# 4-manifold obtained from a ribbon disk exterior by attaching a 2-handle is simply-connected if its boundary is a homology sphere

I am reading Lemma 2.1 of this paper (https://arxiv.org/pdf/2012.12587.pdf) and I can't see why $$W$$ is simply-connected. Here is the situation:

Let $$K$$ be a ribbon knot in $$S^3$$; it bounds a ribbon disk $$D$$ in $$B^4$$. Let $$Y$$ be the 3-manifold obtained by 0-surgery on $$K$$. Let $$\nu D$$ be a closed tubular neighborhood of $$D$$ in $$D^4$$ and let $$X=B^4-\text{int} (\nu D)$$. Then we have $$\partial X=Y$$ (https://math.stackexchange.com/questions/3432597/boundary-of-slice-disk-exterior-is-the-zero-surgery-of-slice-knot). Now let $$J$$ be an arbitrary knot in $$Y$$, let $$Y'$$ be the 3-manifold obtained from $$Y$$ by an integral Dehn surgery on $$J$$, and suppose that $$Y'$$ is an integral homology $$S^3$$.

The lemma is claiming that $$Y'$$ bounds a contractible 4-manifold $$W$$, and in the proof $$W$$ is constructed by attaching a 2-handle to $$X$$ along $$J$$ (with the same framing coefficient with surgery coefficient.) In the second paragraph of the proof, there is the following statement: "$$W$$ must be simply-connected if the resulting 3-manifold is a homology sphere." Here the resulting 3-manifold is $$Y'$$ and it is assumed to be a homology sphere, so this means that $$W$$ is simply-connected, but I can't see why. How can we show this?

What I know is that $$W$$ is a homology 4-ball, and $$X$$ has the homology of $$S^1\times B^3$$. Also from van Kampen's theorem applied to $$B^4=X\cup \nu D$$, $$\pi_1(X)$$ is normally generated by the class of $$\{\text{pt.}\}\times \partial D^2 \subset D\times D^2 =\nu D$$. Finally, by van Kampen's theorem applied to $$W=X\cup h$$ ($$h$$ is the 2-handle), $$\pi_1(W)$$ is the quotient of $$\pi_1(X)$$ by the subgroup normally generated by the class of the knot $$J$$.

This is false in general, I’ll prove the existence of a counterexample.

Since $$D$$ is a ribbon disk, $$X$$ has a handle structure with one $$0$$-handle, $$n$$ $$1$$-handles and $$n-1$$ $$2$$-handles (as described in the paper). In particular, $$X$$ has a 2-complex spine.

Now let $$j$$ be any embedded loop in $$X$$. Then by general position $$j$$ is isotopic to a knot $$J$$ in $$Y =\partial X$$ (since we may isotope it off of the 2-complex spine of $$X$$).

There are ribbon knots whose complements admit loops $$j$$ generating homology such that $$\pi_1(W) \neq 0$$ (where as you describe, $$\pi_1(W)$$ is obtained from $$\pi_1(X)$$ by killing an element representing $$j$$). For example, take any non-trivial knot $$H$$, and let $$K=H\# \overline{H}$$. Then $$K$$ is a ribbon knot bounding a disk $$D$$ (obtained by the spinning construction), and $$\pi_1(X)\cong \pi_1(S^3-K)$$. But for any non-trivial knot $$H$$, $$1/2$$ surgery on it has non-trivial fundamental group (if you like by the knot complement problem and geometrization theorem). Let $$j$$ be a loop representing the loop of slope $$1/2$$ in the peripheral torus of $$K$$ (and transferred to a knot $$J\subset Y$$ as above), then attaching a handle along $$J$$ will give a manifold with non-trivial fundamental group and boundary a homology sphere.

Addendum: I realized in fact in this example that the knot $$J$$ is easy to describe. $$0$$-framed surgery on $$H\#\overline{H}$$ is homeomorphic to the splice of $$S^3-\mathcal{N}(H)$$ and $$S^3-\mathcal{N}(\overline{H})$$, obtained by gluing the peripheral tori so that the meridians and longitudes are identified. Take $$J$$ to be the knot of slope $$1/2$$ in the torus.

• @oguzsavk This appears to find a gap in your paper. Jul 12, 2022 at 17:01