6
$\begingroup$

This question is about the beaviour of 4-genus of knots with respect to connected sum.

Let us indicate with $T(k)$ a Torus knot of type $(2,k)$, $k$ is an odd integer. Fix an orientation for every $T(k)$ so that $T(-k)$ represents the same knot with reversed orientation.

$T(k)\sharp T(-k)$ is a slice knot. More generally if $K^* $ denotes the mirror of $K$ then $K\sharp K^*$ is slice (infact ribbon).

My question is:

  • Is it true that a knot of the form $T(a_1)\sharp\dots\sharp T(a_n)$ is slice if and only if n is even, say $n=2k$, and we can arrange the coefficients so that for every $i\leq n$ we have $a_{k+i}=-a_i$?

Note that for any pair of knots $K_1$ and $K_2$, if both $K_1$ and $K_1\sharp K_2$ are slice then $K_2$ is also slice. Therefore one only needs to show that there exists $i$ and $j$ such that $a_i=-a_j$.

Of course we can generalize this problem:

  • Which connected sums of torus knots are slice?

Here are some links for definitions:

Torus knot: http://en.wikipedia.org/wiki/Torus_knot

Slice knot: http://en.wikipedia.org/wiki/Slice_knot

Slice genus: http://en.wikipedia.org/wiki/Slice_genus

Improve this question
$\endgroup$
5
  • $\begingroup$ I imagine that's a hard problem. A fairly analogous problem of determining which connect-sums of lens spaces bound rational homology balls was completed in 2007 by Lisca. But being in Pisa I imagine you're well aware of that. $\endgroup$
    – Ryan Budney
    Dec 8, 2010 at 19:12
  • $\begingroup$ My first line of attack would be the Alexander polynomial. This is known for torus knots, and it is also known that a slice knot must have Alexander polynomial of the form $f(t)f(t^{-1})$. This should tell you quite a bit about what connect sums could be slice, although it won't be able to distinguish between a knot and its mirror image, so you'd need some other invariant to take care of this. $\endgroup$
    – Jim Conant
    Dec 8, 2010 at 19:19
  • 3
    $\begingroup$ Oh, this problem is easier than I thought. Litherland computed the group structure of the subgroup of the concordance group spanned by the torus knots, in 1979. See Kearton's survey, here: maths.ed.ac.uk/~aar/slides/durham.pdf $\endgroup$
    – Ryan Budney
    Dec 8, 2010 at 19:33
  • $\begingroup$ @ Ryan, yes I am! $\endgroup$
    – Paolo Aceto
    Dec 8, 2010 at 20:58
  • $\begingroup$ @Jim, I have tried using the alexander polynomial, but I didn't go so far. $\endgroup$
    – Paolo Aceto
    Dec 8, 2010 at 21:09

1 Answer 1

Reset to default
10
$\begingroup$

I believe the answer to your last two questions is yes, and it follows from Litherland's (1979) computation of the Tristram-Levine invariants of torus knots. See Kearton's survey here:

http://www.maths.ed.ac.uk/~aar/slides/durham.pdf

i.e. a connect sum of torus knots is slice if and only if the prime summands appear in balancing mirror-reflected pairs.

Improve this answer
$\endgroup$
0

Your Answer

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

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