# Linking number and intersection number

Consider a disjoint union of two circles $$A$$ and $$B$$ smoothly embedded in $$\mathbb{R}^3$$ with linking number more than $$1$$. Suppose we know that there exists a disc $$D$$ in $$\mathbb{R}^3$$ such that $$\partial D=A$$ and $$D$$ intersects $$B$$ at one point. Does that imply that the linking number of $$A$$ and $$B$$ is $$1$$?

I know the answer is yes, if $$D$$ intersects $$B$$ transversely at one point. The answer would be no, if we remove the assumption that $$A$$ and $$B$$ are linked, because we can take the disc to touch $$B$$ at one point.

So, another version of the question is the following: if we perturb the embedding of $$D$$ rel $$A$$ to make it transverse to the circle $$B$$, is it possible that it always increases the number of points in $$D \cap B$$?

$$\DeclareMathOperator\tX{\widetilde{X}}\DeclareMathOperator\tB{\widetilde{B}}\DeclareMathOperator\tD{\widetilde{D}}\DeclareMathOperator\Z{\mathbb{Z}}$$

In fact, $$B$$ must intersect $$D$$ at least $$|\text{link}(A,B)|$$ times. Here, by the way, $$D$$ can be an arbitrary Seifert surface, not just a disk (so $$A$$ need not be the unknot).

The cleanest way to see this is to use covering spaces.

Let $$X = \mathbb{R}^3 \setminus A$$ and let $$\pi\colon \tX \rightarrow X$$ be the (unique) infinite cyclic cover of $$X$$. To calculate the linking number of $$A$$ and $$B$$, regard $$B$$ as a map $$B\colon [0,1] \rightarrow X$$ with $$B(0) = B(1)$$ and let $$\tB\colon [0,1] \rightarrow \tX$$ be a lift of $$B$$ to $$\tX$$. The points $$\tB(0)$$ and $$\tB(1)$$ differ by a deck transformation. Identifying the deck group with $$\Z$$, this deck transformation is an integer $$n$$ that equals the linking number of $$A$$ and $$B$$ (note that I'm being a little sloppy with the sign of the linking number since the above does not depend on the orientation of $$A$$, but this doesn't matter for your question).

The disk $$D$$ lifts to a bunch of disjoint surfaces $$\tD_i$$ indexed by $$i \in \Z$$. Indexing the $$\tD_i$$ correctly, they divide $$\tX$$ up into connected components $$\tX_i$$ such that $$\tX_i$$ contains $$\tD_i$$ and $$\tD_{i+1}$$.

When we lift $$B$$, we have to choose the starting point, and we might as well choose it to lie in $$\tX_0$$. The endpoint of $$\tB$$ will then be in $$\tX_n$$ where $$n$$ is the linking number. Assuming for concreteness that $$n \geq 1$$, we now come to the key point: to get from $$\tX_0$$ to $$\tX_n$$, the path $$\tB$$ must pass through each of $$\tD_1,\ldots,\tD_n$$. Looking downstairs, this means that $$B$$ must intersect $$D$$ a minimum of $$n$$ times, as claimed.

• Interesting. I would have said the right way to think about linking numbers is via Poincare (or Alexander) duality. But that would not have led me to your strong conclusion (unless I had used homology/cohomology with twisted coefficients). And the conclusion does not hold for a linked $p$-sphere and $q$-sphere in $(p+q+1)$-space if $p$ and $q$ are greater than $1$ (in which case covering spaces are not the right way to think). Apr 13 at 19:52
• @TomGoodwillie: Your gentle criticism that I should have been less insistent about the One True Way is well-taken, and I’ll edit it to tone that down (in my defense, I am massively jet lagged right now!). Of course, the right thing to do is to know a bunch of different ways to think about things. Apr 13 at 20:03
• Thank you @AndyPutman for a very nice explanation. Apr 13 at 20:38
• @TomGoodwillie: Do you mean the higher dimensional analog of the fact '$B$ must intersect $D$ at least $|link(A,B)|$ times' does not hold? If yes, can you please give any reference or idea about why this would fail? Apr 13 at 20:38
• @user429294: Yes, that is what I mean. For an example, take $p=1$ and $q\ge 2$. Let $D$ be a $2$-dimensional disk. Choose a map $f:D\to \mathbb R^2$ (for example, the $z\mapsto z^L$ map for complex numbers) such that only $0$ goes to $0$, and such that the winding number of the restricted map $\partial D\to \mathbb R^2\backslash 0$ is $L$. Choose an embedding $e:D\to \mathbb R^q$. Together these give a map $(f,e):D\to \mathbb R^2\times \mathbb R^q$ that is an embedding, and such that its restriction to the boundary of $D$ has linking number $L$ with $0\times \mathbb R^q$. Apr 13 at 21:10