Furstenberg-Katznelson argument: Difference between revisions

From Polymath Wiki
Jump to navigationJump to search
 
(12 intermediate revisions by 3 users not shown)
Line 1: Line 1:
'''Theorem DHJ(3)'''. For every <math>\delta > 0</math> there exists n such that every subset <math>A \subset [3]^n</math> of density at least <math>\delta</math> contains a combinatorial line.
This proof of [[DHJ(3)]] is being pursued in our [http://terrytao.wordpress.com/2009/02/11/a-reading-seminar-on-density-hales-jewett/ reading seminar].  Below is an ongoing "combinatorial" translation of the argument. [Very informal!]


'''Proof'''Suppose for contradiction that the claim failed; then there exists <math>\delta > 0</math>, a sequence of n going to infinity, and a sequence <math>A^{(n)} \subset [3]^n</math> of line-free sets, all with density at least <math>\delta</math>.
We write <math>[3] = \{0,1,2\}</math> for our alphabetWe let <math>Lines(n)</math> be the space of all combinatorial lines <math>\ell: [3] \to [3]^n</math> with "few" wildcards, where we will be vague about what "few" means.  We also have <math>Planes(n)</math>, the space of combinatorial planes <math>\pi: [3]^2 \to [3]^n</math>, and <math>Spaces(n)</math>, the space of combinatorial 3-spaces <math>\sigma: [3]^3 \to [3]^n</math>.  Note that one can think of a combinatorial plane in <math>[3]^n</math> as a combinatorial line in Lines(n), etc.


== Step 1: Random sampling ==
[[DHJ(3)]] is then the claim that if <math>f: [3]^n \to [0,1]</math> has large mean, then


Let <math>m = m^{(n)}</math> be a sufficiently slowly growing integer-valued function of n (e.g. one can take <math>m := \lfloor \log \log \log n \rfloor </math>).  We can then construct a random embedding <math>\phi^{(n)}: [3]^m \to [3]^n</math> as follows:
:<math>{\Bbb E} f(\ell(0)) f(\ell(1)) f(\ell(2))</math> is large,


# Pick a random base point <math>x \in [3]^n</math>.  Note from the law of large numbers that the number of 0s, 1s, and 2s in x will be comparable to n/3 with probability 1-o(1) (here we use o(1) to denote anything that goes to zero in the limit <math>n \to \infty</math>, independently of the choice of m).
where <math>\ell</math> ranges over Lines(n).
# Pick m random indices <math>i_1,\ldots,i_m</math> in the 0-set of x.  These will be distinct with probability 1-o(1), if m is a sufficiently slowly growing function of n.
# Define <math>\phi^{(n)}(w_1 \ldots w_m)</math> to be the word in <math>[3]^n</math> formed by starting with x, and replacing the <math>i_j</math> coefficient of x (which is 0) with <math>w_i</math>, for each <math>i=1,\ldots,m</math>.


For instance, if n=5, m=2, x=01002, and <math>i_1 = 1; i_2 = 4</math>, then <math>\phi(ab) = a10b2</math> for all <math>a,b=0,1,2</math>.
== 12-low influence ==


The above map is well-defined with probability 1-o(1), and maps combinatorial lines to combinatorial lines.  Thus the random set <math>B^{(n)}_m := (\phi^{(n)})^{-1}(A^{(n)}) \subset [3]^m</math> will be line-free.
We say that a function <math>f: [3]^n \to [-1,1]</math> is ''12-low influence'' if


Given any <math>0 \leq j \leq m</math> and <math>w \in [3]^j</math>, let <math>B_{m-j,w} \subset [3]^{m-j}</math> denote the random set
:<math>{\Bbb E} |f(x)-f(y)|^2</math> is small,
: <math>B^{(n)}_{m-j,w} := \{ v \in [3]^{m-j}: vw \in B^{(n)}_m \}</math>
where vw is of course the concatenation of v and w.  A convenient observation is that the <math>B^{(n)}_{m-j,w}</math> are asymptotically ''stationary'' in the sense that the distribution are asymptotically independent of w:


'''Stationarity lemma'''.  Let <math>0 \leq j\leq m</math>, let <math>B \subset [3]^{m-j}</math>, and let <math>w, w' \in [3]^j</math>Then
where x is drawn uniformly from <math>[3]^n</math> and y is formed from x by randomly flipping a 1 to a 2 or vice versaEquivalently, f is 12-low influence if


: <math>{\Bbb P}( B^{(n)}_{m-j,w} = B ) = {\Bbb P}( B^{(n)}_{m-j,w'} = B ) + o(1).</math>
<math>{\Bbb E} |f(\ell(1)) - f(\ell(2))|^2</math> is small,


