Comments for Gowers's Weblog http://gowers.wordpress.com Mathematics related discussions Mon, 16 Nov 2009 22:58:12 +0000 http://wordpress.com/ hourly 1 Comment on The first unknown case of polynomial DHJ by gowers http://gowers.wordpress.com/2009/11/14/the-first-unknown-case-of-polynomial-dhj/#comment-4350 gowers Mon, 16 Nov 2009 22:58:12 +0000 http://gowers.wordpress.com/?p=1134#comment-4350 What other properties are you assuming? For instance, is there some sense in which the graph is "dense"? (E.g. the graph might have n vertices and every vertex might have a distinct label with i and j both at most $latex C\sqrt n$.) Is it possible to say in slightly more detail what kind of statement you are looking for? What other properties are you assuming? For instance, is there some sense in which the graph is “dense”? (E.g. the graph might have n vertices and every vertex might have a distinct label with i and j both at most C\sqrt n.) Is it possible to say in slightly more detail what kind of statement you are looking for?

]]> Comment on The first unknown case of polynomial DHJ by Randall McCutcheon http://gowers.wordpress.com/2009/11/14/the-first-unknown-case-of-polynomial-dhj/#comment-4349 Randall McCutcheon Mon, 16 Nov 2009 22:25:24 +0000 http://gowers.wordpress.com/?p=1134#comment-4349 Sources of optimism, sources of pessimism. I'm going to try to explain why I think DHJ3 with $latex n^2$ wildcards might be easier than QDHJ2. In order to do this, I need to explain a bit about the ergodic theory (actually more functional analysis) formulation of 5 above (albeit with squares instead of upper triangles). I apologize in advance if I skirt too many details to be clear to ergodic theory outsiders...I'll elaborate on any given points if asked. Let $latex (U_{ij})$ be commuting unitary operators on a Hilbert space and let $latex f$ be a vector in the Hilbert space. If $latex \alpha$ is a finite subset of naturals, put $latex V_\alpha = \prod_{i,j\in \alpha} U_{ij}$. Then for any $latex \epsilon>0$ there is an $latex \alpha$ with $latex \langle f, V_\alpha f\rangle > -\epsilon$. A proof of the above would establish 5 from my previous post. Now, the first thing one would try is to take a weak limit of the $latex V_\alpha$ along an "IP ring". (An IP ring is the set of finite unions of a countable family of disjoint finite sets, and the limit is taken as the minimal element of your set tends to infinity.) You can always do this, by Hindman's theorem. Denoting the limit by $latex P$, one has $latex \langle f, V_\alpha f\rangle \rightarrow \langle f, Pf\rangle$. In order to get that this is non-negative you need to know that $latex P$ is positive. We think in fact it is positive but can't prove it (still, this is a very promising avenue), and vexingly there is a counterexample (BFM96) showing that it need not be an orthogonal projection, which is the only way we know how to show that $latex P$ is positive in more benign cases. A further downside is that even if one could show that $latex P$ is positive in the general case, that would be the end of the story...one would have a nice result about two element configurations but no Furstenberg-style "structure theory" to handle more general cases (cases that would generalize Szemeredi's theorem, or even Roth's theorem). Now consider what happens when in DHJ2 you want your wildcard set to have cardinality $latex n^2$, and don't care about the structure of the wildcard set. It suffices to prove the following. Let $latex (U_{i})$ be unitary operators (not necessarily commuting) on a Hilbert space and let $latex f$ be a vector in the Hilbert space. If $latex \alpha$ is a finite subset of naturals, put $latex U_\alpha = \prod_{i\in \alpha} U_{i}$ where the product is in decreasing order of indices. Then for any $latex \epsilon>0$ there is an $latex \alpha$ with $latex |\alpha| =n^2$ and $latex \langle f, U_\alpha f\rangle > -\epsilon$. It turns out (I think...I have it all written down but no one, me included, has checked it carefully) that this can be proved and in a way that promises a (possible...I haven't thought about it too much) full-blown structure theory. It goes like this: first choose a subsequence such that for every $latex k$, the weak operator limit of $latex U_\alpha$ restricted to those $latex \alpha$ with $latex |\alpha|=k$ exists. Call the limit $latex P_k$. Now take any function $latex n$ from subsets of $latex {\bf N}$ to the naturals of the form $latex n(\alpha)=\sum_{i\in \alpha} x_i$ and choose an IP ring such that the weak operator limit of $latex P_{n(\alpha)^2}$ exists. Call it $latex Q$. Now (assuming I didn't cheat) $latex Q$ is an orthogonal projection and everything works. (That orthogonal projections yield structure theories is a meta-theorem, so by "everything" I include that, though I have literally thought about it for less than one minute, in which I surmised that it would be "very very complicated". Also a suitable PET-style induction scheme would have to be found...haven't thought about that either, beyond noting that it might not be completely obvious.) Sources of optimism, sources of pessimism.

