Slice knot

In knot theory, a "knot" means an embedded circle in the 3-sphere

${\displaystyle S^{3}=\{\mathbf {x} \in \mathbb {R} ^{4}\mid |\mathbf {x} |=1\}.}$

The 3-sphere can be thought of as the boundary of the four-dimensional ball

${\displaystyle B^{4}=\{\mathbf {x} \in \mathbb {R} ^{4}\mid |\mathbf {x} |\leq 1\}.}$

A knot ${\displaystyle K\subset S^{3}}$ is slice if it bounds a nicely embedded 2-dimensional disk D in the 4-ball.[1]

What is meant by "nicely embedded" depends on the context: if D is smoothly embedded in B⁴, then K is said to be smoothly slice. If D is only locally flat (which is weaker), then K is said to be topologically slice.

The following is a list of all non-trivial slice knots with 10 or fewer crossings; 6₁, ${\displaystyle 8_{8}}$, ${\displaystyle 8_{9}}$, ${\displaystyle 8_{20}}$, ${\displaystyle 9_{27}}$, ${\displaystyle 9_{41}}$, ${\displaystyle 9_{46}}$, ${\displaystyle 10_{3}}$, ${\displaystyle 10_{22}}$, ${\displaystyle 10_{35}}$, ${\displaystyle 10_{42}}$, ${\displaystyle 10_{48}}$, ${\displaystyle 10_{75}}$, ${\displaystyle 10_{87}}$, ${\displaystyle 10_{99}}$, ${\displaystyle 10_{123}}$, ${\displaystyle 10_{129}}$, ${\displaystyle 10_{137}}$, ${\displaystyle 10_{140}}$, ${\displaystyle 10_{153}}$ and ${\displaystyle 10_{155}}$.[2] All of them are smoothly slice.

Every ribbon knot is smoothly slice. An old question of Fox asks whether every smoothly slice knot is actually a ribbon knot.[3]

The signature of a slice knot is zero.[4]

The Alexander polynomial of a slice knot factors as a product ${\displaystyle f(t)f(t^{-1})}$ where ${\displaystyle f(t)}$ is some integral Laurent polynomial.[4] This is known as the Fox–Milnor condition.[5]

• Slice genus
• Slice link
• Conway knot, a topologically slice knot whose smoothly non-slice status was unproven for 50 years