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Simultaneously vanishing higher derived
limits
(Joint work with Jeffrey Bergfalk)
Chris Lambie-Hanson
Department of Mathematics and Applied Mathematics
Virginia Commonwealth University
Oaxaca
8 August 2019
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I. Homological beginnings
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Additivity
Definition
A homology theory is additive on a class of topological spaces C
if, for every natural number p and every family {Xi | i ∈ J} such
that each Xi and J
Xi are in C, we have
J
Hp(Xi ) ∼
= Hp(
J
Xi )
via the map induced by the inclusions
Xi →
J
Xi .
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Additivity of strong homology
Let Xn denote the n-dimensional Hawaiian earring, i.e., the
one-point compactification of an infinite countable sum of copies
of the n-dimensional open unit ball. Let ¯
Hp(X) denote the pth
strong homology group of X.
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Additivity of strong homology
Let Xn denote the n-dimensional Hawaiian earring, i.e., the
one-point compactification of an infinite countable sum of copies
of the n-dimensional open unit ball. Let ¯
Hp(X) denote the pth
strong homology group of X.
Theorem (Mardeˇ
si´
c-Prasolov, ‘88)
Suppose that 0 < p < n are natural numbers. Then
N
¯
Hp(Xn) = ¯
Hp(
N
Xn)
if and only if limn−p A = 0.
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Additivity of strong homology
Let Xn denote the n-dimensional Hawaiian earring, i.e., the
one-point compactification of an infinite countable sum of copies
of the n-dimensional open unit ball. Let ¯
Hp(X) denote the pth
strong homology group of X.
Theorem (Mardeˇ
si´
c-Prasolov, ‘88)
Suppose that 0 < p < n are natural numbers. Then
N
¯
Hp(Xn) = ¯
Hp(
N
Xn)
if and only if limn−p A = 0.
Consequently, if strong homology is additive on closed subsets of
Euclidean space, then limn A = 0 for all n ≥ 1.
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A Hawaiian earring
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Compact supports
Definition
A homology theory has compact supports on a class of
topological spaces C if, for every natural number p and every
space X in C, Hp(X) is isomorphic to the direct limit of Hp(K)
for compact subspaces K of X via the map induced by the
inclusion maps K → X.
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Compact supports
Definition
A homology theory has compact supports on a class of
topological spaces C if, for every natural number p and every
space X in C, Hp(X) is isomorphic to the direct limit of Hp(K)
for compact subspaces K of X via the map induced by the
inclusion maps K → X.
Mardeˇ
si´
c and Prasolov also showed that if strong homology has
countable supports on closed subsets of Euclidean space, then
limn A = 0 for all n ≥ 1.
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The inverse system A
Given a function f ∈ ωω, let
I(f ) = {(k, m) ∈ ω × ω | m ≤ f (k)} ,
and let
Af =
I(f )
Z.
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The inverse system A
Given a function f ∈ ωω, let
I(f ) = {(k, m) ∈ ω × ω | m ≤ f (k)} ,
and let
Af =
I(f )
Z.
If f ≤ g, we have a projection map pfg : Ag → Af .
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The inverse system A
Given a function f ∈ ωω, let
I(f ) = {(k, m) ∈ ω × ω | m ≤ f (k)} ,
and let
Af =
I(f )
Z.
If f ≤ g, we have a projection map pfg : Ag → Af .
A is the inverse system
Af , pfg f , g ∈ ωω, f ≤ g .
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The inverse system A
Given a function f ∈ ωω, let
I(f ) = {(k, m) ∈ ω × ω | m ≤ f (k)} ,
and let
Af =
I(f )
Z.
If f ≤ g, we have a projection map pfg : Ag → Af .
A is the inverse system
Af , pfg f , g ∈ ωω, f ≤ g .
Its inverse limit, lim0 A, is ω ω Z.
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Condensed mathematics
The question of the consistency of limn A = 0 for n ≥ 1 arose
independently in recent work of Clausen and Scholze on
condensed mathematics, a new approach to doing algebra in
situations in which the algebraic structures carry topologies.
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Condensed mathematics
The question of the consistency of limn A = 0 for n ≥ 1 arose
independently in recent work of Clausen and Scholze on
condensed mathematics, a new approach to doing algebra in
situations in which the algebraic structures carry topologies. They
introduced the category of condensed abelian groups as a
replacement for topological abelian groups.
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Condensed mathematics
The question of the consistency of limn A = 0 for n ≥ 1 arose
independently in recent work of Clausen and Scholze on
condensed mathematics, a new approach to doing algebra in
situations in which the algebraic structures carry topologies. They
introduced the category of condensed abelian groups as a
replacement for topological abelian groups. The natural question
of whether pro-abelian groups embed fully faithfully into
condensed abelian groups is equivalent, in its simplest case, to the
question of whether limn A = 0 for n ≥ 1.