I’m going to try to explain why I think DHJ3 with n^2 wildcards might be easier than QDHJ2. In order to do this, I need to explain a bit about the ergodic theory (actually more functional analysis) formulation of 5 above (albeit with squares instead of upper triangles). I apologize in advance if I skirt too many details to be clear to ergodic theory outsiders…I’ll elaborate on any given points if asked.

Let (U_{ij}) be commuting unitary operators on a Hilbert space and let f be a vector in the Hilbert space. If \alpha is a finite subset of naturals, put V_\alpha = \prod_{i,j\in \alpha} U_{ij}. Then for any \epsilon>0 there is an \alpha with \langle f, V_\alpha f\rangle > -\epsilon.

A proof of the above would establish 5 from my previous post. Now, the first thing one would try is to take a weak limit of the V_\alpha along an “IP ring”. (An IP ring is the set of finite unions of a countable family of disjoint finite sets, and the limit is taken as the minimal element of your set tends to infinity.) You can always do this, by Hindman’s theorem. Denoting the limit by P, one has \langle f, V_\alpha f\rangle \rightarrow \langle f, Pf\rangle. In order to get that this is non-negative you need to know that P is positive. We think in fact it is positive but can’t prove it (still, this is a very promising avenue), and vexingly there is a counterexample (BFM96) showing that it need not be an orthogonal projection, which is the only way we know how to show that P is positive in more benign cases. A further downside is that even if one could show that P is positive in the general case, that would be the end of the story…one would have a nice result about two element configurations but no Furstenberg-style “structure theory” to handle more general cases (cases that would generalize Szemeredi’s theorem, or even Roth’s theorem).

Now consider what happens when in DHJ2 you want your wildcard set to have cardinality n^2, and don’t care about the structure of the wildcard set. It suffices to prove the following.

Let (U_{i}) be unitary operators (not necessarily commuting) on a Hilbert space and let f be a vector in the Hilbert space. If \alpha is a finite subset of naturals, put U_\alpha = \prod_{i\in \alpha} U_{i} where the product is in decreasing order of indices. Then for any \epsilon>0 there is an \alpha with |\alpha| =n^2 and \langle f, U_\alpha f\rangle > -\epsilon.

It turns out (I think…I have it all written down but no one, me included, has checked it carefully) that this can be proved and in a way that promises a (possible…I haven’t thought about it too much) full-blown structure theory. It goes like this: first choose a subsequence such that for every k, the weak operator limit of U_\alpha restricted to those \alpha with |\alpha|=k exists. Call the limit P_k. Now take any function n from subsets of {\bf N} to the naturals of the form n(\alpha)=\sum_{i\in \alpha} x_i and choose an IP ring such that the weak operator limit of P_{n(\alpha)^2} exists. Call it Q. Now (assuming I didn’t cheat) Q is an orthogonal projection and everything works. (That orthogonal projections yield structure theories is a meta-theorem, so by “everything” I include that, though I have literally thought about it for less than one minute, in which I surmised that it would be “very very complicated”. Also a suitable PET-style induction scheme would have to be found…haven’t thought about that either, beyond noting that it might not be completely obvious.)

]]> Comment on The first unknown case of polynomial DHJ by Gil http://gowers.wordpress.com/2009/11/14/the-first-unknown-case-of-polynomial-dhj/#comment-4348 Gil Mon, 16 Nov 2009 22:13:14 +0000 http://gowers.wordpress.com/?p=1134#comment-4348 Is there a regularity lemma for graphs whose vertices are labelled by pairs {i,j}? Is there a regularity lemma for graphs whose vertices are labelled by pairs {i,j}?

]]> Comment on The first unknown case of polynomial DHJ by Randall McCutcheon http://gowers.wordpress.com/2009/11/14/the-first-unknown-case-of-polynomial-dhj/#comment-4347 Randall McCutcheon Mon, 16 Nov 2009 21:04:08 +0000 http://gowers.wordpress.com/?p=1134#comment-4347 Almost right.... It might be useful (to me, if to no one else) to set up a garbage thread for people (maybe just me) to use as a previewer. Conjecture: Let $latex \delta>0, k\in {\bf N}$. There exists $latex M=M(\delta,k)$ having the property that if $latex E\subset \{0,1,\ldots ,k-1\}^M$ with $latex |E|\geq \delta k^M$ then there exist $latex n\in {\bf N}$ and a variable word $latex w(x)$ having $latex n^2$ wildcards such that $latex \big\{w(t):t\in \{0,1,\ldots ,k-1\} \big\}\subset E$. Almost right…. It might be useful (to me, if to no one else) to set up a garbage thread for people (maybe just me) to use as a previewer.