'''Proof'''.  Observe that <math>B_{m-j,w} = (\phi^{(n)}_{m-j,w})^{-1}(A_n)</math>, where <math>\phi^{(n)}_{m-j,w}: [3]^{m-j} \to [3]^n</math> is defined like <math>\phi^{(n)}</math>, but replacing m with m-j, and also replacing x with x', defined by replacing x at j random locations <math>i_{m-j+1},\ldots,i_m</math> in the 0-set by the entries in w.  But a direct computation using Stirling's formula (if m is sufficiently slowly growing) shows that the probability distribution of x' is within o(1) in total variation norm from the distribution of x (intuitively the j changes to x are a "drop in the ocean" compared with the noise level <math>\sqrt{n}</math> of x).  In particular, replacing w with w' only affects the distribution of x' by o(1) in total variation, and the claim follows.
where <math>\ell</math> is drawn randomly from Lines(n).
<math>\Box</math>


== Step 2: Taking limits ==
The following claim is, technically, not true, but is a useful first approximation to the truth:


For fixed n, and <math>m = m^{(n)}</math>, we can construct random sets <math>B^{(n)}_i \subset [3]^i</math> for <math>i=0,\ldots,m</math> by the formula <math>B^{(n)}_i := B^{(n)}_{i,0^{m-i}}</math>; these are all line-free. We then have the asymptotic stationarity property
:'''Lemma 1''' (A lie) Let <math>f: [3]^n \to [-1,1]</math>.  Then the function <math>F(x) := {\Bbb E}_{\ell: \ell(1)=x} f(\ell(2))</math> has 12-low influence, as does the function <math>G(x) := {\Bbb E}_{\ell: \ell(2) = x} f(\ell(1))</math>.


: <math>{\Bbb P}( B^{(n)}_{i,w} = B ) = {\Bbb P}( B^{(n)}_i = B ) + o(1)</math>
Let us say that a function <math>f: [3]^n \to [-1,1]</math> is ''12-uniform'' if <math>{\Bbb E} g(\ell(1)) f(\ell(2))</math> is small for all <math>g: [3]^n \to [-1,1]</math>, where <math>\ell</math> is again ranging over Lines(n).  Assuming Lemma 1, we see that 12-uniform functions are basically orthogonal to 12-low influence functions.  In fact any function can essentially be orthogonally decomposed into 12-uniform and 12-low influence components (see the [[Fourier-analytic proof of Sperner]] for some more discussion).


for all <math>w \in [3]^j</math> with <math>i+j \leq m</math>, where
We can also define 12-uniformity and 12-low influence on functions on Lines(n) rather than <math>[3]^n</math>, where now <math>\ell</math> is ranging over lines in Lines(n) (i.e. in elements of Planes(n)).  Similarly for functions on Planes(n), Spaces(n), etc.


: <math>B^{(n)}_{i,w} := \{ v \in [3]^i: vw \in B_{i+j}^{(n)} \}.</math> (1)
== 01-uniformity relative to 12-low influence ==


Also, observe that <math>{\Bbb P}(B^{(n)}_0 \neq \emptyset)</math> is the density of <math>A_n</math> and is thus at least <math>\delta</math>.
We can define 01-uniformity in exact analogy to 12-uniformity; a function f is 01-uniform if <math>{\Bbb E} g(\ell(0)) f(\ell(1))</math> is small for all <math>g: [3]^n \to [-1,1]</math>.  We need a more sophisticated notion (analogous to <math>U^2({\Bbb Z}/N{\Bbb Z})</math> Gowers uniformity): a function f is 01-uniform relative to 12-low influence if <math>{\Bbb E} g(\ell(0)) f(\ell(1)) h(\ell)</math> is small for all <math>g: [3]^n \to [-1,1]</math> and all 12-low influence <math>h: Lines(n) \to [-1,1]</math>.


Using the Helly selection principle we can take a subsequence limit as <math>n \to \infty</math> in the vague topology. This gives us a new sequence of random sets <math>B_i \subset [3]^i</math> for all <math>i \geq 0</math> obeying the exact stationarity property
'''Example''' Random functions are 01-uniform relative to 12-low influence.


: <math>{\Bbb P}( B_{i,w} = B ) = {\Bbb P}( B_i = B )</math>
:'''Lemma 2''' (von Neumann theorem)  If <math>f: [3]^n \to [-1,1]</math> is 01-uniform relative to 12-low influence and <math>g, h: [3]^n \to [-1,1]</math>, then <math>{\Bbb E} h(\ell(0)) f(\ell(1)) g(\ell(2))</math> is small.


for all <math>w \in [3]^j</math>, where <math>B_{i,w}</math> is defined similarly to (1), and such that <math>B_0 \neq \emptyset</math> with positive probability.  Also, all the <math>B_i</math> are almost surely line-free; by removing an event of probability zero we may assume that they are ''surely'' line-free.
'''Proof''': An "IP van der Corput lemma" allows one to ensure that <math>{\Bbb E} h(\ell(0)) f(\ell(1)) g(\ell(2))</math> is small if we can show that


== Step 3: Creating transformations ==
:<math>{\Bbb E} f(\pi(01)) g(\pi(02)) f(\pi(11)) g(\pi(22))</math> (1)


