BK:Section 3: Difference between revisions

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Rephrasing statement and shortening proof to illustrate link with Sanders's Lemma 2.8 more clearly
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Parent page: [[Improving the bounds for Roth's theorem]]
Parent page: [[Improving the bounds for Roth's theorem]]


One of the take-away results from Section 3 of the Bateman-Katz paper is Proposition 3.1, which is in some places referred to as the "nd-estimate". The rough reason for this terminology is that it says that a set <math>A</math> in <math>\mathbb{F}_3^n</math> of density about <math>1/n</math> either has a `good' density increment on a subspace of codimension <math>d</math>, or else the <math>(1/n)</math>-large spectrum of <math>A</math> intersects any <math>d</math>-dimensional subspace in at most about <math>nd</math> points.
One of the take-away results from Section 3 of the Bateman-Katz paper is Proposition 3.1, an important part of which is in some places referred to as the "nd-estimate". The rough reason for this terminology is that it says that a set <math>A</math> in <math>\mathbb{F}_3^n</math> of density about <math>1/n</math> either has a `good' density increment on a subspace of codimension <math>d</math>, or else the <math>(1/n)</math>-large spectrum of <math>A</math> intersects any <math>d</math>-dimensional subspace in at most about <math>nd</math> points. We shall say later on why this is significant.
 
Here is the precise result, stated in slightly different terms to the paper in order to illustrate how it relates to other results.


:'''Proposition 1''' Let <math>A</math> be a subset of <math>\mathbb{F}_3^n</math> with density <math>\alpha</math>, and let <math>\delta > 0</math> and <math>0 \leq \eta \leq 1</math> be parameters. Set <math>\Delta = \{ \gamma \in \widehat{G} : | \widehat{1_A}(\gamma) | \geq \delta \alpha \} \setminus \{0\}</math>. Then
Here is the precise result, stated in slightly different terms to the paper in order to illustrate how it relates to other results. For a subspace <math>V \leq \mathbb{F}_3^n</math> we write
# either there is a subspace of <math>\mathbb{F}_3^n</math> of codimension <math>d</math> on which <math>A</math> has density at least <math>\alpha(1 + \eta)</math>
:<math>V^{\perp} = \{ \gamma \in \widehat{\mathbb{F}_3^n} : \gamma(x) = 1 \ \forall x \in V \}</math>
# or <math>|\Delta \cap W| \leq \eta \delta^{-2}</math> for each <math>d</math>-dimensional subspace <math>W \leq \widehat{\mathbb{F}_3^n}</math>.
for its annihilator (cf. [[Basic facts about Bohr sets|the section on Bohr sets]]).


'''Proof''' Choose a subspace <math>H</math> such that <math>W</math> is the annihilator of <math>H</math>, and let <math>V</math> be a subspace transverse to <math>H</math>. Then for any <math>\gamma\neq0\in W</math>,
:'''Proposition 1''' Let <math>A \subset \mathbb{F}_3^n</math> be a set with density <math>\alpha</math>, and let <math>0 \leq \delta, \eta \leq 1</math> be parameters. Set
:<math>\widehat{1_A}(\gamma)=3^{-n}\sum_{v\in V}(| A\cap(H+v)|-3^{-d}| A|)\gamma(v)</math>
:<math>\Delta = \{ \gamma \in \widehat{G} : | \widehat{1_A}(\gamma) | \geq \delta \alpha \} \setminus \{ 0_{\widehat{\mathbb{F}_3^n}} \}</math>.
and hence
Suppose <math>V \leq \mathbb{F}_3^n</math> be a subspace. Then
:<math>\sum_{\gamma\neq0\in W}|\widehat{1_A}(\gamma)|^2=3^{d-2n}\sum_{v\in V}(| A\cap(H+v)|-3^{-d}| A|)^2.</math>
* either <math>A</math> has density at least <math>\alpha(1 + \eta)</math> on <math>V</math>,
If we let <math>V^+</math> be the subset of <math>V</math> for which each of the squared summands is positive, then either <math>A</math> has the required density increment on a translate of <math>H</math> (which has codimension <math>d</math>), or
* or <math>|\Delta \cap V^{\perp}| \leq 3\eta \delta^{-2}</math>; in fact <math>\sum_{\gamma \in V^{\perp}} |\widehat{(1_A - \alpha)}(\gamma)|^2 \leq 3\eta \alpha^2</math>.
:<math>|| A\cap(H+v)|-3^{-d}| A||\ll 3^{-d}| A|\eta.</math>
'''Proof''':
Hence
Let us write <math>\mu_V = \frac{|\mathbb{F}_3^n|}{|V|}1_V</math> for the indicator function of <math>V</math> normalized so that <math>\mathbb{E}_x \mu_V(x) = 1</math>. If
:<math>\sum_{v\in V^+}|| A\cap(H+v)|-3^{-d}| A||\ll| A|\eta</math>
:<math>1_A*\mu_V(x) > \alpha(1 + \eta)</math>
and
for some <math>x \in \mathbb{F}_3^n</math> then we are in the first case, so let us assume that <math>1_A*\mu_V \leq \alpha(1+\eta)</math>. Write <math>f = 1_A - \alpha</math> for the balanced function of <math>A</math>. Then
:<math>\sum_{v\in V^+}|| A\cap(H+v)|-3^{-d}| A||^2\ll 3^{-d}| A|^2\eta^2.</math>
:<math> | \Delta \cap V^{\perp} | \delta^2 \alpha^2 \leq \sum_{\gamma \in V^{\perp}} |\widehat{f}(\gamma)|^2 = \sum_{\gamma \in \widehat{\mathbb{F}_3^n}} |\widehat{f}(\gamma)|^2 |\widehat{\mu_V}(\gamma)|^2.</math>
Furthermore, since
By Parseval's identity, this equals
:<math>\sum_{v\in V}|| A\cap(H+v)|-3^{-d}| A||=0</math>
:<math> \mathbb{E}_{x \in \mathbb{F}_3^n} f*\mu_V(x)^2 = \mathbb{E}_{x \in \mathbb{F}_3^n} 1_A*\mu_V(x)^2 - \alpha^2 \leq \alpha^2(2\eta + \eta^2),</math>
defining <math>V^-</math> similarly and combining the trivial estimate
which proves the result.
:<math>|| A\cap(H+v)|-3^{-d}| A||\leq3^{-d}| A|</math>
for <math>v\in V^-</math> with the above gives
:<math>\sum_{v\in V^-}|| A\cap(H+v)|-3^{-d}| A||^2\ll3^{-d}| A|^2\eta.</math>
Combining these sum estimates gives
:<math>\sum_{v\in V}|| A\cap(H+v)|-3^{-d}| A||^2\ll3^{-d}| A|^2\eta</math>
and hence
:<math>\sum_{\gamma\neq0\in W}|\widehat{1_A}(\gamma)|^2\ll \alpha^2\eta.</math>
Recalling the definition of <math>\Delta</math>, we have
:<math>|\Delta\cap W|\delta^2\alpha^2\ll\sum_{\gamma\in\Delta\cap W}|\widehat{1_A}(\gamma)|^2\ll\alpha^2\eta.</math>