Conjecture: Let \delta>0, k\in {\bf N}. There exists M=M(\delta,k) having the property that if E\subset \{0,1,\ldots ,k-1\}^M with |E|\geq \delta k^M then there exist n\in {\bf N} and a variable word w(x) having n^2 wildcards such that \big\{w(t):t\in \{0,1,\ldots ,k-1\} \big\}\subset E.

]]> Comment on The first unknown case of polynomial DHJ by Randall McCutcheon http://gowers.wordpress.com/2009/11/14/the-first-unknown-case-of-polynomial-dhj/#comment-4346 Randall McCutcheon Mon, 16 Nov 2009 20:57:04 +0000 http://gowers.wordpress.com/?p=1134#comment-4346 Hmmm...is the issue that you can't use less than in math mode? Another try: 3a. FVIP Sarkozy (M): If $latex \Omega$ is an infinite commutative semigroup, $latex (y_i)$ a sequence in $latex \Omega$, $latex (n_i)$ is a sequence of integers, and $latex E\subset \Omega$ has positive density, then $latex E$ contains a configuration $latex \{a, a+ \sum_{i,j\in \alpha, j>i} n_i y_j \}$. Next is the first alluded-to proof in the $latex k=2$ case of getting a wildcard set of cardinality $latex n^2$ (hoping for less than 3 latex perplexities). Yes, I do have reason to believe that the $latex k=3$ case of this is easier than $latex {\rm QDHJ}_2$. I can elaborate at some point. (The second, longer proof provides the basis for the optimism.) Conjecture: Let $latex \delta>0, k\in \mathbf{N}$. There exists $latex M=M(\delta,k)$ having the property that if $latex E\subset \{0,1,\ldots ,k-1\}^M$ with $latex |E|\geq \delta k^M$ then there exist $latex n\in {\bf N}$ and a variable word $latex w(x)$ having $latex n^2$ wildcards such that $latex \big\{w(t):t\in \{0,1,\ldots ,k-1\} \big\}\subset E$. Proof for $latex k=2$: Let $latex \delta_0$ be the infimum of the set of $latex \delta$ for which the conclusion holds and assume for contradiction that $latex \delta_0>0$. Choose by Sarkozy-Furstenberg an $latex m$ such that for any $latex A\subset \{1,2,\ldots ,m\}$ with $latex |A|\geq {\delta_0\over 3} m$, $latex A$ contains a configuration $latex \{a, a+n^2\}$, with $latex n>0$. Let $latex \delta=\delta_0-{\delta_0\over 4\cdot 2^m}$ and put $latex M'=M\big(\delta_0+{\delta_0\over 3\cdot 2^m},2\big)$. Finally put $latex M=m+M'$. We claim $latex M$ works as $latex M(\delta,2)$. Suppose then that $latex E\subset \{0,1\}^M$ with $latex |E|\geq \delta 2^M$. Now, for each $latex v\in \{0,1\}^m$, let $latex E_v=\big\{w\in \{0,1\}^{M'}:vw\in E\big\}$.If $latex |E_v|> \big( \delta_0+{\delta_0\over 3\cdot 2^r}\big) 2^{M'}$ for some $latex v$ we are done; $latex E_v$ will by hypothesis contain $latex \{ w(0),w(1)\}$ for some variable word $latex w$ having $latex n^2$ wildcards for some $latex n$, so that $latex \{ vw(0), vw(1) \}\subset E$. (Notice that $latex vw(x)$ is again a variable word having $latex n^2$ wildcards.) We may therefore assume that $latex |E_v|\leq \big( \delta_0+{\delta_0\over 3\cdot 2^m}\big) 2^{M'}$ for every $latex v$. A simple calculation now shows that $latex |E_v|\geq {\delta_0\over 3}2^{M'}$ for all $latex v$ (otherwise, $latex E$ would be too small.) Now for $latex 1\leq i\leq m$, let $latex v_i$ be the word consisting of $latex i$ 0s followed by $latex (m-i)$ 1s. Since $latex \sum_{i=1}^m |E_{v_i}|\geq {m \delta_0\over 3}2^{M'}$, there must be some $latex u\in \{0,1\}^{M'}$ with $latex \big\vert \big\{i: u\in E_{v_i} \big\} \big\vert \geq {\delta_0\over 3}m$. By choice of $latex m$, there are $latex a$ and $latex n>0$ such that $latex u\in E_{v_a}\cap E_{v_{a+n^2}}$. It follows that $latex \{ v_{a}u, v_{a+n^2}u\} \subset E$. This set has the form $latex \{ w(0), w(1)\}$ for a variable word $latex w(x)$ having $latex n^2$ wildcards. Hmmm…is the issue that you can’t use less than in math mode? Another try:

3a. FVIP Sarkozy (M): If \Omega is an infinite commutative semigroup, (y_i) a sequence in \Omega, (n_i) is a sequence of integers, and E\subset \Omega has positive density, then E contains a configuration \{a, a+ \sum_{i,j\in \alpha, j>i} n_i y_j \}.

