# 8.06 Galois correspondence, 2

## Video

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

- Previous video:
**8.05 Galois correspondence, 1**. - Index of all lectures.

## Notes

*(0.00)* In this section, we will put together everything we have seen
about covering spaces to get the *Galois correspondence* between
covering spaces and subgroups of the fundamental group.

### Covering spaces give subgroups of \(\pi_1(X,x)\)

*(0.13)* A covering space \(p\colon Y\to X\) together with a point
\(y\in p^{-1}(x)\) gives a subgroup \(p_*\pi_1(Y,y)\subset
\pi_1(X,x)\). We have also just seen that if there is a
simpy-connected (*universal*) cover then any subgroup arises this
way. This assignment \((Y,y)\mapsto p_*\pi_1(Y,y)\) is called the
*Galois correspondence* between (based) covering spaces and subgroups
of \(\pi_1(X,x)\). There is some more (functorial) structure, which
allows us to read properties of the covering space off from properties
of the subgroup, which we will now explain.

*(1.18)* We know that \(\beta
p_*\pi_1(Y,y)\beta^{-1}=p_*\pi_1(Y,\sigma_\beta(y))\). This tells us
conjugates of subgroups are represented by the *same* covering space
but with a *different* basepoint. More precisely, if \(y,y'\in Y\)
then a path between \(y\) and \(y'\) projects to get a loop \(\beta\)
in \(X\) such that \(y'=\sigma_\beta(y)\) so a single covering space
\(Y\) defines a *conjugacy class* of subgroups of \(\pi_1(X,x)\), and
any subgroup in that conjugacy class arises as \(p_*\pi_1(Y,y)\) by
picking \(y\) suitably.

*(3.27)* We know that there is a covering transformation \(F\colon
Y_1\to Y_2\) with \(F(y_1)=y_2\) if and only if
\[(p_1)_*\pi_1(Y_1,y_1)\subset (p_2)_*\pi_1(Y_2,y_2).\]

Let's draw a picture to represent this fact as follows. For each
(based) covering space \((Y,y)\) of \(X\), we draw a dot labelled
\((Y,y)\). For each covering transformation between covering spaces
\(F\colon Y_1\to Y_2\), \(F(y_1)=y_2\), we draw an arrow from the dot
\((Y_1,y_1)\) to the dot \((Y_2,y_2)\). Now if we replace each dot
\((Y,y)\) by the subgroup \(p_*\pi_1(Y,y)\), we can replace each
covering transformation by an inclusion. This means that the Galois
correspondence is a *functor* from the category of based covering
spaces (with covering transformations) to the category of subgroups
(with inclusions). This is a lot like the Galois theory of field
extensions, in which field extensions correspond to subgroups of the
Galois group.

*(6.20)* Finally, we also saw that \(Deck(Y,p)=N_H/H\), where
\(H=p_*\pi_1(Y,y)\) and \(N_H\) denotes the normaliser of
\(H\subset\pi_1(X,x)\). This is again very reminiscent of the Galois
theory of field extensions.

### Examples

*(7.30)* Take \(X=S^1\) and \(x=1\in S^1\). We have
\(\pi_1(X,x)=\mathbf{Z}\). The subgroups of \(\mathbf{Z}\) are:

- the trivial group \(\{0\}\), which corresponds to the simply-connected covering space \(p\colon\mathbf{R}\to S^1\).
- for any integer \(n\), the subgroup \(n\mathbf{Z}\), which corresponds to the covering space \(p_n\colon S^1\to S^1\), \(p_n(z)=z^n\).

Some of these subgroups are nested, for example \(4\mathbf{Z}\subset 2\mathbf{Z}\), which means that the corresponding covering spaces (say \(Y_4\) and \(Y_2\)) are related by covering transformations \(Y_4\to Y_2\). Inside all of these subgroups we have the trivial group, so the covering space \(\mathbf{R}\) covers all of them.

*(11.32)* Take \(X=S^1\vee S^1\) and \(x\) the cross point. We have
\(\pi_1(X,x)=\mathbf{Z}\star\mathbf{Z}\).

- Corresponding to the trivial subgroup, we have the universal (simply-connected) cover, which is the infinite 4-valent graph we saw in class.
- Corresponding to the subgroup \(\langle a\rangle\) we need to have a covering space with \(\pi_1\cong\mathbf{Z}=\langle a\rangle\) and there needs to be a loop which projects down to \(a\). So we start with one vertex, add a loop (projecting to \(a\)); this vertex needs to have valency 4, so we add two more edges coming out of it (each projecting to \(b\) or \(b^{-1}\)). We are not allowed any more loops, so we now cap these edges off with two contractible infinite 4-valent graphs. This gives a covering space and we can see that \(p_*\pi_1(Y,y)=\langle a\rangle\) by construction.
- Take instead the subgroup \(norm(a)\)
*normally generated by*\(a\) (i.e. generated by \(a\) and all its conjugates). The corresponding covering space is then a normal subgroup and its deck group should be \(\langle a,b|a\rangle\cong\mathbf{Z}\). Therefore, we're looking for a covering space with an action of \(\mathbf{Z}\) and a loop which projects to \(a\). We form such a covering space by putting vertices at each integer point in \(\mathbf{R}\), projecting the intervals between them to \(b\), and then attaching loops (projecting to \(a\)) at each integer point. This clearly has an action of \(\mathbf{Z}\). It is also easy to see that \(p_*\pi_1(Y,y)\) is the group generated by \(b^nab^{-n}\), \(n\in\mathbf{Z}\).

*(17.24)* There is an inclusion \(\langle a\rangle\to
norm(a)\). Under the Galois correspondence, this gives a covering
transformation from the cover \(Y_1\) for \(\langle a\) to the cover
\(Y_2\) for \(norm(a)\). This covering transformation is illustrated
in the video.

Of course, there are many other subgroups of \(\mathbf{Z}\star\mathbf{Z}\), and, correspondingly, many other covering spaces.

## Pre-class questions

## Navigation

- Previous video:
**8.05 Galois correspondence, 1**. - Index of all lectures.