# Map of $\mathbb{R}^3-Knot \to S^1$

Reading Bachman's "A Geometric Approach to Differential Forms", in section 7.8.1 about the Lining Number invariant, I have stumbled upon the following assertion.

Let the knot $$K$$ be defined as a (continuous and differentiable) map from unit circle $$S^1$$ to 3d Eucledian space $$\mathbb{R}^3$$, so:

$$K: S^1 \to \mathbb{R}^3$$

i.e. $$K$$ is a closed curve in 3d space parametrized by a single real variable.

Let $$U=\mathbb{R}^3-K$$, i.e. it is the 3d space without the points associated with the knot (curve).

Then there always exists a map: $$A:U\to S^1$$.

Unfortunatelly author then says that the proof is outside the scope of the book, and I am not sufficiently knowledgeable to know where to find the proof. Can anyone please suggest a good reference for a keen amateur?

Paragraph from the book:

"""

The linking number of a two-component link is precisely the analogous measure that you get when you treat one of the components as the z-axis and the other as the 1-chain over which we integrate. Given any knot $$K$$ in $$\mathbb{R}^3$$, there is a function $$A: \mathbb{R}^3 − K \to S^1$$. (The existence of such a function is, unfortunately, beyond the scope of this book.) If we think of A as a 0-form, then we can differentiate it to get a 1-form on $$\mathbb{R}^3 − K$$. This 1-form is precisely what we can integrate over a second knot to measure how many times it “links” with $$K$$. Interestingly, there is also a point $$p \in S^1$$ such that $$A^{-1}(p)$$ is a surface whose boundary is $$K$$. Just as before, the linking number can also be computed just by appropriately counting the number of intersections with this surface.

"""

• Surely something more is said about $A$ than just that it's a map from $U$ to $S^1$. The mere existence of such a map is trivial, since the map could be constant. Even a surjective map would be easy to produce. It would be reasonable to ask that $A$ not be homotopic to a constant map, and that doesn't look so trivial to me (I'd use Alexander duality and the fact that $S^1$ is a classifying space for $H^1$). Apr 10, 2019 at 0:24
• I have added the relevant paragraph from the book to the main text. Sorry, I didnt do it straight away. $A$ is not trivial. The author starts with two knots $K$ and $K_2$. He represents one of the knots ($K: S^1 \to \mathbb{R}^3$) by differentiating the 0-form $A$ to get the 1-form $dA$, definded everywhere apart from on the actual knot. He then suggests to integrate $dA$ along the 1-chain defined by $K_2$ to get the linking number of the two knots. Unfortunatelly, the author does not go further with it.
– Cryo
Apr 10, 2019 at 0:44
• In this post, I answer a question with the integral that I believe is the author of the book you mention is referring to. See Rolfsen for more details. Apr 10, 2019 at 1:59
• @N. Owad . Thank you!
– Cryo
Apr 10, 2019 at 3:29

The standard reason is homological (Hatcher's book is the usual reference). Alexander duality relates the homology of $$K$$ to the cohomology of $$\mathbb{R}^3-K$$. In particular, $$H^1(\mathbb{R}^3-K;\mathbb{Z})$$ is isomorphic to $$H_1(K;\mathbb{Z})$$ is isomorphic to $$\mathbb{Z}$$. First cohomology with $$\mathbb{Z}$$ coefficients is the same as homotopy classes of maps $$\mathbb{R}^3-K \to S^1$$. If you take the generator $$1\in\mathbb{Z}$$ and run through the isomorphisms, you get a map $$A:\mathbb{R}^3-K\to S^1$$, which we can assume is smooth. Supposing $$S^1$$ is parameterized by $$\theta$$, then $$d\theta$$ is its $$1$$-form. The map $$A^*$$ is the pullback for $$1$$-forms, giving a closed $$1$$-form $$A^*d\theta$$ on $$\mathbb{R}^3-K$$. (I'm a little uncomfortable calling $$dA$$ a $$1$$-form, but it seems $$dA=A^*d\theta$$.)
Given an oriented curve $$C\subset\mathbb{R}^3-K$$, the linking number of $$C$$ with $$K$$ is $$\int_C A^*d\theta.$$
Unfortunately, I don't know of an elementary way to see such an $$A$$ must exist! (Perhaps you could integrate the form in the Gauss linking integral that N. Owad mentions in the comments. The idea would be you fix one knot $$K$$ permanently, then vary the second knot. Fix a point $$x_0\in \mathbb{R}^3-K$$ and define $$A(x)$$ to be the integral along any arc from $$x_0$$ to $$x$$ in $$\mathbb{R}^3-K$$. I haven't checked, but the $$2$$-form ought to be closed, so $$A(x)$$ is well-defined modulo $$1$$, hence $$A$$ can be thought of as a map to a circle.)
A possibly less-mysterious way to get an $$A$$ is through a Seifert surface. There is a triangulation of $$\mathbb{R}^3-K$$ where the Seifert surface is a subcomplex, and then you can define a function to $$S^1$$ by choosing where all the edges in the triangulation go---paths from one side of the Seifert surface to the other should go around $$S^1$$ exactly once---and then there is a way to fill in the triangles and tetrahedra. After this, you have to find a smooth approximation, which is easy in theory.