Next is the first alluded-to proof in the k=2 case of getting a wildcard set of cardinality n^2 (hoping for less than 3 latex perplexities). Yes, I do have reason to believe that the k=3 case of this is easier than {\rm QDHJ}_2. I can elaborate at some point. (The second, longer proof provides the basis for the optimism.)

Conjecture: Let \delta>0, k\in \mathbf{N}. There exists M=M(\delta,k) having the property that if E\subset \{0,1,\ldots ,k-1\}^M with |E|\geq \delta k^M then there exist n\in {\bf N} and a variable word w(x) having n^2 wildcards such that \big\{w(t):t\in \{0,1,\ldots ,k-1\} \big\}\subset E.

Proof for k=2: Let \delta_0 be the infimum of the set of \delta for which the conclusion holds and assume for contradiction that \delta_0>0. Choose by Sarkozy-Furstenberg an m such that for any A\subset \{1,2,\ldots ,m\} with |A|\geq {\delta_0\over 3} m, A contains a configuration \{a, a+n^2\}, with n>0. Let \delta=\delta_0-{\delta_0\over 4\cdot 2^m} and put M'=M\big(\delta_0+{\delta_0\over 3\cdot 2^m},2\big). Finally put M=m+M'. We claim M works as M(\delta,2). Suppose then that E\subset \{0,1\}^M with |E|\geq \delta 2^M.

Now, for each v\in \{0,1\}^m, let E_v=\big\{w\in \{0,1\}^{M'}:vw\in E\big\}.If |E_v|> \big( \delta_0+{\delta_0\over 3\cdot 2^r}\big) 2^{M'} for some v we are done; E_v will by hypothesis contain \{ w(0),w(1)\} for some variable word w having n^2 wildcards for some n, so that \{ vw(0), vw(1) \}\subset E. (Notice that vw(x) is again a variable word having n^2 wildcards.)

We may therefore assume that |E_v|\leq \big( \delta_0+{\delta_0\over 3\cdot 2^m}\big) 2^{M'} for every v. A simple calculation now shows that |E_v|\geq {\delta_0\over 3}2^{M'} for all v (otherwise, E would be too small.)

Now for 1\leq i\leq m, let v_i be the word consisting of i 0s followed by (m-i) 1s. Since \sum_{i=1}^m |E_{v_i}|\geq {m \delta_0\over 3}2^{M'}, there must be some u\in \{0,1\}^{M'} with \big\vert \big\{i: u\in E_{v_i} \big\} \big\vert \geq {\delta_0\over 3}m. By choice of m, there are a and n>0 such that u\in E_{v_a}\cap E_{v_{a+n^2}}. It follows that \{ v_{a}u, v_{a+n^2}u\} \subset E. This set has the form \{ w(0), w(1)\} for a variable word w(x) having n^2 wildcards.

