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I am trying to define a closed simple curve (that is, a smooth curve) on a manifold $M$ such that it belongs to a specific class of $\pi_1(M)$. How can I be sure that such a curve exists?

The representatives of a fixed class are obviously closed, so I am trying to find under what conditions there are simple smooth representatives.

Filippo Bianchi
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  • It should be pretty straightforward to come up with a null element of $\pi_1(M)$ for any $M$: For basepoint $x\in M$ pick a chart around $x$, draw any loop in the chart representation and push it forward to the manifold. Aside from this, it might be near impossible to say anything else without specifying a particular manifold. – Mnifldz May 14 '20 at 23:55
  • Does thi helpful ? Maybe you can pick a continous representatives, making it a piecewise smooth, and then smoothing it using bump function of Whitney approximation theorem. You have to check that in these processes, it is still homotopic to the original loop. – Kelvin Lois May 15 '20 at 00:05
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    This is always possible if has dimension ≥3, but is impossible (in general) if $\dim()\le 2$. – Moishe Kohan May 15 '20 at 04:24
  • I am interested in $3$-manifolds, so this should work. Is there a simple proof? – Filippo Bianchi May 15 '20 at 08:39
  • I'm not sure how to boost a representative up from non-smooth to smooth. I think something like what @Sou said should work. However, once you have a smooth representative, if the dimension of $M$ is at least three, then you can find simple smooth representatives. There is enough ambient space that any two self-intersecting segments can drift apart a little bit, giving you a simple curve. https://en.wikipedia.org/wiki/Transversality_(mathematics) – J. Moeller May 15 '20 at 11:07
  • @Prototank: Do you know how to approximate continuous functions by smooth functions? – Moishe Kohan May 16 '20 at 16:56
  • @MoisheKohan I think so, but the more I think about it the more I think there are lots of ways to do it.. and potentially make a mistake. Imagine a "wild knot" type curve in $S^1\times D^2$ which represents the generator of homology. Maybe it's a sequence of trefoils getting smaller and smaller which converge to a point, or something like that. Surely you can pick smooth representatives which approximate this to arbitrary precision, but most of these won't be isotopic to eachother. How can we be sure that the smooth approximations are representatives of the same homotopy classes? – J. Moeller May 16 '20 at 21:14
  • @Prototank: The standard argument is the following: Embed your manifold smoothly in some $R^N$, $\iota: M\to \iota(M)\subset R^N$. Let $U$ be a tubular neighborhood of $i(M)$ and $p: U\to \iota(M)$ be the "standard smooth retraction." Let $f: S^1\to \iota(M)\subset R^N$ be a continuous loop. Then $f$ is the limit of a sequence of smooth maps $f_i: S^1\to R^N$. Then for all $i\ge i_0$, $f_i(S^1)\subset U$. Thus, $g_i= p\circ f_i, i\ge i_0$, are smooth approximations of $f$. Now, use the fact that if maps of a circle are sufficiently close, they are homotopic. – Moishe Kohan May 22 '20 at 16:15

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