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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 $D \cap B$?

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

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.

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    $\begingroup$ 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). $\endgroup$
    – Tom Goodwillie
    Commented Apr 13 at 19:52
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    $\begingroup$ @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. $\endgroup$
    – Andy Putman
    Commented Apr 13 at 20:03
  • $\begingroup$ Thank you @AndyPutman for a very nice explanation. $\endgroup$
    – user429294
    Commented Apr 13 at 20:38
  • $\begingroup$ @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? $\endgroup$
    – user429294
    Commented Apr 13 at 20:38
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    $\begingroup$ @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 $0\times \mathbb R^q$. $\endgroup$
    – Tom Goodwillie
    Commented Apr 13 at 21:10

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