4
$\begingroup$

It is well-known that the space of "pairs of points in a circle" (also called the symmetric square of $S^1$) can be identified with a Möbius strip. I don't know where the idea originates, but it is a neat elementary result in topology, which has a beautiful application in Vaughan's proof of the rectangular peg problem (3blue1brown gives a really nice account of this). Inspired by a talk I saw on related results, I've been trying to compile a list of proofs that the symmetric square of $S^1$ is a Möbius strip. To be specific, the space $Sym^2(S^1)$ is constructed from $S^1\times S^1$ by identifying $(x,y)\sim (y,x)$ for every $x,y\in S^1$ and taking the quotient topology.

Hence, I am asking this question as community wiki, to try and learn of other peoples' favorite ways to prove this result. A quick ground rule: if someone beats you to the chase on your favorite proof, maybe comment to show your preference for it, rather than posting the same proof again? But if you have a proof that seems like a close variant of one already posted, feel free to post it! Since these are all about the same result, there are bound to be variations on the same idea. Also, if you know where your proof comes from, please give proper credit to its originator. If the proof is your own, be explicit about that fact, so that I can make sure to give you proper credit if I ever cite it elsewhere.

Here are summaries of some proofs that I know of, to get the ball rolling:

  • View the torus $S^1\times S^1$ as square with opposite sides identified. The identification $(x,y)\sim (y,x)$ corresponds to folding the square along its diagonal, so we get a triangle with gluing between two adjacent sides. A brief cut-and-paste argument turns this into the usual presentation of the Möbius strip. (I don't know where this argument originates.)
  • Charles Livingston: To any pair of points $x,y\in S^1$ in the unit circle of $\mathbb R^2$, we associate their perpendicular bisector, which is a line through the origin and thus an element of $\mathbb R\mathbb P^1$ (if $x=y$, we take the line through this point and the origin). This defines a map $Sym^2(S^1)\rightarrow \mathbb R\mathbb P^1$, which is a fiber bundle with the interval as its fiber. By checking that this bundle is not orientable (i.e. going around $\mathbb R\mathbb P^1$ once flips the interval), we can see that it is a Möbius strip. (This argument—and a nice picture of the previous one—are included as §2.2 of this paper by Tuffley.)
  • Hugh Morton (summarized on nLab): This is a case of Morton's more general description of $Sym^n(S^1)$. Consider $S^1\subset \mathbb C$ as the unit circle and define a map $\text{Sym}^2(S^1)\rightarrow S^1$ by sending a pair $(x,y)$ to their product $xy.$ As in the previous proof, we can check that this is a fiber bundle with the interval as its fiber, and that this bundle is not orientable (i.e. going around $S^1$ once flips the interval), so it is a Möbius strip.
  • Étienne Ghys: We can view $\mathbb R\mathbb P^2$ as the plane $\mathbb R^2$ with a copy of $\mathbb R\mathbb P^1$ at infinity. Then the Möbius strip $M$ is formed by cutting the open unit disk out of our projective plane. For any two points $p,q$ on the unit circle $S^1\subset \mathbb R^2$, we can take the tangents $\ell_p$ and $\ell_q$ to the circle at these points. The map which sends $(p,q)\mapsto \ell_p\cap \ell_q$ is a homeomorphism $Sym^2(S^1)\rightarrow M$, where the $\mathbb R\mathbb P^1$ at infinity corresponds to intersections of parallel lines. (I believe this is from “Prolongements des diffeomorphisms de la sphère,” but I am not certain.)
  • Karanbir Sarkaria: We can view $\mathbb R\mathbb P^n$ as the space of non-zero, homogeneous polynomials of degree $n$ in $\mathbb R[x,y]$, up to multiplication by a non-zero scalar. Then we have a map $\mathbb R\mathbb P^1\times \mathbb R\mathbb P^1\rightarrow \mathbb R\mathbb P^2$, given by $$\big((rx+sy),(ux+vy)\big)\mapsto (rx+sy)(ux+vy).$$ This descends to a map $\text{Sym}^2(\mathbb R\mathbb P^1)\rightarrow \mathbb R\mathbb P^2$, which is a homeomorphism onto its image. But $ax^2+bxy+cy^2$ lies in the image of this map if and only $b^2-4ac\geq 0$. Hence, we must cut out from $\mathbb R\mathbb P^2$ the set of quadratics with complex roots, which is homeomorphic to the open disk. But removing a disk from $\mathbb R\mathbb P^2$ yields the Möbius strip, as desired. (See Sarkaria's note for a nice illustration of this fact.)
  • Map a pair of points $\{p,q\}$ on the unit circle to their midpoint $(p+q)/2$. This is a bijection between the punctured disk $D^2-{0}$ and pairs $\{p,q\}$ of non-antipodal points, but every pair of antipodal points gets mapped to $0$. In other words, we can form $Sym^2(S^1)$ from $D^2$ by replacing the origin with a copy of $\mathbb R\mathbb P^1$. This blowup of the disk is precisely a Möbius strip. (Again, I don't know where this argument originates.)

