Deck Transformations and Normal Covers

Deck Transformations

Deck Transformation

A deck transformation of a covering space is a covering space automorphism . That is, the following diagram commutes: In other words, . The set of all deck transformations forms a group under composition, denoted or .

Remark

For a path-connected covering space , a deck transformation is uniquely determined by its action on a single point. (ref. proposition) If for two deck transformations , then .

e.g.

• For the covering given by , the deck transformations are homeomorphisms such that . This implies , which holds if and only if for some integer . The group of deck transformations is therefore isomorphic to the integers under addition: .
• Consider . Then , where .
• Take any , and let . i.e., is just copies of . The deck transformations are simply permutations of the copies, so , the symmetric group on letters.

Proposition

If is PC, LPC, and SLSC. Fix some and , then .

Proof By the lifting criterion, for any , there exists a unique lifting map such that . It is clearly invertible with inverse sending to . Note that is exactly all the homotopy classes of loops at (ref. proof of the fundamental theorem of covering spaces), so .

Normal Covers

A key question is when the deck transformation group is “large enough” to connect any two points in the same fiber. That is, for , when does there exist a deck transformation with ?

Normal Covering Space

is normal if for any with , there exists a deck transformation such that .

Topological Interpretation of Normal Covering Spaces

This means the covering space must “look the same” from the perspective of any two points in the fiber.

e.g. Consider the following covering space of : It is not a normal covering space. Observe that at , is a loop, but at , is just a simple path, so (ref. remark), they are different subgroups of . Hence, by the lifting criterion, there is no deck transformation from the point to the point .

Proposition

Suppose is PC and LPC. Let be a PC covering space (automatically LPC) with basepoint . Let and . Then the following holds:

• is normal iff is a normal subgroup.
• , where is the normaliser of in . In particular, if is normal, then .

Remark

For path-connected, to check if is normal, it suffices to fix some and , and check if for any , there exists a deck transformation with .

Proof

e.g. We can use this to compute the fundamental group of . The antipodal map is a 2-sheeted universal cover. Hence (ref. example). In fact, for all , because is still the double universal cover.