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The deck group of a connected covering acts by a covering-space action

Statement

Let $p : E \to B$ be a covering map with connected total space $E$ . Then the deck group $Deck (p)$ acts on $E$ by a covering-space action: every $e \in E$ has an open neighbourhood $U$ with $h U \cap U = \emptyset$ for every nonidentity $h \in Deck (p)$ .

Facts & Assumptions

Given: A covering map $p : E \to B$ with connected total space $E$ , a point $e \in E$ , and the deck group $Deck (p)$ acting on $E$ by evaluation.

[F1]

A deck transformation is an isomorphism $h : E \to E$ over $B$ , and the deck transformations form the group $Deck (p)$ acting on $E$ by evaluation (Deck transformations and the deck-transformation group of a covering).

[F2]

For a covering with connected total space, two deck transformations agreeing at one point are equal; consequently the deck group acts freely on the total space (On a connected covering space, a deck transformation is determined by one point and the deck action is free).

[F3]

A covering map $p : E \to B$ has for every $b \in B$ an evenly covered open neighbourhood $W$ : the preimage $p^{- 1} (W)$ is a disjoint union of open sheets $V_{j}$ , and each restriction $p ∣_{V_{j}} : V_{j} \to W$ is a homeomorphism (Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings).

[F4]

An action of a group on a space $E$ by homeomorphisms is a covering-space action when every $e \in E$ has an open neighbourhood $U$ with $g U \cap U = \emptyset$ for every nonidentity $g$ (Covering-space actions by disjoint translates of neighbourhoods).

Proof

technique · direct

1.1F3construct

Put $b : = p (e)$ . By [F3] choose an evenly covered open neighbourhood $W$ of $b$ and let $U$ be the sheet of $p^{- 1} (W)$ containing $e$ . Then $U$ is an open neighbourhood of $e$ and $p ∣_{U} : U \to W$ is a homeomorphism, hence injective.

2.1F1F2step 1.1

Suppose $h U \cap U$ is nonempty. Then some $u \in U$ has $h (u) \in U$ , and $p (h (u)) = p (u)$ by [F1]. Injectivity of $p ∣_{U}$ gives $h (u) = u$ . By [F2], a deck transformation fixing any point is the identity. Thus $h U \cap U = \emptyset$ for every nonidentity $h$ . No connectedness of the chosen sheet or of the evenly covered neighborhood is needed.

3.1F1F4step 2.1∎

Since $e$ was arbitrary and every deck transformation is a homeomorphism, [F4] proves that the deck group acts by a covering-space action.

Depends on

Used by

Dependency tree · two levels

13 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

Danny Calegari, Foliations and the Geometry of 3-Manifolds (Oxford Mathematical Monographs) (standard reference, not scraped)
Eckhard Meinrenken, Lie Groupoids and Lie Algebroids, lecture notes (University of Toronto MAT1341, Fall 2017) (standard reference, not scraped)