For each i, Let <math> X := \prod_{n=0}^\infty 2^{[3]^n}</math> denote the space of all sequences of sets <math>B_n \subset [3]^n</math>, with the product sigma-algebraThe probability distribution of the random sequence <math>(B_n)_{n=0}^\infty</math> defined in the previous section is thus a probability measure <math>\mu</math> on X.  The set <math>E := \{ B_0 \neq \emptyset \}</math> has positive measure with respect to <math>\mu</math>. 
is small, where <math>\pi</math> ranges over Planes(n)(Rough sketch of proof: we can rearrange <math>{\Bbb E} h(\ell(0)) f(\ell(1)) g(\ell(2))</math> as


For each i=0,1,2, let <math>U_i: X \to X</math> be the transformation that sends a sequence <math>(B_n)_{n=0}^\infty</math> to <math>(B_{n,i})_{n=0}^\infty</math>, where the <math>B_{n,i}</math> are defined as in (1).  The stationarity property is nothing more than the statement that one has the measure-preserving property <math>\mu(U_i^{-1}(F)) = \mu(F)</math> for any set F depending on only one of the <math>B_n</math>.  Observe that if we define <math>U_w</math> for any word w by the formula
:<math>{\Bbb E}_{\phi,i} h(\phi(0^r)) f(\phi(1^i 0^{r-i})) g(2^i 0^{r-i})</math>


:<math>U_{w_1 \ldots w_n} = U_1 \ldots U_{w_n}</math>
where r is a medium size number, <math>\phi: [3]^r \to [3]^n</math> ranges over r-dimensional subspaces with few wildcards, and i ranges from 1 to r.  Using Cauchy-Schwarz in i, we reduce to bounding


:<math>{\Bbb E}_{\phi,i,j} f(\phi(1^i 0^{r-i})) g(2^i 0^{r-i}) f(\phi(1^j 0^{r-j})) g(2^j 0^{r-j})</math>


...
which can be rearranged to give (1)).
 
We can rearrange (1) further as
 
:<math>{\Bbb E} F(\pi(1)) G(\pi(2))</math>
 
where <math>\pi(1), \pi(2)</math> are thought of as lines and <math>F(\ell) := f(\ell(0)) f(\ell(1))</math> and <math>G(\ell) := g(\ell(0)) g(\ell(2))</math>. But by Lemma 1, we can rewrite this as
 
:<math>{\Bbb E} F(\ell) \overline{G}(\ell)</math>
 
where <math>\overline{G}</math> is 12-low influence.  But as f is 01-uniform relative to 12-low influence, this expression is small.  <math>\Box</math>
 
Thus, when dealing with expressions of the form
 
:<math>{\Bbb E} h(\ell(0)) f(\ell(1)) g(\ell(2))</math>,
 
functions which are 01-uniform relative to 12-low influence can be discarded from the second factor.  For similar reasons, functions which are 20-uniform relative to 12-low influence (defined in the obvious manner) can be discarded from the third factor).
 
== Obstructions to uniformity ==
 
To continue, we need to understand what the obstructions are to 01-uniformity relative to 12-low influence.  The answer turns out to be 01-almost periodicity relative to 12-low influence.  Roughly speaking, a function f is 01-periodic relative to 12-low influence if any shift from 0s to 1s distorts f in a manner which can be described by 12-low influence functions.  More precisely, there exists a bounded number of functions <math>g_h: [3]^n \to [-1,1]</math> and 12-low influence functions <math>c_h: Lines(n) \to [-1,1]</math> for <math>h = 1,\ldots,H</math> such that the equation
 
:<math>f(\ell(1)) = {\Bbb E}_h c_h(\ell) g_h(\ell(0))</math>
 
is approximately true for most lines <math>\ell</math> in Lines(n).
 
'''Example 1''' If f is 01-low influence, then <math>f(\ell(1)) \approx f(\ell(0))</math>, so f is certainly 01-almost periodic relative to 12-low influence.
 
'''Example 2''' Let <math>f(x) \in \{-1,+1\}</math> be the parity of the number of 1s in x, then <math>f(\ell(1)) = c(\ell) f(\ell(0))</math> where <math>c(\ell)</math> depends only on the number of wildcards in <math>\ell</math>, and is in particular 12-low influence.  Thus this function is also 01-almost periodic relative to 12-low influence.
 
'''Example 3''' If <math>f: [3]^n \to [-1,1]</math> is 12-low influence, then trivially <math>f(\ell(1)) = c(\ell)</math> where <math>c(\ell) := f(\ell(1))</math>.  Since c is 12-low influence, we conclude that f is 01-almost periodic relative to 12-low influence.
 
'''Example 4''' Any bounded linear combination of 01-almost periodic functions relative to 12-low influence is also 01-almost periodic relative to 12-low influence; also, any product of 01-almost periodic functions relative to 12-low influence is also 01-almost periodic relative to 12-low influence.
 
These almost periodic functions obstruct uniformity:
 
'''Lemma 2''': If <math>f: [3]^n \to [-1,1]</math> is 01-uniform relative to 12-low influence, and <math>g: [3]^n \to [-1,1]</math> is 01-almost periodic relative to 12-low influence, then f and g have small inner product.
 
