Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As someone who trained as a physicist, I have known for ages that $\operatorname{SU}(2)$ is a double cover of $\operatorname{SO}(3)$, so, during an idle day at the office I decided to look up what this meant. It turned out to be more complicated than I had expected. The definition of a covering space of a topological space seemed to be quite fiddly and left me thinking that these things were born of utility rather than essential beauty.

So, what are they for, and why are they more useful, than say a simple open cover? And why are the pre-images of a point in the covered space called a ‘fibre’ (is it linked to fibre bundles in differential geometry?). And, all of the many definitions I have seen seem to imply that the fibres must be discrete (presumably, countably infinite at the most, does this follow from the definitions?), and beyond that they always seem to be finite and of the same number (i.e., always $2$ at every point in the case of $\operatorname{SU}(2)$ and $\operatorname{SO}(3)$) is this always the case?

share|cite|improve this question
Dear James, The definition only seems fiddly if you don't have experience and a broader context in which to place it. (Just as the definition of group, or normal subgroup, – Matt E Dec 31 '11 at 0:09
I'm certain that that's true. Any chance of an example of this broader context? I am not scared of the fiddly, but then again I prefer elegant, so I hope there was some great need for this. After all, to define the Lesbesgue integral you could take the Caratheodary route or the representation theory route - both fiddly - but at least you know why you were putting the effort in. What wrong with a simple open cover? And why does the 'fiber' make me think of fiber bundles? – James Robson Dec 31 '11 at 0:30
@James: Because a covering space is (locally) a fibre bundle! (The fibres are discrete, however.) – Zhen Lin Dec 31 '11 at 1:12
Ok, so my intuition of link to diff. geom. was not misplaced. However, you say the fibres are discrete (I guess this is in the topological sense, so my countability comment is discounted - but still, is that a requirement or just a consequence of the basic recipe?). Also, my naive notion of a fibre bundle is attaching the same thing to each point of some base manifold. In this case, how do we know that the fibres have the same size - or perhaps they don't? (Apparently some people allow empty fibres, but then what is the point of all this - which is my question!) – James Robson Dec 31 '11 at 1:34
Asking why a covering space is more useful than an open cover is comparing apples to oranges; even though both concepts may fall under the label "cover of a space", they are completely different concepts, just like distributions in the sense of probability theory and in the sense of generalized functions. – KCd Dec 31 '11 at 4:36

Covering spaces naturally occur in the study of analytic continuation (in fact I believe this is where they first appeared). For example, the square root function $\sqrt{z}$ cannot be extended to a holomorphic function on all of $\mathbb{C}$. It can be locally defined on various open sets in $\mathbb{C} - \{ 0 \}$, and by analytic continuation it can be defined, for example, starting from a neighborhood of $z = 1$ (so that $\sqrt{z} = 1$ for example) and counterclockwise around the origin. However, this process is inconsistent: when you get back to $z = 1$ you'll find that $\sqrt{z} = -1$.

The solution is to define $\sqrt{z}$ on a double cover of $\mathbb{C} - \{ 0 \}$; there are two sheets of the cover for each of the two possible values of the square root. Similarly, $\sqrt[n]{z}$ is defined on an $n$-sheeted cover of $\mathbb{C} - \{ 0 \}$, and $\log z$ is defined on a cover of $\mathbb{C} - \{ 0 \}$ with infinitely many sheets.

There is a nice analogy with the theory of field extensions and Galois theory; the double cover mentioned above corresponds in some sense to the field extension $\mathbb{C}(z, \sqrt{z})$ of the field $\mathbb{C}(z)$, and it has Galois group $\mathbb{Z}/2\mathbb{Z}$. In this analogy the fundamental group of a space is analogous to the absolute Galois group of a field, and this has been a very fruitful analogy in mathematics, leading to the theory of the étale fundamental group. These ideas are thoroughly explored in Szamuely's Galois Groups and Fundamental Groups.