]]> Comment on The first unknown case of polynomial DHJ by Randall McCutcheon http://gowers.wordpress.com/2009/11/14/the-first-unknown-case-of-polynomial-dhj/#comment-4345 Randall McCutcheon Mon, 16 Nov 2009 20:42:22 +0000 http://gowers.wordpress.com/?p=1134#comment-4345 Some corrections. (Principal, of course, not principle.) I am finding it difficult to handle the incorporation of latex without encountering baffling anomalies. V is officially for "variant" cf. BFM96 though Hillel chose the name wth "Vitaly" in mind (maybe...I wasn't consulted). The F is for officially for "finitely generated" (the problem with VIP systems is that when you do PET you can possibly get down to infinitely different many IPs, which is the source of all the trouble...if you just stipulate that this is not to happen you can prove things and there are a few natural applications along these lines, such as generalized polynomials and other things), though it is entirely plausible, if unlikely, that I chose it with "Furstenberg" in mind. Here is an attempt to repair 3a. above (in answer to Tim's question $latex \alpha$ is any large finite set): 3a. FVIP Sarkozy (M): If $latex \Omega$ is an infinite commutative semigroup, $latex (y_i)$ a sequence in $latex \Omega$, $latex (n_i)$ is a sequence of integers, and $latex E\subset \Omega$ has positive density, then $latex E$ contains a configuration $latex \{a, a+ \sum_{i,j\in \alpha, j>i} n_iy_j\}$. There exists $latex M=M(\delta,k)$ having the property that if $latex E\subset \{0,1,\ldots ,k-1\}^M$ with $latex |E|\geq \delta k^M$ then there exist $latex n\in {\bf N}$ and a variable word $latex w(x)$ having $latex n^2$ wildcards such that $latex \big\{w(t):t\in \{0,1,\ldots ,k-1\} \big\}\subset E$. Proof for $latex k=2$: Let $latex \delta_0$ be the infimum of the set of $latex \delta$ for which the conclusion holds and assume for contradiction that $latex \delta_0>0$. Choose by Sarkozy-Furstenberg an $latex m$ such that for any $latex A\subset \{1,2,\ldots ,m\}$ with $latex |A|\geq {\delta_0\over 3} m$, $latex A$ contains a configuration $latex \{a, a+n^2\}$, with $latex n>0$. Let $latex \delta=\delta_0-{\delta_0\over 4\cdot 2^m}$ and put $latex M'=M\big(\delta_0+{\delta_0\over 3\cdot 2^m},2\big)$. Finally put $latex M=m+M'$. We claim $latex M$ works as $latex M(\delta,2)$. Suppose then that $latex E\subset \{0,1\}^M$ with $latex |E|\geq \delta 2^M$. Now, for each $latex v\in \{0,1\}^m$, let $latex E_v=\big\{w\in \{0,1\}^{M'}:vw\in E\big\}$.If $latex |E_v|> \big( \delta_0+{\delta_0\over 3\cdot 2^r}\big) 2^{M'}$ for some $latex v$ we are done; $latex E_v$ will by hypothesis contain $latex \{ w(0),w(1)\}$ for some variable word $latex w$ having $latex n^2$ wildcards for some $latex n$, so that $latex \{ vw(0), vw(1) \}\subset E$. (Notice that $latex vw(x)$ is again a variable word having $latex n^2$ wildcards.) We may therefore assume that $latex |E_v|\leq \big( \delta_0+{\delta_0\over 3\cdot 2^m}\big) 2^{M'}$ for every $latex v$. A simple calculation now shows that $latex |E_v|\geq {\delta_0\over 3}2^{M'}$ for all $latex v$ (otherwise, $latex E$ would be too small.) Now for $latex 1\leq i\leq m$, let $latex v_i$ be the word consisting of $latex i$ 0s followed by $latex (m-i)$ 1s. Since $latex \sum_{i=1}^m |E_{v_i}|\geq {m \delta_0\over 3}2^{M'}$, there must be some $latex u\in \{0,1\}^{M'}$ with $latex \big\vert \big\{i: u\in E_{v_i} \big\} \big\vert \geq {\delta_0\over 3}m$. By choice of $latex m$, there are $latex a$ and $latex n>0$ such that $latex u\in E_{v_a}\cap E_{v_{a+n^2}}$. It follows that $latex \{ v_{a}u, v_{a+n^2}u\} \subset E$. This set has the form $latex \{ w(0), w(1)\}$ for a variable word $latex w(x)$ having $latex n^2$ wildcards. Some corrections.

(Principal, of course, not principle.) I am finding it difficult to handle the incorporation of latex without encountering baffling anomalies. V is officially for “variant” cf. BFM96 though Hillel chose the name wth “Vitaly” in mind (maybe…I wasn’t consulted). The F is for officially for “finitely generated” (the problem with VIP systems is that when you do PET you can possibly get down to infinitely different many IPs, which is the source of all the trouble…if you just stipulate that this is not to happen you can prove things and there are a few natural applications along these lines, such as generalized polynomials and other things), though it is entirely plausible, if unlikely, that I chose it with “Furstenberg” in mind.

Here is an attempt to repair 3a. above (in answer to Tim’s question \alpha is any large finite set):

3a. FVIP Sarkozy (M): If \Omega is an infinite commutative semigroup, (y_i) a sequence in \Omega, (n_i) is a sequence of integers, and E\subset \Omega has positive density, then E contains a configuration \{a, a+ \sum_{i,j\in \alpha, j>i} n_iy_j\}. There exists M=M(\delta,k) having the property that if E\subset \{0,1,\ldots ,k-1\}^M with |E|\geq \delta k^M then there exist n\in {\bf N} and a variable word w(x) having n^2 wildcards such that \big\{w(t):t\in \{0,1,\ldots ,k-1\} \big\}\subset E.

Proof for k=2: Let \delta_0 be the infimum of the set of \delta for which the conclusion holds and assume for contradiction that \delta_0>0. Choose by Sarkozy-Furstenberg an m such that for any A\subset \{1,2,\ldots ,m\} with |A|\geq {\delta_0\over 3} m, A contains a configuration \{a, a+n^2\}, with n>0. Let \delta=\delta_0-{\delta_0\over 4\cdot 2^m} and put M'=M\big(\delta_0+{\delta_0\over 3\cdot 2^m},2\big). Finally put M=m+M'. We claim M works as M(\delta,2). Suppose then that E\subset \{0,1\}^M with |E|\geq \delta 2^M.

