Lecture 8
Completion: The final part of Keevash’s proof
Zur Luria
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∗
Sufficient conditions for a triangle decomposition
To recap, we constructed the following SETs:
• An algebraic SET, T = {xyz : π(x) + π(y) + π(z) = 0}, where π : V (G) → F2a is a random
injection. We set G∗ = ∪T , and showed that whp G∗ is ∗(c2 , 16)-typical.
• A random SET N ⊂ K3 (G \ G∗ ), covering most of the remaining edges. We showed that whp
L := G \ G∗ \ ∪N is c4 -bounded.
• We used a random greedy algorithm to construct the SET M C , such that ∪M C = L ∪ S for
some c5 -bounded S ⊂ G∗ .
• In the previous lecture, we constructed SETs M I , M O such that ∪M O is the disjoint union of
∪M I and S, and ∪M O ⊂ G∗ is c6 -bounded. We did this by constructing an integral triangle
set φ such that ∂2 φ = S and ∂2 φ+ is contained in G∗ and is c6 -bounded.
The next step, completion, is to find SETs M1 , M2 , M3 , M4 such that
• ∪M1 is the disjoint union of L and ∪M2 .
• M2 ⊆ M4 .
• ∪M3 = ∪M4 .
• M3 ⊆ T .
Given these SETs, the following is a triangle decomposition of G.
N ∪ M1 ∪ (M4 \ M2 ) ∪ (T \ M3 ).
How can we find M1 , M2 , M3 and M4 ?
We say that a triangle xyz is octahedral if x + y + z 6= 0 (here and later we omit the π, identifying
V (G) with a subset of F2a ) and there is a copy of K2,2,2 in G with the parts {x, y + z}, {y, x + z} and
∗
Institute of Theoretical Studies, ETH, 8092 Zurich, Switzerland. [email protected]. Research supported by
Dr. Max Rössler, the Walter Haefner Foundation and the ETH Foundation.
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{z, x+y}. We call this the associated octahedron of xyz. Note that if the associated octahedron is in G
then it is also in G∗ , because the triangles {x, y, x+y}, {y+z, y, z}, {x, x+z, z} and {y+z, x+z, x+y}
all sum to zero in F2a , and are thus triangles of T . We call this triangle decomposition the algebraic
decomposition of the associated octahedron, noting that the remaining four triangles are also a
triangle decomposition of the associated octahedron.
Now, assuming that the triangles in M O are all octahedral, and that these octahedra are
edge-disjoint, we can construct M1 , M2 , M3 and M4 as follows.
We take M1 = M C ∪ M I , M2 = M O , let M3 be the triangles in the algebraic decomposition of
the associated octahedra of triangles in M O , and let M4 be the other triangles, e.g. {x, y, z}, {y +
z, x + z, z}, {x, x + z, x + y} and {y + z, x + z, z} for every triangle xyz in M O . It is simple to check
that these SETs indeed satisfy our four conditions.
Our task is therefore to modify M O , M I and M C to ensure that the triangles in M O are all
octahedral.
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Modifying M O , M I and M C to get an octahedral M O
This is done using the following two phase random greedy algorithm.
1. We start by setting φ = M C + M I − M O , so ∂2 φ = L. We want to modify φ to ensure that
the triangles in φ− are octahedral.
The first step is to eliminate all of the signed triangles in φ. Given an octahedron Ωf containing
a signed element f = xyz of φ, we say that a triangle f 0 in Ωf if far if |f ∩ f 0 | ≤ 1, and say
that Ωf is valid if (i) none of its triangles are in T , except maybe f ,(ii) all of the new edges
are in G∗ , and (iii) all of its far triangles are octahedral. We denote the union of Ωf and the
associated octahedra of the far triangles by Ω+
f , and call this an extended configuration of f .
0
0
We denote the vertices of Ωf by x, y, z, x , y , z 0 , and then the additional vertices in Ω+
f are
0
0
0
sums of these: x + y , y + z and so on.
We order all of the signed elements of φ, and for each element f we choose a valid Ωf , subject
to the new edges in Ω+
f being distinct from all of the new edges of the extended configurations
that were chosen in previous steps. We add a signed copy of Ωf to φ, where the sign of f is
the opposite of its sign in φ, so that f is eliminated.
2. Let φ0 denote the result of the first phase, if it doesn’t abort. We now eliminate all of the
signed triangles of φ0 , apart from those containing an edge of L, or those that were far in the
first phase.
Let xyz 0 be such a triangle, and assume wlog that φ0 (xyz 0 ) = +1 and that z 0 is the new vertex
of the triangle, i.e. there was a signed element xyz in φ such that φ(xyz) = +1. Since xyz 0
doesn’t contain an edge in L, xy ∈
/ L and since ∂2 φ = L, there was some other signed element,
00
00
xyz with φ(xyz ) = −1. After the first phase, xyz 00 was eliminated, resulting in a triangle
xyz 000 with φ0 (xyz 000 ) = −1. Note that xyz 000 was not far and doesn’t contain an edge of L.
The point of all this is that we can now pair up these triangles into pairs f, f 0 of opposite signs,
such that |f ∩ f 0 | = 2. We eliminate each such pair by subtracting a signed octahedron Ωf,f 0
2
from φ0 . We say that Ωf,f 0 is valid if all of its triangles except for f and f 0 are octahedral, in
which case we denote their union by Ω+
f,f 0 .
As before, we order the pairs (f, f 0 ), and for each pair we choose a valid Ωf,f 0 uniformly at
random, conditioned in the new edges being distinct from all of the new edges of the extended
configurations that were chosen in previous steps.
Let ψ denote the result of the algorithm, if it doesn’t abort. Since the triangles in ψ − are now
all octahedral, we are done. Thus, it only remains to show that whp the algorithm doesn’t abort.
