# Higher homotopy groups of slice disk complement

Let $K \subseteq{\mathbb{S}}^3=∂\mathbb{D}^4$ be a non-trivial slice knot, i.e. $K$ bounds a slice disk $\Delta$ in $\mathbb{D}^4$. Let $N(\Delta)$ be a regular neighborhood of $\Delta$ in $\mathbb{D}^4$.

What is known about $\pi_i(\mathbb{D}^4-N(\Delta))$, for $i\geq2$?

For what it's worth, $\pi_1(\mathbb{D}^4-N(\Delta))$ is known to be normally generated by the meridian of $K$. The knot complement $\mathbb{S}^3 - K$ and $∂(\mathbb{D}^4-N(\Delta))=M_K$ (the zero-framed surgery on the knot $K$) are both known to be aspherical.

• I don't think the non-triviality of $K$ has much impact on your question. For example, connect-sum any slice disc with a $2$-knot such as a Cappell-Shaneson knot. – Ryan Budney Feb 1 '12 at 23:54
• Ah, I wanted $K$ to be non-trivial since I believe that $\mathbb{S}^3-K$ is aspherical only for non-trivial knots. – Aru Ray Feb 2 '12 at 2:00
• You can also do a boundary connect-sum of two slice discs, that would give the connect sum of the two slice knots on the boundary. – Ryan Budney Feb 2 '12 at 2:26
• @Arunima Ray: If $K$ is trivial, its complement is an open solid torus, which is aspherical. – Autumn Kent Feb 2 '12 at 2:54
• oops, my bad, I meant the the zero surgery on the knot $M_K$ is not aspherical for a trivial knot - if I've thought this through correctly, the zero surgery on the trivial knot is $\mathbb{S}^2x\mathbb{S}^1$ – Aru Ray Feb 2 '12 at 3:17

The homotopy groups can be pretty big things. For example, your $D^4 - N(\Delta)$ class of spaces contains the class of all $2$-knot complements -- simply remove a 4-ball neighbourhood of $S^4$ that intersects the $2$-knot in an unknotted disc.

$2$-knot complements have fairly complicated homotopy groups. For example, Cappell-Shaneson knot complements fiber over $S^1$ with fiber a once-punctured $(S^1)^3$. The universal cover of this space is $\mathbb R \times (\mathbb R^3 - \mathbb Z^3)$, so by the Hilton-Milnor theorem, rationally the homotopy groups are a free lie algebra with countably-infinite many generators (up to the action of $\pi_1$ there's just one generator, though).

Off the top of my head I don't know if slice disc groups are any more general than knot groups, they're probably not very far from each other. I think Kawauchi may not cover this but the references in his survey book should mention something.

$2$--knots usually don't have aspherical complements, see Dyer & Vasquez, The sphericity of higher dimensional knots, Canad. J. Math. 25(1973), 1132-1136. This suggests a complicated answer for slice disks in general.