Inverse of a smooth concordance of smooth knots

Question

We say that a smooth concordance of smooth knots C' is inverse to C if the concatenation C•C' is smoothly isotopic to the trivial cylinder. I wonder if there are any known ways of inverting smooth knot concordances? Are all smooth knot concordances invertible? If not, what could be the potential obstructions of that?

Answer 1

I’ll address the second part of the question on obstructions to inverting a concordance. If $C\cdot C’$ is isotopic to the product cylinder, then there is a retract $(S^3\times I, C\cdot C’) \to (S^3, K_1)$, where $C$ is a concordance from $K_1$ to $K_2$, and $C’$ is a concordance from $K_2$ to $K_1$. By restricting this retract, we get a retract $(S^3\times I,C) \to (S^3, K_1)$ and a map $(S^3,K_2)\to (S^3,K_1)$. These also give a retraction $S^3\times I - C\to S^3-K_1$ and a degree 1 map rel boundary $ S^3-K_2\to S^3-K_1$. Hence there is an injective map $\pi_1(S^3-K_1)\to \pi_1(S^3\times I-C)$ and a surjective map $\pi_1(S^3-K_2)\to \pi_1(S^3-K_1)$. Thus a ribbon concordance from $K_1$ to $K_2$ cannot be invertible unless $K_1=K_2$. If one has a degree 1 map between knot complements, then the Seifert genus of $K_2$ must be bigger than that of $K_1$ and the Alexander polynomial of $K_1$ divides that of $K_2$.

In a similar vein, any homological knot invariant (such as Khovanov or knot Floer homology) for which cobordisms act as morphisms must be an embedding. Eg $Id= Kh(C\cdot C’): Kh(K_1)\to Kh(K_1) $ implies $Kh(C): Kh(K_1)\to Kh(K_2)$ is injective.

Answer 2

As far I know, there is no systematic way of inverting a concordance (even assuming you know it is invertible in the first place). The first thing you should try is check whether the double of the concordance works. (By doubling here I mean taking $-(S^3\times[0,1],C)$ and turning it upside down, which gives you a concordance in the other direction.)

There are however obstructions for invertibility. For instance, if $C$ is a concordance from the unknot to $K$, then (essentially by definition) $K$ has to be doubly slice: $K$ has to be a slice of an unknotted embedding of $S^2$ into $S^4$. There are a number of obstructions to this, the simplest of which is probably the fact that the double cover $\Sigma(K)$ (or, indeed, any finite cyclic cover) of $S^3$ branched over $K$ has to embed in $S^4$ if $K$ is doubly slice. A theorem of Hantzsche says that in this case $H_1(\Sigma(K))$ is of the form $G\oplus G$ for some (finite Abelian) group $G$. This gives you plenty of examples of knots that are slice but not doubly slice, for instance all 2-bridge knots $K_{p^2/(p-1)}$ (since the double cover is a lens space, which has cyclic $H_1$). So no concordance to such a knot can be invertible. Something similar happens for concordances where one of the ends has determinant 1 (just because Hantzsche's argument is quite robust). This also works in the topological category, by the way. Further obstructions (in the smooth category) to double-sliceness have been given, for instance, by Jeff Meier here.

From Meier and Livington's paper, I also learnt that the Levine–Tristram signatures vanish on all of $S^1$ for doubly-slice knots (whereas it only needs to vanish away from the roots of the Alexander polynomial for slice knots).