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II. Set theoretic reformulations
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Nontrivial coherent families
Given f , g ∈ ωω, let f ∧ g denote their greatest lower bound in
(ωω, ≤).
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Nontrivial coherent families
Given f , g ∈ ωω, let f ∧ g denote their greatest lower bound in
(ωω, ≤).
Definition
Let Φ = {ϕf : I(f ) → Z | f ∈ ωω} be a family of functions.
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Nontrivial coherent families
Given f , g ∈ ωω, let f ∧ g denote their greatest lower bound in
(ωω, ≤).
Definition
Let Φ = {ϕf : I(f ) → Z | f ∈ ωω} be a family of functions.
1 Φ is coherent if ϕf I(f ∧ g) =∗ ϕg I(f ∧ g) for all
f , g ∈ ωω.
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Nontrivial coherent families
Given f , g ∈ ωω, let f ∧ g denote their greatest lower bound in
(ωω, ≤).
Definition
Let Φ = {ϕf : I(f ) → Z | f ∈ ωω} be a family of functions.
1 Φ is coherent if ϕf I(f ∧ g) =∗ ϕg I(f ∧ g) for all
f , g ∈ ωω.
2 Φ is trivial if there is a function ψ : ω × ω → Z such that
ϕf =∗ ψ I(f ) for all f ∈ ωω.
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Nontrivial coherent families
Given f , g ∈ ωω, let f ∧ g denote their greatest lower bound in
(ωω, ≤).
Definition
Let Φ = {ϕf : I(f ) → Z | f ∈ ωω} be a family of functions.
1 Φ is coherent if ϕf I(f ∧ g) =∗ ϕg I(f ∧ g) for all
f , g ∈ ωω.
2 Φ is trivial if there is a function ψ : ω × ω → Z such that
ϕf =∗ ψ I(f ) for all f ∈ ωω.
Clearly, a trivial family is coherent.
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Nontrivial coherent families
Given f , g ∈ ωω, let f ∧ g denote their greatest lower bound in
(ωω, ≤).
Definition
Let Φ = {ϕf : I(f ) → Z | f ∈ ωω} be a family of functions.
1 Φ is coherent if ϕf I(f ∧ g) =∗ ϕg I(f ∧ g) for all
f , g ∈ ωω.
2 Φ is trivial if there is a function ψ : ω × ω → Z such that
ϕf =∗ ψ I(f ) for all f ∈ ωω.
Clearly, a trivial family is coherent. A nontrivial coherent family Φ
is a clear example of set theoretic incompactness: each local
family {ϕf | f < g} (for a fixed g ∈ ωω) is trivial, as witnessed by
ϕg itself, but the entire family is not.
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2-dimensional nontrivial coherence
Definition
Let Φ = ϕfg : I(f ∧ g) → Z f , g ∈ ωω .
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2-dimensional nontrivial coherence
Definition
Let Φ = ϕfg : I(f ∧ g) → Z f , g ∈ ωω .
1 Φ is alternating if ϕfg = −ϕgf for all f , g ∈ ωω.
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2-dimensional nontrivial coherence
Definition
Let Φ = ϕfg : I(f ∧ g) → Z f , g ∈ ωω .
1 Φ is alternating if ϕfg = −ϕgf for all f , g ∈ ωω.
2 Φ is 2-coherent if it is alternating and ϕfg + ϕgh =∗ ϕfh for
all f , g, h ∈ ωω. (All functions restricted to I(f ∧ g ∧ h).)
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2-dimensional nontrivial coherence
Definition
Let Φ = ϕfg : I(f ∧ g) → Z f , g ∈ ωω .
1 Φ is alternating if ϕfg = −ϕgf for all f , g ∈ ωω.
2 Φ is 2-coherent if it is alternating and ϕfg + ϕgh =∗ ϕfh for
all f , g, h ∈ ωω. (All functions restricted to I(f ∧ g ∧ h).)
3 Φ is 2-trivial if there is a family
Ψ = {ψf : I(f ) → Z | f ∈ ωω}
such that ψg − ψf =∗ ϕfg for all f , g ∈ ωω.
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2-dimensional nontrivial coherence
Definition
Let Φ = ϕfg : I(f ∧ g) → Z f , g ∈ ωω .
1 Φ is alternating if ϕfg = −ϕgf for all f , g ∈ ωω.
2 Φ is 2-coherent if it is alternating and ϕfg + ϕgh =∗ ϕfh for
all f , g, h ∈ ωω. (All functions restricted to I(f ∧ g ∧ h).)
3 Φ is 2-trivial if there is a family
Ψ = {ψf : I(f ) → Z | f ∈ ωω}
such that ψg − ψf =∗ ϕfg for all f , g ∈ ωω.
Again, a 2-trivial family is 2-coherent.
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2-dimensional nontrivial coherence
Definition
Let Φ = ϕfg : I(f ∧ g) → Z f , g ∈ ωω .
