# 8.03 Normal covering spaces

## Video

Below the video you will find accompanying notes and some pre-class questions.

- Previous video:
**8.02 Covering transformations**. - Next video:
**8.04 Deck group**. - Index of all lectures.

## Notes

### Example

*(0.25)* Consider the covering space \(p_n\colon S^1\to S^1\),
\(p_n(z)=z^n\). What is the deck group of this covering space? Deck
transformations are covering isomorphisms \(F\colon(S^1,p_n)\to
(S^1,p_n)\), satisfying \(p\circ F=p\). In other words,
\[(F(z))^n=z^n\]
so \(F(z)=\mu z\) for some \(\mu\) with \(\mu^n=1\). The formula
\(F(z)=\mu z\) defines a deck transformation for each \(n\)th root
of unity, and any deck transformation must have this form. This
tells us that the deck group is \[Deck(S^1,p_n)=\mathbf{Z}/n.\]

*(2.15)* If we take \(\pi_1(S^1,1)=\mathbf{Z}\) and push it forward
along \(p_n\) then we get the subgroup
\[(p_n)_*\pi_1(S^1,1)=n\mathbf{Z}\subset\pi_1(S^1,1)=\mathbf{Z}.\]
Note that, in this example,
\[Deck(S^1,p_n)\cong\mathbf{Z}/(n\mathbf{Z})=\pi_1(X,x)/p_*\pi_1(Y,y)\]
where \(p\colon Y\to X\) is the covering space.

*(3.36)* Consider the covering space \(p\colon\mathbf{R}\to S^1\),
\(p(x)=e^{2\pi ix}\). What is the deck group? It consists of maps
\(F\colon\mathbf{R}\to\mathbf{R}\) such that \(e^{2\pi
iF(x)}=e^{2\pi ix}\), which implies \(F(x)=x+n\). The deck group is
therefore \(Deck(\mathbf{R},p)=\mathbf{Z}\). The pushforward
\(p_*\pi_1(\mathbf{R})\) is trivial as \(\mathbf{R}\) is
simply-connected, so the quotient
\(\pi_1(S^1)/p_*\pi_1(\mathbf{R})\) is isomorphic to \(\mathbf{Z}\),
which is the deck group.

### Normal subgroups

*(5.40)* Is this a general phenomenon? No: the subgroup
\(p_*\pi_1(Y,y)\subset\pi_1(X,x)\) might not be a normal subgroup, in
which case the quotient \(\pi_1(X,x)/p_*\pi_1(Y,y)\) doesn't even make
sense. However, in the case where \(p_*\pi_1(Y,y)\) is a normal
subgroup of \(\pi_1(X,x)\), it will turn out to be true that
\(Deck(Y,p)=\pi_1(X,x)/p_*\pi_1(Y,y)\). We single out these covers
with a definition:

*(6.56)* We say that \(p\colon Y\to X\) is a *normal* (or *regular*,
or *Galois*) cover if \(p_*\pi_1(Y,y)\) is a normal subgroup of
\(\pi_1(X,p(y))\).

Normal covering spaces are maximally symmetric, as we will see from the following lemma.

*(7.34)* A path-connected covering space \(p\colon Y\to X\) is
normal if and only if \(Deck(Y,p)\) acts transitively on
\(p^{-1}(x)\) for some \(x\in X\).

*(8.45)* Note that \(Deck(Y,p)\) always acts on \(p^{-1}(x)\): if
\(F\in Deck(Y,p)\) and \(y\in p^{-1}(x)\) then \(p(F(y))=p(y)=x\), so
\(F(y)\in p^{-1}(x)\).

*(9.56)* Moreover, \(F\in Deck(Y,p)\) is determined uniquely by its
value on a particular point \(y\in p^{-1}(x)\): 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 \(y\in p^{-1}(x)\), and hence would be equal.

*(12.29)* In our earlier examples, the deck group acts transitively.
For \(p_n\colon S^1\to S^1\), the \(n\)th roots of unity act
transitively by rotation on \(p_n^{-1}(1)\), which it itself the set
of \(n\)th roots of unity. For \(p\colon\mathbf{R}\to S^1\), the group
\(\mathbf{Z}\) acts transitively by translation on
\(p^{-1}(1)=\mathbf{Z}\).

*(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 \(\beta\in\pi_1(X,x)\). We want to show that
\[\beta p_*\pi_1(Y,y)\beta^{-1}=p_*\pi_1(Y,y).\] We know from an
exercise in an earlier class that \[\beta
p_*\pi_1(Y,y)\beta^{-1}=p_*\pi_1(Y,\sigma_\beta(y)),\] where
\(\sigma_\beta\colon p^{-1}(x)\to p^{-1}(x)\) is the monodromy
around \(\beta\). Therefore it suffices to prove that
\(p_*\pi_1(Y,y)=p_*\pi_1(Y,\sigma_\beta(y))\).

*(19.34)* By assumption, \(Deck(Y,p)\) acts transitively, so there
exists a deck transformation \(F\colon Y\to Y\) such that
\(F(y)=\sigma_\beta(y)\) and deck transformations are
homeomorphisms, so
[(F_{*}:π_{1}(Y,y)→π_{1}(Y,σ_{β}(y))\] is an
isomorphism. We also know that \(p\circ F=p\), so \(p_*\circ
F_*=p_*\). Putting all this together gives:

Since this holds for all \(\beta\in\pi_1(X,x)\), this shows that \(p_*\pi_1(Y,y)\subset\pi_1(X,x)\) is normal.

*(21.50)* Conversely, suppose that \(p_*\pi_1(Y,y)\) is a normal
subgroup. Let \(y'\) be another point in \(p^{-1}(x)\) and let
\(\tilde{\beta}\) be a path in \(Y\) from \(y\) to \(y'\). Then
\(y'=\sigma_\beta(y)\) and \[p_*\pi_1(Y,y')=\beta
p_*\pi_1(Y,y)\beta^{-1}.\] Since \(p_*\pi_1(Y,y)\) is normal, we get
\(p_*\pi_1(Y,y')=p_*\pi_1(Y,y)\), so the existence theorem for
covering isomorphisms implies that there exists a covering
transformation \(F\colon Y\to Y\) 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 \(p_*\pi_1(Y,y)\subset\pi_1(X,x)\)? Can you see it is not normal without appealing to the lemma we used?

## Navigation

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**8.02 Covering transformations**. - Next video:
**8.04 Deck group**. - Index of all lectures.