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 Z p
How big is:
3
jAj
3
f (a; b; c) 2 A : a + b+ c = X g »
p
If
1cA (x)
is small for
x6
= 0
Character sums
d
dj(p ¡ 1)
Sd = f x : x 2 Fp g
dSd (a) =
1
X
x 2 Sd
jSd j X
e(ax) =
p
jSd j = p
p¡ 1
d
e(ax d )
x 2 Fp
S = f (x d1 ; x d2 ; : : : ; x dn ) : x 2 Fnp g
X
X
jSj
cS (a) =
1
e(a ¢x) =
p
x2 S
x 2 Fp
e(a1 x d1 + a2 x d2 + : : : + an x dn )
Character sums
¯
¯
¯
¯
¯X
¯
d1
d2
dn ¯
1=2
¯
e(a
x
+
a
x
+
:
:
:
+
a
x
)
¿
p
1
2
n
deg(
f
)
¯
¯
¯x 2 Fp
¯
¯
¯
¯
¯
µ
¶
¯X
f (x) ¯
¯
¯ ¿ deg( f ) ;deg( g) p1=2
e
¯
g(x) ¯
¯x 2 Fp
¯
¯
¯
¯
¯
¯X
¯
d1
d2
dr ¯
¯
e(a1 x + a2 x + : : : + ar x ) ¯ ¿
¯
¯x 2 Fp
¯
r
p1¡
±( r )
Weil (1948)
Deligne (1974)
Bourgain (2005)
Lots of Applications:
Distribution of quadratic residues
Distribution / Security of RSA
Gaps between primes
Extractor constructions
(Etc.)
Restriction estimates attempt to understand
exponential sums with arbitrary coefficients
S µ Fn
X
c(x)e(a ¢x)
x2 S
What can we hope to¯ say?
n
For most a 2 F
¯
¯X
¯
¯
¯
¯ c(x)e(a ¢x) ¯
¯
¯
is small.
x2 S
Ã
Estimate:
X
x 2 Fn
¯
¯p !
¯X
¯
¯
¯
c(x)e(a
¢x)
¯
¯
¯
¯
x2 S
Ã
1=p
in terms of:
X
x2 S
!
jc(x)j q
1=q
Let us reformulate the goal:
S µ Fn
measure on S
d¾ surface
0
1
jjf jj L q ( S;d¾)
f :S! C
X
@
:=
»2 S
1=q
jf (»)j q A
jSj
X
1
(f d¾) _ (x) :=
f (»)e(x ¢»)
jSj
»2 S
jj(f d¾) _ jj L p (Fn ) · R(q ! p)jjf jj L q (S;d¾)
R(1 ! 1 ) = 1
jj(f d¾) _ jj L 1
(Fn )
· jjf jj L 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 ¡ 1 g µ Fn
_
(f d¾) (x) :=
X
1
jFj n ¡
1
f (»)e(x 1 »1 + x 2 »2 + : : : + x n (»12 + »22 + : : : + »n2 ))
1
»2 Fn ¡
jPj = Fn ¡
1
Classify (p,q) such that
jj(f d¾) _ jj L p (Fn ) · R(q ! p)jjf jj L q (S;d¾)
holds with R(q ! p) independent of the field size.
Finite Field Restriction Conjecture for the
Paraboloid, II
• Extension Estimate:
jj(f d¾) _ jj L p (Fn ) · R(q ! p)jjf jj L q (S;d¾)
_
(f d¾) (x) :=
X
1
jFj n ¡ 1
»2 Fn ¡
f (»)e(x 1 »1 + x 2 »2 + : : : + x n (»12 + »22 + : : : + »n2 ))
1
• Restriction Estimate:
jj^
gjj L q 0(P ;d¾) · R(q ! p)jjgjj L p 0(Fn )
Finite Field Restriction: Motivation
• Understanding Exponential sums with coefficients
• Model problem for Euclidean harmonic analysis
The Fourier Transform
fb(») :=
R
Rn
2¼i x ¢»
f (x)e
dx
The Fourier Restriction Problem:
Given a surface S with measure d¾ can we define:
fb(»)
for a.e.
»2 S?
The Fourier Restriction Problem I
(3-d Paraboloid)
(3-d Sphere)
f 2 L1
fb(») is continuous
f 2 L2
2
b
arbitrary
in
f (»)
L
d¾
d¾
What about 1 < p < 2 ?
