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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.

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  • $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented Feb 1, 2012 at 23:54
  • $\begingroup$ Ah, I wanted $K$ to be non-trivial since I believe that $\mathbb{S}^3-K$ is aspherical only for non-trivial knots. $\endgroup$
    – Aru Ray
    Commented Feb 2, 2012 at 2:00
  • $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented Feb 2, 2012 at 2:26
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    $\begingroup$ @Arunima Ray: If $K$ is trivial, its complement is an open solid torus, which is aspherical. $\endgroup$
    – Autumn Kent
    Commented Feb 2, 2012 at 2:54
  • $\begingroup$ 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 $\mathbb{S}^2x\mathbb{S}^1$ $\endgroup$
    – Aru Ray
    Commented Feb 2, 2012 at 3:17

2 Answers 2

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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.

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To add to Ryan's answer,

$2$--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.

On the other hand, if the disk is ribbon, then the complement is aspherical, see Asano, Marumoto, and Yanagawa, T, Ribbon knots and ribbon disks, Osaka J. Math. 18 (1981), 161-174

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    $\begingroup$ I've been thinking about this problem a bit more recently and learned that there are some issues with the second reference. See: On the asphericity of ribbon disc complements ams.org/journals/tran/1985-289-01/S0002-9947-1985-0779064-8/… and Some remarks on a problem of JHC Whitehead doi.org/10.1016/0040-9383(83)90038-1, both by James Howie. The asphericity of ribbon disc complements appears to be open and is related to the Whitehead conjecture. $\endgroup$
    – Aru Ray
    Commented Nov 15, 2018 at 16:31
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    $\begingroup$ Oh wow interesting. $\endgroup$
    – Autumn Kent
    Commented Nov 27, 2018 at 21:05

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