'''Proof''': The inner product of f and g can be expressed as
 
:<math> {\Bbb E} f(\ell(1)) g(\ell(1))</math>
 
where <math>\ell</math> ranges over Lines(n).  On the other hand, from the almost periodicity of g we can write
 
:<math>g(\ell(1)) = {\Bbb E}_h c_h(\ell) G_h(\ell(0))</math>
 
for some 12-low influence <math>c_h</math> and some bounded <math>G_h</math>.  So we can reexpress the inner product as
 
:<math> {\Bbb E}_h {\Bbb E} f(\ell(1)) G_h(\ell(0)) c_h(\ell).</math>
 
But as f is 01-uniform relative to 12-low influence, the inner expectation is small, and the claim follows.  <math>\Box</math>
 
Conversely, these are the only obstructions to uniformity:
 
'''Lemma 3''': If <math>f: [3]^n \to [-1,1]</math> is not 01-uniform relative to 12-low influence, then there exists a <math>g: [3]^n \to [-1,1]</math> is 01-almost periodic relative to 12-low influence, such that f and g have large inner product.
 
'''Proof''': By hypothesis, we can find a 12-low influence c and a bounded G such that
 
:<math>{\Bbb E} f(\ell(1)) G(\ell(0)) c(\ell)</math>
 
is large.  We then take
 
:<math> g(x) := {\Bbb E}_{\ell: \ell(1) = x} G(\ell(0)) c(\ell)</math>.
 
Clearly f correlates with g.
 
If the expression inside the expectation was constant, then we would have
:<math> g(\ell(1)) = G(\ell(0)) c(\ell)</math>
and g would be 01-almost periodic relative to 12-low influence.  Of course, the expression here is not necessarily constant, but if it oscillates too much, then g would be very small and the claim would be easy.  I think we can use some sort of statistical sampling argument to handle the general case (I did this in [http://front.math.ucdavis.edu/math.CO/0405251 my finitisation of Furstenberg's proof of Szemeredi's theorem]); I'll come back to this point later.  <math>\Box</math>
 
In view of this lemma, I believe we have
 
'''Corollary 4''': Any function <math>f: [3]^n \to [0,1]</math> can be decomposed into a function <math>f_{01}: [3]^n \to [0,1]</math> with the same mean as f which is 01-almost periodic relative to 12-low influence, plus an error <math>f-f_{01}</math> which is 01-uniform relative to 12-low influence.
 
We have a similar decomposition <math>f = f_{20} + (f-f_{20})</math> into a function <math>f_{20}: [3]^n \to [0,1]</math> with the same mean as f which is 20-almost periodic relative to 12-low influence, plus an error <math>f-f_{20}</math> which is 01-uniform relative to 12-low influence.  Using the von Neumann theorem, we can thus replace
 
:<math>{\Bbb E} f(\ell(0)) f(\ell(1)) f(\ell(2))</math>
 
by
 
:<math>{\Bbb E} f(\ell(0)) f_{01}(\ell(1)) f_{20}(\ell(2))</math>.  (2)
 
== Conclusion of proof ==
 
Let <math>f_{12}</math> be the 12-low influence component of f, then <math>f_{12}</math> is non-negative, has large density, and correlates with f.  If we then let E be the set where <math>f_{12}</math> is large, then E is also 12-low influence, and f is large on E.
 
By [[DHJ(2.5)]], the 12-low influence set E contains large combinatorial subspaces.  Passing to such a subspace, we may assume that E is everything, i.e. <math>f_{12}</math> is large everywhere.  To put it another way, this implies that f has large density inside any 12-low influence set.
 
Having done this, we now perform the reductions in the previous section to get to the expression (2).  From the almost periodicity of <math>f_{01}, f_{20}</math>, we have
 
:<math>f_{01}(\ell(1)) = {\Bbb E}_h c_h(\ell) g_{01,h}(\ell(0))</math> (3)
 
where <math>c_h</math> is 12-low influence, and similarly
 
:<math>f_{20}(\ell(2)) = {\Bbb E}_h c'_h(\ell) g_{20,h}(\ell(0))</math> (4)
where <math>c'_h</math> is 12-low influence.
 
Direct substitution of (3) and (4) into (2) does not quite work; we have to do something a bit sneakier.  Let r be a medium size number and consider an r-dimensional subspace <math>\phi: [3]^r \to [3]^n</math>.  By the [[Graham-Rothschild theorem]], if r is large enough, one can find a three-dimensional subspace <math>\sigma: [3]^3 \to [3]^n</math> of <math>\phi</math> on which <math>c_h, c'_h</math> are essentially constant.  Call such a subspace ''monochromatic''; thus the collection of monochromatic 3-dimensional subspaces has positive density in Spaces(n).  Furthermore, since this collection is defined using the 12-low influence functions <math>c_h, c'_h</math>, the collection of 3-dimensional monochromatic spaces is itself 12-low influence.
 
We now rewrite (2) as
 
:<math>{\Bbb E} f(\sigma(012)) f_{01}(\sigma(112)) f_{20}(\sigma(212))</math>
 
where <math>\sigma</math> ranges over Spaces(n).
 