1 Φ is alternating if ϕfg = −ϕgf for all f , g ∈ ωω.
2 Φ is 2-coherent if it is alternating and ϕfg + ϕgh =∗ ϕfh for
all f , g, h ∈ ωω. (All functions restricted to I(f ∧ g ∧ h).)
3 Φ is 2-trivial if there is a family
Ψ = {ψf : I(f ) → Z | f ∈ ωω}
such that ψg − ψf =∗ ϕfg for all f , g ∈ ωω.
Again, a 2-trivial family is 2-coherent. A non-2-trivial 2-coherent
family Φ is an example of incompactness: each local family
{ϕfg | f , g < h} (for a fixed h ∈ ωω) is 2-trivial, as witnessed by
the family {−ϕfh | f < h}, but the entire family is not.
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n-dimensional nontrivial coherence
Given a sequence f = (f0, . . . , fn−1) and i < n, f i is the sequence
of length n − 1 formed by removing fi from f .
Definition
Fix n ≥ 2, and let Φ = ϕ
f
: I(∧f ) → Z f ∈ (ωω)n .
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n-dimensional nontrivial coherence
Given a sequence f = (f0, . . . , fn−1) and i < n, f i is the sequence
of length n − 1 formed by removing fi from f .
Definition
Fix n ≥ 2, and let Φ = ϕ
f
: I(∧f ) → Z f ∈ (ωω)n .
1 Φ is alternating if ϕ
f
= sgn(σ)ϕ
σ(f )
for all f ∈ (ωω)n and all
permutations σ.
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n-dimensional nontrivial coherence
Given a sequence f = (f0, . . . , fn−1) and i < n, f i is the sequence
of length n − 1 formed by removing fi from f .
Definition
Fix n ≥ 2, and let Φ = ϕ
f
: I(∧f ) → Z f ∈ (ωω)n .
1 Φ is alternating if ϕ
f
= sgn(σ)ϕ
σ(f )
for all f ∈ (ωω)n and all
permutations σ.
2 Φ is n-coherent if it is alternating and n
i=0
(−1)i ϕ
f i
=∗ 0
for all f ∈ (ωω)n+1.
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n-dimensional nontrivial coherence
Given a sequence f = (f0, . . . , fn−1) and i < n, f i is the sequence
of length n − 1 formed by removing fi from f .
Definition
Fix n ≥ 2, and let Φ = ϕ
f
: I(∧f ) → Z f ∈ (ωω)n .
1 Φ is alternating if ϕ
f
= sgn(σ)ϕ
σ(f )
for all f ∈ (ωω)n and all
permutations σ.
2 Φ is n-coherent if it is alternating and n
i=0
(−1)i ϕ
f i
=∗ 0
for all f ∈ (ωω)n+1.
3 Φ is n-trivial if there is an alternating family
ψ
f
: I(∧f ) → Z f ∈ (ωω)n−1
such that n−1
i=0
(−1)i ψ
f i
=∗ ϕ
f
for all f ∈ (ωω)n.
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limn A and nontrivial coherence
Coherent and trivial can now be thought of as 1-coherent and
1-trivial.
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limn A and nontrivial coherence
Coherent and trivial can now be thought of as 1-coherent and
1-trivial.
Theorem (Mardeˇ
si´
c-Prasolov (n = 1), Bergfalk (n ≥ 2))
Fix n ≥ 1. Then limn A = 0 if and only if every n-coherent family
Φ = ϕ
f
f ∈ (ωω)n
is n-trivial.
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limn A and nontrivial coherence
Coherent and trivial can now be thought of as 1-coherent and
1-trivial.
Theorem (Mardeˇ
si´
c-Prasolov (n = 1), Bergfalk (n ≥ 2))
Fix n ≥ 1. Then limn A = 0 if and only if every n-coherent family
Φ = ϕ
f
f ∈ (ωω)n
is n-trivial.
Thus, to prove that limn A = 0, it suffices to show that every
n-coherent family is n-trivial.
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lim1 A
Theorem (Mardeˇ
si´
c-Prasolov, Simon, ’88)
CH ⇒ lim1 A = 0.
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lim1 A
Theorem (Mardeˇ
si´
c-Prasolov, Simon, ’88)
CH ⇒ lim1 A = 0.
Theorem (Dow-Simon-Vaughan, ‘89)
d = ℵ1 ⇒ lim1 A = 0.
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lim1 A
Theorem (Mardeˇ
si´
c-Prasolov, Simon, ’88)
CH ⇒ lim1 A = 0.
Theorem (Dow-Simon-Vaughan, ‘89)
d = ℵ1 ⇒ lim1 A = 0.
PFA ⇒ lim1 A = 0.
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lim1 A
Theorem (Mardeˇ
si´
c-Prasolov, Simon, ’88)
CH ⇒ lim1 A = 0.