Lemma 2.1. With high probability the algorithm doesn’t abort, the edges of φ0 form a c7 -bounded
+ form a c -bounded graph.
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graph for c7 = d(G2c
8
∗ )30 γ 9 , and the edges of ψ
Proof. We will only analyze the first phase. The analysis of the second phase follows similar lines.
The proof is similar to the one in cover, where we showed that whp the random greedy algorithm
doesn’t abort. As before, the main technical part of the proof is to show that for a fixed signed
triangle f = xyz in φ, there are many valid octahedra Ωf . In other words, we are counting extended
∗
configurations Ω+
f in G , conditioned on the event that the extra octahedra that are sitting on the
far triangles of Ωf are their associated octahedra.
∗ 30 3 9
We claim that whp the number of choices for Ω+
f is at least (1 ± 32c2 )d(G ) n γ . Let X denote
+
the number of choices for Ωf , viewed as a random variable depending on the injection π. We will
bound E[X], and then use Azuma’s inequality to show that whp X is close to its expectation.
Note that E[X] is the sum over all potential extended configurations, regardless of π, of the
probability that the values assigned by π imply that the octahedra of the far triangles are indeed
their associated octahedra.
Now, by the c2 -typicality of G∗ , the number of summands is (1 ± 30c2 )d(G∗ )30 n12 , since an extended configuration has 30 new edges (by my count) and 12 new vertices. There are 9 vertices whose
π value is constrained by previous choices, so the probability that a potential extended configuration
is good is (1/(2a − O(1))9 ≈ γ 9 /n9 . Thus, E[X] = (1 ± 31c2 )d(G∗ )30 n3 γ 9 .
We recall Azuma’s inequality applied to the symmetric group.
Theorem 2.2. Let f : Sn → R such that f is b-Lipschitz, e.g. for any permutation τ ∈ Sn and
transposition σ, |f (σ ◦ τ ) − f (τ )| ≤ b. Then if we choose a permutation for f uniformly at random,
a2
Pr(|f − E[f ]| > a) < 2 exp −
.
2nb2
In order to apply this theorem, we have to show that X is O(n2 )-Lipschitz, e.g., that for any
given π, changing the values of two vertices not contained in f cannot change the value of X by more
than O(n2 ). We show that the number of extended configurations Ω+
f that use a given vertex v is
2
O(n ). Roughly speaking, this holds because we have three degrees of freedom when constructing
2
Ω+
f , and so fixing one of the vertices leaves us two degrees of freedom, or at most n possible choices.
0 0 0
Indeed, assume that v is one of the 12 new vertices in Ω+
f . If v ∈ {x , y , z }, then there are
at most n2 ways to choose the other two new vertices of Ωf , and then there is at most one way to
complete Ωf to an extended configuration, so this accounts for at most 3n2 choices for Ω+
f . If v is one
of the other vertices, say x0 + y 0 , then we get the constraint x0 + y 0 = v, so there are at most n ways
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to choose them both and n2 ways to choose Ωf . Altogether we have shown that X is 12n2 -Lipschitz,
so by Azuma:
√
(c2 d(G∗ )30 n3 γ 9 )2
∗ 30 3 9
< 2 exp(− n).
Pr(|X − E[X]| > c2 d(G ) n γ ) < 2 exp −
2
2
2n(12n )
Applying the union bound, we get that whp the number of choices is at least (1±32c2 )d(G∗ )30 n3 γ 9
for every signed triangle f in φ. Therefore, as long as the union of the extended configurations is
c7 -bounded, the algorithm doesn’t abort, because the total number of Ω+
f that can be ruled out for
a given triangle f is at most 30c7 n, a small fraction of the total number.
Therefore, it remains to show that whp φ0+ is c7 -bounded. As in the analysis of cover, let t
denote the number of signed triangles in φ, which is the number of steps in the greedy algorithm, let
Si denote the union of the extended configurations that were chosen up to time i, and let τ denote
the first time that Si is not c7 -bounded, or ∞ otherwise. We will show that whp τ = ∞. There
holds
t
X
Pr(τ < ∞) =
Pr(τ = i|τ ≥ i).
i=1
The next step is to show that for a fixed vertex v and time t0 , given that τ ≥ t0 , whp the degree
of v in St0 −1 is at most (c7 /2)n. By applying a union bound over all n vertices, we will show that
whp St0 −1 is (c7 /2)-bounded, implying that τ > τ0 . By applying a union bound for all times t0 , we
conclude that whp τ = ∞.
We have degSt −1 (v) ≤ 6 degφ+ (v) + 8Y , where Y is the number of times that v appeared in an
0
extended configuration as a new vertex up to time t0 . This holds because the maximal degree of a
vertex in an extended configuration is 8, and the degrees of the three original vertices are 6.
Now we get a concentration bound on Y , by representing Y as the sum of indicator variables Yi ,
which indicate whether v appeared as a new vertex in the extended configuration that was chosen
at time i. The probability that v appears in a given extended configuration Ω+
f is the number of
extended configurations of f that use v as a new vertex divided by the total number, which, as we
12n2
calculated above, is at most (1±32c2 )d(G
∗ )30 n3 γ 9 , and so
E[Y ] ≤
t
γ 9 d(G∗ )30 n
≤
c6
n
9
2γ d(G∗ )30
= (c7 /4)n.
Chernoff’s bound shows that whp Y ≤ (c7 /2)n, completing the proof.
References
[1] P. Keevash, The existence of designs, arXiv preprint arXiv:1401.3665 (2014).
[2] P. Keevash, Counting designs, arXiv preprint arXiv:1504.02909 (2015).
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