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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