## When k looks and smells like the unknot…

February 14, 2011

Valentine’s day special issue~ ^_^

Professor Gabai decided to ‘do some classical topology before getting into the fancy stuff’ in his course on Heegaard structures on 3-manifolds. So we covered the ‘loop theorem’ by Papakyriakopoulos last week. I find it pretty cool~ (So I started applying it to everything regardless of whether a much simpler argument exists >.<)

Let $M$ be a three dimensional manifold with (non-empty) boundary. In what follows everything is assumed to be in the smooth category.

Theorem: (Papakyriakopoulos, ’58)
If $f: \mathbb{D}^2 \rightarrow M$ extends continuously to $\partial \mathbb{D}$ and the image $f(\partial \mathbb{D}) \subseteq \partial M$ is homotopically non-trivial in $\partial M$. Then in any neighborhood $N(f(\mathbb{D}))$ we can find embedded disc $D \subseteq M$ such that $\partial D$ is still homotopically non-trivial in $\partial M$.

i.e. this means that if we have a loop on $\partial M$ that is non-trivial in $\partial M$ but trivial in $M$, then in any neighborhood of it we can find a simple loop that’s still non-trivial in $\partial M$ and bounds an embedded disc in $M$.

We apply this to the following:

Corollary: If a knot $k \subseteq \mathbb{S}^3$ has $\pi_1(\mathbb{S}^3 \backslash k) = \mathbb{Z}$ then $k$ is the unknot.

Proof: Take tubular neighborhood $N_\varepsilon(k)$, consider $M=\mathbb{S}^3 \backslash \overline{N_\varepsilon(k)}$, boundary of $M$ is a torus.

By assumption we have $\pi_1(M) = \pi_1(\mathbb{S}^3 \backslash k) = \mathbb{Z}$.

Let $k' \subseteq \partial M$ be a loop homotopic to $k$ in $N_\varepsilon(k)$.

Since $\pi_1(M) = \mathbb{Z}$ and any loop in $M$ is homotopic to a loop in $\partial M = \mathbb{T}^2$. Hence the inclusion map $i: \pi_1(\mathbb{T}^2) \rightarrow \pi_1(M)$ is surjective.

Let $l \subseteq \partial M$ be the little loop winding around $k$.

It’s easy to see that $i(l)$ generates $\pi_1(M)$. Hence there exists $n$ s.t. $k'-n \cdot l = 0$ in $\pi_1(M)$. In other words, after $n$ Dehn twists around $l$, $k'$ is homotopically trivial in $M$ i.e. bounds a disk in $M$. Denote the resulting curve $k''$.

Since $k''$ is simple, there is small neighborhood of $k''$ s.t. any homotopically non-trivial simple curve in the neighborhood is homotopic to $k''$. The loop theorem now implies $k''$ bounds an embedded disc in $M$.

By taking a union with the embedded collar from $k$ to $k''$ in $N_\varepsilon(k)$:

We conclude that $k$ bounds an embedded disc in $\mathbb{S}^3 \backslash k$ hence $k$ is the unknot.

Establishes the claim.

Happy Valentine’s Day, Everyone! ^_^

### 5 Responses to “When k looks and smells like the unknot…”

1. guangbox Says:

It’s better to draw a heart-shaped picture today~

2. Pengfei Says:

OH! I love this!

3. […] When k looks and smells like the unknot… […]