# What happens if we generalize the fundamental group to make knotted loops distinct?

The definition of an element of the fundamental group of a space $X$ based at point $p$ is $$f:[0,1]\rightarrow X,\quad f(0)=p=f(1),$$ defined up to homotopy. This homotopy allows self-intersection, so that there is no sense in thinking about images of loops as knots. Suppose we were to constrain the homotopy class by disallowing self-intersection. Then for each element of the ordinary fundamental group, we would have a copy of every possible knot.

My feeling is that this would get us nowhere, but I can't help but ask if anyone has thought about this and if there might be an interesting application. For instance, how about the complement of a sphere that has some infinite set of knots drilled out of it?

• If you want the set you obtain to be a group, you need to allow «self-intersections» (although you used this term without defining it properly, so I can only guess what you mean by this). – Loïc Teyssier Dec 1 '17 at 14:35
• If you just take $X= S^1$ (a circle), then your "group" would have only two elements: go around the circle in one direction, or in the other direction. You could not compose these elements, and there would be no neutral element. – Goldstern Dec 1 '17 at 14:38
• And if $X$ is a single point, the "group" would be empty. shudder – Goldstern Dec 1 '17 at 14:39
• @Goldstern Good point. – j0equ1nn Dec 1 '17 at 14:47
• I think the reasonable thing would be to look at the space of embeddings of $S^1$ up to isotopy. This is reasonable in dimension $3$, but for higher dimensions this is probably just the space of homotopy classes of maps from $S^1$ into $X$ by general position arguments. I do not see how you could define compositions on this space. – Thomas Rot Dec 1 '17 at 16:07