Theorem (Dow-Simon-Vaughan, ‘89)
d = ℵ1 ⇒ lim1 A = 0.
PFA ⇒ lim1 A = 0.
Theorem (Todorcevic, ‘98)
OCA ⇒ lim1 A = 0.
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lim1 A
Theorem (Mardeˇ
si´
c-Prasolov, Simon, ’88)
CH ⇒ lim1 A = 0.
Theorem (Dow-Simon-Vaughan, ‘89)
d = ℵ1 ⇒ lim1 A = 0.
PFA ⇒ lim1 A = 0.
Theorem (Todorcevic, ‘98)
OCA ⇒ lim1 A = 0.
MAℵ1
⇒ lim1 A = 0.
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lim1 A
Theorem (Mardeˇ
si´
c-Prasolov, Simon, ’88)
CH ⇒ lim1 A = 0.
Theorem (Dow-Simon-Vaughan, ‘89)
d = ℵ1 ⇒ lim1 A = 0.
PFA ⇒ lim1 A = 0.
Theorem (Todorcevic, ‘98)
OCA ⇒ lim1 A = 0.
MAℵ1
⇒ lim1 A = 0.
Theorem (Kamo, ‘93)
V Add(ω,ω2) |= “ lim1 A = 0”.
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Higher limits
Theorem (Bergfalk, ‘17)
PFA ⇒ lim2 A = 0.
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Higher limits
Theorem (Bergfalk, ‘17)
PFA ⇒ lim2 A = 0.
Theorem (Bergfalk-LH, ‘19)
Suppose that κ is a measurable cardinal and that P is a length-κ
finite support iteration of Hechler forcings. Then
V P |= “ lim nA = 0 for all n ≥ 1”.
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III. A strong ∆-system lemma
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Hechler forcing
Recall that conditions of Hechler forcing are of the form
p = (sp, f p), where sp ∈ <ωω and f p ∈ ωω.
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Hechler forcing
Recall that conditions of Hechler forcing are of the form
p = (sp, f p), where sp ∈ <ωω and f p ∈ ωω. If p and q are
Hechler conditions, then q ≤ p if and only if
• sq end-extends sp;
• f q ≥ f p;
• sq(k) > f p(k) for all k ∈ dom(sq) \ dom(qp).
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Hechler forcing
Recall that conditions of Hechler forcing are of the form
p = (sp, f p), where sp ∈ <ωω and f p ∈ ωω. If p and q are
Hechler conditions, then q ≤ p if and only if
• sq end-extends sp;
• f q ≥ f p;
• sq(k) > f p(k) for all k ∈ dom(sq) \ dom(qp).
In what follows, P denotes a length-κ finite support iteration of
Hechler forcings, where κ is a fixed regular uncountable cardinal.
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Hechler forcing
Recall that conditions of Hechler forcing are of the form
p = (sp, f p), where sp ∈ <ωω and f p ∈ ωω. If p and q are
Hechler conditions, then q ≤ p if and only if
• sq end-extends sp;
• f q ≥ f p;
• sq(k) > f p(k) for all k ∈ dom(sq) \ dom(qp).
In what follows, P denotes a length-κ finite support iteration of
Hechler forcings, where κ is a fixed regular uncountable cardinal.
As a result of a simple ∆-system argument, we know that, if
pα | α < κ is a sequence of conditions in P, then there is an
unbounded set A ⊆ κ such that pα | α ∈ A consists of pairwise
compatible conditions.
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A first attempt
In our proof, we will be working with families of conditions
indexed by finite increasing sequences from κ. We would like to
use a multi-dimensional ∆-system argument to perform a similar
sort of uniformization. A na¨
ıve first attempt might be:
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A first attempt
In our proof, we will be working with families of conditions
indexed by finite increasing sequences from κ. We would like to
use a multi-dimensional ∆-system argument to perform a similar
sort of uniformization. A na¨
ıve first attempt might be:
Lemma?
Suppose that κ is sufficiently large (e.g., measurable) and fix
n ≥ 1. Given any family pα | α ∈ [κ]n from P, there is an
unbounded A ⊆ κ such that pα | α ∈ [A]n consists of pairwise
compatible conditions.
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A first attempt
In our proof, we will be working with families of conditions
indexed by finite increasing sequences from κ. We would like to
use a multi-dimensional ∆-system argument to perform a similar
sort of uniformization. A na¨
ıve first attempt might be:
Lemma?
Suppose that κ is sufficiently large (e.g., measurable) and fix
n ≥ 1. Given any family pα | α ∈ [κ]n from P, there is an
unbounded A ⊆ κ such that pα | α ∈ [A]n consists of pairwise
compatible conditions.
This is obviously false. If n = 2, for example, we can define
pαβ | α < β < κ such that pαβ(α) is (a name for) ( 0 , 0) and
pαβ(β) is ( 1 , 0). Then, for any α < β < γ, we have pαβ ⊥ pβγ.