Suppose that <math>\sigma</math> is monochromatic.  Then from (3) we have
:<math>f_{01}(\sigma(112)) = {\Bbb E}_h c_h(\sigma(**2)) g_{01,h}(\sigma(002))</math>
and
:<math>f_{01}(\sigma(012)) = {\Bbb E}_h c_h(\sigma(0*2)) g_{01,h}(\sigma(002))</math>
and hence by monochromaticity
:<math>f_{01}(\sigma(112)) = f_{01}(\sigma(012))</math>
and similarly
:<math>f_{20}(\sigma(212)) = f_{20}(\sigma(012))</math>.
Thus we can bound (2) from below by
:<math>{\Bbb E} 1_{mono}(\sigma) [f f_{01} f_{20}](\sigma(012))</math>
where <math>1_{mono}</math> is the indicator function on the class of monochromatic spaces.
 
Now, we claim that <math>f_{01}</math> is large on almost all of the support of f.  Indeed, if there was a large set in the support of f on which <math>f_{01}</math> was small, then denoting f' by the restriction of f to that set, <math>{\Bbb E} f'</math> would be large but <math> {\Bbb E} f' f_{01}</math> would be small.  But if <math>f'_{01}</math> denotes the component of f' which is 01-almost periodic relative to 12-low influence, then <math>f'_{01}</math> is dominated by <math>f_{01}</math> (since f' is dominated by f), and so <math> {\Bbb E} f' f'_{01}</math> would be small.  But by orthogonality of f' and <math>f'-f'_{01}</math>, this is equal to <math> {\Bbb E} (f'_{01})^2 </math>.  Since <math>f'_{01}</math> has the same mean as f', which is large, this is a contradiction.  Thus <math>f_{01}</math> is large on almost all the support of f, and similarly for <math>f_{20}</math>.  Thus we can bound (2) from below by some multiple of
 
:<math>{\Bbb E} 1_{mono}(\sigma) 1_{supp(f)}(\sigma(012))</math>.
 
But because f has positive density inside any 12-low influence set, this expression is large, and we are done.

Latest revision as of 17:43, 24 February 2009

This proof of DHJ(3) is being pursued in our reading seminar. Below is an ongoing "combinatorial" translation of the argument. [Very informal!]

We write [math]\displaystyle{ [3] = \{0,1,2\} }[/math] for our alphabet. We let [math]\displaystyle{ Lines(n) }[/math] be the space of all combinatorial lines [math]\displaystyle{ \ell: [3] \to [3]^n }[/math] with "few" wildcards, where we will be vague about what "few" means. We also have [math]\displaystyle{ Planes(n) }[/math], the space of combinatorial planes [math]\displaystyle{ \pi: [3]^2 \to [3]^n }[/math], and [math]\displaystyle{ Spaces(n) }[/math], the space of combinatorial 3-spaces [math]\displaystyle{ \sigma: [3]^3 \to [3]^n }[/math]. Note that one can think of a combinatorial plane in [math]\displaystyle{ [3]^n }[/math] as a combinatorial line in Lines(n), etc.

DHJ(3) is then the claim that if [math]\displaystyle{ f: [3]^n \to [0,1] }[/math] has large mean, then

[math]\displaystyle{ {\Bbb E} f(\ell(0)) f(\ell(1)) f(\ell(2)) }[/math] is large,

where [math]\displaystyle{ \ell }[/math] ranges over Lines(n).

12-low influence

We say that a function [math]\displaystyle{ f: [3]^n \to [-1,1] }[/math] is 12-low influence if

[math]\displaystyle{ {\Bbb E} |f(x)-f(y)|^2 }[/math] is small,

where x is drawn uniformly from [math]\displaystyle{ [3]^n }[/math] and y is formed from x by randomly flipping a 1 to a 2 or vice versa. Equivalently, f is 12-low influence if

[math]\displaystyle{ {\Bbb E} |f(\ell(1)) - f(\ell(2))|^2 }[/math] is small,

where [math]\displaystyle{ \ell }[/math] is drawn randomly from Lines(n).

The following claim is, technically, not true, but is a useful first approximation to the truth:

Lemma 1 (A lie) Let [math]\displaystyle{ f: [3]^n \to [-1,1] }[/math]. Then the function [math]\displaystyle{ F(x) := {\Bbb E}_{\ell: \ell(1)=x} f(\ell(2)) }[/math] has 12-low influence, as does the function [math]\displaystyle{ G(x) := {\Bbb E}_{\ell: \ell(2) = x} f(\ell(1)) }[/math].

Let us say that a function [math]\displaystyle{ f: [3]^n \to [-1,1] }[/math] is 12-uniform if [math]\displaystyle{ {\Bbb E} g(\ell(1)) f(\ell(2)) }[/math] is small for all [math]\displaystyle{ g: [3]^n \to [-1,1] }[/math], where [math]\displaystyle{ \ell }[/math] is again ranging over Lines(n). Assuming Lemma 1, we see that 12-uniform functions are basically orthogonal to 12-low influence functions. In fact any function can essentially be orthogonally decomposed into 12-uniform and 12-low influence components (see the Fourier-analytic proof of Sperner for some more discussion).

