Any more thoughts?

]]>Dear Mark, But the Sperner version should suffice to imply Sarkozy -Furstenberg’s theorem, right? Can we try the Cauchy-Swartz argument from http://terrytao.wordpress.com/2013/02/28/a-fourier-free-proof-of-the-furstenberg-sarkozy-theorem/ ?

]]>I think the Sperner version for forbidding differences (for all t) is like a variant of conjecture 1 where rather than requiring the wildcard set to be of the form we just require it to be of the form where $X$ and $Y$ have the same size. I think (from experience with the colouring case) that this is a significantly weaker statement (and is less applicable). However, that weakness might make it an excellent first step.

In the symmetric difference version then even if we only forbid a for a single fixed then we cannot have a positive proportion of the graphs on vertices. (Forbidding symmetric differences of odd size is trivial: hence the constraint that one part has even size in the above.)

Let us prove this. Suppose is a collection of an proportion of the graphs on vertices. Fix a set of vertices and let be disjoint sets (from and each other) of vertices. For any graph on vertices we can map into the space of graphs on vertices by mapping a point in viewed as a subset of to the graph

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By an averaging argument there must be some graph for which this image contains at least points and thus, for this , if we take the preimage of we have an proportion of and so, as in an earlier comment, we must have two points with symmetric difference of size 2, say . But this corresponds under our mapping to two graphs with symmetric difference : a as required.

]]>I don't see how to extend it to complete graphs (but perhaps complete bipartite graphs can be regarded as the first case of PHJ as well). We can try an application of the polynomial method a la Frankl Wilson or the remarkable argument for Frankl-Rodl theorem.

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What is for an arbitrary graph on vertices? It is

.

We can interpret this as follows: we define a random variable by picking a random clique of size 4 and letting take the value 1 if is even and 0 otherwise. Then is the expectation of .

For example, if is a single edge, then with probability , so . If is a complete bipartite graph with vertex sets of density and (so ), then we get that (since the number of edges in the joining the two vertex sets is even if the vertices of the are split into two even-sized sets and odd otherwise), which equals . If is a random graph, then the probability that is even looks close to 1/2, so will be small.

It looks hard to give some useful expression for , so the next step would probably be to try to work out what one hopes to extract from the Fourier information, just so we have a target to aim at.

]]>I did do something with Fourier things but I never really had a feel for the Fourier techniques so I could have easily overlooked something obvious. In other words it is very definitely worth trying them again.

]]>I asked the question because I was thinking about the problem. Here’s an idea that doesn’t work but that might conceivably be a model for a cleverer idea that does.

I’ll give a false proof for the problem. Define the –*ball* of a graph to be the set of all such that is a path of length 3. If the -balls of and intersect, then we have a graph that differs from each of and in a . If those two s are and , then putting them together we get the clique with vertex set .

Of course, that’s rubbish because there is no reason for the two s to be related in such a helpful way. So we have to make do with a much weaker condition: that the -ball of does not intersect the -ball of via two s of the form and . That’s probably too weak to be of use.

Did you at some point try to work out the Fourier transform of the set of s? I’m curious to know whether one can get anywhere with a density-increment argument.

]]>Oops. I know that argument well, and now can’t see why I didn’t see that it worked here.

]]>I think this one is easy. Split the cube into 2 bits: one of even parity and one of odd parity. Working just inside one we are asking for a set with no 2 points at distance at most 2: therefore the 1 balls miust be disjoint. These each contain n points o you cant have more than 2^n/n points in total.(and 2^n/n in the other part).

]]>Right, Mark, for subspaces the density and coloring questions are the same…good point.

]]>For the symmetric difference K_4 case I can’t even see if the colouring case is true (and I don’t remember thinking about it). Just to be explicit: suppose you r colour all graphs on n vertices. Must there be two graphs G and H with the same colour and symmetric difference a K_4?

]]>Doesn’t this last question (vector space of bounded codimension) follow from the colouring PHJ: colour the space of all graphs by which coset of the vector space it lies in? Or am I missing something?