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A first attempt
In our proof, we will be working with families of conditions
indexed by finite increasing sequences from κ. We would like to
use a multi-dimensional ∆-system argument to perform a similar
sort of uniformization. A na¨
ıve first attempt might be:
Lemma?
Suppose that κ is sufficiently large (e.g., measurable) and fix
n ≥ 1. Given any family pα | α ∈ [κ]n from P, there is an
unbounded A ⊆ κ such that pα | α ∈ [A]n consists of pairwise
compatible conditions.
This is obviously false. If n = 2, for example, we can define
pαβ | α < β < κ such that pαβ(α) is (a name for) ( 0 , 0) and
pαβ(β) is ( 1 , 0). Then, for any α < β < γ, we have pαβ ⊥ pβγ.
This turns out to be the only obstacle, though.
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Aligned sets
Definition
Fix n ≥ 1 and suppose that α = α0, . . . , αn−1 and
β = β0, . . . , βn−1 are in [κ]n (and hence are increasing). We say
that α and β are aligned if, for all i, j < n,
αi = βj ⇒ i = j.
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Aligned sets
Definition
Fix n ≥ 1 and suppose that α = α0, . . . , αn−1 and
β = β0, . . . , βn−1 are in [κ]n (and hence are increasing). We say
that α and β are aligned if, for all i, j < n,
αi = βj ⇒ i = j.
Lemma
Suppose that κ is weakly compact, and fix n ≥ 1. Given any
family pα | α ∈ [κ]n , there is an unbounded A ⊆ κ such that,
for all α, β ∈ [κ]n, if α and β are aligned, then pα
and p
β
are
compatible.
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Aligned sets
Definition
Fix n ≥ 1 and suppose that α = α0, . . . , αn−1 and
β = β0, . . . , βn−1 are in [κ]n (and hence are increasing). We say
that α and β are aligned if, for all i, j < n,
αi = βj ⇒ i = j.
Lemma
Suppose that κ is weakly compact, and fix n ≥ 1. Given any
family pα | α ∈ [κ]n , there is an unbounded A ⊆ κ such that,
for all α, β ∈ [κ]n, if α and β are aligned, then pα
and p
β
are
compatible.
We actually require a bit more than this, for which we seem to
need a measurable cardinal.
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The ∆-system lemma
Lemma
Suppose that κ is measurable, with normal measure U, and fix
n ≥ 1. Let P be a length-κ finite support iteration of Hechler
forcings, and let {pα | α ∈ [κ]n} be a family of conditions in P.
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The ∆-system lemma
Lemma
Suppose that κ is measurable, with normal measure U, and fix
n ≥ 1. Let P be a length-κ finite support iteration of Hechler
forcings, and let {pα | α ∈ [κ]n} be a family of conditions in P.
Then there is a set A ∈ U and conditions {pα | α ∈ [A]
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The ∆-system lemma
Lemma
Suppose that κ is measurable, with normal measure U, and fix
n ≥ 1. Let P be a length-κ finite support iteration of Hechler
forcings, and let {pα | α ∈ [κ]n} be a family of conditions in P.
Then there is a set A ∈ U and conditions {pα | α ∈ [A]
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The ∆-system lemma
Lemma
Suppose that κ is measurable, with normal measure U, and fix
n ≥ 1. Let P be a length-κ finite support iteration of Hechler
forcings, and let {pα | α ∈ [κ]n} be a family of conditions in P.
Then there is a set A ∈ U and conditions {pα | α ∈ [A]
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The ∆-system lemma
Lemma
Suppose that κ is measurable, with normal measure U, and fix
n ≥ 1. Let P be a length-κ finite support iteration of Hechler
forcings, and let {pα | α ∈ [κ]n} be a family of conditions in P.
Then there is a set A ∈ U and conditions {pα | α ∈ [A]
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IV. Vanishing higher limits
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Proof sketch (n = 2)
Main Theorem
Suppose that κ is a measurable cardinal and that P is a length-κ
finite support iteration of Hechler forcings. In V P, for every
n ≥ 1, every n-coherent family ϕ
f
f ∈ (ωω)n is n-trivial.
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Proof sketch (n = 2)
Main Theorem
Suppose that κ is a measurable cardinal and that P is a length-κ
finite support iteration of Hechler forcings. In V P, for every
n ≥ 1, every n-coherent family ϕ
f
f ∈ (ωω)n is n-trivial.
Proof sketch for n = 2: We start by making some simplifying
assumptions.
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Proof sketch (n = 2)
Main Theorem
Suppose that κ is a measurable cardinal and that P is a length-κ
finite support iteration of Hechler forcings. In V P, for every
n ≥ 1, every n-coherent family ϕ
f
f ∈ (ωω)n is n-trivial.
Proof sketch for n = 2: We start by making some simplifying
assumptions.
• To show that a 2-coherent family of functions is trivial, it
suffices to find a trivialization indexed by some ≤∗-cofinal
family of functions.