We can also define 12-uniformity and 12-low influence on functions on Lines(n) rather than [math]\displaystyle{ [3]^n }[/math], where now [math]\displaystyle{ \ell }[/math] is ranging over lines in Lines(n) (i.e. in elements of Planes(n)). Similarly for functions on Planes(n), Spaces(n), etc.

01-uniformity relative to 12-low influence

We can define 01-uniformity in exact analogy to 12-uniformity; a function f is 01-uniform if [math]\displaystyle{ {\Bbb E} g(\ell(0)) f(\ell(1)) }[/math] is small for all [math]\displaystyle{ g: [3]^n \to [-1,1] }[/math]. We need a more sophisticated notion (analogous to [math]\displaystyle{ U^2({\Bbb Z}/N{\Bbb Z}) }[/math] Gowers uniformity): a function f is 01-uniform relative to 12-low influence if [math]\displaystyle{ {\Bbb E} g(\ell(0)) f(\ell(1)) h(\ell) }[/math] is small for all [math]\displaystyle{ g: [3]^n \to [-1,1] }[/math] and all 12-low influence [math]\displaystyle{ h: Lines(n) \to [-1,1] }[/math].

Example Random functions are 01-uniform relative to 12-low influence.

Lemma 2 (von Neumann theorem) If [math]\displaystyle{ f: [3]^n \to [-1,1] }[/math] is 01-uniform relative to 12-low influence and [math]\displaystyle{ g, h: [3]^n \to [-1,1] }[/math], then [math]\displaystyle{ {\Bbb E} h(\ell(0)) f(\ell(1)) g(\ell(2)) }[/math] is small.

Proof: An "IP van der Corput lemma" allows one to ensure that [math]\displaystyle{ {\Bbb E} h(\ell(0)) f(\ell(1)) g(\ell(2)) }[/math] is small if we can show that

[math]\displaystyle{ {\Bbb E} f(\pi(01)) g(\pi(02)) f(\pi(11)) g(\pi(22)) }[/math] (1)

is small, where [math]\displaystyle{ \pi }[/math] ranges over Planes(n). (Rough sketch of proof: we can rearrange [math]\displaystyle{ {\Bbb E} h(\ell(0)) f(\ell(1)) g(\ell(2)) }[/math] as

[math]\displaystyle{ {\Bbb E}_{\phi,i} h(\phi(0^r)) f(\phi(1^i 0^{r-i})) g(2^i 0^{r-i}) }[/math]

where r is a medium size number, [math]\displaystyle{ \phi: [3]^r \to [3]^n }[/math] ranges over r-dimensional subspaces with few wildcards, and i ranges from 1 to r. Using Cauchy-Schwarz in i, we reduce to bounding

[math]\displaystyle{ {\Bbb E}_{\phi,i,j} f(\phi(1^i 0^{r-i})) g(2^i 0^{r-i}) f(\phi(1^j 0^{r-j})) g(2^j 0^{r-j}) }[/math]

which can be rearranged to give (1)).

We can rearrange (1) further as

[math]\displaystyle{ {\Bbb E} F(\pi(1)) G(\pi(2)) }[/math]

where [math]\displaystyle{ \pi(1), \pi(2) }[/math] are thought of as lines and [math]\displaystyle{ F(\ell) := f(\ell(0)) f(\ell(1)) }[/math] and [math]\displaystyle{ G(\ell) := g(\ell(0)) g(\ell(2)) }[/math]. But by Lemma 1, we can rewrite this as

[math]\displaystyle{ {\Bbb E} F(\ell) \overline{G}(\ell) }[/math]

where [math]\displaystyle{ \overline{G} }[/math] is 12-low influence. But as f is 01-uniform relative to 12-low influence, this expression is small. [math]\displaystyle{ \Box }[/math]

Thus, when dealing with expressions of the form

[math]\displaystyle{ {\Bbb E} h(\ell(0)) f(\ell(1)) g(\ell(2)) }[/math],

functions which are 01-uniform relative to 12-low influence can be discarded from the second factor. For similar reasons, functions which are 20-uniform relative to 12-low influence (defined in the obvious manner) can be discarded from the third factor).

Obstructions to uniformity

To continue, we need to understand what the obstructions are to 01-uniformity relative to 12-low influence. The answer turns out to be 01-almost periodicity relative to 12-low influence. Roughly speaking, a function f is 01-periodic relative to 12-low influence if any shift from 0s to 1s distorts f in a manner which can be described by 12-low influence functions. More precisely, there exists a bounded number of functions [math]\displaystyle{ g_h: [3]^n \to [-1,1] }[/math] and 12-low influence functions [math]\displaystyle{ c_h: Lines(n) \to [-1,1] }[/math] for [math]\displaystyle{ h = 1,\ldots,H }[/math] such that the equation

[math]\displaystyle{ f(\ell(1)) = {\Bbb E}_h c_h(\ell) g_h(\ell(0)) }[/math]

is approximately true for most lines [math]\displaystyle{ \ell }[/math] in Lines(n).