The Fourier Restriction Problem II
We want an inequality of the form:
R
b(»)jd¾· Cjjf jj L p (R n )
j
f
S
jj fbjj L 1 (S;d¾) · Cjjf jj L p ( R n )
This is equivalent to the `extension’ inequality:
jj(gd¾) _ jj L p 0 (R n ) · Cjjgjj L 1
(S;d¾)
Score Board
p0 = 1
p0 ¸ 6
p0 > 4
p0 ¸ 4
(3-d Sphere/Paraboloid):
(trivial)
Stein
1968
Tomas
1975
Stein/Sjolin 1975
p0 ¸ 3:866 Bourgain 1991
_
p0 ¸
p0 ¸
p0 ¸
p0 ¸
p0 ¸
p0 ¸
3:818
3:777
3:715
3:333
3:3
3:27 *
Wolff
1995
Tao, Vargas, Vega 1998
Tao, Vargas
2000
Tao
2002
Bourgain, Guth
Bourgain, Guth
jj(gd¾) jj L p 0(R 3 ) · Cjjgjj L 1
(S;d¾)
2010
2010
Geometric Properties
_
jj(gd¾) jj L p 0(R 3 ) · Cjjgjj L 1
g-
(g- d¾)
(S;d¾)
_
¿
2¼i ¿¢»
e
g- (»)
(e2¼i ¿¢»g- (»)d¾) _
Geometric Properties II
_
jj(gd¾) jj L p 0(R 3 ) · Cjjgjj L 1
Overlap is the Enemy!
(S;d¾)
How much overlap can tubes have?
Kakeya Maximal Conjecture
jj
Restriction Conjecture
P
¿i (x)jj L p (R 3 ) ¿
Kakeya Maximal Conjecture
Kakeya Set Conjecture
If a set E µ R3 contains a line in every direction, how small
can its dimension be?
E²
E
Restriction Conjecture
Kakeya Maximal
Kakeya set Conjecture
3-d Kakeya Set Score Board
di m(E) ¸ 2
di m(E) ¸ 2:333
di m(E) ¸ 2:5
di m(E) ¸ 2:5 + 10¡
Drury
1983
Bourgain 1991
Wolff
1995
10
Tao, Katz, Laba 1999
Back to Finite Fields
So what is the 3-d finite field restriction conjecture:
jj(f d¾) _ jj L p (F3 ) · R(q ! p)jjf jj L q (P ;d¾)
P := f (! ; ! ¢! ) : ! 2 Fn ¡ 1 g µ F2
¡ 1 is a square
jj(f d¾) _ jj L 3 (F3 ) ¿ jjf jj L 3 (P ;d¾)
p¸ 4
p ¸ 3:6 ¡ ±
Stein-Tomas
L
2013
¡ 1 is not a square
jj(f d¾) _ jj L 3 (F3 ) ¿ jjf jj L 2 (P ;d¾)
p ¸ 4 Stein-Tomas
2002
p > 3:6 Mockenhaupt, Tao
p ¸ 3:6 Bennett, Carbery, Garrigos, and
Wright / Lewko-L
2010
p ¸ 3:6 ¡ ± L
2013
L
2013
p > 3:5*
The Stein-Tomas method
(doesn’t care if -1 is a square)
Want to prove:
jj(f d¾) _ jj L 4 (F3 ) ¿ jjf jj L 2 (P ;d¾)
jj^
gjj L 2 (P ;d¾) ¿ jjgjj L 4= 3 (F3 )
jj^
gjj L 2 (P ;d¾) ¿ jFj 1=2 jjgjj L 2 (F3 )
_
jj^
gjj L 2 (P ;d¾) = jhg; g ¤ (d¾) i j
max j(d¾) _ j ¿ jFj ¡
x6
=0
1
1=2
(Parseval)
» jjgjj 1 max j(d¾) _ j 1=2
x6
=0
(via Gauss Sums)
The Stein-Tomas method, I
X
1
(d¾) _ (x 1 ; x 2 ; x 3 ) =
jFj 2
e(x 1 »1 + x 2 »2 + x 3 (»12 + »22 ))
»2 F2
(d¾) _ (0; 0; 0) = 1
= 0
If x 3 6
0
1 @X
(d¾) (x 1 ; x 2 ; x 3 ) =
jFj 2
1 0
»2 Fp
@
X
e(x 1 »1 +
e(x 1 »2 + x 3 »22 ) A
»1 2 F20
»2 2 F2
1
