# Are there surfaces with more than two sides?

I'm watching a naive introduction to the Möbius band, the lecturer asks if it's possible to construct a one sided surface and then she says that there is one of these surfaces, namely the Möbius band. Then she mentions that some surfaces have two sides.

I've also had this doubt when reading Flegg's From Geometry to Topology. So, is it possible to have a surface with more than two sides? My intuition says no, but perhaps someone made some magic trick and made it somehow. I've looked at some wikipedia articles and I've seen no mention to something with more than two surfaces (I've used my browser search tool), unless such surfaces have other names.

I guess that the tags should be geometry and topology, if you think there is something more, please edit.

• This depends on precisely defining what a "side" is. There is a standard formal definition, under which the answer is "no", but the definition is complicated and might not satisfy you as capturing the intuitive notion of a side. – Alex Becker Apr 12 '14 at 0:53
• @AlexBecker You can point me to it anyway. I'm studying on a math major, someday I'll understand it. – Billy Rubina Apr 12 '14 at 0:54
• An assignment of inside and outside is called an "orientation" of a manifold, which is how we formalize the notion of a side. – Alex Becker Apr 12 '14 at 1:01
• It is not orientation but coorientation. – Moishe Kohan Apr 12 '14 at 1:11
• Do you allow disconnected surfaces? In this case, the disjoint union of, say, a sphere and a mobius band might qualify in your mind as an example of a surface with "3 sides". – Mike F Apr 12 '14 at 18:20

Here's an intuitive explanation (and I am writing this under the assumption that the surface is connected).

The key lies in understanding the difference between the number of sides "locally" and the number of sides "globally".

For any surface depicted in 3-D space, the number of sides locally is exactly two. The intuitive reason for this is that if you train a microscope on a point of the surface and zoom way in, you see a local piece of the surface which looks very, very similar to a flat plane in 3-dimensional space. A flat plane in 3-D space always has exactly two sides. So from a local perspective, a surface has exactly two sides. Not one, not three or more, but exactly two.

Now we switch to the global perspective. Imagine in your microscope you see a tiny creature walking along one side of the local piece of the surface. Is it possible for the creature to travel along some path, one which is allowed to exit the view of the microscope but which returns at a later time, so that when the path returns the creature is on the opposite side of the piece of the surface in the microscope? If so, if such a path exists, then from a global perspective the surface is one-sided. If not, if no matter what path the creature walks around it always comes back to the same side of the surface in the microscope, then the surface is globally two-sided.

As @MoisheCohen says, this is the concept of "co-orientation", also called "transverse orientation". At every point on every surface in 3-D space there are exactly two local transverse orientations. The surface is globally one-sided if there exists a path which switches the two local transverse orientations, and the surface is globally two sided if no path switches the two local transverse orientations.

To summarize: the reason there are not "3 or more sides" is that there are exactly two sides locally, and those two sides are either globally equivalent or globally inequivalent. When you have an equivalence relation on a set of two elements, the number of equivalence classes is either one or two.

• What is the proof that there is not a global 3rd side? I could use the categorization of compact surfaces to show all of them have 1 or 2 sides. But that does not include non-compact surfaces and it seems to me that a hidden assumption in the categorization proof is that the surfaces have 1 or 2 sides... – Michaela Light Apr 15 '14 at 22:27

Getting a true answer to your question requires some familiarity with differential topology (or algebraic topology), which I assume in what follows. My favorite reference is do Carmo's "Riemannian Geometry".

The definitions below are (intentionally) formal (you had enough informal input from other answers). To gain an intuition, one should work out specific examples of surfaces in $R^3$.

In order to keep things simple, I will assume in what follows that $S$ is a smooth hypersurface in an $n$-dimensional Riemannian manifold $(M,g)$, i.e., $S$ is a smooth submanifold in $M$ and $S$ has dimension $n-1$. (The assumptions of smoothness and presence of a Riemannian metric, simplify definition and proofs but can be avoided by doing a bit more work. To avoid using Riemannian metric I would have to define the normal bundle of $S$ in $M$. To eliminate smoothness assumptions, I would have to invoke some heavy machinery from the theory of topological manifolds.) For concreteness, you can consider the case when $M=R^n$ with the standard metric.

Definition. A unit normal vector to $S$ at a point $x\in S$ is a unit vector in $T_x(M)$ (the tangent space of $M$ at $x$) which is orthogonal to $T_x(S)$ (the tangent space of $S$ at $x$).

