I am working on a presentation about the complexity of determining whether a knot is trivial, and while reading about some topological stuff I somehow managed to "show" that there are no non-trivial knots. Obviously, there has to be something wrong with my "proof", but I can't see where I did something wrong, or maybe which of the things I read about are inaccurate.

Here are the definitions and propositions I've picked up from various papers.

Definition 1: An embedding $N\hookrightarrow M$ is proper, if $\partial N=N\cap\partial M$. And a properly embedded hypersurface $S\subset M$ is essential in $M$, if it cannot be homotoped into $\partial M$ while keeping $\partial S$ fixed in place, and if the fundamental group $\pi_1(S)$ of $S$ injects into $\pi_1(M)$.

Proposition 2: Let $K$ be a tame knot embedded in $\mathbb S^3$, and let $T_K\subseteq\mathbb S^3$ be an open tubular neighborhood of $K$. The knot $K$ is trivial, if and only if the knot complement $M_K:=\mathbb S^3\setminus T_K$ of $K$ contains an essential disk (whose boundary certainly is a longitude of $\overline{T_K}$).

(as far as I know this is a corollary from the fact that a knot embedded in $\mathbb R^3$ is trivial, if and only if it has a spanning disk, that is, a disk embedded in $\mathbb R^3$ whose boundary is the knot traversed once.)

Proposition 3: The complement $T'$ of the interior of a solid torus $T$ in $\mathbb S^3$ is homeomorphic to a solid torus, the longitudes of $T$ are the meridians of $T'$ and the meridians of $T$ are the longitudes of $T'$.

Proposition 4: Every solid torus $T$ contains an essential disk, and every such disk has a meridian of $T$ as boundary.

And here is my corollary from those propositions.

My corollary: Every tame knot is trivial. Proof. Let $K\subset\mathbb S^3$ be any tame knot. Since $K$ is an embedding of the circle $\mathbb S^1$ into $\mathbb S^3$, any open tubular neighborhood $T_K\subseteq\mathbb S^3$ of $K$ is homeomorphic to $\mathbb S^1\times(D^2)^\circ$, i.e. the interior of a solid torus. By proposition 3 the knot complement $M_K=\mathbb S^3\setminus T_K$ is homeomorphic to a solid torus which, by proposition 4, contains an essential disk (with a meridian of $M_K$ or a longitude of $\overline{T_K}$ as boundary). Hence, $K$ is trivial by proposition 1.

Can you see where I've made a mistake or which propositions are inaccurate?


1 Answer 1


Proposition 3 is false, or at least phrased in a misleading way: the complement of a knotted solid torus is S³ is certainly not a solid torus: this happens if and only if the solid torus is unknotted, essentially because of your argument.

Actually, more is true: a theorem of Gordon and Luecke says that the homeomorphism type of the complement of a knotted solid torus determines the type of the knot.

What's true is that the complement of a standard, revolution solid torus in S³ (which certainly is unknotted) is a solid torus: this doesn't apply to knoted solid tori.

  • $\begingroup$ Thank you for your answer :) I initially thought that complements of homeomorphic subspaces are homeomorphic as well (in this case the unknotted torus and knotted torus were homeomorphic, so I thought that their complements are homeomorphic as well). Obviously, this is not the case for unconnected ambient space, but then I found a counter-example even for simply connected spaces (X=[0,2], A={0}, B={1}). Thanks again for your answer :) $\endgroup$
    – Cubi73
    Commented Jan 28, 2021 at 20:35

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