Now, for each v\in \{0,1\}^m, let E_v=\big\{w\in \{0,1\}^{M'}:vw\in E\big\}.If |E_v|> \big( \delta_0+{\delta_0\over 3\cdot 2^r}\big) 2^{M'} for some v we are done; E_v will by hypothesis contain \{ w(0),w(1)\} for some variable word w having n^2 wildcards for some n, so that \{ vw(0), vw(1) \}\subset E. (Notice that vw(x) is again a variable word having n^2 wildcards.)

We may therefore assume that |E_v|\leq \big( \delta_0+{\delta_0\over 3\cdot 2^m}\big) 2^{M'} for every v. A simple calculation now shows that |E_v|\geq {\delta_0\over 3}2^{M'} for all v (otherwise, E would be too small.)

Now for 1\leq i\leq m, let v_i be the word consisting of i 0s followed by (m-i) 1s. Since \sum_{i=1}^m |E_{v_i}|\geq {m \delta_0\over 3}2^{M'}, there must be some u\in \{0,1\}^{M'} with \big\vert \big\{i: u\in E_{v_i} \big\} \big\vert \geq {\delta_0\over 3}m. By choice of m, there are a and n>0 such that u\in E_{v_a}\cap E_{v_{a+n^2}}. It follows that \{ v_{a}u, v_{a+n^2}u\} \subset E. This set has the form \{ w(0), w(1)\} for a variable word w(x) having n^2 wildcards.

]]> Comment on The first unknown case of polynomial DHJ by gowers http://gowers.wordpress.com/2009/11/14/the-first-unknown-case-of-polynomial-dhj/#comment-4344 gowers Mon, 16 Nov 2009 19:14:50 +0000 http://gowers.wordpress.com/?p=1134#comment-4344 Randall, Many thanks for this interesting collection of further problems (not all of which I yet understand -- e.g. is alpha an infinite set in 3-5 or is it just large and finite?). I haven't yet managed to guess what V and FV stand for ... One small remark is that the problem about getting a wildcard set of perfect-square size in DHJ3 looks to me unlikely to be easier than quadratic DHJ, since it implies that you can find (x,x+y^2,x+2y^2) in a dense set of integers, which is strictly harder than finding (x,x+y^2), which seems to be all you can get out of QDHJ. But perhaps you see things differently because you know more about the currently available proof techniques: I would be interested to hear your thoughts on this. Randall, Many thanks for this interesting collection of further problems (not all of which I yet understand — e.g. is alpha an infinite set in 3-5 or is it just large and finite?). I haven’t yet managed to guess what V and FV stand for …

One small remark is that the problem about getting a wildcard set of perfect-square size in DHJ3 looks to me unlikely to be easier than quadratic DHJ, since it implies that you can find (x,x+y^2,x+2y^2) in a dense set of integers, which is strictly harder than finding (x,x+y^2), which seems to be all you can get out of QDHJ. But perhaps you see things differently because you know more about the currently available proof techniques: I would be interested to hear your thoughts on this.

]]> Comment on The first unknown case of polynomial DHJ by Randall McCutcheon http://gowers.wordpress.com/2009/11/14/the-first-unknown-case-of-polynomial-dhj/#comment-4343 Randall McCutcheon Mon, 16 Nov 2009 17:01:24 +0000 http://gowers.wordpress.com/?p=1134#comment-4343 Somehow this got eaten....I mention it only because it's the one instance in which the principle difficulty has been overcome, though unfortunately in a way that cannot possibly generalize too far: 3b. DVIP Sarkozy (Balister-Bergelson-M): If $latex E$ is a set of natural numbers having positive upper density and $latex (n_{ij})$ is an infinite upper triangular matrix whose entries are natural numbers with (various conditions here having a rather complicated and not easy to formulate common theme, including $latex n_{ij}=j^i$ and $latex n_{ij}= o (\sqrt{i\over \log i})$ as notable special cases) then $latex E$ contains a configuration of the form $latex \{a, a+\sum_{i,j\in \alpha, i>j} n_{ij} \}$. Somehow this got eaten….I mention it only because it’s the one instance in which the principle difficulty has been overcome, though unfortunately in a way that cannot possibly generalize too far:

3b. DVIP Sarkozy (Balister-Bergelson-M): If E is a set of natural numbers having positive upper density and (n_{ij}) is an infinite upper triangular matrix whose entries are natural numbers with (various conditions here having a rather complicated and not easy to formulate common theme, including n_{ij}=j^i and n_{ij}= o (\sqrt{i\over \log i}) as notable special cases) then E contains a configuration of the form \{a, a+\sum_{i,j\in \alpha, i>j} n_{ij} \}.

