Consider two quadratics in four variables whose only common zero is the zero vector. They lead to a solution (w,x,y,z,Q1,Q2) with points, and the same density as Seva’s original example.

I don’t know if they include new examples or not.

The only example I can think of is and . where is not a square.

Unfortunately, that is just two solutions multiplied together.

The Chevally-Warning Theorem (1936) says that any set of polynomials over a finite field, with more variables than the total of the degrees of the polynomials, has simultaneous zeros.

In our case, that means two quadratics in five variables have simultaneous non-trivial zeros, so A can’t have points.

Or three quadratics in seven variables have simultaneous non-trivial zeros, and again we can’t do better than your first solution.

Perhaps two quadratics in eight variables could have just q simultaneous zeros?

I did a Monte Carlo with q=9, and I think found two quadratics whose zero-sets shared q points, all on a line through zero. (That is not surprising for homogeneous quadratics.)

If that happened in other fields, it would allow points in A, out of a total points, which is not as good as your other solution.

The only better result for this method would be if there are two quadratics whose only zero-point in common is the zero vector itself. In that case, the size of A would be and you get your density

Any arithmetic progression in is of the form

where and for . The latter condition implies , whence . But now the former condition along with the assumption that contains no arithmetic progressions with the difference in show that ! Thus, is a cap, and we have . Can we choose , and so that (as )? (What gives a hope that this is possible is that we do allow progressions in a vast majority of directions!) If so, we have constructed a cap in of size which, as far as I remember (I may well be wrong about it though), beats the best known constructions. ]]>

About that problem 4.13. Let’s continue this conversation off of this blog. Send me an email.

I guess I was being silly about the 3APs and incidences question, now that I think about it. I should have thought about the artist’s drawing of a horizon — that transformation preserves lines and incidences, but not 3APs.

But funnily enough I had asked this problem about 3APs to several people (including a geometer), and not one of them thought of using these projective transformations.

]]>Dear Ernie,

it seems to me that this kind of map is by far to general to imply anything interesting about the inverse question associated with the Szemerédi-Trotter thereom, except for that it suffices to classify the relevant configurations up to “projective isomorphisms”. By this I mean: These maps can be applied not only to grids, but to all configurations of points and lines, so in particular to those maximizing the number of incidences. By the way, I think I can do Problem 4.13 from your list, but probably that’s not new given that the list seems to be a few years old.

Best regards,

Christian.

]]>Dear Christian,

Wow! That’s fantastic. I had thought about various kinds of maps like you had described, but had somehow overlooked this particular possibility. I suppose it sinks my hope of encoding 3APs by incidences, but at the same time it means that problems about 3APs really are fundamentally different from incidence-type problems (at least superficially).

By the way, there is an old problem posed by T. Tao (and I think also J. Solymosi and others) appearing in the following (problem 4.6)

about classifying collections of points and lines with almost the maximal number of incidences. The fact that there are maps such as you describe suggests that the answer will not simply be “dense subsets of uniform grids”, but will instead be much more complicated.

]]>Oups, apart from that I failed to press the reply button, a crucial formula did not parse, so here is another attempt to write it down with slightly more English: The point is mapped to .

]]>your argument involving a large number of quadratic equations sound compelling, but I still don’t see why the following rules out that, e.g. a grid, can enforce 3APs. The map preserves collinearity of points, so if we apply it to all points with integers ranging from to we get a copy of our grid. (I assume here that and are linearly independent over the rationals, so that we do not divide by zero. It is also useful to assume these numbers to be algebraically independent for the sake of discussion. If one does not want to rely on that, one can replace them by two other “sufficiently general” real numbers). Now in the new drawing no point seems to be the midpoint of two other points (basically because this would tell us something non-trivial about and .)

So it may be that the quadratic equations you produce are not very independent from one another. A possible reason for that, apart from getting only insted of equations for any $ latex k$ collinear points might be hidden in results such as Pappus’ or Desargues’ theorem, which seem to tell us that sometimes one of these equations is entailed by eight or nine others in a non-trivial way. Hilbert’s Nullstellensatz suggests that these implication can be made rather explicit in single equations, which would incidentally yield – probably rather ugly – one-line-proofs of these two theorems.)

]]>Christian,

“but do you mean by three points forming a 3AP that one of them is the midpoint…?” Yes, that is what I mean.

“Can you indicate a… ?” I don’t know of a single one! That’s my question — prove that one exists.

Assuming that the point-line incidence restrictions for all the lines associated to some grid pin down the grid structure up to affine transformations, it should be possible to decide this algebraically: in order to bypass the problems of multiple configurations due to affine transformations (which preserve incidences), one can anchor three of the points to be and . Then, one can represent triples of collinear points by quadratic questions, which reflect the fact that the slopes of the segments and are equal. In principle, then, one can represent all the point-line incidences (using only the lines meeting three or more of the points) by a large number of these quadratic equations, which *should* have a unique solution (since the anchoring prevents multiple solutions coming from affine transformations). Perhaps one of these computer algebra packages (like MAGMA) can then verify that the equations indeed have a unique solution in the reals.

Assuming this is true (the incidence data for sufficiently large grids uniquely determines the structure of the grid up to affine transformations), a bigger and more interesting challenge would be to produce a set of point-line pairs where every line is incident to EXACTLY three points (i.e. not four or more), such that for any drawing of the incidence the points must ALWAYS contain a 3AP. I happen to believe it is possible to find such a configuration; in fact, I believe that projecting down a grid (and the associated lines hitting three points each) to the 2D plane (bijectively) will give such a configuration when the number of dimensions is large enough.

Why consider point configurations where every line meets exactly three points? Well, that has to do with ellitpic curves… another time, perhaps.

