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A finite coordinate-bump map embeds a compact manifold in some Euclidean space

Statement

Let $M^{n}$ be a compact smooth manifold. Then there are finitely many coordinate charts $(U_{i}, x_{i})$ , open sets $V_{i} \subseteq U_{i}$ covering $M$ , and smooth bump functions $ϕ_{i} : M \to [0, 1]$ supported in $U_{i}$ and equal to $1$ on $V_{i}$ such that $F : = (ϕ_{1}, ϕ_{1} x_{1}^{1}, \dots, ϕ_{1} x_{1}^{n}, \dots, ϕ_{m}, ϕ_{m} x_{m}^{1}, \dots, ϕ_{m} x_{m}^{n}) : M \to R^{m (n + 1)}$ is a smooth embedding.

Facts & Assumptions

Given: A compact smooth $n$ -manifold $M$ .

[L1]

For every point of a smooth manifold, the construction in A chart bump at a point with prescribed support, Proof 2.1–5.1, gives a chart $(U, x)$ , an open neighbourhood $V ⋐ U$ of the point, and a smooth bump supported in $U$ and equal to $1$ on $V$ .

[L2]

An injective immersion from a compact manifold is an embedding (An injective immersion from a compact manifold is an embedding).

Proof

technique · direct

1.1L1given

Let $T$ be the family of all triples $((U, x), V, ϕ)$ where $(U, x)$ is a coordinate chart, $V ⋐ U$ is open, and $ϕ : M \to [0, 1]$ is a smooth bump supported in $U$ and equal to $1$ on $V$ . By [L1], the family of their open sets $V$ covers $M$ . Compactness gives finitely many such sets $V_{i}$ covering $M$ ; retain a witnessing chart $(U_{i}, x_{i})$ and bump $ϕ_{i}$ for each of these finitely many sets. This uses only finite choices after the subcover is found.

2.1step 1.1construct

Define the coordinate-bump blocks $B_{i} (q) : = (ϕ_{i} (q), ϕ_{i} (q) x_{i}^{1} (q), \dots, ϕ_{i} (q) x_{i}^{n} (q)) \in R^{n + 1},$ and let $F : = (B_{1}, \dots, B_{m})$ . The map is smooth because each block is smooth on $U_{i}$ and vanishes off $U_{i}$ .

2.2step 1.1algebra

To prove immersion, fix $p \in M$ and choose $i$ with $p \in V_{i}$ . Because $ϕ_{i}$ is identically $1$ on the open set $V_{i}$ , its differential vanishes there. Thus on $V_{i}$ the last $n$ coordinates of $B_{i}$ are just the chart coordinates $x_{i}^{1}, \dots, x_{i}^{n}$ , whose differentials form an isomorphism $T_{p} M \to R^{n}$ . Hence $d F_{p}$ is injective.

3.1step 1.1step 2.1algebra

To prove injectivity, suppose $F (p) = F (q)$ . Choose $i$ with $p \in V_{i}$ . Then $ϕ_{i} (p) = 1$ , hence $ϕ_{i} (q) = 1$ as well because the first coordinates of $B_{i} (p)$ and $B_{i} (q)$ agree. Therefore $q \in U_{i}$ , and the equalities $ϕ_{i} (p) x_{i}^{a} (p) = ϕ_{i} (q) x_{i}^{a} (q)$ give $x_{i}^{a} (p) = x_{i}^{a} (q)$ for every $a$ . Since $x_{i}$ is injective on $U_{i}$ , one gets $p = q$ .

4.1L2step 3.1step 2.2∎

Steps 3.1 and 2.2 show that $F$ is an injective immersion. By [L2], $F$ is a smooth embedding.

Depends on

Used by

A coordinate-bump embedding of the circle in Euclidean space Example

Dependency tree · two levels

19 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

Marco Gualtieri, Topology I: Smooth Manifolds, Part 11 (standard reference, not scraped)
John M. Lee, Introduction to Smooth Manifolds, 2nd ed., Chapter 6 (standard reference, not scraped)