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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.

"""

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    $\begingroup$ 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 $H^1$). $\endgroup$ Apr 10, 2019 at 0:24
  • $\begingroup$ 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. $\endgroup$
    – Cryo
    Apr 10, 2019 at 0:44
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    $\begingroup$ 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. $\endgroup$
    – N. Owad
    Apr 10, 2019 at 1:59
  • $\begingroup$ @N. Owad . Thank you! $\endgroup$
    – Cryo
    Apr 10, 2019 at 3:29

1 Answer 1

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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.

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  • $\begingroup$ Thank you! About the book that you mentioned, is it A. Hatcher "Algebraic Topology"? $\endgroup$
    – Cryo
    Apr 10, 2019 at 9:30
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    $\begingroup$ @Cryo Yes, that's the one. (For learning algebraic topology, some people also recommend Rotman's Introduction, Lee's two books on manifolds, Bredon's Topology and Geometry, May's Concise Course, or Spanier.) For knots themselves, you can look at Rolfsen's Knots and Links or Lickorish's Introduction to Knot Theory. $\endgroup$ Apr 10, 2019 at 17:07

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