1 Y @X
_
2 A
(d¾) (x 1 ; x 2 ; x 3 ) =
e(»
»
=4x
)e(x
(»
+
x
=2x
)
)
i
i
3
3
1
1
3
2
jFj i = 1;2
»1 2 F2
1
2
(d¾) _ (x 1 ; x 2 ; x 3 ) =
e(x
¢x=4x
)(S(x
))
n
n
jFj 2
X
S(x n ) =
e(x»2 )
j(d¾) _ (x ; x ; x )j ¿ jFj ¡ 1
_
x 3 »12 ) A
1
1
2
3
The Stein-Tomas method, II
X
g=
g1E i
g(x) » 2¡
i
x 2 Ei
1· i · 10 log( F)
jj 1cE jj L 2 ( P ;d¾) ¿ jFj 1=2 jj1E jj L 2 (F3 )
¿ jFj (1+ ° )=2
jE j = jFj °
if ° ¸ 2
¿ jFj 3° =4 = jj1E jj L 4 = 3 (F3 )
The Stein-Tomas method, III
Consider:
jE j = jFj °
for ° · 2
_ 1=2
c
jj 1E jj L 2 (P ;d¾) = jh1E ; 1E ¤ (d¾) i j
_
(d¾) (x) = ±(x) + K (x)
jK (x)j ¿ jFj ¡
1
1=2
c
jj 1E jj L 2 (P ;d¾) ¿ jE j + j h1E ; 1E ¤ K i j 1=2
X
j1E ¤ K (x)j = j
1E (t)K (x ¡ t)j ¿ jE jjFj ¡
t
j h1E ; 1E ¤ K i j 1=2 ¿ jEjjFj ¡
1=2
1
The Stein-Tomas method, IV
1=2
c
jj 1E jj L 2 ( P ;d¾) ¿ jE j + jE jjFj ¡
° =2
°¡
c
jj 1E jj L 2 (P ;d¾) ¿ jFj
+ jFj
1=2
1=2
for ° · 2
¿ jFj 3° =4 = jj1E jj L 4= 3 (F3 )
We have proven:
jj 1cE jj L 2 (P ;d¾) ¿ jj1E jj L 4= 3 ( F3 )
jj(f d¾) _ jj L 4 (F3 ) ¿ jjf jj L 2 (P ;d¾)
How did Mockenhaupt-Tao go beyond Stein-Tomas?
(-1 not a square)
Extension estimate
jj(gd¾) _ jj L p (F3 ;dx ) · Cp jjgjj L 2 (P ;d¾)
Restriction estimate
jj f^jj L 2 (P ;d¾) · Cjjf jj L p 0 (F3 ;dx )
jj
Eµ
3
F
P
f = 1E
dE s jj L 2 (P ;d¾) ¿
1
s
P
dE s jj L 2 (P ;d¾)
jj
1
s
P
dE s jj L 2 ( P ;d¾)
jj
1
s
dE s jj L 4 ( P ;d¾)
jj 1
Mockenhaupt-Tao
Es
Ls
jj(gd¾) _ jj L p (F1 8= 5 ;dx ) ¿ jjgjj L 2 (P ;d¾)
N points N lines
F2
# incidences* ¿ N 3=2
Detour: Sum-product Estimates
A ½R
jAj · jA + Aj · jAj
arithmetic progression
2
jAj · jA ¢Aj · jAj
2
geometric progression
Erdős and Szemerédi’s sum-product conjecture:
max(jA + Aj; jA ¢Aj) ¸ jAj 2+ o(1)
1+ ± Erdős and Szemerédi’s (1983)
¸ jAj
….
¸ jAj
4=3+ o(1)
Solymosi (2008)
Sum-product estimates (finite fields)
(*A not `near’ a subfield)
A ½F
max(jA + Aj; jA ¢Aj) ¸ jAj
1+ ±
Bourgain, Katz, Tao (2002)
Szemerédi-Trotter Incidence Problem (finite fields)
N points
N lines
# incidences ¿ N 3=2 (Cauchy-Schwarz)
# incidences* ¿ N 3=2¡ ±
2
F
(Bourgain, Katz, Tao)
Beyond Mockenhaupt-Tao
P
Es
Ls
N points N lines
# incidences* ¿ N 3=2¡
dE s jj L 2 ( P ;d¾)
jj
1
s
dE s jj L 4 ( P ;d¾)
jj 1
F2
±
A) The Stein-Tomas / Mockenhaupt-Tao method
isn’t sharp.
B) Each slice E s contains the same number of
points, and is far from being contained in a subfield.
