Math 216A Notes, Week 8
Scribe: Jesse Benavides
Disclaimer: These notes are not nearly as polished (and quite possibly not nearly as correct) as a
published paper. Please use them at your own risk.
1. Van der Waerden’s Theorem
Last Time:
If we color [1, n] ({1, . . . , n}), with k colors, and n ≥ n0 (k) some color has solution to x + y = z.
x + y = 2z, Progressions {a, a + 2, . . . , a + (k − 1)d}
Theorem (Van der Waerden) For all k, r, there exists W (r, k) such that if n ≥ W (r, k), you color [1, n] with
r colors. Some color has a k-term progression.
Example 1 W (2, 3) = 9 because
RBRBBRBR
has no 3 term progression, and this sequence no longer has this property.
Claim: W (2, 3) ≤ 325
Proof. We write 365 = 5 · 65.
Then there are 65 blocks of size 5. Consider the first 33 blocks. Notice two are colored the same
Without loss of generality say a1 , a1 + e are the color red. Suppose I’m trying to avoid a monochromatic
progression. Since a1 , a1 + e are both red, it must be that a1 + 2e is blue. Since a1 + 5d, a1 + 5d + e are
both red, then a1 + 5d + 2e has to be blue. Now we no matter how we color a1 + 10d + 2e either have
the monochromatic red progression a1 , a1 + 5d + e, a1 + 10d + 2e or the monochromatic blue progression
a1 + 2e, a1 + 5d + 2e, a1 + 10d + 2e.
1
The following object will play a role in our proof of the general theorem.
Definition A sunflower with m petals of size k is a collection of m-arithmetic progressions of the form
a + d1
a + 2d1
...
a + (k − 1)d1
a + d2
..
.
a + 2d2
...
a + (k − 1)d2
..
.
a + dm
a + 2dm
...
a + (k − 1)dm
such that:
(i) Each progression is monochromatic.
(ii) All petals are different colors.
In the W (2, 3) example from before, the key step was to find a sunflower with 2 petals of size 2, with
the idea being that then we’d be done no matter how we colored the center of the flower. Our proof of the
general case will go along similar lines.
Proof. (of Van der Waerden) Double induction:
The base case for the outer loop is k = 2. W (r, 2) = r + 1.
Now we suppose W (r, k − 1) exists for all r. We want to show W (r, k) exists for all r.
Lemma Suppose W (r, k − 1) exists for all r. For all r, m ≥ 1 there’s an W (r, m, k − 1) such that one of the
following holds for all colorings [1, n]:
(i) There is a sunflower of m petals of length k − 1 having center in [1, n] OR
(ii) There is a monochromatic progression of length k.
Note that if we take m = r + 1 then (i) cannot occur, as we just don’t have enough colors for all the
petals. This gives us our longer progression.
Proof. (induct on m)
If m = 1, we are done by our assumption. Else, let N1 = W (r, m − 1, k − 1), N2 = 2W (rN1 , k − 1). Consider
a coloring of [1, N1 N2 ] Think of it as N2 blocks of size N1 . rN1 ways of coloring block. Applying the Van der
2
Waerden Theorem for k − 1-progressions, we get a monochromatic arithmetic progression of k − 1 blocks.
The start of each block forms an arithmetic progression, and furthermore, each block is colored the same.
Consider the colorings within a block. A block of size N1 , has either a monochromatic progression of length
k OR a sunflower with m − 1 petals of size k − 1. If the block has a monochromatic progression, then we
are done. Else consider the following:
Color 1:
a + d1
a + 2d1
...
a + (k − 1)d1
a + c + d1
..
.
a + c + 2d1
...
a + c + (k − 1)d1
..
.
a + (k − 2)c + d1
a + (k − 2)c + 2d1
...
a + (k − 2)c + (k − 1)d1
a + d2
a + 2d2
...
a + (k − 1)d2
a + c + d2
..
.
a + c + 2d2
...
a + c + (k − 1)d2
..
.
a + (k − 2)c + d2
a + (k − 2)c + 2d2
...
a + (k − 2)c + (k − 1)d2
Color 2:
..
.
