Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Edit: I've really enjoyed everyone's examples (especially the pictures!), but I was mostly looking for a general theorem. For instance, a similar statement to mine is, Can the mapping cylinder of every finite-sheeted covering map of a finite graph be embedded in $\mathbb{R}^3$? This is the level of generality I'd be most interested in. On second thought, and after reading the comments, I think a community wiki list of interesting examples (like those below) would be more helpful to other instructors. If you have any more, please post them!

I recently taught my students about covering maps of topological spaces, using the classic example of the real line winding onto to the circle. This covering map can be exhibited in real life (well, not all of it, but a representative chunk). All other covering maps of the circle really can be exhibited with string. Other things that can be exhibited include winding the upper half plane onto the punctured plane, although this is equivalent to the real line covering the circle.

After some thought, it seems that some graph coverings (like a double cover of the figure eight) can be realized in real life, while it seems that nontrivial surface coverings cannot be exhibited unless the circle has boundary. My question is,

What is the class of covering maps of graphs or two-dimensional CW-complexes that can be realized in 3-space, i.e. so that there is a homotopy of the covering space in $\mathbb{R}^3$ onto the base space which is an isotopy for $0\leq t< 1$ and is the covering map for $t=1$?

(This is my working definition of "real life homotopy", since we allow things to touch but not pass through each other. There may be a better definition.)

share|improve this question
The orientable double cover of the mobius strip can be realized in $3$-space using a non-standard embedding of the cylinder. –  Will Sawin Dec 6 '12 at 2:25
@Brian Rushton: Think of cutting a paper Möbius strip (aka. paper-strip-glued-with-a-half-twist) along the midline. You get the paper-strip-glued-with-a-twist, which is the required nonstandard cylinder embedding. –  Ketil Tveiten Dec 6 '12 at 13:02
I like the one where the unit quaternions double cover $SO_3.$ The other idea is to roll a student in carpet. When the carpet just begins to touch, that it a single cover. Roll him over again and that is a double cover. Very real life. –  Will Jagy Dec 7 '12 at 2:30
@Ketil Tveiten: Not quite. You actually get a paper-strip-glued-with-two-full-twists that way. The construction you want is to make a Mobius strip, but with two strips of paper stacked on top of each other (so that the gluing glues one end of each strip to the opposite end of the other). –  Alison Miller Dec 8 '12 at 21:12
Will Sawin's example can be realized in a very pleasing manner which I learned recently at MoMath. Modify the usual construction of the Mobius band, using three strips of paper stacked on top of each other. Do a single half-twist of this triple stack, and then carefully stick the required pieces of tape into this contraption. You will get a paper model of the double covering of the Mobius band (the glued inner strip) by the annulus (the glued pair of outer strips). –  Lee Mosher Mar 14 '13 at 14:57

9 Answers 9

The fact that the fundamental group SO(3) is Z/2 is sometimes shown by a funny movement of the arm, twisting it once while keeping the hand palm up, and detwisting it by twisting it once more. See also this thread on Stackexchange.

share|improve this answer

For classroom purposes, you can tell your students they probably already studied covering maps in kindergarten when folding paper, cutting out a figure and then unfolding the paper to see a paper dolls chain:

alt text

share|improve this answer
Nice example! I think this covering is branched though (as the fiber at a point on the doll's foot, say, has twice as many points as the fiber at a point right between her eyes). –  Mark Grant Dec 6 '12 at 7:01
Mark, I think Guntram's covering map wasn't to the folded sheet of paper, but to a single doll holding its own hand. But I agree--otherwise this is not at all an example of a covering map: It wouldn't even be a local homeomorphism. (Showing such an example to students could really hurt their understanding.) –  Hiro Lee Tanaka Dec 6 '12 at 16:00
Counterexamples which just barely miss being examples can often enhance understanding. Furthermore this example, while not a covering space in the usual topological category, is indeed a covering space in the orbifold category, and it can be used to exhibit the isomorphism between the infinite dihedral group (the deck transformation group of the orbifold universal covering space) with the fundamental group of the orbifold $[0,1] \times \mathbb{R}$ with mirror reflectors on the boundary. –  Lee Mosher Mar 14 '13 at 15:02

