Deck transformation

A covering is regular iff its deck transformation group acts transitively on fibres

Let 𝑝 :˜𝑋 →𝑋 be a connected and locally path-connected covering and Γ =Aut𝖢𝗈𝗏𝑋⁡(𝑝) be its deck transformation group. Then 𝑝 is a regular covering iff Γ acts on one (and therefore every) fibre 𝑝−1{𝑥0} transitively. ¹ homotopy Equivalently, the orbit of each ˜𝑥0 is a whole fibre.

Proof

Let 𝑝 :˜𝑋 →𝑋 be a regular covering and 𝑥0 ∈𝑋. Let ˜𝑥0,˜𝑥′0 ∈𝑝−1{𝑥0}, and let 𝐻 and 𝐻′ denote the characteristic subgroups with respect to ˜𝑥0 and ˜𝑥′0 respectively. Since 𝑝 is regular, 𝐻 =𝐻′, and by equivalence of coverings criterion the coverings with either base point are equivalent. Hence there exists 𝛾 ∈Γ such that 𝛾(˜𝑥0) =˜𝑥′0.

For the converse, assume Γ acts transitively on 𝑝−1{𝑥0}, i.e. the following diagram commutes for any ˜𝑥0,˜𝑥′0 ∈𝑝−1{𝑥0}:

Applying the Fundamental group functor 𝜋1 to this diagram it is clear that the characteristic subgroups is basepoint-invariant. Therefore 𝑝 is a regular covering.

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Footnotes

1. 2010, Algebraische Topologie, ¶2.3.36, p. 96 ↩