The fact in question also relates to a fascinating result of Bott and Shchepin. Let $\exp_n(S^1)$ consist of all subsets of $S^1$ containing between 1 and $n$ points. This is topologized via the Hausdorff metric on compact subspaces of $S^1$. Bott showed that $\exp_3(S^1)=S^3$ ("On the third symmetric potency of $S^1$") and Schepin showed that the natural embedding $S^1=\exp_1(S^1)\hookrightarrow \exp_3(S^1)=S^3$ is a trefoil knot (unpublished). Identifying pairs of equal elements with singleton sets, we get $Sym^2(S^1)=\exp_2(S^1)$, so the Möbius band $\exp_2(S^1)$ fits into this picture as a (non-orientable) Seifert surface of the trefoil. This theorem is refined in the aforementioned article by Tuffley, and has also been proven by: Nakandakari and Tsukuda via elementary pasting techniques (they animate their construction here), Mostovoy via lattices in $\mathbb C$, and Rose via isometries of the hyperbolic plane. I am not describing these in detail because they are lengthier and I haven't taken the time to fully understand them yet, but if you know of another description of $Sym^2(S^1)$ inside $\exp_3(S^1)=S^3$, that would also be a great thing to contribute.

$\endgroup$

1 Answer 1

1
$\begingroup$

Lemma 1: The set of lines on the plane is the open Möbius band.

Proof: The set of lines on $\rm\Bbb RP^2\cong S^2/{-1}$ is is continuous bijection with the points of $\rm\Bbb RP^2$: associate each great circle with pair of points located perpendicular to that circle. Deleting the line at infinity (the equator) gets us to the lines on the plane, through the gnomonic projection. This leaves us with $\rm\Bbb RP^2\setminus\{pt\}$, which is the open Möbius band.

(To see that the projective plane minus a point is the Möbius band, relate them both to a cylinder mod antipodes.)

Lemma 2: The set of lines on the plane that intersect the unit circle is the closed Möbius band.

Proof: This is the previous construction minus an open neighborhood of the line at infinity.

Finally, to relate this construction to the set of pairs of points on $S^1$, intersect these lines with the unit circle.


(Incidentally, the set of infinite rows of unit-spaced parallel lines (example element) is topologically a Klein bottle. We can decompose it into two Möbius bands by drawing a diameter $\frac12$ circle and asking whether a given element intersects it or not.)

$\endgroup$
2
  • $\begingroup$ Another remark: we can find the oriented cover of the first two constructions by orienting the lines (either by drawing an arrow on the lines or shading in one of the half-planes to either side of it). That is, a 180-degree rotation is an orientation-reversing path. $\endgroup$ Commented Mar 2, 2021 at 0:12
  • $\begingroup$ (Also, apparently the fact that the configuration space of a plane of stripes is a Klein bottle is relevant to condensed matter physics and liquid crystals) $\endgroup$ Commented Mar 2, 2021 at 0:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .