# 8.06 Galois correspondence, 2

## Video

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

## 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.