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Proof sketch (n = 2)
Main Theorem
Suppose that κ is a measurable cardinal and that P is a length-κ
finite support iteration of Hechler forcings. In V P, for every
n ≥ 1, every n-coherent family ϕ
f
f ∈ (ωω)n is n-trivial.
Proof sketch for n = 2: We start by making some simplifying
assumptions.
• To show that a 2-coherent family of functions is trivial, it
suffices to find a trivialization indexed by some ≤∗-cofinal
family of functions.
• Since the Hechler reals fα | α < κ are ≤∗-increasing and
cofinal, it suffices to find a trivialization indexed by
fα | α ∈ B for some cofinal B ⊆ κ.
Slide 67
Slide 67 text
• The 2-triviality of a 2-coherent family ϕfg f , g ∈ ωω is
equivalent to the following statement:
Slide 68
Slide 68 text
• The 2-triviality of a 2-coherent family ϕfg f , g ∈ ωω is
equivalent to the following statement:
There is an alternating family ψfg f , g ∈ ωω such that
each ψfg : I(f ∧ g) → Z is finitely supported and, for all
f , g, h ∈ ωω, (ϕgh − ψgh) − (ϕfh − ψfh) + (ϕfg − ψfg) = 0.
Slide 69
Slide 69 text
• The 2-triviality of a 2-coherent family ϕfg f , g ∈ ωω is
equivalent to the following statement:
There is an alternating family ψfg f , g ∈ ωω such that
each ψfg : I(f ∧ g) → Z is finitely supported and, for all
f , g, h ∈ ωω, (ϕgh − ψgh) − (ϕfh − ψfh) + (ϕfg − ψfg) = 0.
In V , fix a condition p ∈ P and a P-name ˙
Φ = ˙
ϕαβ α, β < κ
for a 2-coherent family indexed by pairs of Hechler reals.
Slide 70
Slide 70 text
• The 2-triviality of a 2-coherent family ϕfg f , g ∈ ωω is
equivalent to the following statement:
There is an alternating family ψfg f , g ∈ ωω such that
each ψfg : I(f ∧ g) → Z is finitely supported and, for all
f , g, h ∈ ωω, (ϕgh − ψgh) − (ϕfh − ψfh) + (ϕfg − ψfg) = 0.
In V , fix a condition p ∈ P and a P-name ˙
Φ = ˙
ϕαβ α, β < κ
for a 2-coherent family indexed by pairs of Hechler reals. For
α < β < γ < κ above the domain of p, let ˙
e(α, β, γ) be a P
name for ϕβγ − ϕαγ + ϕαβ, let ˙
¯
e(α, β, γ) be a P-name for its
restriction to its (finite) support, and fix pαβγ ≤ p such that
Slide 71
Slide 71 text
• The 2-triviality of a 2-coherent family ϕfg f , g ∈ ωω is
equivalent to the following statement:
There is an alternating family ψfg f , g ∈ ωω such that
each ψfg : I(f ∧ g) → Z is finitely supported and, for all
f , g, h ∈ ωω, (ϕgh − ψgh) − (ϕfh − ψfh) + (ϕfg − ψfg) = 0.
In V , fix a condition p ∈ P and a P-name ˙
Φ = ˙
ϕαβ α, β < κ
for a 2-coherent family indexed by pairs of Hechler reals. For
α < β < γ < κ above the domain of p, let ˙
e(α, β, γ) be a P
name for ϕβγ − ϕαγ + ϕαβ, let ˙
¯
e(α, β, γ) be a P-name for its
restriction to its (finite) support, and fix pαβγ ≤ p such that
• pαβγ fα ≤ fβ ≤ fγ;
Slide 72
Slide 72 text
• The 2-triviality of a 2-coherent family ϕfg f , g ∈ ωω is
equivalent to the following statement:
There is an alternating family ψfg f , g ∈ ωω such that
each ψfg : I(f ∧ g) → Z is finitely supported and, for all
f , g, h ∈ ωω, (ϕgh − ψgh) − (ϕfh − ψfh) + (ϕfg − ψfg) = 0.
In V , fix a condition p ∈ P and a P-name ˙
Φ = ˙
ϕαβ α, β < κ
for a 2-coherent family indexed by pairs of Hechler reals. For
α < β < γ < κ above the domain of p, let ˙
e(α, β, γ) be a P
name for ϕβγ − ϕαγ + ϕαβ, let ˙
¯
e(α, β, γ) be a P-name for its
restriction to its (finite) support, and fix pαβγ ≤ p such that
• pαβγ fα ≤ fβ ≤ fγ;
• pαβγ decides the value of ˙
¯
e(α, β, γ).
