# 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.