10
$\begingroup$

Isotopy classes of oriented knots in $S^3$ form a commutative monoid with respect to connected sum. Smoothly slice knots, i.e. knots that are the boundary of a smooth properly embedded disk in $B^4$, form a submonoid. The quotient of the monoid of all knots by the submonoid of slice knots is the smooth knot concordance group $\mathcal{C}$. It is a group because every knot has in inverse: its mirror image with reversed orientation.

My question is: what is known about the isomorphism type of the group $\mathcal{C}$?

Of course, much research has gone into studying $\mathcal{C}$, but here I'm only interested in purely group theoretic properties of $\mathcal{C}$, which can be formulated without making reference to the geometric meaning of $\mathcal{C}$. For example, satellite operations promise to reveal much about $\mathcal{C}$, but (as far as I know) nothing about the isomorphism type of $\mathcal{C}$.

Here are the properties of $\mathcal{C}$ that I can think of:

  1. $\mathcal{C}$ is countable and abelian.
  2. There is an epimorphism $\mathcal{C}\to \mathbb{Z}^{\infty}$ (the sum of countably many copies of $\mathbb{Z}$).
  3. There is a split epimorphism $p\colon\mathcal{C}\to (\mathbb{Z/2})^{\infty}$ (i.e. there exists $i\colon (\mathbb{Z/2})^{\infty}\to \mathcal{C}$ such that $p\circ i$ is the identity of $(\mathbb{Z/2})^{\infty}$).

Of course, any epimorphism as in 2. is automatically split, since $\mathbb{Z}^{\infty}$ is free.

Are these three properties all that is known?

Here's how the three properties can be proven:

  1. Every knot has a diagram, and there are clearly only countably many isotopy types of diagrams, so only countably many isotopy types of knots. Connected sum may be seen to be abelian by waving your hands around.
  2. Such an epimorphism is given by (half of) Levine-Tristram signatures at $e^{2\pi i/m}$ for $m$ ranging over the prime numbers.
  3. Such an epimorphism can e.g. be constructed by choosing amphichiral knots $K_1, K_2, ...$ whose Alexander polynomials $\Delta(K_j)$ are irreducible and pairwise not related by multiplication by a unit. Take $i$ to be the inclusion of the subgroup of $\mathcal{C}$ generated by the classes $[K_j]$. Let $p$ send a knot concordance class represented by a knot $J$ to the sum of those $[K_j]$ for which the maximal $e$ such that $\Delta(K_j)^e$ divides $\Delta(J)$ is odd.

Instead of smoothly slice knots, one may quotient by topologically slice knots (boundaries of locally flat embedded disks), which gives the topological knot concordance group $\mathcal{C}_{t}$. The same question can be asked about $\mathcal{C}_{t}$.

$\endgroup$
2
  • 2
    $\begingroup$ On the devil's advocate side of the fence, what questions could you answer if you knew the isomorphism type? $\endgroup$ Apr 6, 2022 at 16:52
  • 2
    $\begingroup$ There's of course lots known about individual knots and classes of knots, but I don't think anyone knows anything more than your 1/2/3 about the group structure. Perhaps the most important group theoretic properties that might one might ask for are the presence of torsion other than 2-torsion, and whether C contains any infinitely divisible elements. Levine's algebraic concordance group surjects onto an infinite sum of Z/4's, but it's not known if this is split. Lots of (all known?) knots that hit those Z/4's are known to be of infinite order. $\endgroup$ Apr 6, 2022 at 16:57

2 Answers 2

3
$\begingroup$

The followings are not purely algebraic results but contain crucial information about the smooth concordance group $\mathcal C$:

  1. Every knot is concordant to a prime knot.

Kirby, R. C., Lickorish, W. R. (1979). Prime knots and concordance. In Mathematical Proceedings of the Cambridge Philosophical Society (Vol. 86, No. 3, pp. 437-441). Cambridge University Press.

  1. Every concordance class admits a hyperbolic representative.

Myers, R. (1983). Homology cobordisms, link concordances, and hyperbolic 3-manifolds. Transactions of the American Mathematical Society, 278(1), 271-288.

  1. Montesinos knots cannot generate $\mathcal C$.

Hendricks, K., Hom, J., Stoffregen, M., Zemke, I. (2020). Surgery exact triangles in involutive Heegaard Floer homology. arXiv preprint arXiv:2011.00113.

$\endgroup$
2
  • 1
    $\begingroup$ This has to do with higher algebraic structures on the concordence classes of knots, i.e. satellite and arborescent structures. So it is only loosely associated with the question at hand. $\endgroup$ Apr 9, 2022 at 18:40
  • 1
    $\begingroup$ That is why I wrote “not” in italics. $\endgroup$ Apr 9, 2022 at 20:42
1
$\begingroup$

The kernel of the product of the split surjections in 2 and 3, namely $\mathcal{C} \to \mathbb{Z}^{\infty} \oplus (\mathbb{Z}/2)^{\infty}$, has a subgroup isomorphic to $\mathbb{Z}^{\infty} \oplus (\mathbb{Z}/2)^{\infty}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.