Quail Entry: Deck and Cov. Sp. Action

this happened december 17th

Deck Transformations

rep. by commutative triangles (only one covering space involved!)

a deck transformation D \colon X' \to X' is a lift of the projection X' \to X.

via. lifting property if X' is path conn. then D determined by a point in the sense that if D,D' agree on a point then they are equal.

normal means essentially transitive, any two points in a fiber can be mapped to each other via a deck transfomration.

A covering space is normal, ‘normal’ is not applied to deck transformations.

Reason for ‘normal’ : X' \to X a normal covering space iff \pi_1(X') is normal subgroup in \pi_1(X).

hypotheses: to prove need lifting criterion which requires X' path connected and locally path conn. (but by general properties of path conn. cov. space I guess loc. path. is necessarily inherited by X) so require X loc. path connected and restrict to path connected cov. spaces to apply result.

regard pf: let y,y' be points in the fiber of x \in X want to know when there is comm. triangle with corners (X',y),\ (X',y'),\ (X,x) by lift crit this happens when \pi_1 X',y = \pi_1 X',y' in \pi_1 X,x.  Upstairs these are related by conjugating by a path y \to y'; downstairs this is conjugation by a loop.

H is image of \pi_1 X' downstairs.  If f \in N(H), then above implies \exists deck tranf. taking a pt in fiber y to f'(1) where f' is a lift starting at $y$.  In fact for G(X) gp of deck tranf

And isomorphism: G(X) = N(H)/H

pf: uses image of \pi_1 X' are just loops downstairs that lift to loops upstairs.

Covering Space Actions

require: G acts on Y and \forall p \in Y has nbd U s.t. gU \cap U \ne \emptyset iff g = 1.

In particular the collection \cup_{g \in G} gU are disjoint and Y \to Y/G becomes a covering space.

1. Its always a normal covering space.

Y path conn. implies deck transf. det. by image of point

2. G = G(Y \to Y/G) = Deck(Y \to Y/G)

To use iso G(X) = N(H)/H need Y to be path. conn and loc. path conn to get

3. G = G( Y \to Y/G) = \pi_1(Y/G)/H

where H is as above and using normality here.

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