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.

  • $\begingroup$ 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 $\epsilon(K)=0$. $\endgroup$ Oct 24, 2018 at 21:13
  • $\begingroup$ But all those results are for $S^3$ $\endgroup$ Oct 24, 2018 at 21:24
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    $\begingroup$ @Anubhav I also don't know if anything from HFK can be considered too elementary. $\endgroup$
    – mme
    Oct 24, 2018 at 21:30
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    $\begingroup$ 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 $Y^{\#r}$. $\endgroup$ Oct 24, 2018 at 23:22
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    $\begingroup$ Also, for topological concordance there is a nice invariant due to Schneiderman (Algebraic linking numbers of knots in 3–manifolds, AG&T, 2003). $\endgroup$ Oct 24, 2018 at 23:24

1 Answer 1


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 $.


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