# 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

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).