Color m − 1:
a + dm−1
a + 2dm−1
...
a + (k − 1)dm−1
a + c + dm−1
..
.
a + c + 2dm−1
...
a + c + (k − 1)dm−1
..
.
a + (k − 2)c + dm−1
a + (k − 2)c + 2dm−1
...
a + (k − 2)c + (k − 1)dm−1
If a is any of the first m − 1 colors, we are done. If a has color m, then a + c, a + 2c, . . . , a + (k − 2)c are
also color m. For each i, a + (k − 1)di , a + c(k − 2)di , . . . , a + (k − 2)c + di are color i. Then, this gives a
sunflower centered at a + (k − 1)c.
The bounds shown on W (2, k) by this argument is worse than k k
.
..
k
(k’s tower of exponents). W (r, k)’s
bound from this argument is related to the Ackerman function. Both of these are far from optimal.
3
Gowers: (2001)
2k+9
r2
W (r, k) ≤ 22
Ron Graham offers a $1000 prize to anyone showing
W (2, k) < 2k
2
2. Better bounds on W (r, 3) (Ron Graham, Solymosi; 2006)
Theorem For any r, there is a N (r) such that if we color the two dimensional grid [1, N ] × [1, N ] with r
colors and N > N (r), the coloring has a monochromatic isosceles right triangle.
Proof. (somewhat of a sketch) Let S1 be the points on the line x + y = N + 1. By the Pigeonhole Principle,
there must be at least
N
r
of some color. Let T1 ⊆ S1 a monochromatic subset of this size and having color 1.
Let S2 be the set of points which form a triangle with two points in T1 (note that |S2 | =
the Pigeonhole Principle again, we get that some diagonal has ≥
|S2 |
N
|T1 |
2
). Applying
points. If any point in S2 has color 1,
we are done. Else, by the Pigeonhole Principle, there has to be a monochromatic subset T2 of size ≥
4
|S2 |
(r−1)N
on some diagonal. Let S3 be the set of points forming a triangle with points in T2 .
If any of these points are color 2, then by definition there is an isosceles triangle of that color. The key point
here is that each point in S3 also forms an isosceles right triangle two points in T1 (see the above diagram),
so we are also done if any point in S3 has color 1. Otherwise by pigeonhole we can choose a monochromatic
T3 ⊆ S3 of size at least
|S3 |
(r−2)N
on the same diagonal. Now just keep on repeating the process.
If at the end |Tr | ≥ 1, then we have a point which forms an isosceles triangle no matter which of the r
available colors it is assigned. Now it just becomes a matter of working through and seeing how large N has
to be for this to hold. It turns out that if N ≥ 22
cr
for some c, this works.
cr
Corollary W (r, 3) ≤ 22
Proof. Given a coloring of [−N, N ], we can color the grid by having (x, y) given the color of y − x. We have
a set (x, y), (x + d, y), (x, y + d) that is monochromatic in the new coloring. y − x, y − (x + d), (y + d) − x is
monochromatic in the old coloring.
3. Miscellaneous side notes:
- Van der Waerden: W (2, 3) = 9
If N > 9, color [1, N ] with 2 colors, then there has to be a 3-term monochromatic progression.
Suppose N is large. We want to color [1, N ] with two colors. By Van der Waerden, there will be
(many) monochromatic arithmetic progressions of length 3. So instead of completely avoiding monochromatic
progressions, suppose instead we want to create as few as possible.
Conjecture: (Parillo / Robertson / Saracino, and Butler / Costello / Graham) Divide N into 548 blocks of
approximately equal size. Color the first 28 blocks red, the next 6 blue, the next 28 red, the next 37 blue,
the next 59 red, the next 118 blue, then color the remaining 274 blocks to make everything antisymmetric.
This coloring is best.
5
This particular coloring is known to be locally optimal (changing the coloring by only a small amount
increases the number of monochromatic progressions), but the hard part is showing it is globally optimal.
4. Rado’s Theorem
Question: Is there a coloring of Z with a finite number of colors with no monochromatic solution x + 2y −
4z = 0?
Example 2 (attempted) If we color all odd integers red, there is no monochromatic red solution. But we get
many solutions with x, y, z all even, so just using two colors and following this pattern doesn’t quite work.
