, go on. After steps, we have planes whose sum is of dimension 2k and for which there is a subspace of codimension at most such that is in . Either we never find injectivity before steps, in which case we basically end up with a rectangle made up from two subspaces of dimension , or we find injectivity on a space of dimension at least . Then we have a bilinear set for some bilinear form .

**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. Suppose for instance that while . Thus for , we have where . I can’t quite prove it’s not a bilinear set though.

Anyway it contains a large rectangle. Indeed, one of the must have size at least .

## Commentaires récents