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The notion of a knot concordance is a rich subject in low-dimensional topology, see Livingston's survey. More precisely:

For $i=0,1$, let $K_i$ be knots in $S^3$. A knot concordance from $K_0$ to $K_1$ is a smooth annulus $A=S^1 \times [0,1]$ in $S^3 \times [0,1]$ such that $\partial A= -(K_0) \cup K_1$ where $-$ denotes the reversed orientation.

Using this relation, we can form a group structure on the set of oriented knots in $S^3$, denoted by $\mathcal{C}$.

Let $K$ be a slice knot in $S^3$, that is, $K$ bounds a smooth disk $D$ embedded in $B^4$. We can also show that $K$ is slice if and only if $K$ is concordant to the unknot in $S^3$.

Let $M$ be a closed oriented $3$-manifold. I wonder:

  1. Can we talk about a knot $M$ behaves like the unknot in $S^3$?
  2. Can we generalize the notion of knot concordance to the oriented knots in $M$?
  3. (Extra) Can we define inverses of knots in $M$ as in the case of $S^3$?
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    $\begingroup$ The unknot in any connected 3-manifold is the (unique up to isotopy) knot that bounds an embedded disk. And sure - we can generalize concordance and study it in any 3-manifold we like using the same definition with $M$ in place of $S^3$. $\endgroup$
    – user101010
    Commented Apr 1, 2022 at 20:42
  • $\begingroup$ How do you justify the uniqueness result up to isotopy? $\endgroup$ Commented Apr 1, 2022 at 21:18
  • $\begingroup$ Intuitively, you can use the disk to shrink the knot down to be really small and move it around however you like. $\endgroup$
    – user101010
    Commented Apr 1, 2022 at 23:29
  • $\begingroup$ The space of smooth embeddings of a disc $D^j$ in a smooth $n$-manifold $N$ has the homotopy-type of the bundle of orthonormal $j$-frames in the tangent bundle to $N$. This is connected provided $j<n$ and $N$ is connected. This is basically just a version of the classification of tubular neighbourhoods theorem, and uses a shrinking argument as the anonymous user suggests. $\endgroup$ Commented Apr 2, 2022 at 7:11
  • $\begingroup$ Thanks for your comments. I have added one more question. Finding inverses might be an issue, is it true in general? $\endgroup$ Commented Apr 2, 2022 at 8:08

1 Answer 1

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One can certainly define concordance by cylinders in $M \times I$ as you suggest. Making the set of concordance classes into a group is problematic, however. Most of the troubles come from the observation that for two knots to be concordant, they must be freely homotopic. This was explored in the Indiana U. PhD thesis of Prudence Heck, Knot concordance in non-simply connected manifolds. I don't think you get a group in any obvious way, for much the same reason that you don't get a group out of the set of free homotopy classes of loops.

There is a notion of homology concordance of oriented pairs $(M,K)$ where $M$ is a homology sphere. The equivalence relation is then concordance in a homology cobordism between $M_0$ and $M_1$. This becomes a group under pairwise connected sum. The unknot in $S^3$ is the 0 element, and pairwise orientation reversal provides inverses.

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  • $\begingroup$ Thanks in advance! I have two questions: 1) is it possible to find a copy of his thesis, online? 2) I heard the homology concordances. Even if cannot we define usual concordances in homology spheres without homology cobordisms? $\endgroup$ Commented Apr 2, 2022 at 16:57
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    $\begingroup$ You can find her thesis (Knot concordance in non -simply connected manifolds) via Mathscinet if you have access. If you don't have access, then send me an email; I'm not that hard to find. I don't understand your question 2). $\endgroup$ Commented Apr 2, 2022 at 20:42
  • $\begingroup$ In $S^3$, the inverse of $K$ is $-\overline{K}$. My second question is can we find inverses of knots in homology spheres explicitly like the case in $S^3$. Is it again, problematic? $\endgroup$ Commented Apr 3, 2022 at 21:18
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    $\begingroup$ Before you can ask about inverses, you have to have a well-defined notion of addition. My point above was that in (most) non-simply connected 3-manifolds, there isn't an obvious notion of addition. So yes, it seems a bit problematic to me. But maybe you can figure it out! $\endgroup$ Commented Apr 3, 2022 at 23:42
  • $\begingroup$ Always there is a hope! :) $\endgroup$ Commented Apr 4, 2022 at 22:36

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