In quantum mechanics, covering spaces of topological groups naturally occur for the following reason. Because a wave function $\psi$ and any multiple $e^{i \theta} \psi$ of it represent the same physical state, to say that a group $G$ acts as a group of symmetries of a quantum system whose states lie in a Hilbert space $H$ is not to say that there is a representation $G \to \text{U}(H)$ (the unitary group of $H$) but rather a projective representation $G \to \text{PU}(H)$. For Lie groups $G$ such representations can be analyzed using representations of a covering group $\tilde{G}$ of $G$. This is why quantum systems with $\text{SO}(3)$-symmetry, e.g. an electron orbiting a proton, are naturally analyzed using the representation theory of $\text{SU}(2)$.

share|cite|improve this answer
Many thanks. So covering spaces were devised to provide spaces that functions C could have complex values in? Sounds good to me. If that's correct, I think it should be mentioned more often as a pedagogical motivation? – James Robson Dec 31 '11 at 2:17
@James: values on. I agree, but at the same time the modern perspective recognizes that the theory of covering spaces is topological and not analytic in nature. – Qiaochu Yuan Dec 31 '11 at 2:28
When you say "on" do you mean I should have said "functions ON C", which was a typo, or do you mean something mathematical that I got wrong? Probably the latter, but words have many meanings in maths? – James Robson Dec 31 '11 at 3:16
@JamesRobson: My guess is he meant that the functions are to be defined on the covering spaces, i.e., their input values come from the covering spaces. When we say a function has values in $Y$ it usually means that the output values are in $Y$. Explicitly, if $f:X\to Y$, then $f$ is defined on $X$, and the values of $f$ are in $Y$. – Jonas Meyer Dec 31 '11 at 6:51

The definition probably only seems fiddly if you haven't seen it (or related) definitions before. What is says is the following: a map $p: Y \to X$ is a covering map if $p$ locally looks like the projection from $$X \times \text{ a discrete space} \to X.$$

A little more precisely: each point $x \in X$ has a neighbourhood $U$ such that the map $$p^{-1}(U) \to U$$ is isomorphic to a projection $$U \times \text{ a discrete space} \to U.$$

This kind of property of a map --- that locally on the target it looks like the projection from a certain kind of product --- is very common in topology and geometry, and underlies the fundamental notion of a fibre bundle. Covering spaces are perhaps the simplest example, since they are fibre bundles with discrete fibres. Fibre bundles of all kinds appear everywhere, and so it is not so much a question of asking what they are useful for, but rather, of identifying a ubiquitous property and giving it a name.

Another, more global, way to describe covering spaces is as follows: if a discrete group $\Gamma$ acts on a space $Y$ in such a way that each point has a neighbourhood $U$ such that the orbits $\gamma U$ are distinct for distinct elements $\gamma \in \Gamma$, then the quotient map $Y \to Y/\Gamma$ is a covering map (i.e. $Y$ is a covering space of $Y/\Gamma$).

Since group actions on spaces are pretty ubiquitous, this gives some indiction of why covering maps might be commonly encountered in topology. (The basic example is $\Gamma = \mathbb Z$ acting by translation on $Y = \mathbb R$, with the cover being $\mathbb R/\mathbb Z$, which is a circle.)

Finally, if you begin with space $X$, in order to construct covers of $X$, you have to "unwind" certain directions in $X$. Thus investigating covering spaces of $X$ is the same as investigating the extent to which the various directions in $X$ are "wound up".

E.g. in the circle there is just one direction, and unwinding it, you get the covering space $\mathbb R$. In $SO(3)$ there is one direction which is wound up, and unwinding it gives $SU(2)$. Often this "unwinding" can be thought of in a physical way: e.g. imagine that you are walking around a stadium, and measuring the distance you have travelled as you walk. When you get all the way around, you are back where you are started (the stadium is a circle), but your distance travelled isn't at zero (it's at 400 metres, say). The numerical distance travelled "unwinds" the circle of the stadium into the line $\mathbb R$.

E.g. in $SO(3)$, the "belt trick" mentioned by Georges allows you to "unwind" a rotation to get an element of $SU(2)$. (And when you do it twice, you get back where you started --- unlike in the case of the stadium, where your distance travelled never resets to zero; so here you see the difference between a double cover, like $SU(2)$ over $SO(3)$, and an infinite cover, like $\mathbb R$ over $\mathbb Z$.)

share|cite|improve this answer

First of all, your technical questions, such as why the number of sheets (the cardinality of the fiber) is constant (for a connected covering space, of course) are answered in any algebraic topology text. I'll point you to Hatcher's text, available here because one, it is free, and two, it includes some less formal discussion that you might find illuminating. (EDIT: Having looked at the relevant section in Hatcher, he only asserts that the number of sheets is locally constant. It follows from the requirement that about any point, there is a neighborhood that is evenly covered, as this same neighborhood will work for nearby points, giving locally constant and hence globally constant if your space is connected.)