]]>It looks that the easiest question we mentioned but did not unswer is this: Is there a vector space of graphs with n vertices of bounded codimension which does not contain any clique.

]]>@Gil: Yes, I’m pretty sure Conjecture 4 is completely open.

]]>Ah, that’s interesting. My first reaction to the fixed-graph suggestion was that it must surely be false, but then I realized that I was confusing what happens in the integers (where there are only two numbers at distance from ) with what happens here (where there are many graphs at “distance” from ).

That suggests two nice problems. One is to see what can be said about a set of graphs that doesn’t have any symmetric differences. Another is perhaps to ask for more with general cliques: for example, can we say that a set of graphs with no clique differences has exponentially small density? (I’d be happy with an exponential bound in the number of vertices, but probably this is a very hard problem — or false.)

]]>I don’t think I ever had anything non-trivial for this question but I will list a few of the things I remember.

I thought briefly about the symmetric difference being a clique and then wondered whether forbidding a single fixed size clique (eg ) was sufficient to rule out a positive density set . I was unable to solve this and did not return to the case of forbidding all clique symmetric differences. I have some recollection that I thought was more likely to rule out a positive proportion as it has an even number of edges (obviously necessary) and each vertex has even degree. I do not know whether the latter has any relevance.

I think it is also easy to show that forbidding as a symmetric difference is sufficient to rule out a positive proportion. It essentially follows from an averaging argument followed by an application of the fact that a positive proportion of the cube contains 2 points with symmetric difference of size 2.

Indeed suppose is an proportion of all graphs on vertices. Fix two vertices and let be the union of all the paths of length 2 between them. With an easy (but slightly tedious to write out) argument we can find and and two graphs with symmetric difference .

]]>Tim, just to be sure: Conjecture 4 and the others are completely open, right? it is not that they have known but extremely complex ergodic theoretic proof like was the case for DHJ?

On the code side, here is a nice question: Is there a family of graphs on n vertices of size 2**(alpha n^2) so that the symmetric difference between every two can be covered by ten (say) cliques.

]]>To judge from his web page, I think they are not published, so you’d have to ask him directly and hope that he still has his thesis in electronic form.

]]>Interesting… Is the special case where the collection of graphs itself is a subspace (namely closed under taking symmetric differences) known? Are Mark’s results in this area published?

]]>My former research student Mark Walters worked on the first question you ask during his PhD. He didn’t solve it (but did prove some theorems for other families of graphs, such as cycles), and as far as I know it is still not known. Of course, the obvious thing to try for this variant is Fourier analysis, and Mark did. I can’t remember what the difficulty was. It might be interesting to have another go.

]]>Is this known?

While at it when we consider families of graphs we can replace various notions of error-correcting codes by refine notions. E.g. families of graphs that the symmetric difference between any two has cromatic number larger than k, is connected, etc. Where these graph codes studied?

]]>How many edges can a graph G_i to G_j:s with j>i ? If G_i has edges to some set graphs with larger indices it follows that all edges added to G_i must form a sequence of nested cliques for those values of j. This means that G_0 will typically only have G_1 and the final complete graphs as neighbours.

For a G_i with i close to half the total number of edges all cliques for higher j will come from picking the edges of a sequence of nested cliques in the complement of G_i. This puts an upper bound of about log(n) on the degree, but again it is more likely to be constant.

Yet another bound comes from the fact that since all added edges have to be included in the cliques the size of the clique is controlled by the number of vertices included in the edges. For a uniform random graph this means hat as soon as the graph becomes connected the only possible clique is the whole K_n.

This means that for preferential attachment models where vertices of very high degree are created quickly the sizes of the cliques adding to the degree of G_i will have to grow quickly and we are again forced towards lower, or even bounded, degree for the clique-difference graph.

This is of course only informal reasoning but I don’t think it would be too hard to turn it into a proper result saying that for nested sequences of random graphs the chromatic number will be bounded.

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