Definition Let p:E \rightarrow X be a covering. A
deck transformation is a homeomorphism d:E \rightarrow E, such that the diagram of continuous maps commutes. Together with the composition of maps, the set of deck transformation forms a
group \operatorname{Deck}(p), which is the same as \operatorname{Aut}(p). Now suppose p:C \to X is a covering map and C (and therefore also X) is connected and locally path connected. The action of \operatorname{Aut}(p) on each fiber is
free. If this action is
transitive on some fiber, then it is transitive on all fibers, and we call the cover
regular (or
normal or
Galois). Every such regular cover is a
principal, where G = \operatorname{Aut}(p) is considered as a discrete topological group. Every universal cover p:D \to X is regular, with deck transformation group being isomorphic to the
fundamental group Examples • Let q : S^1 \to S^1 be the covering q(z)=z^{n} for some n \in \mathbb{N} , then the map d_k:S^1 \rightarrow S^1 : z \mapsto z \, e^{2\pi ik/n} for k \in \mathbb{Z} is a deck transformation and \operatorname{Deck}(q)\cong \mathbb{Z}/ n\mathbb{Z}. • Let r : \mathbb{R} \to S^1 be the covering r(t)=(\cos(2 \pi t), \sin(2 \pi t)), then the map d_k:\mathbb{R} \rightarrow \mathbb{R} : t \mapsto t + k for k \in \mathbb{Z} is a deck transformation and \operatorname{Deck}(r)\cong \mathbb{Z}. • As another important example, consider \Complex the
complex plane and \Complex^{\times} the complex plane minus the origin. Then the map p: \Complex^{\times} \to \Complex^{\times} with p(z) = z^{n} is a regular cover. The deck transformations are multiplications with n-th
roots of unity and the deck transformation group is therefore isomorphic to the
cyclic group \Z/n\Z. Likewise, the map \exp : \Complex \to \Complex^{\times} with \exp(z) = e^{z} is the universal cover.
Properties Let X be a path-connected space and p:E \rightarrow X be a connected covering. Since a deck transformation d:E \rightarrow E is
bijective, it permutes the elements of a fiber p^{-1}(x) with x \in X and is uniquely determined by where it sends a single point. In particular, only the identity map fixes a point in the fiber. Because of this property every deck transformation defines a
group action on E, i.e. let U \subset X be an open neighborhood of a x \in X and \tilde U \subset E an open neighborhood of an e \in p^{-1}(x), then \operatorname{Deck}(p) \times E \rightarrow E: (d,\tilde U)\mapsto d(\tilde U) is a
group action.
Normal coverings Definition A covering p:E \rightarrow X is called normal, if \operatorname{Deck}(p) \backslash E \cong X. This means, that for every x \in X and any two e_0,e_1 \in p^{-1}(x) there exists a deck transformation d:E \rightarrow E, such that d(e_0)=e_1.
Properties Let X be a path-connected space and p:E \rightarrow X be a connected covering. Let H=p_{\#}(\pi_1(E)) be a
subgroup of \pi_1(X), then p is a normal covering iff H is a
normal subgroup of \pi_1(X). If p:E \rightarrow X is a normal covering and H=p_{\#}(\pi_1(E)), then \operatorname{Deck}(p) \cong \pi_1(X)/H. If p:E \rightarrow X is a path-connected covering and H=p_{\#}(\pi_1(E)), then \operatorname{Deck}(p) \cong N(H)/H, whereby N(H) is the
normaliser of H. Let E be a topological space. A group \Gamma acts
discontinuously on E, if every e \in E has an open neighborhood V \subset E with V \neq \empty, such that for every d_1, d_2 \in \Gamma with d_1 V \cap d_2 V \neq \empty one has d_1 = d_2. If a group \Gamma acts discontinuously on a topological space E, then the
quotient map q: E \rightarrow \Gamma \backslash E with q(e)=\Gamma e is a normal covering. Hereby \Gamma \backslash E = \{\Gamma e: e \in E\} is the
quotient space and \Gamma e = \{\gamma(e):\gamma \in \Gamma\} is the
orbit of the group action.
Examples • The covering q : S^1 \to S^1 with q(z)=z^{n} is a normal coverings for every n \in \mathbb{N}. • Every simply connected covering is a normal covering.
Calculation Let \Gamma be a group, which acts discontinuously on a topological space E and let q: E \rightarrow \Gamma \backslash E be the normal covering. • If E is path-connected, then \operatorname{Deck}(q) \cong \Gamma. • If E is simply connected, then \operatorname{Deck}(q)\cong \pi_1(\Gamma \backslash E).
Examples • Let n \in \mathbb{N}. The antipodal map g:S^n \rightarrow S^n with g(x)=-x generates, together with the composition of maps, a group D(g) \cong \mathbb{Z/2Z} and induces a group action D(g) \times S^n \rightarrow S^n, (g,x)\mapsto g(x), which acts discontinuously on S^n. Because of \mathbb{Z_2} \backslash S^n \cong \mathbb{R}P^n it follows, that the quotient map q : S^n \rightarrow \mathbb{Z_2}\backslash S^n \cong \mathbb{R}P^n is a normal covering and for n > 1 a universal covering, hence \operatorname{Deck}(q)\cong \mathbb{Z/2Z}\cong \pi_1({\mathbb{R}P^n}) for n > 1. • Let \mathrm{SO}(3) be the
special orthogonal group, then the map f : \mathrm{SU}(2) \rightarrow \mathrm{SO}(3) \cong \mathbb{Z_2} \backslash \mathrm{SU}(2) is a normal covering and because of \mathrm{SU}(2) \cong S^3, it is the universal covering, hence \operatorname{Deck}(f) \cong \mathbb{Z/2Z} \cong \pi_1(\mathrm{SO}(3)). • With the group action (z_1,z_2)*(x,y)=(z_1+(-1)^{z_2}x,z_2+y) of \mathbb{Z^2} on \mathbb{R^2}, whereby (\mathbb{Z^2},*) is the
semidirect product \mathbb{Z} \rtimes \mathbb{Z} , one gets the universal covering f: \mathbb{R^2} \rightarrow (\mathbb{Z} \rtimes \mathbb{Z}) \backslash \mathbb{R^2} \cong K of the
klein bottle K, hence \operatorname{Deck}(f) \cong \mathbb{Z} \rtimes \mathbb{Z} \cong \pi_1(K). • Let T = S^1 \times S^1 be the
torus which is embedded in the \mathbb{C^2}. Then one gets a homeomorphism \alpha: T \rightarrow T: (e^{ix},e^{iy}) \mapsto (e^{i(x+\pi)},e^{-iy}), which induces a discontinuous group action G_{\alpha} \times T \rightarrow T, whereby G_{\alpha} \cong \mathbb{Z/2Z}. It follows, that the map f: T \rightarrow G_{\alpha} \backslash T \cong K is a normal covering of the klein bottle, hence \operatorname{Deck}(f) \cong \mathbb{Z/2Z}. • Let S^3 be embedded in the \mathbb{C^2}. Since the group action S^3 \times \mathbb{Z/pZ} \rightarrow S^3: ((z_1,z_2),[k]) \mapsto (e^{2 \pi i k/p}z_1,e^{2 \pi i k q/p}z_2) is discontinuously, whereby p,q \in \mathbb{N} are
coprime, the map f:S^3 \rightarrow \mathbb{Z_p} \backslash S^3 =: L_{p,q} is the universal covering of the
lens space L_{p,q}, hence \operatorname{Deck}(f) \cong \mathbb{Z/pZ} \cong \pi_1(L_{p,q}). == Galois correspondence ==