# Sliceness of knots

For a subring $$R⊂ \mathbb Q$$, a knot $$K⊂S^3$$ is called $$R$$-slice if there exists an embedded disk $$D$$ in an $$R$$-homology $$4$$-ball $$B$$ such that $$∂(B,D) = (S^3,K)$$, see [Definition 1.3, KW16]. We say $$K$$ is rationally (resp. integrally) slice if $$R= \mathbb Q$$ (resp. $$= \mathbb Z$$).

In terms of crossing, the minimal example of rationally slice knot seems (probably is) figure-eight knot $$4_1$$, see [Theorem 4.16, Cha07].

A knot $$K ⊂ S^3$$ is slice if it bounds a smoothly embedded disk $$D^2$$ in the $$4$$-ball $$B^4$$. Again in terms of crossing, the minimal example of slice knot is unknot $$0_1$$.

My question is that is there any minimal example of integrally slice knot?

## 1 Answer

Both $$6_1$$ and $$3_1 \# m(3_1)$$ are smoothly slice (as is the unknot), and I claim that all other knots of at most seven crossings are not integrally slice. This will follow from two claims: first, if $$K$$ is integrally slice then $$\frac{\Delta_K''(1)}{2}$$ is even; and second, if $$K$$ is alternating and rationally slice then its signature is zero. (Note that $$8_3$$ satisfies both of these but is not smoothly slice; I don't know whether it is integrally slice.)

The key observation is that if $$K$$ is integrally slice, then $$S^3_1(K)$$ must be integrally homology cobordant to $$S^3$$. To see this, we take a slice disk bounded by $$K$$ in a homology ball, and we remove a ball about some point on the disk to get a concordance from $$U$$ to $$K$$ inside a homology cobordism from $$S^3$$ to itself. Performing a 1-surgery along this cylinder gives us the desired homology cobordism.

From here, filling in the $$S^3$$ end with a ball gives us a smooth homology ball bounded by $$S^3_1(K)$$. Thus $$S^3_1(K)$$ has vanishing Rohklin invariant, or equivalently its Casson invariant is even, and the surgery formula for the latter says that $$\frac{\Delta_K''(1)}{2}$$ must be even.

For the second claim, $$S^3_1(K)$$ is rationally homology cobordant to $$S^3$$, so its Heegaard Floer d-invariant must be zero. When $$K$$ is alternating, this was computed to be $$2\min(0,-\lceil -\sigma(K)/4\rceil)$$ for alternating $$K$$ by Ozsváth and Szabó (arXiv:0209149, corollary 1.5), so we must have $$\sigma(K) \geq 0$$. But the same argument applies to the mirror $$m(K)$$, with signature $$-\sigma(K)$$, so in fact $$K$$ must have signature zero.