]]> Comment on The first unknown case of polynomial DHJ by Randall McCutcheon http://gowers.wordpress.com/2009/11/14/the-first-unknown-case-of-polynomial-dhj/#comment-4342 Randall McCutcheon Mon, 16 Nov 2009 16:55:12 +0000 http://gowers.wordpress.com/?p=1134#comment-4342 In this post I will suggest a couple of (ostensibly easier) problems. Perhaps they should be taken with a grain of salt in that I was also keen on going for easier cases than DHJ$latex _3$ on the polymath1 blog. In that case it seems I was unduly pessimistic, however here there may be a better case for pessimism. Here are some progressively more general statements. 1. Sarkozy's theorem: every set of natural numbers having positive upper density contains a configuration $latex \{a,a+n^2\}$. 2. IP Sarkozy (Bergelson-Furstenberg): If $latex E$ is a set of natural numbers having positive upper density and $latex R$ is an IP set (that is, the set of finite sums without repeats of some infinite sequence) then $latex E$ contains a configuration of the form $latex \{a, a+r^2\}$ with $latex r\in R$. 3a. FVIP Sarkozy (M): If $latex \Omega$ is an infinite commutative semigroup, $latex (y_i)$ a sequence in $latex \Omega$, $latex (n_i)$ a sequence of integers, and $latex E\subset \Omega$ has positive density, then $latex E$ contains a configuration $latex \{a, a+ \sum_{i,j\in \alpha, ij} n_{ij} \}$. 4. VIP Sarkozy in $latex {\bf N}$ (open). If $latex E$ is a set of natural numbers having positive upper density and $latex (n_{ij})$ is an infinite upper triangular matrix whose entries are natural numbers then $latex E$ contains a configuration of the form $latex \{a, a+\sum_{i,j\in \alpha, i>j} n_{ij} \}$. 5. VIP Sarkozy (open). If $latex \Omega$ is an infinite commutative semigroup, $latex E\subset \Omega$ is a set of positive upper density and $latex (n_{ij})$ is an infinite upper triangular matrix whose entries come from $latex \Omega$ then $latex E$ contains a configuration of the form $latex \{a, a+\sum_{i,j\in \alpha, i>j} n_{ij} \}$. (It's fully general here to take $latex \Omega ={\bf N}^{{\bf N}\times {\bf N}}$ and let $latex n_{ij}$ be the coordinate-wise basis, i.e. $latex n_{ij}(i,j)=1$ and $latex n_{ij}=0$ elsewhere.) 6. QDHJ$latex _2$ (Conjecture 1 above). In the ergodic theory setting, we have worked on 6 virtually not at all...some attention has been given the less ambitious 5 (2, 3a and 3b constituting some miniscule progress), where one has a very natural ergodic formulation that everyone believes should be true and looks like it might be amenable to attack. It's not really even ergodic theory, but straight functional analysis, so even Tim might like this angle...I can explain it more fully in a separate reply. For 6, we don't even have a decent ergodic theory formulation I think one can use Tao's ideas from Polymath1 to give a lousy one, but my somewhat strong (and slightly principled) hunch is that ergodic theory will prove useless in attacking 6. This is one respect in which the problem appeals to me polymathically...since I don't have any idea how to do it using methods familiar to me, I am very interested in what people would come up with from different angles. On the other hand, there are often (perhaps even always) strong parallels in the methods, which is some cause for pessimism. Another strong plus is that the problem has what I see as at least two weak formulations that are themselves very good problems; a solution to either would amount to a great success, I think...they are 5 above and DHJ$latex _3$ with $latex n^2$ wildcards, which I mentioned in my previous reply. From the ergodic theory perspective, 5 is almost surely cleaner to think about and already exhibits the crucial source of all the pessimism, though the graph/clique formulation may be nicer from a strictly combinatorial perspective. In this post I will suggest a couple of (ostensibly easier) problems.
Perhaps they should be taken with a grain of salt in that I was also
keen on going for easier cases than DHJ_3 on the polymath1 blog. In
that case it seems I was unduly pessimistic, however here there may
be a better case for pessimism. Here are some progressively more
general statements.

1. Sarkozy’s theorem: every set of natural numbers having positive
upper density contains a configuration \{a,a+n^2\}.

2. IP Sarkozy (Bergelson-Furstenberg): If E is a
set of natural numbers having positive upper density and R
is an IP set (that is, the set of finite sums without repeats of
some infinite sequence) then E contains a configuration of
the form \{a, a+r^2\} with r\in R.

3a. FVIP Sarkozy (M): If \Omega is an infinite commutative
semigroup, (y_i) a sequence in \Omega, (n_i)
a sequence of integers, and E\subset \Omega has positive
density, then E contains a configuration \{a, a+ \sum_{i,j\in \alpha, ij} n_{ij} \}.

4. VIP Sarkozy in {\bf N} (open). If E is a set of
natural numbers having positive upper density and (n_{ij})
is an infinite upper triangular matrix whose entries are natural
numbers then E contains a configuration of the form \{a, a+\sum_{i,j\in \alpha, i>j} n_{ij} \}.