Slide 73
Slide 73 text
Apply our ∆-system lemma to obtain A ∈ U and conditions
{pα | α ∈ [A]<3} such that
Slide 74
Slide 74 text
Apply our ∆-system lemma to obtain A ∈ U and conditions
{pα | α ∈ [A]<3} such that
• for all α β in [A]≤3, we have dom(pα
) dom(p
β
) and
pα
= p
β
dom(pα
);
Slide 75
Slide 75 text
Apply our ∆-system lemma to obtain A ∈ U and conditions
{pα | α ∈ [A]<3} such that
• for all α β in [A]≤3, we have dom(pα
) dom(p
β
) and
pα
= p
β
dom(pα
);
• for all α ∈ [A]
Slide 76
Slide 76 text
Apply our ∆-system lemma to obtain A ∈ U and conditions
{pα | α ∈ [A]<3} such that
• for all α β in [A]≤3, we have dom(pα
) dom(p
β
) and
pα
= p
β
dom(pα
);
• for all α ∈ [A]
Slide 77
Slide 77 text
Apply our ∆-system lemma to obtain A ∈ U and conditions
{pα | α ∈ [A]<3} such that
• for all α β in [A]≤3, we have dom(pα
) dom(p
β
) and
pα
= p
β
dom(pα
);
• for all α ∈ [A]
Slide 78
Slide 78 text
We claim that p∅
forces ˙
Φ to be trivial.
Slide 79
Slide 79 text
We claim that p∅
forces ˙
Φ to be trivial. In V , partition A into
unbounded sets Γ0, Γ1, Γ2.
Slide 80
Slide 80 text
We claim that p∅
forces ˙
Φ to be trivial. In V , partition A into
unbounded sets Γ0, Γ1, Γ2. Let G be P-generic with p∅
∈ G. The
following facts, which hold in V [G], are the point of our
∆-system lemma:
Slide 81
Slide 81 text
We claim that p∅
forces ˙
Φ to be trivial. In V , partition A into
unbounded sets Γ0, Γ1, Γ2. Let G be P-generic with p∅
∈ G. The
following facts, which hold in V [G], are the point of our
∆-system lemma:
• B := {α ∈ Γ0 | pα ∈ G} is unbounded.
Slide 82
Slide 82 text
We claim that p∅
forces ˙
Φ to be trivial. In V , partition A into
unbounded sets Γ0, Γ1, Γ2. Let G be P-generic with p∅
∈ G. The
following facts, which hold in V [G], are the point of our
∆-system lemma:
• B := {α ∈ Γ0 | pα ∈ G} is unbounded.
• For all α < β in B, there is αβ ∈ Γ1 \ β such that
pα αβ
, pβ αβ
∈ G.
Slide 83
Slide 83 text
We claim that p∅
forces ˙
Φ to be trivial. In V , partition A into
unbounded sets Γ0, Γ1, Γ2. Let G be P-generic with p∅
∈ G. The
following facts, which hold in V [G], are the point of our
∆-system lemma:
• B := {α ∈ Γ0 | pα ∈ G} is unbounded.
• For all α < β in B, there is αβ ∈ Γ1 \ β such that
pα αβ
, pβ αβ
∈ G.
• For all α < β < γ in B, there is
αβγ ∈ Γ2 \ max{ αβ, αγ, βγ}
such that
pα αβ αβγ
, pα αγ αβγ
, pβ αβ αβγ
, pβ βγ αβγ
, pγ αγ, αβγ
, pγ βγ, αβγ
∈ G.
Slide 84
Slide 84 text
For α < β in B, let ψαβ := e(α, β, αβ) = ϕβ αβ
− ϕα αβ
+ ϕαβ.
Slide 85
Slide 85 text
For α < β in B, let ψαβ := e(α, β, αβ) = ϕβ αβ
− ϕα αβ
+ ϕαβ.
We claim that this works. To see this, fix α < β < γ in B. We
must show that e(α, β, γ) = ψβγ − ψαγ + ψαβ.
Slide 86
Slide 86 text
For α < β in B, let ψαβ := e(α, β, αβ) = ϕβ αβ
− ϕα αβ
+ ϕαβ.
We claim that this works. To see this, fix α < β < γ in B. We
must show that e(α, β, γ) = ψβγ − ψαγ + ψαβ.
NOTE: For any α < β < γ < δ < κ, we have
e(β, γ, δ) − e(α, γ, δ) + e(α, β, δ) − e(α, β, γ) = 0.
Slide 87
Slide 87 text
For α < β in B, let ψαβ := e(α, β, αβ) = ϕβ αβ
− ϕα αβ
+ ϕαβ.
We claim that this works. To see this, fix α < β < γ in B. We
must show that e(α, β, γ) = ψβγ − ψαγ + ψαβ.
NOTE: For any α < β < γ < δ < κ, we have
e(β, γ, δ) − e(α, γ, δ) + e(α, β, δ) − e(α, β, γ) = 0.
Thus,
e(α, β, γ) = e(β, γ, αβγ) − e(α, γ, αβγ) + e(α, β, αβγ).
