Galois Correspondence of Covering Spaces

We interrupt our regularly scheduled series on the fundamental group of the circle to bring you this beautiful connection between Galois theory and the theory of covering spaces. In this post, I will assume the reader has some familiarity with both Galois theory and the material on covering spaces found in section 1.3 of Hatcher's Algebraic Topology.


One of the things about math that thrills me the most is that occasionally a result/idea known in one area will reappear - perhaps unexpectedly! - in a different, seemingly unrelated area. So when I recently learned that the Galois correspondence between groups and fields has almost an exact analogue in the correspondence between fundamental groups and covering spaces, I was tickled pink! Interestingly enough, Hatcher devotes only a single sentence to this fact in the introduction to section 1.3 of his book. In my opinion, the resemblance is far too perfect not to say a little more.

So my goal for today is to show you the similarities side-by-side.  I'm still in the process of learning about covering spaces, and I'm certainly no expert in Galois theory, so forgive me for any glosses or for not getting too deep here. (Although, the resemblance itself is pretty deep!) But I believe I know enough to at least convey the general idea, even if we're just scratching the surface!


Definitions & Preliminary Remarks

Let's start by recalling a few definitions.

Suppose $X$ is a path-connected, locally path-connected, and semilocally simply-connected space. Given a covering space $(\tilde{X},\tilde{x_0})\overset{p}{\to}(X,x_0)$, the set of all covering space isomorphisms $f:(\tilde{X},\tilde{x_0})\to(\tilde{X},\tilde{x_0})$ forms the group of deck transformations. Notice from the diagram on the right that we must have $p\circ f = p$ for all $f\in G(\tilde{X})$. In other words, a deck transformation $f$ is a homeomorphism of $\tilde{X}$ such that the shadows/projections of both $\tilde{X}$ and $f(\tilde{X})$ onto the base space $X$ look the same.
The cover is called regular (or normal) if for any two points $\tilde{x_0}$ and $\tilde{x_1}$ in $p^{-1}(x)$ for any $x\in X$, there exists $f\in G(\tilde{X})$ such that $f(\tilde{x_0})$ = $\tilde{x_1}$. So we might say the cover is regular if $G(\tilde{X})$ acts transitively on $p^{-1}(x_0)$. But it's really more than that since there's an action on the entire space $\tilde{X}$, not just $p^{-1}(x)$. (Each $f\in G(\tilde{X})$ is a homeomorphism of $\tilde{X}$, not just a permutation of $p^{-1}(x)$.) So regular covers are extra special because they have, in Hatcher's words, maximal symmetry*.

Turning now to Galois theory, let $F$ be a field and suppose $K/F$ is a finite degree extension. (Recall, K is a field that contains $F$, but we can also view it as an $F$-vector space. So to say $K/F$ is a finite degree extension means $K$ is a finite dimensional vector space over $F$.) We say that $K/F$ is a Galois extension if $K/F$ is the splitting field of a separable polynomial (i.e. a polynomial whose roots are all distinct) in $F[x]$. Equivalently, we say $K/F$ is Galois if the degree of $K/F$ equals the number of automorphisms of $K$ that fix $F$.

An observation: From Hatcher's Proposition 1.39, we know that the group of deck transformations $G(\tilde{X})$ is isomorphic to $\pi_1(X)$ precisely when the cover $\tilde{X}\overset{p}{\to}X$ is the universal cover (and this is always regular). And as we know, the elements of $G(\tilde{X})$ act on $\tilde{X}$ while permuting the elements of $p^{-1}(x)$. In other words, the deck transformations of $\tilde{X}$ are the automorphisms of $\tilde{X}$. And analogously, whenever $K$ is the splitting field of some separable polynomial $p\in F[x]$, the automorphisms in the Galois group Gal$(K/F)$ act on $K$ while permuting all the roots of $p$ in $K$.

But the analogue doesn't stop there! It gets better.

Galois Theory

The Fundamental Theorem of Galois Theory tells us the following:**

Let $K/F$ be a Galois extension and let $G=\text{Gal}(K/F)$ be the group of automorphisms of $K$ which fix $F$. Then there is a bijection

given by the correspondences
And we can summarize some of the other results in the following diagram:

And here's what's so cool: The above has a direct analogue in the relationship between groups and covering spaces!

Covering Space Theory

Let $X$ be a path-connected, locally path-connected, and semilocally simply-conencted space with universal cover $\tilde{X}\overset{p}{\to}X$. Then there is a bijection
(where basepoints are preserved, although to save ink I've omitted them) given by the correspondences
where $\tilde{X}/H$ is the orbit space obtained by letting $H$ act on $\tilde{X}$ where the action must be a covering space action (i.e. for each $x\in \tilde{X}$ there is a neighborhood $U$ of $x$ so that $U\cap hU=\varnothing$ for all $h\in H\smallsetminus\{e\}$.) And just like above, we can summarize some of the other results in the following diagram:
(Perhaps we should call this 'The Fundamental Theorem of Covering Spaces'?)

Let's close by looking at a quick example to summarize what we've just seen.

A Side-by-Side Example

Consider the polynomial $p(x)=(x^2-2)(x^2-3)$ over $\mathbb{Q}$ which has splitting field $\mathbb{Q}(\sqrt{2},\sqrt{3})$. And let $X=\mathbb{R}P^2\times\mathbb{R}P^3$ (where $\mathbb{R}P^n$ is the real projective space of dimension $n$) which has universal cover $S^2\times S^3$.*** Then $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ is both the Galois group of $p$ and the fundamental group of $X$. In particular, we have the following diagrams (click to expand):


Pretty cool, huh?

I'm sure that none of this is a coincidence. It's much too nice! So it's got me wondering - what is the larger context in which these two correspondences (in Galois theory and covering spaces) coalesce?



*Indeed, for any $x\in X$ and for any $\tilde{x_0},\tilde{x_1}\in p^{-1}(x)$, there is a homeomorphism $f:\tilde{X}\to\tilde{X}$ that will not only take $\tilde{x_0}$ to $\tilde{x_1}$ but will also 'move' the entire space $\tilde{X}$ in such a way that its shadow/projection on to $X$ still matches $X$ perfectly! This is maximal symmetry indeed.

**For instance, see Theorem 14, ch. 14 of Dummit and Foote's Abstract Algebra.

*** Recall, $\mathbb{R}P^n$ has universal cover $S^n$ via the (2-fold) quotient map $S^n\overset{\times 2}{\to} \mathbb{R}P^n$ where antipodal points are identified. We also have $\pi_1(\mathbb{R}P^n)=\mathbb{Z}/2\mathbb{Z}$ for all $n$.