# What is the best way explain to undergraduates that all 1-dimensional manifolds are orientable?

Let's suppose that $M$ is a connected $1$-dimensional smooth manifold (Haussdorf and paracompact). We know that there are exactly two types, up to diffeomorphism (even up to homeomorphism), namely $\mathbb R$ and $S^1$. These are clearly both orientable, and one "high-powered" way to show this it is to show that the top exterior power $\Lambda^n (T^*M)$ of its cotangent bundle, which is the cotangent bundle itself since $n=1$, is trivial, because each of these manifolds is a Lie group. Of course, if you already know that the only ones are $\mathbb R^1$ and $S^1$, you can also directly show from the "oriented cover" definition of orientability that they are indeed both orientable.

I am teaching a third-year curves and surfaces course, and what I am looking for is the following. Suppose you don't know what all the connected $1$-dimensional smooth manifolds are. (In my course we can suppose they are embedded submanifolds of $\mathbb R^n$, but that's not very important.) How can one show, using elementary ideas, that any connected curve has to be orientable?

I believe that one way to do this is the following: show that any connected curve can be expressed as the image of a single regular parametrized curve. Once we have this, we're done. It is clear that the argument should use the fact that we can always "reparametrize by arc-length", and by measuring the arc-length from a fixed point in a fixed initial direction, one either gets a diffeomorphism with $\mathbb R$ or with $S^1$, depending on whether or not the curve is closed. Is there an easy way to justify this to students in a curves and surfaces class? Is there an easier argument that I haven't noticed?

The reason I am thinking about this is because all the undergraduate curves and surfaces texts spend a lot of time explaining why surfaces need not be orientable, but never discuss why orientability is never an issue for curves. They also spend a lot of time talking about "covering a surface with multiple coordinate charts" but never discuss (except for $S^1$, sometimes) the need to cover a curve by more than one chart. It would be nice to be able to give my students an easy (but rigorous) explanation of the orientability of any connected curve.

• It seems to me that your argument that any connected curve can be expressed as the image of a single regular parametrized curve does not fall very short of proving that $\mathbb{R}$ and $S^1$ are the only $1$-dimensional varieties. – Andrea Ferretti Feb 8 '11 at 10:07
• Have you looked at Milnor's Topology from the differentiable viewpoint where he has an elementary proof of classification of one manifolds using the arclength parametrization approach – Mohan Ramachandran Feb 17 '11 at 19:57

I think that it's easiest to model an orientable manifold as one whose gluing maps are in $\mathrm{Diff}^+$. Actually the following proof also works in the topological category.

You can assume that all of the charts of the 1-manifold are open intervals, and that any two charts also intersect in an interval. Then, once you orient one of the intervals, the orientation spreads to its neighbors. Now, there is a principle in combinatorics that the orientations will all be consistent unless there is a finite obstruction. What would this obstruction look like? Using the Hausdorff condition, and throwing away redundant charts, you can clean up any finite collection of charts until you either have a sequence of charts or a cyclic sequence of charts chained together at the ends. Then it's clear in either case that there is no finite obstruction.

Note that there are non-Hausdorff 1-manifolds that are not orientable; the Hausdorff condition is thus essential to the proof.

Here are some extra remarks about the argument. First, this combinatorial principle. Another illustration of the same principle is the fact that a graph is $k$-colorable if and only if every finite subgraph is $k$-colorable. If the graph only has countably many vertices, then there is a standard proof by induction; it is only the uncountable case that requires something fancier such as the ultrafilter lemma or Tychonoff's theorem. In fact, the case that's needed is almost the same as the $k=2$ case of colorability. This case, and the orientability argument, is even easier than the general case because the local coloring or orientation is essentially unique.

Second, cleaning up the atlas so that every chart is an interval and a non-empty intersection of any two charts is an interval. The first condition is sometimes part of the definition of an atlas. But if not, every open set in $\mathbb{R}$ is a countable union of intervals (or every open set in $\mathbb{R}^n$ is a countable union of balls) and you can just make them separate charts. As for the second condition, using the intermediate value theorem and the Hausdorff condition, two interval charts can only intersect at one end or at both ends. If they intersect at both ends, then all of the other charts are redundant and the manifold is a circle.

