An RBG construction of integral surgery homeomorphisms

We define $n$-special RBG links, for which the associated knot pairs are easier to find diagrammatically and the associated knots have the same $n$-surgery.

As in [12], one can experiment on parametrized families of $n$-special RBG links to look for knot pairs that share the same $n$-surgery, such that one of the knots is $n$-slice in some $\#^{l}\overline{\mathbb{CP}^{2}}$ and the other is not $n$-slice in any $\#^{m}\overline{\mathbb{CP}^{2}}$.

In [5], Nakamura showed that for special RBG links such that $R=U$ and $K_{B}$ is $H$-slice in $W$, Rasmussen’s $s$-invariant cannot be used to obstruct $K_{G}$ from being $H$-slice. We generalize Nakamura’s theorem to $n$-special RBG links.

If $n>0$, the above theorem leaves open the possibility of using the $s$-invariant to detect exotic pairs of definite $4$-manifolds from $n$-special RBG links where $R$ is the unknot. For example, we have the following:

Theorem 1.5 also gives a new way of obstructing knots from being $n$-slice, by finding another knot with the same $n$-surgery and checking its $s$-invariant. (This generalizes the $n=0$ method, which was used by Piccirillo in her proof that the Conway knot is not slice in [17], and then extended by Nakamura [13].) For example, we prove the following theorem.

We will also define a different class of $|n|$-RBG links called $n$-peculiar RBG links (see Section 5), for which the red components $R$ are rationally framed and the associated knot pairs can be obtained diagrammatically.

This gives a new construction of RBG links when $n=0$, for which Nakamura’s obstruction in [13] does not immediately apply; so, in principle, they can potentially produce exotic $4$-spheres.

Organization of the paper. In Section 2, we generalize the RBG construction of zero surgeries to integral surgeries. We discuss the construction of a potential exotic pair by cutting and pasting of $n$-traces. In Section 3, we introduce $n$-special RBG links, for which the associated knot pairs can be obtained diagrammatically. In Section 4, we generalize Nakamura’s sliceness obstruction to an $n$-sliceness obstruction using $n$-special RBG links, and we give examples where the $n$-sliceness of knots in $\#^{m}\overline{\mathbb{CP}^{2}}$ is obstructed by this new method. In Section 5, we discuss $n$-peculiar RBG links.

Acknowledgements. The author is grateful to her advisor, Ciprian Manolescu, for his suggestion to work on this project and his continued guidance and support. We would also like to thank Anthony Conway, Nathan Dunfield, Kyle Hayden, Maggie Miller, Lisa Piccirillo, Qiuyu Ren, Ali Naseri Sadr and Charles Stine for helpful suggestions and conversations.

Let $X$ be a smooth, closed, oriented 4-manifold, and let $X^{\circ}=X\backslash int(B^{4})$. Let $K$ be a knot in $\partial X^{\circ}\cong S^{3}$. Suppose $K$ bounds a properly embedded disk $D$ in $X^{\circ}$. There exists a tubular neighborhood $\nu(D)\cong D^{2}\times D^{2}$, where $D$ is identified with $D^{2}\times\{0\}$. Pick a point $p\in\partial D^{2}$ and denote $S^{1}\times\{p\}$ by $K_{D}$. Following Section 2.2 of [13], we make the following definition.

Denote $X^{\circ}\backslash\nu(D)$ by $E(D)$ and denote the trace of the $n$-surgery along $K$ by $X_{n}(K)$. By the trace embedding lemma [9, Lemma 3.3], if $K$ is $n$-slice in $X$, then $-X_{n}(K)$ is smoothly embedded in $X$. In particular, we have that $[D]\cdot[D]=-n$.

Now, given an $n$-surgery homeomorphism $\phi:S_{n}^{3}(K)\rightarrow S^{3}_{n}(J)$, we define

For $n=0$, if the disk $D$ is not null-homologous, then it is possible that $X_{(D,\phi)}$ is not homeomorphic to $X$. Note that if $X$ is definite, such as $\#^{m}\mathbb{CP}^{2}$, every $0$-slice disk is null-homologous.

However, for $X=\#^{m}\overline{\mathbb{CP}^{2}}$, we have that $X_{(D,\phi)}$ is homeomorphic to $X$.

For $(X,X_{(D,\phi)})$ to be a potential exotic pair, we need that the $n$-surgery homeomorphism $\phi:S_{n}^{3}(K)\rightarrow S_{n}^{3}(J)$ does not extend smoothly to an $n$-trace diffeomorphism $\Phi:X_{n}(K)\rightarrow X_{n}(J)$. In some cases, one can see that $\phi$ actually extends smoothly over $n$-traces.

Generalizing Definition 3.12 in [12], we say that an $n$-surgery homeomorphism $\phi:S_{n}^{3}(K)\rightarrow S_{n}^{3}(J)$ has property U, if there exists a choice of surgery diagrams of $S_{n}(K)$ and $S_{n}(J)$, such that $\phi$ sends a $0$-framed meridian of $K$ to a $0$-framed curve $\gamma$ which appears unknotted in the diagram of $S_{n}(J)$.

For example, all links in Example 2.9 have property U.

If the mapping class group of the $n$-surgeries is trivial, then if there exists some $\phi:S_{n}^{3}(K)\rightarrow S_{n}^{3}(J)$ which does not extend over the trace, then the traces $X_{n}(K)$ and $X_{n}(J)$ are not diffeomorphic. In general, it is hard to obstruct extensibility smoothly, but we have obstructions for extending homeomorphically over the traces.

For example, when $n=1,2,4,p^{k}$ or $2p^{k}$ with $p$ an odd prime, every $n$-surgery homeomorphism extends over the traces homeomorphically. For other $n$, it is possible that $\phi:S_{n}^{3}(K)\rightarrow S_{n}^{3}(J)$ does not map meridian to meridian homologically; in such a case, if we also have that the mapping class group of $S_{n}^{3}(K)$ is trivial, then we conclude that their $n$-traces $X_{n}(K)$ and $X_{n}(J)$ are not homeomorphic.