I don't understand the definition of a knotted curve. It says,

A simple closed space curve is knotted if we cannot fill it in with a disk.

But how can we "fill it in with a disk" and not have intersections? Why else would we be unable to fill it in?

  • $\begingroup$ Look at the simplest knotted curve, a trefoil knot, and see if you can "fill it in with a disk". $\endgroup$
    – Somos
    May 13, 2021 at 17:50
  • 2
    $\begingroup$ Perhaps I don't understand your question "But how can we 'fill it in with a disk' and not have intersections?", because it seems to me that this question can be answered by the simplest possible example: the circle $\{(x,y,0) \in \mathbb R^3 \mid x^2 + y^2 = 1\}$ can be filled in with the disk $\{(x,y,0) \in \mathbb R^3 \mid x^2 + y^2 \le 1\}$ and that disk does not have intersections. $\endgroup$
    – Lee Mosher
    May 13, 2021 at 18:06

1 Answer 1


Here's a precise statement: An unknotted curve is the boundary of an embedded disk in 3-space. A knotted curve is a curve that is not an unknotted curve.

By this definition, we can say a curve can be filled in with a disk if it is an unknotted curve (i.e., there exists an embedded disk whose boundary is the given curve).

The function of the disk is that it records a way to "trivialize" the curve. Imagine that the disk is given polar coordinates $(r,\theta)$. The $r=1$ curve is the boundary, and for $r=\epsilon$ for $\epsilon>0$ very small, the curve is obviously unknotted in an intuitive sense. The family of curves as $r$ changes gives an isotopy from the original curve to the intuitively trivial curve.

If you allow self-intersecting disks (immersed disks rather than embedded disks), then every knot is the boundary of a disk. This corresponds to the fact that every knotted curve can be unknotted by dragging the curve through itself some number of times (see unknotting number). Technical note: there's a straightforward way to make a disk that also has Whitney umbrella singularities, which is a failure of the disk to be an immersion, but I think you can drag them to the boundary and eliminate them. Another technical note: this is different from the notion of immersed genus (which Gabai proved is equal to Seifert genus) because it allows the interior of the disk to intersect its boundary.


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