]]>I’m not sure whether I understand the problem correctly, but do you mean by three points forming an 3AP that one of them is the midpoint of the line segment joining the two other ones? If so (or also otherwise) could you, please, indicate a point/line-configuration that enforces the presence of a 3AP? At the moment I cannot think of any, mainly because there seem to be far too many projective maps that preserve incidence but tend to destroy the midpoint property. (To be more precise: Take a drawing of a given configuration, colour all points which are fourth harmonic points of collinear triples of points you have used yellow, consider a line avoing all these yellow points and send it to infinity.)

]]>Problem. Can one encode three-term progressions in terms of point-line incidences in ? More specifically: suppose you are given a set of pairs , which represents the fact that the point is on the line . Furthermore, suppose that this list of incidences may not be exhaustive in that there could be other point-line incidences in the drawing not accounted for by the list. Then, no matter how you draw the configuration of points and lines, so long as the incidences are respected, must the set of points ALWAYS contain a 3AP?

Take, for example, the grid of points, with lines, each incident with three of those points. Some points are incident with of those lines; some are incident with only ; and one point is incident with lines. It turns out that there are many OTHER ways of drawing the points and lines (uncountably many, even after accounting for affine transformations), besides the lattice grid, such that the points do not contain a three-term progression. (It is a nice exercise to work out some examples!)

But what about grids, or , etc.?

….

The original motivation of this problem was to find a way to convert problems about 3APs into incidence questions that could then be approached algebraically (e.g. using elliptic curves).

]]>Thanks for fixing my latex. Yes, the terminology is supposed to evoke the geometry of although it may not be necessarily true even there that a set with no 3 collinear points looks like a hat.

]]>Many thanks for this interesting comment. I have to confess that the hattishness of a set with no three collinear points doesn’t jump out at me, unless you are talking about subsets of — is that the point?

I was about to ask why the last statement doesn’t solve the cap-set problem, but suddenly realized that not containing three collinear points in is much stronger than not containing a line in But that is an interesting result you mention.

In case you are wondering why your latex didn’t initially come out, it is because you put backslashes in front of “latex”.

]]>See section 1.4 of the note. Basically, Blokhuis’s idea is an algebraic method that gives good bounds on the “union of subspaces” problem in that section (sec. 1.4).

Perhaps there is a way to combine Tom’s Bogolyubov theorem (or my work with Olof) with this algebraic approach to give a strong solution to the problem at the beginning of that note (since Tom’s result implies that sumsets contain 99 percent of the elements of a large coset progression, one almost has that contains the union , where is some large subspace/subgroup with the element removed). If so, then, as I comment, one would get strong upper bounds on the size of subsets in without 3APs, under a certain “uniformity assumption” (which perhaps can be removed) mentioned in the note.

Assuming this could be done, it makes me wonder whether one could say something similar in — can algebraic and combinatorial/analytic methods be combined to attack the CAP set problem?

]]>It’s clear that what I have just said *must* be true, since one of the key facts driving what Nets was writing about is that you can’t have more than Fourier coefficients of size in a subspace of dimension whereas the function I discussed in my previous comment had Fourier coefficients lying in a subspace of dimension This isn’t a contradiction, because the function does not take values in though it does take values in But let us see in more detail why it isn’t a contradiction.

The argument for the bound goes like this. (I have already given it, and Nets has given it in a different form.) Suppose we can find a subspace of dimension such that for values of the Fourier coefficient has modulus at least Then (The initial comes from the fact that Now the restriction of to is the Fourier transform of where is the subspace annihilated by (so has codimension ). Therefore, by Parseval, If is substantially less than 1, then this roughly tells us that and therefore that there is an affine subspace of codimension (a translate of ) where takes at least the value which is the same as saying that the density of in that affine subspace is at least It is here that I have used positivity: if I didn’t know that was positive, then its norm might be accounted for by some large negative values instead.

Now if we did Roth-Meshulam iterations, we would obtain a density increment of whereas actually we have obtained So the assumption that we cannot improve on Roth-Meshulam implies that cannot be substantially bigger than as claimed.

]]>Suppose we want to try to prove that a set of density with no APs of length 3 must have a substantial density increase on some subspace of not very high codimension. It is tempting to try to prove a more general assertion, namely that if is any bounded function that averages and is significantly smaller than then averages significantly more than on some low-codimensional subspace.

However, I think the “killer example” shows that this statement is false. It gives us an example of a set with density and with AP-density roughly Now consider the function This again has density Its 3AP count is However, the fact that it is hard to obtain a density *decrease* for makes it hard to obtain a density *increase* for

This shows that any proof that hopes to use a density-increment method to obtain a substantial improvement to the Roth-Meshulam bound must use in a crucial way the fact that the characteristic function of takes values in rather than, say, in More suggestively, the balanced function of takes values in and we must use the “one-sided” fact that its negative values cannot have large modulus.

Since this is potentially an important point, let me try to be more precise about what kinds of arguments the example rules out. To do that, I’ll bound from below the density of in an -codimensional affine subspace Let be the annihilator of and let be the annihilator of Then the density of in is unless contains a point of In the second case, contains a vector capable of distinguishing from Whether or not it does so depends on whether we choose the right translate of but the point is that the density of in is roughly where is the number of the that are intersected by Since the are assumed to be in general position, the density of in is at least roughly Thus, to get a density increase proportional to the number of dimensions we must lose is also proportional to

If this is correct, it gives a test that should be applied to any attempted proof that tries to use a density-increment argument: does the argument generalize in a natural way to -valued functions? If it does, then the attempt will not improve on the Roth-Meshulam bound.

As with the post, this comment comes with a health warning: I haven’t written things up carefully so it is possible that what I am saying is nonsense.

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