# Is Rasmussen's s-invariant of a knot an invariant of a 4-manifold?

Let $K$ be a knot in the 3-sphere $S^3$.

Here we denote by $s(K)$ Rasmussen's s-invariant for $K$, and by $X_{K}(n)$ the 4-manifold obtained from the standard 4-ball $B^4$ by attaching a $2$-handle along $K$ with framing $n$.

My question is the following.

For two knots $K$ and $K'$ such that $X_{K}(0)$ and $X_{K'}(0)$ are diffeomorphic, is it true that $s(K)=s(K')$?

I am also interested in the same question for Ozsváth-Szabó's $\tau$-invariant.

Note that it is known that there exist knots $J$ and $J'$ such that $\partial X_{J}(0)$ and $\partial X_{J'}(0)$ are diffeomorphic, and $s(J) \neq s(J')$, see Yasui's paper Corks, exotic 4-manifolds and knot concordance.

• I think that the question is true if $K'$ is the unknot. If $X_K(0)$ is diffeomorphic to $S^2\times D^2=X_U(0)$ where $U$ is the unknot, then $K$ is slice in homotopy 4-ball. By Kronheimer-Mrowka, $s(K)=0$. Similar argument works for $\tau$. Aug 11, 2015 at 17:44
• @user156937 Thank you very much! If one of them is the unknot, the other is also the unknot by Property R theorem by Gabai. If one of them is slice, then the other, say $K'$, is slice in a homotopy 4-ball, therefore $s(K')=0$.The problem is when $K$ and $K'$ are not slice. Aug 11, 2015 at 17:52

Lisa Piccirillo just posted a few days ago this paper on the arXiv. Corollary 1.3 asserts exactly that $s$ is not a 0-trace invariant of the knot.