Revision as of 11:48, 6 February 2011

Parent page: Improving the bounds for Roth's theorem

One of the take-away results from Section 3 of the Bateman-Katz paper is Proposition 3.1, an important part of which is in some places referred to as the "nd-estimate". The rough reason for this terminology is that it says that a set [math]\displaystyle{ A }[/math] in [math]\displaystyle{ \mathbb{F}_3^n }[/math] of density about [math]\displaystyle{ 1/n }[/math] either has a `good' density increment on a subspace of codimension [math]\displaystyle{ d }[/math], or else the [math]\displaystyle{ (1/n) }[/math]-large spectrum of [math]\displaystyle{ A }[/math] intersects any [math]\displaystyle{ d }[/math]-dimensional subspace in at most about [math]\displaystyle{ nd }[/math] points. We shall say later on why this is significant.

Here is the precise result, stated in slightly different terms to the paper in order to illustrate how it relates to other results. For a subspace [math]\displaystyle{ V \leq \mathbb{F}_3^n }[/math] we write

[math]\displaystyle{ V^{\perp} = \{ \gamma \in \widehat{\mathbb{F}_3^n} : \gamma(x) = 1 \ \forall x \in V \} }[/math]

for its annihilator (cf. the section on Bohr sets).

Proposition 1 Let [math]\displaystyle{ A \subset \mathbb{F}_3^n }[/math] be a set with density [math]\displaystyle{ \alpha }[/math], and let [math]\displaystyle{ 0 \leq \delta, \eta \leq 1 }[/math] be parameters. Set
[math]\displaystyle{ \Delta = \{ \gamma \in \widehat{G} : | \widehat{1_A}(\gamma) | \geq \delta \alpha \} \setminus \{ 0_{\widehat{\mathbb{F}_3^n}} \} }[/math].

Suppose [math]\displaystyle{ V \leq \mathbb{F}_3^n }[/math] be a subspace. Then

  • either [math]\displaystyle{ A }[/math] has density at least [math]\displaystyle{ \alpha(1 + \eta) }[/math] on [math]\displaystyle{ V }[/math],
  • or [math]\displaystyle{ |\Delta \cap V^{\perp}| \leq 3\eta \delta^{-2} }[/math]; in fact [math]\displaystyle{ \sum_{\gamma \in V^{\perp}} |\widehat{(1_A - \alpha)}(\gamma)|^2 \leq 3\eta \alpha^2 }[/math].

Proof: Let us write [math]\displaystyle{ \mu_V = \frac{|\mathbb{F}_3^n|}{|V|}1_V }[/math] for the indicator function of [math]\displaystyle{ V }[/math] normalized so that [math]\displaystyle{ \mathbb{E}_x \mu_V(x) = 1 }[/math]. If

[math]\displaystyle{ 1_A*\mu_V(x) \gt \alpha(1 + \eta) }[/math]

for some [math]\displaystyle{ x \in \mathbb{F}_3^n }[/math] then we are in the first case, so let us assume that [math]\displaystyle{ 1_A*\mu_V \leq \alpha(1+\eta) }[/math]. Write [math]\displaystyle{ f = 1_A - \alpha }[/math] for the balanced function of [math]\displaystyle{ A }[/math]. Then

[math]\displaystyle{ | \Delta \cap V^{\perp} | \delta^2 \alpha^2 \leq \sum_{\gamma \in V^{\perp}} |\widehat{f}(\gamma)|^2 = \sum_{\gamma \in \widehat{\mathbb{F}_3^n}} |\widehat{f}(\gamma)|^2 |\widehat{\mu_V}(\gamma)|^2. }[/math]

By Parseval's identity, this equals

[math]\displaystyle{ \mathbb{E}_{x \in \mathbb{F}_3^n} f*\mu_V(x)^2 = \mathbb{E}_{x \in \mathbb{F}_3^n} 1_A*\mu_V(x)^2 - \alpha^2 \leq \alpha^2(2\eta + \eta^2), }[/math]

which proves the result.


To be added:

  • Statement of size bound on [math]\displaystyle{ \Delta }[/math] from Parseval alone
  • Statement of Chang's theorem
  • Relation to Lemma 2.8 in Sanders's paper