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Positive intermediate cohomology of compactified Euclidean space vanishes

Statement

Assume AC. For $N \geq 2$ , $0 < q < N$ , and $R = Z$ or $F_{2}$ , the one-point compactification $(R^{N})^{+}$ of $R^{N}$ (The one-point (Alexandroff) compactification $X^{*} = X \cup {\infty}$ , whose open sets are the open sets of $X$ together with the complements in $X^{*}$ of the closed compact subsets of $X$ ) is homeomorphic to $S^{N}$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ ) and has $H^{q} ((R^{N})^{+}; R) = 0$ (Topological universal coefficient short exact sequence for cohomology, The Axiom of Choice).

Facts & Assumptions

Given: integers $N \geq 2$ and $0 < q < N$ , a coefficient ring $R = Z$ or $F_{2}$ , and the one-point compactification $(R^{N})^{+} = (R^{N})^{*}$ with added point $\infty$ (The one-point (Alexandroff) compactification $X^{*} = X \cup {\infty}$ , whose open sets are the open sets of $X$ together with the complements in $X^{*}$ of the closed compact subsets of $X$ ).

[F1]

The topology $T^{*}$ of $X^{*} = X \cup {\infty}$ consists of the open sets of $X$ together with the sets $X^{*} ∖ C$ for $C \subseteq X$ closed in $X$ and a compact subset of $X$ ; $X^{*}$ is compact, $X$ is an open subspace with its original topology, and $X^{*}$ is Hausdorff exactly when $X$ is locally compact and Hausdorff (The one-point (Alexandroff) compactification $X^{*} = X \cup {\infty}$ , whose open sets are the open sets of $X$ together with the complements in $X^{*}$ of the closed compact subsets of $X$ , $X^{*}$ is compact and contains $X$ as an open subspace; $X$ is dense in $X^{*}$ exactly when $X$ is not compact; and $X^{*}$ is Hausdorff exactly when $X$ is locally compact and Hausdorff).

[F2]

$R^{N}$ is locally compact, its topology is the Euclidean metric topology and the product topology, and $S^{N} = {y \in R^{N + 1} : ∥ y ∥_{2} = 1}$ carries the subspace topology, hence is metrizable and Hausdorff ( $R^{n}$ is locally compact and $σ$ -compact, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it, For $n \geq 1$ the product topology on $n$ copies of the usual topology of $R$ is the metric topology of $d_{\infty}$ on $R^{n}$ , and hence also of $d_{1}$ and $d_{2}$ , so $R^{n}$ as a product and $R^{n}$ as a metric space are one space, Euclidean spheres and closed balls as subspaces of $R^{n}$ , Every subspace of a metrizable space is metrizable and every subspace of a first countable space is first countable, the metric case being the subspace metric already identified with the subspace topology, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[F3]

A map into a finite product is continuous exactly when its components are; sums, products and quotients of continuous real-valued functions with nowhere-vanishing denominator are continuous, and $x \mapsto x_{i}$ and $x \mapsto ∥ x ∥_{2}^{2}$ are continuous on $R^{N}$ (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, Sums, products, absolute values, finite maxima and minima, and quotients of continuous real-valued maps on a topological space are continuous where defined, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it, Continuity of a map of topological spaces at a point and globally).

[F4]

A closed bounded subset of $R^{N}$ is compact (A subset of $R^{n}$ with the product topology is compact exactly when it is closed and bounded, the product topology being the Euclidean metric topology).

[F5]

A continuous bijection from a compact space onto a Hausdorff space is a homeomorphism, and a continuous map into a subspace that corestricts to the image is continuous as a map onto that image (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

[F6]

Under AC, for every space $X$ , abelian group $G$ and $n \geq 0$ the evaluation sequence is natural and exact (Topological universal coefficient short exact sequence for cohomology, The Axiom of Choice).

[F7]

For the sphere $S^{N}$ with $N \geq 2$ one has $H_{0} (S^{N}; Z) ≅ Z$ and $H_{j} (S^{N}; Z) = 0$ for $0 < j < N$ and for $j > N$ ; this is the case $N \geq 1$ of the reduced-homology computation of Homology of spheres. For every abelian group $G$ the explicit free resolution $0 \to 0 \to Z \overset{1}{\to} Z \to 0$ computes ${Ext}_{Z}^{1} (Z, G) = 0$ by the definition of $Ext$ (the same resolution computation performed for $G = Z$ in Integral cohomology detects adjacent homology torsion), and ${Ext}_{Z}^{1} (0, G) = {Hom}_{Z} (0, G) = 0$ holds trivially (Ext via a projective resolution of the first variable).

[F8]

A homeomorphism induces isomorphisms on singular cohomology, contravariantly in the map (Singular cohomology is contravariantly functorial).