I like this one :

alt text

It is (part of) a cyclic covering of $S^2$ minus $5$ points :

alt text

share|improve this answer
The homotopy that the question asks for is not obvious here; could you give some more detail? –  j.c. Dec 6 '12 at 15:55
Well if I understant the question, it's just like for the real line winding around the circle, the map $H : \mathbf R^3 \times [0,1]\to\mathbf R^3$ defined by (say) $H((x_1,x_2,x_3),t)=(x_1,x_2,(1-t).x_3)$ ... it furnishes the projection onto the surface I will add in a few minutes in the answer. –  G.C. Dec 6 '12 at 16:09

Here's an example I learned from John Franks. This is a nice example because it's used to produce an example of Smale's sphere eversion problem. It also generalizes to include (for example) Will Sawin's comment above.

Consider the Boy's surface. This is an immersion of $\mathbb{R}P^2$ into $\mathbb{R}^3$. If you look at its normal bundle, there's no sense of +1 or -1 (because it's non-orientable) but you can look at its associated unit sphere bundle. This is the orientation double cover--a.k.a. the sphere--immersed into $\mathbb{R}^3$. By scaling the fibers of the normal bundle from 1 to 0, you see the "covering homotopy" of $S^2$ onto the Boy's surface, as you ask for.

So that's the thing you're looking for, but let's go further--instead of just scaling from 1 to 0, scale from 1 to -1. This is a homotopy, through immersions, of $S^2$ to itself, and it leads to one way in which you can evert the sphere (i.e., turn it inside out). I think this strategy originally came from Shapiro, though any historical corrections are welcome!

More generally, if you immerse any 2-dimensional object (i.e., an un/orientable surface with or without boundary) you can perform the same trick, examining the unit-length elements of its normal bundle. In Will Sawin's example, take an embedded Mobius band and examine its unit normal bundle--this gives the non-standard embedding of the cylinder you're asking for, and scaling the normal bundle to the zero section gives you the "covering homotopy" you seek.

share|improve this answer
Yes, Shapiro. Check out the 1980 Intelligencer article by B. Morin and G. Francis on sphere eversion. –  Igor Rivin Dec 6 '12 at 15:25
See also Apéry, François La surface de Boy. (French) [Boy's surface] Adv. in Math. 61 (1986), no. 3, 185–266. As an undergraduate in Krakow in late 1980s I saw a movie depicting some things described in this article (shown by Apery during his visit), but it does not seem to be available over the internet. –  Margaret Friedland Dec 6 '12 at 16:36

As far as I can tell, you can cover any graph embedded in $\mathbb{R}^3$ with its universal cover using a homotopy that satisfies your conditions. This is because trees are contractible, and can snake themselves into tight spaces without intersecting themselves.

share|improve this answer

My favourite example of a real-life covering space is a multi-level parking garage, along with all up-down ramps. Wander around for a while, and you're on a different deck.

share|improve this answer


I like the Cayley graph of the symmetric group $S_3$ for the presentation $P$ with generators $x,y$ and relations $r=x^3,s=y^2,t=xyxy$. This gives a $2$-complex $K(P)$ with one $0$-cell, two $1$-cells, labelled $x,y$, and three $2$-cells labelled $r,s,t$. The Cayley graph of $P$, i.e. the $1$-skeleton of the universal cover of $K(P)$, is shown above (part of Fig 10.6 of Topology and Groupoids), where the solid lines are mapped to $x$, and the broken lines to $y$. Also shown is a choice of maximal tree. You can also see the lifts of the $2$-cells, but they can't all really be drawn since the corresponding universal cover is a $6$-fold cover, and there are not enough $2$-cells in the above diagram. The dots show where a relation starts.

Note that in group theory, a so called Schreier transversal for a subgroup $H$ of a free group $F$ with generating set $X$ can be seen as a maximal tree in the covering graph of the $1$-complex $K$ determined by the generating set $X$ of $F$, the covering being determined by the subgroup $H$ of $F$.

