Finite Field Restriction Estimates
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Finite Field Restriction Estimates
Mark Lewko
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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
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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)
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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.)
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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
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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.
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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))
jj(f d¾)_ jjL p (Fn ) · R (q! p)jjf jjL q(S;d¾)
Classify (p,q) such that
holds with independent of the field size.R(q! p)
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Finite Field Restriction Conjecture for the Paraboloid, II• Extension Estimate:
jj(f d¾)_ jjL p (Fn ) · R (q! p)jjf jjL q(S;d¾)
(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 )
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Finite Field Restriction: Motivation
• Understanding Exponential sums with coefficients
• Model problem for Euclidean harmonic analysis
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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. ?
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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
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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)
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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
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Geometric Propertiesjj(gd¾)_ jjL p0(R 3) · CjjgjjL 1 (S;d¾)
g
(g d¾)_
e2¼i¿¢»g (»)
¿
(e2¼i¿¢»g (»)d¾)_
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Geometric Properties IIjj(gd¾)_ jjL p0(R 3) · CjjgjjL 1 (S;d¾)
Overlap is the Enemy!
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Kakeya Maximal Conjecture
How much overlap can tubes have?
Restriction Conjecture Kakeya Maximal Conjecture
jjP¿i (x)jjL p (R 3) ¿
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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
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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
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Back to Finite Fields
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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¡ ±
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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)
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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
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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)
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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
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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)
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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)
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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
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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)
….
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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
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Ps jjd1E s jjL 2(P ;d¾)
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
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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
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-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)
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Let’s run the Mockenhaupt-Tao argument even though it can’t work
jjP
sd1E s jjL 2(P ;d¾) ¿
Ps jjd1E 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:
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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±
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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)_
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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!
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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
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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.
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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
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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.
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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
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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
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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
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Thank You!