# Why do we often consider knots to be embedded in $S^3$ instead of $\mathbb{R}^3$?

When doing knot theory toward the end of my algebraic topology course, we often defined knots as embeddings of $S^1$ in $S^3$ instead of $\mathbb{R}^3$. My professor justified this by saying that $S^3$ is just $\mathbb{R}^3$ with a point added at infinity, which I agree with, but it didn't help with any of the proofs we did. He said $S^3$ is compact which makes it nice but I didn't see this 'niceness' show up anywhere. I can see that almost every knot in $S^3$ corresponds to a knot in $\mathbb{R}^3$, except for knots that pass through the point of infinity. Could someone give me an example of where having knots embedded in a compact surface was more useful or convenient than embedding them in $\mathbb{R}^3$?

• $\mathbb{R}^3$ is $S^3$ with a point deleted, and you can choose that particular point arbitrarily (since the group of homeomorphisms of $S^3$ in whatever reasonably category you're interested in acts transitively on it). A knot $S^1 \to S^3$ can't be surjective, so there's no difference between knots in $\mathbb{R}^3$ and $S^3$. – anomaly Jun 20 '17 at 7:05
• As to the specific question: There's no real difference between $\mathbb{R}^3$ and $S^3$ for the reason mentioned above, but $S^3$ is compact and not contractible, which makes topological invariants easier to find. – anomaly Jun 20 '17 at 7:08
• Which kind of topological invariants would be easier to find? When using Seifert-van Kampen to derive the Wirtinger presentation for a knot $X$, we considered $\pi_1(\mathbb{R}^3-X)$, so $\mathbb{R}^3$ was fine for knot groups. – Osama Ghani Jun 20 '17 at 7:10
• One example that comes to mind: The Lickorish-Wallace theorem states that any closed, orientable, connected $3$-manifold can be obtained from surgery on a link in $S^3$, and that seems to be the natural framework for that sort of result. But ultimately, the choice of working with $S^3$ or $\mathbb{R}^3$ is irrelevant for the reason above. – anomaly Jun 20 '17 at 7:16
• I think Poincare duality is a little more convenient in the compact case (and really homology is in general as you don't have to say any nonsense about compactly supported). The manifolds of a knot associated to a knot (e.g. branched covers and surgeries) are much easier to define and work with as compact manifolds (and this way they appear as natural boundaries of 4-manifolds without having to talk about different pieces of the boundary). – PVAL-inactive Jun 20 '17 at 7:24

## 1 Answer

Knots are commonly studied nowadays using geometry.

In particular, it is important to know whether a knot is hyperbolic, meaning that its complement in $S^3$ has a complete hyperbolic metric of finite volume (this property would be meaningless for the complement of knot in $\mathbb{R}^3$, which never has a complete hyperbolic metric of finite volume).

In the 1970's, Troels Jorgensen and Bob Riley produced examples of hyperbolic knots.

In the late 1970's, William Thurston completely characterized hyperbolic knots in simple topological terms, which was part of a revolution in 3-dimensional geometric topology that he started.