Definition. A unit normal vector field to $S$ is a vector field along $S$ consisting of unit normal vectors, i.e., a smooth map $\nu: S\to TM$ (the tangent bundle of $M$) which sends each $x\in S$ to a unit normal vector $\nu_x$ to $S$ at the point $x$.

Definition. A coorientation of $S$ is a choice of a unit normal vector field $\nu$ to $S$. A submanifold $S\subset M$ is called coorientable if it admits a coorientation.

Clearly, if $\nu$ is a coorientation of $S$, then the vector field $-\nu$ is also a coorientation of $S$, and $-\nu\ne \nu$.

Exercise. Coorientability of $M$ is independent of the choice of a Riemannian metric on $M$: $S$ is coorientable if and only if there exists a continuous vector field $\xi$ along $S$ such that $\xi_x\notin T_xS$ for every $x\in S$. Hint: Use the orthogonal projection of $T_xM$ to the orthogonal complement of $T_xS$ in $T_xM$.

Note that coorientability is independent of orientability of $S$ and $M$. It is also independent of the number of components of $M\setminus S$. However:

Exercise. a. If both $M$ and $S$ are orientable, then $S$ is coorientable.

b. If $M$ and $S$ are connected and $M\setminus S$ is not connected, then $S$ is cooriented.

Definition. Suppose that $S$ is connected. Then $S$ is called 1-sided if it does not admit a coorientation. (Such $S$ is said to have one side.)

Lemma. Suppose that $S$ is connected and coorientable. Then $S$ admits exactly two coorientations.

Proof. Let $\nu, \mu$ be coorientations of $S$. I claim that either $\nu=\mu$ or $\mu=-\nu$. Consider a point $x\in S$. Then either $\nu_x=\mu_x$ or $\nu_x=-\mu_x$ since these are unit normal vectors to $S$ and $T_x M= T_x S\oplus {\mathbb R}$ is the orthogonal decomposition. Therefore, we obtain a partition $U\sqcup V$ of $S$ where $$U=\{x\in M: \nu_x=\mu_x\}, V=\{x\in M: \nu_x=-\mu_x\}.$$ Both sets are closed in $S$ since $\nu_x, \mu_x$ are continuous vector fields. If both sets are nonempty then $S$ is not connected, which contradicts our assumption. Hence, either $\nu_x=\mu_x$ for every $x\in S$ or $\nu_x=-\mu_x$ for every $x\in S$. qed

Definition. A side of a connected hypersurface $S$ is a choice of coorientation of $S$.

Corollary. Every connected hypersurface either is 1-sided or has exactly two sides. In other words, the number of sides, $\sigma(S)$, of a connected hypersurface $S$ is either 1 or 2.

Suppose now that $S$ is not necessarily connected and $$S= \coprod_{j\in J} S_j$$ is the decomposition of $S$ in its connected components. Then the number of sides of $S$ is defined as $$\sigma(S):= \sum_{j\in J} \sigma(S_j).$$ For instance, if $S$ is the disjoint union of the Moebius band and an annulus in $R^3$, then $\sigma(S)=1+2=3$, that is, $S$ has three sides. If $M=RP^2\times S^1$ and $S=RP^2\times \{p\}\subset M$, then $\sigma(S)=2$, that is, $S$ has two sides.

This answers your question on the "number of sides" of a hypersurface. Note that if $S$ is not a hypersurface in $M$ then its "number of sides" is not defined.

This is a great question. In 1-dimensions - the degenerate case where you slice your surface appropriately - the Jordan Curve Theorem (YouTube video) says the curve will have an "inside" and an outside.

This result sounds tautological - but if the curve is very complicated, it gets harder to check whether you are inside your not. Is the red dot inside the curve or not. In the second example, the boundary is very complicated but it still has 2 sides.

[][1] (source: [oberlin.edu](http://www.oberlin.edu/math/faculty/bosch/smiley2006-yellow-red-dot-web.png))
• Ok, tell the answer! Is it inside or not? – Sawarnik Apr 16 '14 at 8:41
• The red dot is inside the black face curve, because no matter how you flee the yellow region, you have to cross the black line (which is the inside/outside boundary) an odd number of times. – Steve Kass Apr 16 '14 at 22:20