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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 the $D$ intersectintersects $B$ 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 questionthe 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 $D \cap B$?

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 the $D$ intersect $B$ 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 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 $D \cap B$?

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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user429294
user429294
  • 125
  • 4

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 the $D$ intersect $B$ 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 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 $D \cap B$?