• That example can be given in terms of an embedding in $\R^2$: take a line segment AB, and glue B to a point C in the middle in a way such that the loop based at C forms a drop shape. Then the non-Hausdorff manifold is not orientable at C. To my knowledge the proof that the real line and $S^1$ are the only two orientable 1-manifolds essentially reduces to the Hausdorff condition. But maybe one does not need countable basis to prove orientability? – John Jiang Feb 7 '11 at 17:16
• The principle of a finite obstruction follows from a special case of Tychnoff's theorem, that the product of any collection of finite sets is compact. So it works for any cardinality, so I am not using a countable basis. Note then that there are two Hausdorff 1-manifolds that don't have a countable basis: A line that is Alexandroff-long in one direction, and a line that is Alexandroff-long in both directions. – Greg Kuperberg Feb 7 '11 at 17:36
• @Greg: not that I don't believe you, but could you elaborate on how you are applying this finite obstruction principle? I see that it is secretly a form of the compactness theorem / ultrafilter lemma but I am not sure how you are using it. – Qiaochu Yuan Feb 7 '11 at 17:44
• @Qiaochu I'm saying that if you can consistently orient any finite collection of interval charts, then you can consistently orient all of them. That is exactly using the ultrafilter lemma, which in my other remark I call Tychonoff's theorem. – Greg Kuperberg Feb 7 '11 at 18:03
• @Spiro If you are only interested in the countable or paracompact case, then the ultrafilter lemma is an overly technical name for a fairly obvious proof by induction. The manifold has a countable atlas of charts; enumerate them somehow. If there is no finite obstruction to an orientation, then by induction you've oriented everything. – Greg Kuperberg Feb 8 '11 at 6:45

I have an idea, but it might not be simpler than proving directly that there only two diffeomorphism classes of 1-dimensional manifolds...

A preliminary result is that any two distinct points can be joined by a regular curve. Assume now that $M$ is non-orientable. Then its orientation cover, say $N$, has an orientation-reversing involution $s$ without fixed points. Take a regular curve $c:[0,1]\to N$ joining $x$ and $s(x)$ for some $x\in N$ (i.e. $c(0)=x$ and $c(1)=s(x)$).

Since $s$ reverses the orientation of $N$, we must have $s_*(\dot c(0))=a\dot c(1)$ for some negative $a$.

By working in local charts, this shows that $s$ actually defines an involution $\sigma$ of the interval $[0,1]$ with $c\circ \sigma=s\circ c$, and thus $s$ has a fixed point, contradiction.

• An interesting idea, for sure, since it does come down to properties of the interval $[0,1]$ in the end. However, talking about the oriented double cover is too much for my particular course. – Spiro Karigiannis Feb 8 '11 at 4:06

I'm struck by the idea that one ought to prove first that there is a Morse function, and work out orientability and other things as corollaries.

Elementary? Let M be a curve. Can we find a smooth, non-vanishing vector field defined everywhere on M? This proves the theorem.

How to explain? Calling whatever direction the vector points "ahead" is a choice of orientation for the curve.

• Well, this is indeed the right idea. But how can you find such a smooth, non-vanishing vector field defined everywhere on $M$? If one can show, for example, that the entire curve can be parametrized by arc-length (which is the idea I discuss in the actual question), then of course its velocity vector field would be everywhere non-zero. But how does one make rigorous that any connected $1$-dimensional smooth embedded submanifold of $\mathbb R^n$ can always be globally parametrized by arc-length, not just locally? – Spiro Karigiannis Feb 8 '11 at 4:19
• Can you not adapt the same argument that shows that any locally path-connected, connected space is globally path-connected? (I.e., define two points to be equivalent if there is a local parametrization hitting them both; show that this is, in fact, an equivalence relation; then the curve is partitioned into disjoint, open equivalence classes; since it is connected, there is only one equivalence class. So any two points can be connected by an arclength parametrization. Use Zorn's lemma to show there exists a maximal parametrization, then show (using Hausdorff?) it contains entire equiv class. – Charles Staats Feb 8 '11 at 15:12