Slide 88
Slide 88 text
For α < β in B, let ψαβ := e(α, β, αβ) = ϕβ αβ
− ϕα αβ
+ ϕαβ.
We claim that this works. To see this, fix α < β < γ in B. We
must show that e(α, β, γ) = ψβγ − ψαγ + ψαβ.
NOTE: For any α < β < γ < δ < κ, we have
e(β, γ, δ) − e(α, γ, δ) + e(α, β, δ) − e(α, β, γ) = 0.
Thus,
e(α, β, γ) = e(β, γ, αβγ) − e(α, γ, αβγ) + e(α, β, αβγ).
e(β, γ, αβγ) = −e(γ, βγ, αβγ) + e(β, βγ, αβγ) + e(β, γ, βγ)
= −e + e + ψβγ = ψβγ
Slide 89
Slide 89 text
For α < β in B, let ψαβ := e(α, β, αβ) = ϕβ αβ
− ϕα αβ
+ ϕαβ.
We claim that this works. To see this, fix α < β < γ in B. We
must show that e(α, β, γ) = ψβγ − ψαγ + ψαβ.
NOTE: For any α < β < γ < δ < κ, we have
e(β, γ, δ) − e(α, γ, δ) + e(α, β, δ) − e(α, β, γ) = 0.
Thus,
e(α, β, γ) = e(β, γ, αβγ) − e(α, γ, αβγ) + e(α, β, αβγ).
e(β, γ, αβγ) = −e(γ, βγ, αβγ) + e(β, βγ, αβγ) + e(β, γ, βγ)
= −e + e + ψβγ = ψβγ
Similarly for e(α, γ, αβγ) and e(α, β, αβγ).
Slide 90
Slide 90 text
For α < β in B, let ψαβ := e(α, β, αβ) = ϕβ αβ
− ϕα αβ
+ ϕαβ.
We claim that this works. To see this, fix α < β < γ in B. We
must show that e(α, β, γ) = ψβγ − ψαγ + ψαβ.
NOTE: For any α < β < γ < δ < κ, we have
e(β, γ, δ) − e(α, γ, δ) + e(α, β, δ) − e(α, β, γ) = 0.
Thus,
e(α, β, γ) = e(β, γ, αβγ) − e(α, γ, αβγ) + e(α, β, αβγ).
e(β, γ, αβγ) = −e(γ, βγ, αβγ) + e(β, βγ, αβγ) + e(β, γ, βγ)
= −e + e + ψβγ = ψβγ
Similarly for e(α, γ, αβγ) and e(α, β, αβγ).
This reduces to e(α, β, γ) = ψβγ − ψαγ + ψαβ, as desired.
Slide 91
Slide 91 text
V. Further directions
Slide 92
Slide 92 text
Optimality
There are a number of questions one can ask about the
optimality of our theorem.
Slide 93
Slide 93 text
Optimality
There are a number of questions one can ask about the
optimality of our theorem.
Question
What is the consistency strength of the statement
“ limn A = 0 for all n ≥ 1”?
Slide 94
Slide 94 text
Optimality
There are a number of questions one can ask about the
optimality of our theorem.
Question
What is the consistency strength of the statement
“ limn A = 0 for all n ≥ 1”?
Question
How small can the continuum be in a model of
“ limn A = 0 for all n ≥ 1”?
Slide 95
Slide 95 text
Optimality
There are a number of questions one can ask about the
optimality of our theorem.
Question
What is the consistency strength of the statement
“ limn A = 0 for all n ≥ 1”?
Question
How small can the continuum be in a model of
“ limn A = 0 for all n ≥ 1”?
We conjecture that it must be greater than ℵω. A natural place
to look is the model obtained from adding ℵω+1 Cohen reals to a
model of CH.
Slide 96
Slide 96 text
Strong homology
Question
Is it consistent that strong homology is additive on some nice
class of topological spaces, e.g., locally compact metric spaces?
Slide 97
Slide 97 text
Strong homology
Question
Is it consistent that strong homology is additive on some nice
class of topological spaces, e.g., locally compact metric spaces?
Question
Is it consistent that strong homology has compact supports on
some nice class of topological spaces, e.g., locally compact metric
spaces?
Slide 98
Slide 98 text
Strong homology
Question
Is it consistent that strong homology is additive on some nice
class of topological spaces, e.g., locally compact metric spaces?
Question
Is it consistent that strong homology has compact supports on
some nice class of topological spaces, e.g., locally compact metric
spaces?
The model for our Main Theorem is a natural first place to look
for these questions.
Slide 99
Slide 99 text
Results contained in “Simultaneously vanishing higher derived
limits,” joint with Jeffrey Bergfalk.
Available at: https://arxiv.org/abs/1907.11744
All artwork by Josef Albers, inspired by visits to Monte Alban
Slide 100
Slide 100 text
Thank you!