Your question is a big one, and the answers already posted give different insights into why covering spaces are as important a notion as they are. To expand on Lopsy's answer a bit, I think it is worth stressing that covering spaces and fundamental groups are intimately linked; this is one basic reason for their ubiquity. An essential fact about covering spaces is that for nice enough spaces $X$, the various connected covering spaces $\tilde{X}$ for $X$ are in one-one correspondence with subgroups of the fundamental group $\pi_1(X)$. That is, for each such subgroup $H \le \pi_1(X)$, there is a covering space $\tilde{X}$ with $\pi_1(\tilde{X}) = H$. Algebraic concepts like the index of a subgroup, or whether or not $H$ is normal, have elegant interpretations in the covering space setting: the index equals the number of sheets, and if the subgroup is normal, then the covering space is "homogeneous", loosely speaking (see Hatcher for precise statement + compelling pictures). So one answer to your question of why they are more useful than more general covers is that they provide exactly the right setting for realizing spaces with fundamental groups that are subgroups of your original space.

So why should we care about this? Again, this is a huge question. One answer is that it provides a wealth of group actions, which will allow us to study the structure of groups by understanding properties of the actions that they have on spaces. If we are interested in a group $G$ and we happen to have a space $X$ with fundamental group $G$, then $G$ acts on the universal cover $\tilde{X}$ of $X$ (which is just the simply-connected covering space for $X$) by deck transformations. A deck transformation is a map $\tilde{X} \rightarrow \tilde{X}$ that preserves the fibers. An example would be as follows: as Lopsy said, the universal cover of $S^1$ is $\mathbb{R}$, and $\pi_1(S^1) = \mathbb{Z}$. The action of $\mathbb{Z}$ on $\mathbb{R}$ is just translation; i.e. $n \in \mathbb{Z}$ acts by sending $x$ to $x+n$. Since the fiber of a point $x \in S^1$ is the set $\{x + n, n \in \mathbb{Z}\}$, we can see that only translation by an integer will preserve this set. Using this setting, you can prove, for instance, that any subgroup of a free group is free: thus you can derive purely algebraic facts by appealing to covering space theory.

share|cite|improve this answer
Thanks NKS. I feel I now have achieved semiotic closure. I studied quite a few sources in the last two days, and I can honestly say, this site has been pretty amazing. I now have closure - acting on one idea will only lead to a combination of the others. So, I can now make the effort to learn this stuff (well,it is new year, so the time for resolutions) or be content that they are are contained and can't bite me any more?. Happy New Year. – James Robson Dec 31 '11 at 4:50
I have argued for the last 44 years in books and papers that 1-dimensional homotopy theory is really about groupoids rather than groups, leading to more powerful theorems with simpler proofs. In particular, covering maps of spaces are nicely modelled by covering morphisms of groupoids, and I feel this makes the theory easier to follow: lifting morphisms is to me clearer than lifting actions, even if they are equivalent. This also leads the way to thinking of higher dimensional groupoids, which of course are even more non-abelian. – Ronnie Brown Apr 23 '12 at 13:20

You can dazzle your friends at tomorrow's New Year party with Dirac's belt trick ( or this variant. See also here)

share|cite|improve this answer

In topology, they're used to prove results about fundamental groups: for example, the fact that $R$ covers $S_1$ is used to prove that $\pi_1(S_1)=Z$. Check out this picture:

Say you're walking around the ring in the middle. There's exactly one way to project your journey up/down onto the spiral staircase so that it stays continuous. When you get back to where you started on the ring, the winding number of your journey is going to be the number of flights of the spiral staircase your journey's projection ascended.

In this example, the spiral staircase is $R$, the ring is $S_1$, and the winding number is the corresponding element in the fundamental group. In general, this sort of argument works for any covering space!

share|cite|improve this answer
And now you risk getting me started on other questions - such as the proof that SO(3) is not simply connected because of the "loop" you can take from from one pole to the other. But just to finish this off - can you tell me anything about the fibres? Do they have to be finite sets, and if so are they required to have the same size for each point? – James Robson Dec 31 '11 at 2:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.