What is a slice knot? so I've recently come across Lisa Piccirillo's proof on the Conway Knot Problem, and I want to learn more about it, but I can't really understand what is meant by whether or not a knot is "slice", and why this is useful.
I have tried looking at other articles, but I still don't understand what is meant by the 4 dimensional space disk.
Any help would be greatly appreciated :)
 A: The underlying question about sliceness of knots is whether a circle always bounds a disk. A knot is an embedding of a circle into $S^3$: note that $S^3=\{x\in\mathbb{R}^4:|x|=1\}$ is the boundary of $D^4=\{x\in\mathbb{R}^4:|x|\leq 1\}$. $D^4$ (the four-dimensional 'disk' or 'ball') and its boundary $S^3$ are the analog of a tennis ball ($D^3$) and its boundary sphere ($S^2$) in one dimension higher. In general, the $n$-dimensional 'disk' is $D^n=\{x\in\mathbb{R}^n:|x|\leq 1\}$ and its boundary is the $(n-1)$-dimensional sphere $S^{n-1}=\{x\in\mathbb{R}^n:|x|=1\}$.
So embedding a circle into the boundary $S^3$, it is a natural question to ask whether that circle (a copy of $S^1$) bounds a 'nicely embedded' $D^2$ inside $D^4$. Clearly the standard embedding of $S^1$ into $\mathbb{R}^2$ (i.e. its definition as an analog to the ones above) bounds a disk, namely the standard embedding of $D^2$. In fact, due to the Jordan-Schoenflies theorem, any embedding of $S^1$ into $\mathbb{R}^2$ bounds a disk.
However, in higher dimensions we might not always have this result. Our particular choice of embedding of $S^1$ into $S^3$ might mean that our $S^1$ no longer bounds a 'nicely embedded' $D^2$. A knot being 'slice' means that it does, and is 'non-slice' if it does not. In particular, Lisa Piccirillo showed that the Conway knot is not slice, so does not bound a nicely embedded disk.
Why this is a 'useful' result is up to opinion. But there is a lot of study in geometric topology surrounding whether certain things bound other things. Problems such as the 'generalized Schoenflies problem' and objects such as 'Alexander's Horned Sphere' may be of interest.
Hope this helps! Feel free to let me know if you don't understand any of what I've written. :)
