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Vanishing integral middle cohomology of a closed oriented seven-manifold implies real vanishing

Statement

Assume the Axiom of Choice. Let $M$ be a closed oriented seven-manifold. If $H^{3} (M; Z) = 0$ and $H^{4} (M; Z) = 0$ , then $H^{3} (M; R) = 0$ and $H^{4} (M; R) = 0$ .

Facts & Assumptions

Given: A closed oriented seven-manifold $M$ with $H^{3} (M; Z) = H^{4} (M; Z) = 0$ .

[A1]

The Axiom of Choice is assumed (The Axiom of Choice).

[L1]

Assume AC. For a closed $Z$ -oriented manifold all integral homology groups are finitely generated (Finite generation from cap with a finite fundamental cycle).

[L2]

Every finitely generated abelian group is a direct sum $Z^{r} \oplus T$ with $T$ finite (The fundamental theorem of finitely generated abelian groups from PID modules).

[L3]

Assume AC. For every space $X$ , abelian group $G$ and $n \geq 0$ there is a natural short exact sequence $0 \to {Ext}_{Z}^{1} (H_{n - 1} (X; Z), G) \to H^{n} (X; G) \to {Hom}_{Z} (H_{n} (X; Z), G) \to 0.$ (Topological universal coefficient short exact sequence for cohomology).

[L4]

Assume DC. For $n > 1$ , ${Ext}_{Z}^{1} (Z / n, Z) ≅ Z / n$ , so it is nonzero (Ext one of Z modulo n by Z is Z modulo n).

[L5]

In ZF, $A C \Rightarrow D C$ (AC implies DC implies countable choice).

[L6]

$R$ is an ordered field: it has no nonzero elements of finite order, and for every $m > 0$ and every $y \in R$ there is $x \in R$ with $m x = y$ (The reals form a totally ordered field).

Proof

technique · direct

1.1A1L1given

By [L1] and [A1] every homology group $H_{k} (M; Z)$ is finitely generated.

2.1step 1.1L3given

The universal coefficient sequence [L3] in degree three is $0 \to Ext (H_{2} (M; Z), Z) \to H^{3} (M; Z) \to Hom (H_{3} (M; Z), Z) \to 0$ ; since the middle term vanishes, the two outer groups vanish, so $Ext (H_{2}, Z) = 0$ and $Hom (H_{3}, Z) = 0$ .

3.1step 2.1L3given

The universal coefficient sequence [L3] in degree four is $0 \to Ext (H_{3} (M; Z), Z) \to H^{4} (M; Z) \to Hom (H_{4} (M; Z), Z) \to 0$ ; since the middle term vanishes, $Hom (H_{4}, Z) = 0$ .

4.1step 2.1step 3.1A1L2L4L5

By [L2] and [L4] with [A1], [L5], write $H_{k} (M; Z) = Z^{r_{k}} \oplus T_{k}$ with $T_{k}$ finite: the vanishing $Hom (H_{3}, Z) = Hom (H_{4}, Z) = 0$ and $Hom (Z^{r} \oplus T, Z) ≅ Z^{r}$ force $r_{3} = r_{4} = 0$ , so $H_{3}$ and $H_{4}$ are finite; moreover $Ext (Z^{r}, Z) = 0$ and additivity of Ext together with $Ext (Z / n, Z) ≅ Z / n \neq 0$ show that $Ext (H_{2}, Z) = 0$ forces $T_{2} = 0$ , so $H_{2}$ is a finitely generated free abelian group.

5.1step 4.1L3L6

The real universal coefficient sequence in degree three is $0 \to Ext (H_{2}, R) \to H^{3} (M; R) \to Hom (H_{3}, R) \to 0$ ; here $Ext (H_{2}, R) = 0$ because $H_{2}$ is free, and $Hom (H_{3}, R) = 0$ because $H_{3}$ is finite while $R$ has no nonzero elements of finite order by [L6], so exactness gives $H^{3} (M; R) = 0$ .

6.1step 5.1L3L6

The real universal coefficient sequence in degree four is $0 \to Ext (H_{3}, R) \to H^{4} (M; R) \to Hom (H_{4}, R) \to 0$ ; here $Ext (H_{3}, R) = 0$ because $H_{3}$ is finite and every $m > 0$ acts surjectively on $R$ by [L6], and $Hom (H_{4}, R) = 0$ because $H_{4}$ is finite, so exactness gives $H^{4} (M; R) = 0$ .

7.1step 5.1step 6.1∎

Therefore $H^{3} (M; R) = 0$ by step 5.1 and $H^{4} (M; R) = 0$ by step 6.1, as asserted.

Depends on

Used by

The boundary middle form is well defined and glues Lemma

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Allen Hatcher, Algebraic Topology, Cambridge University Press 2002 (complete book) (standard reference, not scraped)
Charles A. Weibel, An Introduction to Homological Algebra, Chapter 3, Tor and Ext (complete chapter) (standard reference, not scraped)