# When is a connected sum of torus knots a slice knot?

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

• 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. – Ryan Budney Dec 8 '10 at 19:12
• 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. – Jim Conant Dec 8 '10 at 19:19
• 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 – Ryan Budney Dec 8 '10 at 19:33
• @ Ryan, yes I am! – Paolo Aceto Dec 8 '10 at 20:58
• @Jim, I have tried using the alexander polynomial, but I didn't go so far. – Paolo Aceto Dec 8 '10 at 21:09