(0.25) Consider the covering space pn:S1→S1,
pn(z)=zn. What is the deck group of this covering space? Deck
transformations are covering isomorphisms F:(S1,pn)→(S1,pn), satisfying pn∘F=pn. In other words,
(F(z))n=zn
so F(z)=μz for some μ with
μn=1. The formula F(z)=μz defines a deck
transformation for each nth root of unity, and any deck
transformation must have this form. This tells us that the deck
group is Deck(S1,pn)=Z/n.
(2.15) If we take π1(S1,1)=Z and push it
forward along pn then we get the subgroup
(pn)∗π1(S1,1)=nZ⊂π1(S1,1)=Z.
Note that, in this example,
Deck(S1,pn)≅Z/(nZ)=π1(X,x)/p∗π1(Y,y)
where p:Y→X is the covering space.
(3.36) Consider the covering space p:R→S1,
p(x)=e2πix. What is the deck group? It consists of maps
F:R→R such that e2πiF(x)=e2πix, which implies F(x)=x+n. The deck group is
therefore Deck(R,p)=Z. The pushforward
p∗π1(R) is trivial as R is
simply-connected, so the quotient
π1(S1)/p∗π1(R) is isomorphic to Z,
which is the deck group.
Normal subgroups
(5.40) Is this a general phenomenon? No: the subgroup
p∗π1(Y,y)⊂π1(X,x) might not be a normal subgroup, in
which case the quotient π1(X,x)/p∗π1(Y,y) doesn't even make
sense. However, in the case where p∗π1(Y,y) is a normal
subgroup of π1(X,x), it will turn out to be true that
Deck(Y,p)=π1(X,x)/p∗π1(Y,y). We single out these covers
with a definition:
(6.56) We say that p:Y→X is a normal (or
regular, or Galois) cover if p∗π1(Y,y) is a
normal subgroup of π1(X,p(y)).
Normal covering spaces are maximally symmetric, as we will see from
the following lemma.
(7.34) A path-connected covering space p:Y→X is
normal if and only if Deck(Y,p) acts transitively on
p−1(x) for some x∈X.
(8.45) Note that Deck(Y,p) always acts on p−1(x): if
F∈Deck(Y,p) and y∈p−1(x) then p(F(y))=p(y)=x, so
F(y)∈p−1(x).
(9.56) Moreover, F∈Deck(Y,p) is determined uniquely by its
value on a particular point y∈p−1(x): indeed, any covering
transformation is determined uniquely by its value at a
point. Therefore, if the action of Deck(Y,p) is transitive on
p−1(x), then Deck(Y,p) is as big as it could possibly be:
if it were any bigger, two of its elements would necessarily agree at
some point y∈p−1(x), and hence would be equal.
(12.29) In our earlier examples, the deck group acts transitively.
For pn:S1→S1, the nth roots of unity act
transitively by rotation on p−1n(1), which it itself the set
of nth roots of unity. For p:R→S1, the group
Z acts transitively by translation on
p−1(1)=Z.
(14.03) Below is a covering space of the figure 8 which is not
normal.
Let x be the cross-point of the figure 8. It has three pre-images,
but only one of these (say y) is the endpoint of a blue
loop. Since deck transformations are symmetries of the covering space,
they must preserve the unique blue loop, and hence fix this point
y. If a deck transformation fixes a point then it is equal to the
identity, by uniqueness. Therefore the deck group in this case is
trivial.
Proof of lemma
(16.57) Assume that Deck(Y,p) acts transitively on
p−1(x). Pick β∈π1(X,x). We want to show that
βp∗π1(Y,y)β−1=p∗π1(Y,y).
We know from an
exercise in an earlier class that βp∗π1(Y,y)β−1=p∗π1(Y,σβ(y)),
where
σβ:p−1(x)→p−1(x) is the monodromy
around β. Therefore it suffices to prove that
p∗π1(Y,y)=p∗π1(Y,σβ(y)).
(19.34) By assumption, Deck(Y,p) acts transitively, so
there exists a deck transformation F:Y→Y such that
F(y)=σβ(y) and deck transformations are
homeomorphisms, so
(F∗:π1(Y,y)→π1(Y,σβ(y))
is an
isomorphism. We also know that p∘F=p, so p∗∘F∗=p∗. Putting all this together gives:
βp∗π1(Y,y)β−1=p∗π1(Y,σβ(y))=p∗(F∗(π1(Y,y)))=p∗π1(Y,y).
Since this holds for all β∈π1(X,x), this shows that
p∗π1(Y,y)⊂π1(X,x) is normal.
(21.50) Conversely, suppose that p∗π1(Y,y) is a
normal subgroup. Let y′ be another point in p−1(x) and
let ˜β be a path in Y from y to y′. Then
y′=σβ(y) and p∗π1(Y,y′)=βp∗π1(Y,y)β−1.
Since p∗π1(Y,y) is normal, we get
p∗π1(Y,y′)=p∗π1(Y,y), so the existence theorem for
covering isomorphisms implies that there exists a covering
transformation F:Y→Y such that F(y)=y′. Therefore
Deck(Y,p) acts transitively on p−1(x).
Pre-class questions
Consider the non-normal covering space example. What is the subgroup
p∗π1(Y,y)⊂π1(X,x)? Can you see it is not normal
without appealing to the lemma we used?