5. VIP Sarkozy (open). If \Omega is an infinite commutative
semigroup, E\subset \Omega is a set of positive upper
density and (n_{ij}) is an infinite upper triangular matrix
whose entries come from \Omega then E contains a
configuration of the form \{a, a+\sum_{i,j\in \alpha, i>j} n_{ij} \}. (It’s fully general here to take \Omega ={\bf N}^{{\bf N}\times {\bf N}} and let n_{ij} be the
coordinate-wise basis, i.e. n_{ij}(i,j)=1 and n_{ij}=0
elsewhere.)

6. QDHJ_2 (Conjecture 1 above).

In the ergodic theory setting, we have worked on 6 virtually not at
all…some attention has been given the less ambitious 5 (2, 3a and
3b constituting some miniscule progress), where one has a very
natural ergodic formulation that everyone believes should be true and
looks like it might be amenable to attack. It’s not really even
ergodic theory, but straight functional analysis, so even Tim might
like this angle…I can explain it more fully in a separate reply.

For 6, we don’t even have a decent ergodic theory formulation I think
one can use Tao’s ideas from Polymath1 to give a lousy one, but my
somewhat strong (and slightly principled) hunch is that ergodic theory
will prove useless in attacking 6. This is one respect in which the
problem appeals to me polymathically…since I don’t have any idea
how to do it using methods familiar to me, I am very interested in
what people would come up with from different angles. On the other
hand, there are often (perhaps even always) strong parallels in the
methods, which is some cause for pessimism. Another strong plus is
that the problem has what I see as at least two weak formulations
that are themselves very good problems; a solution to either would
amount to a great success, I think…they are 5 above and
DHJ_3 with n^2 wildcards, which I mentioned in my
previous reply. From the ergodic theory perspective, 5 is almost
surely cleaner to think about and already exhibits the crucial
source of all the pessimism, though the graph/clique formulation
may be nicer from a strictly combinatorial perspective.

]]> Comment on The first unknown case of polynomial DHJ by Randall McCutcheon http://gowers.wordpress.com/2009/11/14/the-first-unknown-case-of-polynomial-dhj/#comment-4341 Randall McCutcheon Mon, 16 Nov 2009 14:55:01 +0000 http://gowers.wordpress.com/?p=1134#comment-4341 I will have some more detailed comments but, very quickly, the coloristic version is due to Bergelson and Leibman...I did publish a proof in my springer lecture notes but it's not even independent...it is in fact their proof precisely, but with the ergodic facade removed (their proof is not actually dynamical...they set it up as a recurrence theorem for continuous maps of compact metric spaces, but nowhere in the proof did they use continuity of the maps or completeness of the space...all I did was remove the nonsense and make the proof read as a purely combinatorial one). A second point I would like to make is that there is one joint extension of the Sarkozy-Furstenberg theorem and the density Hales-Jewett theorem that is known to obtain...and that is that in DHJ_2, you can choose your wildcard set to have cardinality n^2. I have two proofs of this fact. The first is one paragraph, completely elementary but uses SF as a lemma; it is unlikely to generalize. (I'll try to dig it up and post it later.) The second is rather long (quite a few pages), uses ergodic theory and contains a lot of ideas that might prove useful in attacking more general problems of this type (like, for example, that one can choose a wildcard set of cardinality n^2 in DHJ_3). In fact, this seems to be a good problem to make explicit...it might be more accessible than the problem you discuss here and people might already have some good ideas about it, since we so recently were heavily into DHJ. Question: In DHJ_3, can one always find a combinatorial line with wildcard set of cardinality n^2? I will have some more detailed comments but, very quickly, the coloristic version is due to Bergelson and Leibman…I did publish a proof in my springer lecture notes but it’s not even independent…it is in fact their proof precisely, but with the ergodic facade removed (their proof is not actually dynamical…they set it up as a recurrence theorem for continuous maps of compact metric spaces, but nowhere in the proof did they use continuity of the maps or completeness of the space…all I did was remove the nonsense and make the proof read as a purely combinatorial one).

A second point I would like to make is that there is one joint extension of the Sarkozy-Furstenberg theorem and the density Hales-Jewett theorem that is known to obtain…and that is that in DHJ_2, you can choose your wildcard set to have cardinality n^2. I have two proofs of this fact. The first is one paragraph, completely elementary but uses SF as a lemma; it is unlikely to generalize. (I’ll try to dig it up and post it later.) The second is rather long (quite a few pages), uses ergodic theory and contains a lot of ideas that might prove useful in attacking more general problems of this type (like, for example, that one can choose a wildcard set of cardinality n^2 in DHJ_3).

In fact, this seems to be a good problem to make explicit…it might be more accessible than the problem you discuss here and people might already have some good ideas about it, since we so recently were heavily into DHJ.

Question: In DHJ_3, can one always find a combinatorial line with wildcard set of cardinality n^2?

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