Timeline for What is the best way explain to undergraduates that all 1-dimensional manifolds are orientable?
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| Feb 8, 2011 at 15:12 | comment | added | Charles Staats | 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. | |
| Feb 8, 2011 at 4:19 | comment | added | Spiro Karigiannis | 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? | |
| Feb 7, 2011 at 21:57 | history | answered | jasomill | CC BY-SA 2.5 |