Finite Field Restriction Estimates

Finite Field Restriction Estimates

Mark Lewko

What is Fourier analysis good for?Quantifying pseudo-randomness with respect to linear objects (equations/subspaces/subgroups/etc).

Let A µ ZpHow big is:

f (a;b;c) 2 A3 : a+b+c= X g »jAj3

p

c1A (x) x 6= 0If is small for

Character sums Sd = fxd : x 2 Fpg

=jSdjp

X

x2Fp

e(axd)d1Sd (a) =X

x2Sd

e(ax)

dj(p¡ 1) jSdj = pp¡ 1d

S = f (xd1 ;xd2 ; : : : ;xdn ) : x 2 Fnpg

=jSjp

X

x2Fp

e(a1xd1 +a2xd2 +:::+anxdn )c1S (a) =X

x2S

e(a¢x)

Character sums ¯¯¯¯¯¯

X

x2Fp

e(a1xd1 +a2xd2 +:::+anxdn )

¯¯¯¯¯¯¿ deg(f ) p

1=2

¯¯¯¯¯¯

X

x2Fp

eµf (x)g(x)

¶¯¯¯¯¯¯¿ deg(f );deg(g) p

1=2

Weil (1948)

Deligne (1974)

¯¯¯¯¯¯

X

x2Fp

e(a1xd1 +a2xd2 +:::+arxdr )

¯¯¯¯¯¯¿ r p1¡ ±(r ) Bourgain

(2005)Lots of Applications:

Distribution of quadratic residues

Gaps between primes

Distribution / Security of RSA

Extractor constructions

(Etc.)

Restriction estimates attempt to understand exponential sums with arbitrary coefficients

X

x2S

c(x)e(a¢x)

S µ Fn

What can we hope to say?¯¯¯¯¯

X

x2S

c(x)e(a¢x)

¯¯¯¯¯

is small.a 2 FnFor most

ÃX

x2Fn

¯¯¯¯¯

X

x2S

c(x)e(a¢x)

¯¯¯¯¯

p! 1=p

Estimate: in terms of:

ÃX

x2S

jc(x)jq! 1=q

Let us reformulate the goal:S µ Fn

(f d¾)_ (x) :=1jSj

X

»2S

f (»)e(x ¢»)

d¾surface measure on S

f : S ! C jjf jjL q (S ;d¾) :=

0

@X

»2S

jf (»)jq

jSj

1

A

1=q

jj(f d¾)_ jjL p (Fn ) · R (q! p)jjf jjL q(S;d¾)

R(1! 1 ) = 1 jj(f d¾)_ jjL 1 (Fn ) · jjf jjL 1(S;d¾)

Estimates get better (harder to prove) as p decreases and q increases.

Finite Field Restriction Conjecture for the Paraboloid

P := f (! ;! ¢! ) : ! 2 Fn¡ 1gµ Fn jP j = Fn¡ 1

(f d¾)_ (x) :=1

jFjn¡ 1X

»2Fn ¡ 1

f (»)e(x1»1+x2»2+:::+xn(»21 +»22 +:::+»2n))


Classify (p,q) such that

holds with independent of the field size.R(q! p)

Finite Field Restriction Conjecture for the Paraboloid, II• Extension Estimate:


(f d¾)_ (x) :=1

jFjn¡ 1X

»2Fn ¡ 1

f (»)e(x1»1+x2»2+:::+xn(»21 +»22 +:::+»2n))

• Restriction Estimate:

jjgjjL q0(P ;d¾) · R (q! p)jjgjjL p0(Fn )

Finite Field Restriction: Motivation

• Understanding Exponential sums with coefficients

• Model problem for Euclidean harmonic analysis

The Fourier Transform

bf (») :=RR n f (x)e2¼ix¢»dx

The Fourier Restriction Problem:

bf (») »2 S

Given a surface with measure can we define: S d¾

for a.e. ?

