Question

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

Answer 1

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.