Example 1 If f is 01-low influence, then [math]\displaystyle{ f(\ell(1)) \approx f(\ell(0)) }[/math], so f is certainly 01-almost periodic relative to 12-low influence.

Example 2 Let [math]\displaystyle{ f(x) \in \{-1,+1\} }[/math] be the parity of the number of 1s in x, then [math]\displaystyle{ f(\ell(1)) = c(\ell) f(\ell(0)) }[/math] where [math]\displaystyle{ c(\ell) }[/math] depends only on the number of wildcards in [math]\displaystyle{ \ell }[/math], and is in particular 12-low influence. Thus this function is also 01-almost periodic relative to 12-low influence.

Example 3 If [math]\displaystyle{ f: [3]^n \to [-1,1] }[/math] is 12-low influence, then trivially [math]\displaystyle{ f(\ell(1)) = c(\ell) }[/math] where [math]\displaystyle{ c(\ell) := f(\ell(1)) }[/math]. Since c is 12-low influence, we conclude that f is 01-almost periodic relative to 12-low influence.

Example 4 Any bounded linear combination of 01-almost periodic functions relative to 12-low influence is also 01-almost periodic relative to 12-low influence; also, any product of 01-almost periodic functions relative to 12-low influence is also 01-almost periodic relative to 12-low influence.

These almost periodic functions obstruct uniformity:

Lemma 2: If [math]\displaystyle{ f: [3]^n \to [-1,1] }[/math] is 01-uniform relative to 12-low influence, and [math]\displaystyle{ g: [3]^n \to [-1,1] }[/math] is 01-almost periodic relative to 12-low influence, then f and g have small inner product.

Proof: The inner product of f and g can be expressed as

[math]\displaystyle{ {\Bbb E} f(\ell(1)) g(\ell(1)) }[/math]

where [math]\displaystyle{ \ell }[/math] ranges over Lines(n). On the other hand, from the almost periodicity of g we can write

[math]\displaystyle{ g(\ell(1)) = {\Bbb E}_h c_h(\ell) G_h(\ell(0)) }[/math]

for some 12-low influence [math]\displaystyle{ c_h }[/math] and some bounded [math]\displaystyle{ G_h }[/math]. So we can reexpress the inner product as

[math]\displaystyle{ {\Bbb E}_h {\Bbb E} f(\ell(1)) G_h(\ell(0)) c_h(\ell). }[/math]

But as f is 01-uniform relative to 12-low influence, the inner expectation is small, and the claim follows. [math]\displaystyle{ \Box }[/math]

Conversely, these are the only obstructions to uniformity:

Lemma 3: If [math]\displaystyle{ f: [3]^n \to [-1,1] }[/math] is not 01-uniform relative to 12-low influence, then there exists a [math]\displaystyle{ g: [3]^n \to [-1,1] }[/math] is 01-almost periodic relative to 12-low influence, such that f and g have large inner product.

Proof: By hypothesis, we can find a 12-low influence c and a bounded G such that

[math]\displaystyle{ {\Bbb E} f(\ell(1)) G(\ell(0)) c(\ell) }[/math]

is large. We then take

[math]\displaystyle{ g(x) := {\Bbb E}_{\ell: \ell(1) = x} G(\ell(0)) c(\ell) }[/math].

Clearly f correlates with g.

If the expression inside the expectation was constant, then we would have

[math]\displaystyle{ g(\ell(1)) = G(\ell(0)) c(\ell) }[/math]

and g would be 01-almost periodic relative to 12-low influence. Of course, the expression here is not necessarily constant, but if it oscillates too much, then g would be very small and the claim would be easy. I think we can use some sort of statistical sampling argument to handle the general case (I did this in my finitisation of Furstenberg's proof of Szemeredi's theorem); I'll come back to this point later. [math]\displaystyle{ \Box }[/math]

In view of this lemma, I believe we have

Corollary 4: Any function [math]\displaystyle{ f: [3]^n \to [0,1] }[/math] can be decomposed into a function [math]\displaystyle{ f_{01}: [3]^n \to [0,1] }[/math] with the same mean as f which is 01-almost periodic relative to 12-low influence, plus an error [math]\displaystyle{ f-f_{01} }[/math] which is 01-uniform relative to 12-low influence.

We have a similar decomposition [math]\displaystyle{ f = f_{20} + (f-f_{20}) }[/math] into a function [math]\displaystyle{ f_{20}: [3]^n \to [0,1] }[/math] with the same mean as f which is 20-almost periodic relative to 12-low influence, plus an error [math]\displaystyle{ f-f_{20} }[/math] which is 01-uniform relative to 12-low influence. Using the von Neumann theorem, we can thus replace

[math]\displaystyle{ {\Bbb E} f(\ell(0)) f(\ell(1)) f(\ell(2)) }[/math]

by

[math]\displaystyle{ {\Bbb E} f(\ell(0)) f_{01}(\ell(1)) f_{20}(\ell(2)) }[/math]. (2)

Conclusion of proof

Let [math]\displaystyle{ f_{12} }[/math] be the 12-low influence component of f, then [math]\displaystyle{ f_{12} }[/math] is non-negative, has large density, and correlates with f. If we then let E be the set where [math]\displaystyle{ f_{12} }[/math] is large, then E is also 12-low influence, and f is large on E.