The Fourier Restriction Problem I

d¾

(3-d Paraboloid) (3-d Sphere) f 2 L1

bf (»)is continuous

f 2 L2

bf (») arbitrary inL2

d¾What about ? 1< p< 2

The Fourier Restriction Problem II

RS jbf (»)jd¾· Cjjf jjL p (R n )

We want an inequality of the form:

This is equivalent to the `extension’ inequality:

jj bf jjL 1(S;d¾) · Cjjf jjL p (R n )

jj(gd¾)_ jjL p0(R n ) · CjjgjjL 1 (S;d¾)

jj(gd¾)_ jjL p0(R 3) · CjjgjjL 1 (S;d¾)

Score Board (3-d Sphere/Paraboloid):

(trivial)

Stein 1968

p0=1p0¸ 6p0>4 Tomas

1975p0¸ 4 Stein/Sjolin 1975

p0¸ 3:866

Wolff 1995p0¸ 3:818

Bourgain 1991

Tao, Vargas, Vega 1998 Tao, Vargas 2000

Tao 2002

Bourgain, Guth 2010

p0¸ 3:777p0¸ 3:715p0¸ 3:333p0¸ 3:3p0¸ 3:27 * Bourgain, Guth

2010

Geometric Propertiesjj(gd¾)_ jjL p0(R 3) · CjjgjjL 1 (S;d¾)

g

(g d¾)_

e2¼i¿¢»g (»)

¿

(e2¼i¿¢»g (»)d¾)_

Geometric Properties IIjj(gd¾)_ jjL p0(R 3) · CjjgjjL 1 (S;d¾)

Overlap is the Enemy!

Kakeya Maximal Conjecture

How much overlap can tubes have?

Restriction Conjecture Kakeya Maximal Conjecture

jjP¿i (x)jjL p (R 3) ¿

If a set contains a line in every direction, how small can its dimension be?

E µ R3

Kakeya Set Conjecture

E E²

Kakeya Maximal

Restriction Conjecture

Kakeya set Conjecture

3-d Kakeya Set Score Board

dim(E ) ¸ 2dim(E ) ¸ 2:333dim(E ) ¸ 2:5dim(E ) ¸ 2:5+10¡ 10

Drury 1983

Bourgain 1991

Wolff 1995

Tao, Katz, Laba 1999

Back to Finite Fields

So what is the 3-d finite field restriction conjecture:

jj(f d¾)_ jjL p (F3) · R (q! p)jjf jjL q(P ;d¾)

¡ 1 is a square ¡ 1 is not a square

P := f (! ;! ¢! ) : ! 2 Fn¡ 1gµ F2

jj(f d¾)_ jjL 3(F3) ¿ jjf jjL 2(P ;d¾)jj(f d¾)_ jjL 3(F3) ¿ jjf jjL 3(P ;d¾)

Mockenhaupt, Tao 2002

p¸ 4

L 2013

p> 3:6Bennett, Carbery, Garrigos, and Wright / Lewko-L 2010

p¸ 3:6p¸ 3:6¡ ±p> 3:5 L 2013*

Stein-Tomasp¸ 4 Stein-Tomas

L 2013p¸ 3:6¡ ±

The Stein-Tomas method

jj(f d¾)_ jjL 4(F3) ¿ jjf jjL 2(P ;d¾)

(doesn’t care if -1 is a square)

Want to prove:

jjgjjL 2(P ;d¾) ¿ jjgjjL 4=3(F3)

jjgjjL 2(P ;d¾) ¿ jFj1=2jjgjjL 2(F3) (Parseval)

jjgjjL 2(P ;d¾) = jhg;g¤(d¾)_ i j1=2 » jjgjj1maxx6=0

j(d¾)_ j1=2

maxx6=0

j(d¾)_ j ¿ jFj¡ 1 (via Gauss Sums)

The Stein-Tomas method, I(d¾)_ (x1;x2;x3) =

1jFj2

X

»2F2

e(x1»1+x2»2+x3(»21 +»22))

(d¾)_ (0;0;0) = 1x3 6= 0If

(d¾)_ (x1;x2;x3) =1jFj2

0

@X

»12F2

e(x1»1+x3»21)

1

A

0

@X

»22F2

e(x1»2+x3»22)

1

A

(d¾)_ (x1;x2;x3) =1jFj2

Y

i=1;2

0

@X

»12F2

e(»i»i=4x3)e(x3(»1+x1=2x3)2)