Another nice exercise for students is to ask them to construct two $3$-fold covers of $S^1 \vee S^1$ such that one is regular and the other is not. For the convenience of readers here is picture of the two graphs, but with unlabelled arrows! It nicely shows how one graph has 3-fold symmetry and the other does not.


share|improve this answer
Ronnie, you beat me to it!;-) –  Tim Porter Dec 6 '12 at 15:33
I'll add something which is not exactly an answer to the question. I got the impression at Oxford in the late 1950s under Henry Whitehead that algebraic topology was about modelling topology by algebra, and any algebra was OK as long as it modelled the geometry. So when I found that most aspects of 1-dimensional homotopy theory were better modelled by groupoids rather than groups, I expected people to be pleased. But most algebraic topology texts model covering spaces by group actions, rather than by covering morphisms of groupoids, which gives a base point free approach, published in 1968! –  Ronnie Brown Mar 28 '13 at 16:51

Embed the Cayley graph of the free group $F_2$ in $\mathbb{R}^2 \subset \mathbb{R}^3$ (note that the edges must shrink geometrically to avoid self-intersection). Increase the vertical component of each vertex $v$ by an amount $h_v$ to be described below. This is your isotopy at $t=0$. Now, as $t \to 1$, stretch the edges to unit length and scale the vertical height of vertices by $(1-t)h_v$ so the full thing comes to rest back on the plane. In the end you covered the infinite square grid that represents the Cayley graph of $\mathbb{Z} \times \mathbb{Z}$. To ensure that this gives an embedding for every $t$, write $v = x_1 \ldots x_r$ with $x_i \in \{ a,b,a^{-1},b^{-1} \}$ (generators of $F_2$), and let $h_v = \sum_{x_j \in \{a,a^{-1}\}} 2^{-j}$. In other words, read $x_1, \ldots, x_r$ as a binary fraction replacing $a,a^{-1}$ by 1 and $b,b^{-1}$ by 0. This way any pair of edges that run the risk of overlapping will have disjoint heights and not intersect after all.

For a second example, take the standard Cantor set in $[-1,1]$. It consists of two copies of itself scaled down by $1/3$ in $\big[-1,-\frac13\big] \cup \big[\frac13,1\big]$. As $t \to 1$, rotate $[-1,0]$ by $\pi t$ about 0, while rescaling by $3t$ and translating the fold point to $-t$. The Cantor set covered itself !

share|improve this answer
For your first construction, how do you know that your homotopy doesn't intersect itself at some $t \in (0,1)$? –  S. Carnahan Dec 6 '12 at 3:31
@Scott: Thanks. I tweaked the construction. –  Rodrigo A. Pérez Dec 6 '12 at 4:14

Not only a covering, but every piecewise linear map $f$ between finite graphs can be realized in $\Bbb R^3$ in your sense. To see this, pick a triangulation $T$ of the mapping cylinder of $f$ that projects simplicially to some triangulation of $I=[0,1]$. Then pick a generic level-preserving map $g$ of the mapping cylinder in $\Bbb R^3\times I$ that is linear on each simplex of $T$. (Generic here means that the images of any $k+1$ vertices span a $k$-dimensional affine subspace in $\Bbb R^4$ as long as either $k\le 3$, or $k\le 4$ and not all of the $k+1$ vertices have the same $I$-coordinate.)

Then $g$ has only finitely many double points, all in $\Bbb R^3\times (0,1)$. (Say, if a pair of 2-simplices with at most one vertex in common have a 1-dimensional intersection, then their 5 or 6 vertices aren't affinely independent.) If $s$ is the minimal $I$-coordinate of a double point, then the part of $g$ that lands in $\Bbb R^3\times [0,s/2]$ yields the desired homotopy.

I'm quite amazed that this rather standard argument didn't occur to anyone having seen this question in 3 months. Maybe a tag like "geometric topology" could have helped. There is some literature on realizing maps in your sense, which goes under "isotopic realization".

share|improve this answer

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.