Proof

1.1F3algebra

Define $h : (R^{N})^{*} \to S^{N}$ by $h (x) = (\frac{2 x}{1 + ∥ x ∥_{2}^{2}}, \frac{∥ x ∥_{2}^{2} - 1}{∥ x ∥_{2}^{2} + 1}), h (\infty) = (0, \dots, 0, 1),$ writing $∥ x ∥_{2}^{2}$ for the squared Euclidean norm. For $x \in R^{N}$ the identity $(2 ∥ x ∥_{2})^{2} + (∥ x ∥_{2}^{2} - 1)^{2} = (∥ x ∥_{2}^{2} + 1)^{2}$ shows $∥ h (x) ∥_{2} = 1$ , so $h (x) \in S^{N}$ ; and $h (x) \neq (0, \dots, 0, 1)$ since $∥ x ∥_{2}^{2} - 1 = ∥ x ∥_{2}^{2} + 1$ is impossible. For $y = (y_{0}, \dots, y_{N}) \in S^{N}$ with $y_{N} \neq 1$ put $ψ (y) : = (y_{0}, \dots, y_{N - 1}) / (1 - y_{N}) \in R^{N}$ . If $r = ∥ ψ (y) ∥_{2}$ then $r^{2} = (1 - y_{N}^{2}) / (1 - y_{N})^{2} = (1 + y_{N}) / (1 - y_{N})$ , so $1 + r^{2} = 2 / (1 - y_{N})$ and $r^{2} - 1 = 2 y_{N} / (1 - y_{N})$ ; hence $h (ψ (y)) = y$ . Conversely, if $h (x) = (z, t)$ then $1 - t = 2 / (1 + ∥ x ∥_{2}^{2})$ and $ψ (h (x)) = 2 x / (1 + ∥ x ∥_{2}^{2}) \cdot (1 + ∥ x ∥_{2}^{2}) / 2 = x$ . So $h$ is a bijection with inverse $ψ$ on $S^{N} ∖ {(0, \dots, 0, 1)}$ and $\infty \mapsto (0, \dots, 0, 1)$ .

2.1F1F2F3F4F5step 1.1

The map $h$ is continuous. On $R^{N}$ its components $x \mapsto 2 x_{i} / (1 + ∥ x ∥_{2}^{2})$ and $x \mapsto (∥ x ∥_{2}^{2} - 1) / (∥ x ∥_{2}^{2} + 1)$ are quotients with denominator $1 + ∥ x ∥_{2}^{2} \geq 1$ never zero, hence continuous by [F3]; the components are continuous, so $h ∣_{R^{N}}$ is continuous into $R^{N + 1}$ and, since its image lies in $S^{N}$ , continuous into the subspace $S^{N}$ by [F5]. At $\infty$ , let $V \subseteq S^{N}$ be open with $(0, \dots, 0, 1) \in V$ . The subspace topology on $S^{N}$ is the metric topology of the maximum metric, so there is $r > 0$ such that every $y \in S^{N}$ with $∥ y - (0, \dots, 0, 1) ∥_{\infty} < r$ lies in $V$ ; choose $M \geq 1$ with $1 / M < r / 2$ and put $C : = {x \in R^{N} : ∥ x ∥_{2} \leq M}$ , which is closed and bounded, hence compact by [F4]. For $x \notin C$ one has $∥ x ∥_{2} > M \geq 1$ , so $∣ 2 x_{i} / (1 + ∥ x ∥_{2}^{2}) ∣ \leq 2 ∥ x ∥_{2} / ∥ x ∥_{2}^{2} = 2 / ∥ x ∥_{2} < 2 / M < r$ and $∣ (∥ x ∥_{2}^{2} - 1) / (∥ x ∥_{2}^{2} + 1) - 1 ∣ = 2 / (∥ x ∥_{2}^{2} + 1) \leq 2 / ∥ x ∥_{2}^{2} < 2 / M < r$ ; hence $h (x) \in V$ . Therefore $W : = (R^{N})^{*} ∖ C$ is open in $(R^{N})^{*}$ by [F1], contains $\infty$ , and satisfies $h (W) \subseteq V$ .

3.1F1F2F5F8step 1.1step 2.1

Hence $h$ is a homeomorphism. Indeed $R^{N}$ is locally compact and Hausdorff by [F2], so $(R^{N})^{*}$ is compact and Hausdorff by [F1], while $S^{N}$ is Hausdorff by [F2]; a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism by [F5]. Consequently $h^{*} : H^{q} (S^{N}; R) \to H^{q} ((R^{N})^{*}; R)$ is an isomorphism for every $q$ by [F8].

4.1F6F7step 3.1algebra

By the homeomorphism of step 3.1 it suffices to compute $H^{q} (S^{N}; R)$ . By [F7], $H_{q} (S^{N}; Z) = 0$ for $0 < q < N$ , while $H_{0} (S^{N}; Z) ≅ Z$ . Apply the universal coefficient sequence of [F6] in degree $q$ with $X = S^{N}$ and $G = R$ : $0 ⟶ {Ext}_{Z}^{1} (H_{q - 1} (S^{N}; Z), R) ⟶ H^{q} (S^{N}; R) ⟶ {Hom}_{Z} (H_{q} (S^{N}; Z), R) ⟶ 0.$ If $q \geq 2$ then $1 \leq q - 1 < q \leq N - 1$ , so both $H_{q - 1}$ and $H_{q}$ vanish and both outer terms are zero by [F7]. If $q = 1$ (so $N \geq 2$ ) then $H_{1} (S^{N}; Z) = 0$ and $H_{0} (S^{N}; Z) ≅ Z$ , so the right term is ${Hom}_{Z} (0, R) = 0$ and the left term is ${Ext}_{Z}^{1} (Z, R) = 0$ by [F7]. In both cases exactness forces $H^{q} (S^{N}; R) = 0$ , for $R = Z$ and for $R = F_{2}$ alike.

5.1F6step 3.1step 4.1∎

Combining steps 3.1 and 4.1, $H^{q} ((R^{N})^{*}; R) ≅ H^{q} (S^{N}; R) = 0$ for $0 < q < N$ , which is the claimed vanishing for the one-point compactification. The case $N = 2$ , $q = 1$ , both coefficient rings, and both orders of the two outer terms in the universal coefficient sequence are covered by the case distinction of step 4.1; the empty coefficient ring and negative $q$ are excluded by the hypotheses, and no further choice beyond AC, used through [F6] and [F7], enters.

Depends on

Used by

Top normal classes vanish for Euclidean embeddings Corollary

Dependency tree · two levels

108 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

John W. Milnor and James D. Stasheff, Characteristic Classes (standard reference, not scraped)