1

A

(d¾)_ (x1;x2;x3) =1jFj2

e(x ¢x=4xn)(S(xn))2

S(xn) =X

»2Fp

e(x»2) j(d¾)_ (x1;x2;x3)j ¿ jFj¡ 1

The Stein-Tomas method, II

jjc1E jjL 2(P ;d¾) ¿ jFj1=2jj1E jjL 2(F3)

g=X

1· i · 10 log(F)

g1E i g(x) » 2¡ i x 2 E i

¿ jFj(1+° )=2

jE j = jFj°

¿ jFj3°=4if ° ¸ 2

= jj1E jjL 4=3(F3)

The Stein-Tomas method, III

jjc1E jjL 2(P ;d¾) = jh1E ;1E ¤(d¾)_ i j1=2

jE j = jFj° for ° · 2Consider:

(d¾)_ (x) = ±(x) +K (x) jK (x)j ¿ jFj¡ 1

jjc1E jjL 2(P ;d¾) ¿ jE j1=2+jh1E ;1E ¤K i j1=2

j1E ¤K (x)j = jX

t

1E (t)K (x ¡ t)j ¿ jE jjFj¡ 1

j h1E ;1E ¤K i j1=2 ¿ jE jjFj¡ 1=2

The Stein-Tomas method, IV

jjc1E jjL 2(P ;d¾) ¿ jE j1=2+jE jjFj¡ 1=2

jjc1E jjL 2(P ;d¾) ¿ jFj°=2+jFj° ¡ 1=2

¿ jFj3°=4

for ° · 2

= jj1E jjL 4=3(F3)

jjc1E jjL 2(P ;d¾) ¿ jj1E jjL 4=3(F3)We have proven:

jj(f d¾)_ jjL 4(F3) ¿ jjf jjL 2(P ;d¾)

jj(gd¾)_ jjL p (F3 ;dx) · CpjjgjjL 2(P ;d¾)

jjf jjL 2(P ;d¾) · Cjjf jjL p0(F3 ;dx)

Restriction estimate

Extension estimate

F3

f = 1E

Eµ

How did Mockenhaupt-Tao go beyond Stein-Tomas? (-1 not a square)

jjP

sd1E s jjL 2(P ;d¾) ¿

Ps jjd1E s jjL 2(P ;d¾)


jjd1E s jjL 4(P ;d¾)

F2pointsN linesN# incidences*

¿ N 3=2

Es Ls

jj(gd¾)_ jjL p (F18=5 ;dx) ¿ jjgjjL 2 (P ;d¾)

Mockenhaupt-Tao

Detour: Sum-product Estimates

A ½RjAj · jA +Aj · jAj2 jAj · jA ¢Aj · jAj2

arithmetic progression geometric progression

max(jA +Aj; jA ¢Aj) ¸ jAj2+o(1)Erdős and Szemerédi’s sum-product conjecture:

¸ jAj1+±

¸ jAj4=3+o(1)

Erdős and Szemerédi’s (1983)

Solymosi (2008)

….

Sum-product estimates (finite fields) A ½F

max(jA +Aj; jA ¢Aj) ¸ jAj1+±A(* not `near’ a

subfield)

Bourgain, Katz, Tao (2002)

Szemerédi-Trotter Incidence Problem (finite fields)

F2

pointslines

# incidences ¿ N 3=2(Cauchy-Schwarz)

# incidences* ¿ N 3=2¡ ±

(Bourgain, Katz, Tao)

NN


jjd1E s jjL 4(P ;d¾)

F2pointsN linesN# incidences*

¿ N 3=2¡ ±(Bourgain, Katz, Tao)

A) The Stein-Tomas / Mockenhaupt-Tao method isn’t sharp.

B) Each slice contains the same number of points, and is far from being contained in a subfield.