More generally, let A be the set of integers congruent to a modulo k. Then for any x, y, z ∈ A, x+y −4z ≡
−a. This isn’t zero unless a = 0. So this gives us a nice way of coloring all the integers not divisible by k.
The question becomes what to do with the rest.
Suppose x, y, z ≡ 0 mod k. Then x + 2y − 4z = 0 if and only
by k, we should look at
x
k
+
2y
k
−
4z
k
= 0. So maybe if x is divisible
x
k.
This motivates our proposed coloring with k − 1 colors: Each x is colored according to its last nonzero
digit in base k.
Suppose we’re checking color i, so that x, y, z all have last digit i and we want to show x + 2y − 4z 6= 0. If
we’re not careful, this can still happen. A bad example is if k = 3. Then
x = 4, (1, 1); y = 4, (1, 1); z = 3, (1, 0) all have color 1 and x + 2y − 4z = 0.
To show that this does not happen, it’s enough to just find one digit which of x + 2y − 4z which is nonzero.
The digit we will consider is the right most place where at least one of the x, y, z have a nonzero digit. The
digit in that place of x + 2y − 4z is determined by which of x, y, z have zeros in that place.
Only x nonzero: place has digit i
Only y nonzero: place has digit 2i
Only z nonzero: place has digit −4i
Only x, y nonzero: place has digit 3i
Only x, z nonzero: place has digit −3i
Only y, z nonzero: place has digit −2i
x, y, z nonzero: place has digit −i
If we pick k so 1, 2, −4, 3, −3, −2, −1 are all invertible modulo k then this certainly can’t happen, so
x + 2y − 4z 6= 0.
Notice that this argument falls apart for the equations x+y −2z = 0, x+y −z = 0 (as it better, given that
we’ve already proven Schur’s theorem and Van der Waerden’s theorem). In both cases one of the numbers
that has to be invertible mod k is 0. As it turns out, this is what matters in general.
6
Theorem (Rado) For a set {a1 , . . . , an } of nonzero integers, there is a coloring of Z with finitely many
colors and no monochromatic solutions to a1 x1 + . . . an xn = 0 if and only if no nonempty set of ai add up
to zero.
Proof. Take m to be a prime larger than |a1 | + |a2 | + . . . |ak |. Color x according to the last digit modulo m.
P
For any nonempty S = {a1 , . . . , ak }, −m < i∈S ai < m, and by assumption this sum is nonzero. So all the
equivalent expressions to the seven above are invertible, and we can color by the last nonzero digit modulo m.
Conversely, assume
P
i∈S, ai 6=0
ai = 0. Fix i0 ∈ S. We will look for a solution of the following form: for
some x, y, z of the same color,
xi =



 x
if
i 6= i0
y
if
i∈
/S



z
if
i ∈ S and i 6= i0
We first check what needs to happen for our choice of xi to solve the equation.
P
a i xi = 0
P
?
P
ai0 x +
i6=i0 ,i∈S z +
i∈S
/ ai y = 0
?
P
ai0 x +
i∈S
/ ai y = ai0 z
P
( i∈S
/ ai )
x+
y = z.
ai
i
0
To show we can pick an appropriate x, y, z, it thus suffices to show the following:
Lemma For any λ ∈ Q, and any k (number of colors), there’s an N such that if n > N and if I color [1, n]
with k colors there is a monochromatic solution to x + λy = z.
Proof. (Induct on k)
Let λ = rs . For k = 1, the result is trivial. Now let’s assume Nk−1 , which works for k − 1 colors,
Nk = W (k, rNk−1 + 1) (if I color with k colors, I get a progression of length rNk−1 + 1. Color [1, Nk ] with
some a, a + d1 , a + 2d1 , . . . , a + Nk−1 rd. Look at {ds, 2ds, . . . , 2dNk−1 s} If any jds is the same color as the
progression, then we can construct a solution: x = a, y = jds, z = a+jdr. If a, a+d1 , a+2d1 , . . . , a+nk−1 rd
has nothing the same as {ds, 2ds, . . . , 2dNk−1 s}, then by the induction hypothesis, we are done. x + λy = z
if and only if dsx + λdsy = dsz.
