# Simple invariants to detect concordance in general 3-manifolds

Let $$Y$$ be a closed, connected, orientable 3-manifold. We call to oriented knots $$K_1, K_2$$ in $$Y$$ (smoothly) concordant if there is a smoothly, properly embedded annulus in $$Y \times I$$ such that the boundary components of the annulus are in $$Y \times \{0\}$$ and $$Y \times \{1\}$$ and are $$K_1$$ and $$K_2$$ (with there respective orientations).

If you are walking along and explicitly given a 3-manifold (say by a framed link) and a pair of knots $$K_1$$ and $$K_2$$ and you are asked if they are concordant, after checking that they are the same class in $$\pi_1$$, what do you do? Specifically what if $$K_2$$ is just the unknot - what sorts of simple invariants can I try and compute to see if $$K_1$$ is not null concordant?

I've seen a couple of papers floating around but I was wondering what kinds of "classical" invariants there are for this sort of problem. For example if $$Y = S^3$$, the first things I might try would be the Arf invariant and the signature.

As an example, maybe take $$Y = S^1 \times S^2$$ and take $$K_1$$ as one of the components of the Whitehead link (where the other component is given 0-framing thus giving $$S^1 \times S^2$$), and take $$K_2$$ to be the unknot.

• Jen Hom has quite a few papers on knot concordant and slice knot related stuffs. You can have a look at those. there are some invariant like $\tao$-invariant, $\epsilon$-invariant etc. If I am not wrong, Jen Hom proved that $K$ is slice(smoothly) then $\epsilon(K)=0$. – Anubhav Mukherjee Oct 24 '18 at 21:13
• But all those results are for $S^3$ – Anubhav Mukherjee Oct 24 '18 at 21:24
• @Anubhav I also don't know if anything from HFK can be considered too elementary. – Mike Miller Eismeier Oct 24 '18 at 21:30
• In the specific case of null-homologous knots, you can use any invariant of (rational or integral, say) homology cobordism for 3-manifold, by doing surgeries and branched covers. Specifically, say $K_1 \subset Y$ is the unknot; then $-p/q$-surgery along $K_2$ is integrally homology cobordant to $Y\# L(p,q)$; likewise, if $\Sigma_r(K_2)$ is rationally homology cobordant to $Y^{\#r}$. – Marco Golla Oct 24 '18 at 23:22
• Also, for topological concordance there is a nice invariant due to Schneiderman (Algebraic linking numbers of knots in 3–manifolds, AG&T, 2003). – Marco Golla Oct 24 '18 at 23:24

If you allow the relaxation of the closed condition so that $$\partial Y$$ is permitted to be a surface, there is a very simple concordance obstruction for knots in a thickening of the boundary, due to Kauffman.
Let $$\partial Y = \Sigma_g$$, a closed orientable surface of genus g. Given a knot $$K \hookrightarrow \Sigma_g \times I$$, take its regular projection to $$\Sigma_g$$: the result is a knot diagram on $$\Sigma_g$$, denoted $$D$$.
Pick a crossing, $$c$$, of $$D$$. Leave $$c$$ from any outgoing arc, and traverse $$D$$ until you return to $$c$$, counting the number of other crossing you passed through. If you passed through an even number of crossings, declare $$c$$ to be even, otherwise delcare it odd. Repeat this for the other crossings of $$D$$.
Denote by $$J ( D )$$ the sum of the signs of the odd crossings. Kauffman shows this is an invariant of $$K$$, and we may define $$J ( K ) = J ( D )$$ [1]. It is shown in [2,3] that if $$J ( K ) \neq 0$$ then $$K$$ is not concordant in $$Y \times I$$ to the trivial knot. Equivalently, $$K$$ does not bound a disc in $$Y \times I$$ (it does not bound a disc in any $$3$$-manifold with boundary $$\Sigma_g$$, in fact).
This extends automatically to obstructing concordances between two non-trivial knots $$K_1$$ and $$K_2$$, in the case when $$\partial Y = \Sigma_g \sqcup \Sigma_{g'}$$, $$K_1 \hookrightarrow \Sigma_g$$, $$K_2 \hookrightarrow \Sigma_{g'}$$. If $$J ( K_1 ) \neq J ( K_2 )$$ then $$K_1$$ is not concordant in $$Y \times I$$ to $$K_2$$.