By DHJ(2.5), the 12-low influence set E contains large combinatorial subspaces. Passing to such a subspace, we may assume that E is everything, i.e. [math]\displaystyle{ f_{12} }[/math] is large everywhere. To put it another way, this implies that f has large density inside any 12-low influence set.

Having done this, we now perform the reductions in the previous section to get to the expression (2). From the almost periodicity of [math]\displaystyle{ f_{01}, f_{20} }[/math], we have

[math]\displaystyle{ f_{01}(\ell(1)) = {\Bbb E}_h c_h(\ell) g_{01,h}(\ell(0)) }[/math] (3)

where [math]\displaystyle{ c_h }[/math] is 12-low influence, and similarly

[math]\displaystyle{ f_{20}(\ell(2)) = {\Bbb E}_h c'_h(\ell) g_{20,h}(\ell(0)) }[/math] (4)

where [math]\displaystyle{ c'_h }[/math] is 12-low influence.

Direct substitution of (3) and (4) into (2) does not quite work; we have to do something a bit sneakier. Let r be a medium size number and consider an r-dimensional subspace [math]\displaystyle{ \phi: [3]^r \to [3]^n }[/math]. By the Graham-Rothschild theorem, if r is large enough, one can find a three-dimensional subspace [math]\displaystyle{ \sigma: [3]^3 \to [3]^n }[/math] of [math]\displaystyle{ \phi }[/math] on which [math]\displaystyle{ c_h, c'_h }[/math] are essentially constant. Call such a subspace monochromatic; thus the collection of monochromatic 3-dimensional subspaces has positive density in Spaces(n). Furthermore, since this collection is defined using the 12-low influence functions [math]\displaystyle{ c_h, c'_h }[/math], the collection of 3-dimensional monochromatic spaces is itself 12-low influence.

We now rewrite (2) as

[math]\displaystyle{ {\Bbb E} f(\sigma(012)) f_{01}(\sigma(112)) f_{20}(\sigma(212)) }[/math]

where [math]\displaystyle{ \sigma }[/math] ranges over Spaces(n).

Suppose that [math]\displaystyle{ \sigma }[/math] is monochromatic. Then from (3) we have

[math]\displaystyle{ f_{01}(\sigma(112)) = {\Bbb E}_h c_h(\sigma(**2)) g_{01,h}(\sigma(002)) }[/math]

and

[math]\displaystyle{ f_{01}(\sigma(012)) = {\Bbb E}_h c_h(\sigma(0*2)) g_{01,h}(\sigma(002)) }[/math]

and hence by monochromaticity

[math]\displaystyle{ f_{01}(\sigma(112)) = f_{01}(\sigma(012)) }[/math]

and similarly

[math]\displaystyle{ f_{20}(\sigma(212)) = f_{20}(\sigma(012)) }[/math].

Thus we can bound (2) from below by

[math]\displaystyle{ {\Bbb E} 1_{mono}(\sigma) [f f_{01} f_{20}](\sigma(012)) }[/math]

where [math]\displaystyle{ 1_{mono} }[/math] is the indicator function on the class of monochromatic spaces.

Now, we claim that [math]\displaystyle{ f_{01} }[/math] is large on almost all of the support of f. Indeed, if there was a large set in the support of f on which [math]\displaystyle{ f_{01} }[/math] was small, then denoting f' by the restriction of f to that set, [math]\displaystyle{ {\Bbb E} f' }[/math] would be large but [math]\displaystyle{ {\Bbb E} f' f_{01} }[/math] would be small. But if [math]\displaystyle{ f'_{01} }[/math] denotes the component of f' which is 01-almost periodic relative to 12-low influence, then [math]\displaystyle{ f'_{01} }[/math] is dominated by [math]\displaystyle{ f_{01} }[/math] (since f' is dominated by f), and so [math]\displaystyle{ {\Bbb E} f' f'_{01} }[/math] would be small. But by orthogonality of f' and [math]\displaystyle{ f'-f'_{01} }[/math], this is equal to [math]\displaystyle{ {\Bbb E} (f'_{01})^2 }[/math]. Since [math]\displaystyle{ f'_{01} }[/math] has the same mean as f', which is large, this is a contradiction. Thus [math]\displaystyle{ f_{01} }[/math] is large on almost all the support of f, and similarly for [math]\displaystyle{ f_{20} }[/math]. Thus we can bound (2) from below by some multiple of

[math]\displaystyle{ {\Bbb E} 1_{mono}(\sigma) 1_{supp(f)}(\sigma(012)) }[/math].

But because f has positive density inside any 12-low influence set, this expression is large, and we are done.