Es

p¸ 3:6¡ ±The finite field restriction conjecture holds for:

Es Ls

p> 3:5

Beyond Mockenhaupt-Tao

What happens if -1 is a square?

f (! ;! ¢! ) : ! 2 F2gf (x ¡ iy);(x+ iy);x2 ¡ (iy)2) : x;y 2 Fg

f (! 1; ! 2;! 1! 2) : ! 1;! 2 2 Fg

(1 d¾)_ (x) :=1jFj2

X

»12F

e(x1»1)

` := ((»;0;0) : »2 F)

=1jFj

±(x1)

jj(1 d¾)_ jjL 3(F3) =µ(1jFj

)3jFj2¶1=3

= jFj¡ 1=3

jj1 jjL p (P ;d¾) =µ

1jFj2

jFj¶1=p

= jFj¡ 1=p

-1 is a square, what goes wrong with the Mockenhaupt-Tao argument?

jjf jjL 2(P ;d¾) · Cjjf jjL 4=3 (F3;dx)

f = 1E

Want to go beyond S-T: Increase this exponent

But you need to decrease this exponent(and M-T needs to use the 2 for Parseval)

Let’s run the Mockenhaupt-Tao argument even though it can’t work

jjP

sd1E s jjL 2(P ;d¾) ¿


If the slices of E do not concentrate on lines then one can get some improvementF3Eµ

Unless jE j » F2one can get more out of the Stein-Tomas method

Consistent with the known problematic case:

If E concentrates on a plane:

E c1E

jjc1E jjL 3=2(P ;d¾)

We can then geometrically understand

It is here were we have to (and do) avoid methodsL2

Being more careful, we can handle sets contained in planes jFj±

Last Case: Every slice of E is a line but E isn’t contained in a small number of planes.

jjf jjL 2(P ;d¾) = jhf ;f ¤(d¾)_ i j1=2 · jjf jj2jjf ¤(d¾)_ jj2

Tf := f ¤(d¾)_

f s Tf i := f i ¤(d¾)_

jjf ¤(d¾)_ jj2Planes correspond to 1-d Fourier coefficients off sf

Only potential problem is if all the planes stack up

…but this can’t happen since we have assumed that the slices (green lines) don’t lie in small number of planes!

Stein-Tomas does better

F3Eµ

jE j ¿ jFj2Summary of cases 1.

2. Most vertical slices don’t concentrate on lines

Mockenhaupt-Tao argument3. E is contained in a small number of planes

Direct computation using geometry of paraboloid 4. Slices of E are contained

in lines, but E isn’t contained in a small number of planes

Geometric estimate for the BR operator

jj(gd¾)_ jjL 3:6(F3 ;dx) · CpjjgjjL 3(P ;d¾)

Can do better with sum-product

M-T still bottleneck

Finite Field Kakeya conjectureF finite field

E µ F3

wesay E has diminesion ®if jE j ¸ CjFj®

is a Kakeya set if it contains a line in every direction

a line is a set of the form ` := fx + tv : t 2 Fgwherex;v 2 F3

Finite Field Kakeya conjecture (Wolff):A Kakeya set has dimension 3.

Finite Field Kakeya

E

F3How big must E be?

d¸ 2:5Wolff ~1995

d¸ 2:5+±Bourgain, Katz, Tao 2002

d¸ 3 Dvir 2008

(elementary combinatorics)

(sum-product estimates)

(Polynomial method)

jE j À jFjd

What’s the relation between finite field restriction and Kakeya?

f

(3-d Euclidean Paraboloid)

(f d¾)_One can’t do this in a finite field!

Kakeya and restriction thought to be less connected over finite fields.

They are connected.Restriction for hyperbolic paraboloid in 2n-1 dimensions implies n dimensional Kakeya

In odd dimensions with -1 a square this is equivalent to the standard paraboloid.

f (! 1; ! 2; ! 1 ¢! 2) : ! 1;! 2 2 Fng

f (! 1; ! 2; ! 1 ¢! 2) : ! 1; ! 2 2 F2gConsider

Hµ := f (µ;! 2;µ¢! 2) : ! 2 2 F2g

(Hµd¾)_ (x1;x2;x3;x4;x5)x3

x4

x5µ

(e(¡ b1»1; ¡ b2»2)Hµd¾)_ (x1;x2;x3;x4;x5)

b

If we had a 3-d Kakeya set

X

µ

(e(¡ b1(µ)»1; ¡ b2(µ)»2)Hµd¾)_ (x1;x2;x3;x4;x5)

x3x4

x5

Thank You!

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Finite Field Restriction Estimates