# Homologically zero circles in smooth manifolds

Let $M$ be a $n$-dimensional smooth manifold, $\gamma:S^1 \to M$ be a smoothly embedded circle such that the homology class it represents, $[\gamma]\in H_1(M;\mathbb{Z})$, is zero. Picking a triangulation on $M$ and using simplicial homology, one can see that $\gamma$ bounds a topological surface $F$ in $M$. However, is it true that $\gamma$ must bound a smoothly embedded surface?

One can always put a smooth structure on $F$, since "smooth=topological" in dimension 2, but the inclusion $F \hookrightarrow M$ may not be smooth then. This can be fixed by using smoothing lemma to get a smooth $F \rightarrow M$, but then the boundary of (the image of) $F$ may not be $\gamma$ anymore, but in general only a smooth curve homotopic to $\gamma$. Also such a smooth map need not be a smooth embedding, and one cannot apply Whitney embedding theorem to fix this if $n=3,4$.

This question is motivated after noticing the book 'Lectures of the Topology of 3-Manifolds' by Saveliev tacitly uses this fact for $n=3,4$, and so I am particularly interested in these two dimensions. Any remarks are appreciated!

• In 3d this results goes back to Seifert, see also rybu.org/node/46. In 4d, use general position to get self-intersections which are isolated points, then get rid of them via self-connected sum of your surface. In higher dimensions use general position. – Moishe Kohan Jan 22 '18 at 14:58