(Bourgain, Katz, Tao)
The finite field restriction conjecture holds for:
p ¸ 3:6 ¡ ±
p > 3:5
What happens if -1 is a square?
f (! ; ! ¢! ) : ! 2 F2 g
f (x ¡ i y); (x + i y); x 2 ¡ (iy) 2 ) : x; y 2 Fg
f (! 1 ; ! 2 ; ! 1 ! 2 ) : ! 1 ; !
2
2 Fg
` := ((»; 0; 0) : » 2 F)
1 X
(1` d¾) (x) :=
jFj 2
_
e(x 1 »1 )
»1 2 F
_
jj(1` d¾) jj L 3 (F3 )
jj1` jj L p (P ;d¾)
µ
=
µ
=
=
1
±(x 1 )
jFj
1 3 2
( ) jFj
jFj
1
jFj
2
jFj
¶ 1=3
= jFj ¡
1=3
= jFj ¡
1=p
¶ 1=p
-1 is a square, what goes wrong with the
Mockenhaupt-Tao argument?
Want to go beyond S-T:
Increase this exponent
jj f^jj L 2 (P ;d¾) · Cjjf jj L 4= 3 (F3 ;dx )
But you need to decrease this exponent
(and M-T needs to use the 2 for Parseval)
f = 1E
Let’s run the Mockenhaupt-Tao argument even though it can’t work
jj
Eµ
P
d
s 1E s jj L 2 (P ;d¾)
¿
P
s
dE s jj L 2 (P ;d¾)
jj 1
3
F
If the slices of E do not concentrate on lines then one can get some improvement
Unless
jE j »
2 one can get more out of the Stein-Tomas method
F
Consistent with the known problematic case:
If E concentrates on a plane:
We can then geometrically understand
jj 1cE jj L 3= 2 ( P ;d¾)
E
1cE
It is here were we have to (and do) avoid
L2
Being more careful, we can handle sets contained injFj
methods
±
planes
Last Case: Every slice of E is a line but E isn’t
contained in a small number of planes.
jj f^jj L 2 (P ;d¾) = j hf ; f ¤ (d¾) _ i j 1=2 · jjf jj 2 jjf ¤ (d¾) _ jj 2
Tf := f ¤ (d¾) _
fs
Tf i := f i ¤ (d¾) _
Planes correspond to 1-d Fourier coefficients of f s
f
jjf ¤ (d¾) _ jj 2
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!
Summary of cases
1. jE j ¿ jFj2
Stein-Tomas does better
2. Most vertical slices don’t
concentrate on lines
Mockenhaupt-Tao argument
Eµ
3
F
jj(gd¾) _ jj L 3: 6 (F3 ;dx ) · Cp jjgjj L 3 (P ;d¾)
M-T still bottleneck
Can do better with sum-product
3. 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
Finite Field Kakeya conjecture
F finite field
3
Eµ F
is a Kakeya set if it contains a line in every direction
a line is a set of the form ` := f x + tv : t 2 Fg where x; v 2 F3
we say E has diminesion ® if jEj ¸ CjFj
Finite Field Kakeya conjecture (Wolff ):
A Kakeya set has dimension 3.
®
Finite Field Kakeya
3
F
How big must E be?
jEj À jFj d
d ¸ 2:5 Wolff
~1995
(elementary combinatorics)
2002
d ¸ 2:5 + ± Bourgain, Katz, Tao
(sum-product estimates)
d¸ 3
E
Dvir
2008
(Polynomial method)
What’s the relation between finite field
restriction and Kakeya?
(3-d Euclidean Paraboloid)
f
(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
f (! 1 ; ! 2; ! 1 ¢! 2 ) : ! 1 ; ! 2 2 Fn g
In odd dimensions with -1 a square this is equivalent to the standard
paraboloid.
Consider f (! 1 ; ! 2 ; !
1 ¢! 2 ) : ! 1 ; !
2
2
F
g
2
H µ := f (µ; ! 2 ; µ ¢! 2 ) : !
µ
x3
(H µd¾) _ (x 1 ; x 2 ; x 3; x 4 ; x 5 )
2
2 F2 g
x5
b
x4
(e(¡ b1 »1 ; ¡ b2 »2 )H µd¾) _ (x 1 ; x 2 ; x 3 ; x 4; x 5 )
If we had a 3-d Kakeya set
x5
x3
X
µ
x4
(e(¡ b1 (µ)»1 ; ¡ b2 (µ)»2 )H µ d¾) _ (x 1 ; x 2 ; x 3 ; x 4 ; x 5 )
Thank You!