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Seifert surfaces

The standard definition of a Seifert surface for a link in $S^3$ is an oriented, compact surface embedded in $S^3$, bounding the link. Often, it is assumed to be connected, but given a disconnected Seifert surface, we can always connect it by surgery.

I want to go the opposite direction: What's the most disconnected Seifert surface for a given link?

Boundary links and disconnected Seifert surfaces.

In the following, assume that $L$ is a link with $m$ components.

Definition $L$ is a boundary link if there exists a Seifert surface consisting of $m$ disjoint connected components, each bounding one circle.

Boundary links are rather rare. As Jim Conant says below, all Milnor invariants must vanish for boundary links. Let us relax the definition a bit then.

Definition An $n$-disconnected Seifert surface for a link $L$ is a surface $\Sigma = \Sigma_1 \sqcup \dots \sqcup \Sigma_n$, where:

  • $\partial \Sigma = L$
  • $\partial \Sigma_i \neq \emptyset$
  • $\Sigma_i$ is connected

Clearly, $n$ is at most the number of components of the link, but I believe that e.g. the Hopf link does not have a 2-disconnected Seifert surface. On the other hand, replace the unknots in the Hopf link by annuli. This is a 2-disconnected Seifert surface for the "doubled" Hopf link, and I believe you can't disconnect it any further. Naturally, we can ask now how disconnected a Seifert surface for a link can be.

Link partitions

Definition An $n$-disconnected Seifert surface for a link $L$ gives a partition of the link into $n$ sublinks.

Question Given a partition of the link into sublinks, can the partition be extended to a Seifert surface for the whole link?

(A partitioned Seifert surface gives Seifert surfaces for each of the sublinks, which are disconnected from each other.)

For example, $n$ unlinked unknots of course admit a partitioned Seifert surface for any partition. But my conjecture is that the finest partition for the doubled Hopf link is the one into two annuli. It is only natural to ask then:

Question For a given link, what's the maximally disconnected Seifert surface? How unique is it?

I can formalise this question a bit. Partitions of the link naturally form a partial order, indeed a lattice. If we restrict this partial order to partitions that arise from disconnected Seifert surface, it still has a minimum (coming from any connected Seifert surface), but does it also has a maximum? Which disconnected surface represents it?

Examples and my research so far

  • Every disconnected Seifert surface can be connected by surgery/connected sum, so we could start with a completely connected Seifert surface and see how far we can disconnect it with (reverse) surgery. The two problems with this approach are:
    1. We don't know a priori with which surface to start.
    2. Surgeries don't commute here, since they're embedded.
  • The Hopf link does not have a disconnected Seifert surface, since the two parts of the link have nonzero linking number.
  • The Whitehead link is not a boundary link (as per Jim Conant's answer) since its Milnor invariants are nontrivial. Since the only other partition is the trivial (indiscrete) partition, this is already the maximum.
  • The Borromean rings aren't a boundary link either. But maybe it's an interesting question whether there exists a 2-disconnected Seifert surface! The linking numbers of two circles are all 0, and indeed there is a surface bounding one circle that doesn't intersect the other two circles. Is there a surface bounding the other two circles not intersecting the first surface?
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A boundary link is a link where each component bounds a connected Seifert surface and all of these surfaces are disjoint. Milnor invariants have to vanish for boundary links, which eliminates the Whitehead link and the Borromean Rings as possibilities.

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  • $\begingroup$ Thanks, that's a great start! I edited my question, since I was completely oblivious of boundary links. $\endgroup$ – Manuel Bärenz Jun 15 '18 at 12:59
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    $\begingroup$ Hi Manuel, I'm pretty sure you can't find a disconnected surface for the Borromean Rings, but I can't quite see how to prove it. $\endgroup$ – Jim Conant Jun 15 '18 at 15:10
  • $\begingroup$ I'm also wondering about a more general criterion, like some invariants that obstruct a disconnected surface. $\endgroup$ – Manuel Bärenz Jun 16 '18 at 16:04

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