5. Beginnings of Roth’s Theorem, and a Crash Course in Fourier Analysis on ZN
Conjecture: (Erdös - Turán, 1936)
Fix δ > 0 and any k ∈ Z, k > 0, then for large N , any A, a subset satisfying A ⊂ {1, . . . , N }, then |A| ≥ δN ,
has a k-term progression.
7
Note that this is stronger than Van der Waerden’s theorem – if we color with a finite number m of colors,
one color has density at least
1
m.
Also, it is false if “progression” is replaced by “x + y = z” (e.g. the odd
integers have positive density, no solutions to x + y = z).
(Roth, 1953): Fourier Analysis proof. Conjecture is true if k = 3.
(Szemeredi, 1969-1975): Combinatorial proof. Conjecture is true for all k.
(Furstenberg, 1977): Ergodic Theory proof.
(Gouwers, 2001): Generalized Fourier Analysis proof.
We’ll see Roth’s proof in more detail next week, but for now, we’ll just give a brief idea of the proof and
then start with some of the Fourier analysis we’ll need to make the argument work.
(N )
Idea of Roth’s Proof: (k = 3 case) Note that {1, . . . , N } contains ≈ 22 three term arithmetic progressions
(for about half the choices of x and y the middle z =
|A| = δN , we would expect ≈
δ3 N 2
4
x+y
2
is an integer). If A were a random set with
2
progressions in A. If δ >> N − 3 , then A probably has many progressions
(e.g. by a second moment method argument).
The trouble is that A is not random, and we have no control over it. The key idea in Roth’s argument is
to come up with the right idea of ”pseudo-random”, so that
-If A is pseudo-random, then A has the right number of progressions (≈
δ3 N 2
4 ).
-If A is not pseudo-random, then we can use this fact to find a “long” subprogression on which
T
|A progression|
>δ+
progression
You then just keep on passing to more and more subprogressions until your density becomes so large you
have to have an arithmetic progression.
We’re going to work in Z/N Z = ZN instead of Z. (Note: Not all arithmetic progressions in ZN are also
progressions in Z. This will cause us a big headache when we do the full proof!).
Definition The Fourier transform of f : ZN → C is given by
fˆ(r) =
N
−1
X
f (k)e
−2πirk
N
k=0
Properties:
1) Orthogonality

N −1
 1
1 X −2πirk
e N =
 0
N
k=0
x ≡ 0(
if
otherwise
2) Plancherel’s Theorem:
N
−1
X
|f (k)|2 =
k=0
3) If h(x) =
P
k
N −1
1 X ˆ 2
|f (r)|
N
k=0
f (x)g(k − x), then ĥ(r) = fˆ(r)ĝ(r)
8
mod N )
Consequence of 3):
|fˆ(r)||ĝ(r)| = |ĥ(r)|
!
X X
2πirx f (k)g(k − x) e N =
x
k
X X
f (k)g(k − x)
≤
x
k
Suppose LHS is large. The Fourier Transform’s of f, g are large in the same place. (e.g. |fˆ(r)||ĝ(r)| ≥ cN ,
which implies LHS ≥ c2 N 2 . Then RHS ≥ c2 N 2 . By the Pigeonhole principle, for some k,
X
f (k)g(k − x) ≥ c2 N
k
Fact/Lemma: For any A, we can use the Fourier Transform to count solutions to x + y − 2z = 0 in A.

 1
Let 1A (x) =
 0
if
x∈A
if
x∈
/A
The number of solutions to x + y − 2z = 0 in ZN can be written as
P
x,y,z∈A
χ(x + y − 2z = 0). Using
the orthogonality observation above, we can rewrite this as
X
x,y,z∈A
N −1
1
1 X −2πik(x+y−2z)
N
e
=
N
N
k=0
X X
e
−2πikx
N
e
−2πiyk
N
e
4πikz
N
x,y,z∈A k
1 X
=
N
k
!
X
e
−2πikx
N
e
−2πiyk
N
x∈A
1 X
=
1̂A (k)1̂A (k)1̂A (−2k)
N
k
9
4πizk e N