Why does the Gluck twist on a spun knot give the standard $S^4$?

Question

Given a knotted arc $A \subset D^3$ (whose endpoints are, say, at $(\pm 1,0,0)$), the spun knot on this arc is $$\partial\left((D^3, A) \times D^2\right), = (\partial(D^3,A) \times D^2) \cup ((D^3,A) \times \partial D^2),$$ a smoothly embedded 2-sphere in $S^4$.

The Gluck twist on a smoothly embedded 2-sphere $S^2 \hookrightarrow S^4$ deletes a tubular neighborhood $\nu(S^2)$, giving a closed manifold with boundary $S^2 \times S^1$. Reglue $\nu(S^2)$ via the diffeomorphism $S^2 \times S^1 \to S^2 \times S^1, (x,t) \mapsto (\alpha(t)x, t)$, where $\alpha$ is the nontrivial element of $\pi_1(SO(3))$ (ie, $\alpha(t)$ is rotation by the angle $t$). This gives a homotopy 4-sphere.

Gluck claims, in "Embeddings of two-spheres in the four-sphere", that on a spun knot, this construction results in $S^4$. I don't understand his argument, which is (essentially) as follows.

"If the two-sphere $S^2 \times 0$ on the boundary of $S^4 \setminus \nu(S^2)$ is itself the boundary of a three-sphere with handles, "nicely" situated in $S^4 \setminus \nu(S^2)$, then $M^4$ is homeomorphic to $S^4$." He then claims spun knots as a particular case.

(It should be possible to replace homeomorphic here with diffeomorphic, if we can see this geometrically - i.e., without invoking Freedman's theorem. References to this particular case agree that the resulting smooth manifold is the standard $S^4$.)

What he calls $S^2 \times 0$, I assume, just means $S^2 \times \{pt\}$, the latter some point in $S^1$. I do not understand what he means by a "three-sphere with handles" - and this term does not appear elsewhere - nor do I see why he should conclude that $M^4$ is homeomorphic to $S^4$.

(I know there are more general results involving twist-spun knots and 0-concordant knots, but I'd like to understand this argument.)

Answer 1

This was shown by Gluck, in the cited paper; see section 22. The basic point is explained in section 17. (What we now call) the Gluck twist will produce an equivalent knot if the circle action on the 2-sphere extends over some 3-manifold that the knot bounds; this is Theorem 17.1 but I recommend that you try to prove it yourself before looking. For the spin of a knot K in $S^3$, you can cook up such a 3-manifold by `spinning' any Seifert surface that K bounds.

I didn't see the phrase you mention (three-sphere with handles) but I presume that it means a punctured connected sum of copies of $S^1\times S^2$. Note that you can readily find the extension of the circle action over such a manifold. Indeed, $S^1\times S^2$ has a circle action with two circle fixed points. Forming equivariant connected sums gives you such an action on $\#^n (S^1\times S^2)$, and you can just remove an invariant neighborhood of any point on a circle of fixed points.

Finally, a spun knot always bounds a punctured $\#^n (S^1\times S^2)$.