# Minimum required crossings in a link diagram for a $k$-component Brunnian link

What is the minimum number, $s$, of crossings in a link diagram for $k$ (component) links fully knotted together such that cutting any single link frees all individual component links--becomes an unlink? (We say such links are Brunnian.)

Clearly if $k=2$ (Hopf link), then $s=2$.

If $k=3$, the Borromean rings show that $s \leq 6$.

Might there be an approach based on roots of some knot polynomial of order $k$?

I believe you are asking how many crossings a Brunnian link must have. It is possible for a Brunnian link with $k$ components to have $8k-8$ crossings but $6k$ is possible using a rubber band chain that closes up. I don't know if this is asymptotically the best possible.

• For $k=6$ (en.wikipedia.org/wiki/Brunnian_link#/media/…) I count $s = 92$, not $8 \cdot 6 - 8 = 40$. Is that figure simply a non-minimal link diagram? – David G. Stork Jan 25 '18 at 2:12
• For $k=3$, the L10a140 link has just $s=10$ crossing, less than either of your formulas (en.wikipedia.org/wiki/L10a140_link), and the Borromean link has just $6$. – David G. Stork Jan 25 '18 at 2:23
• And of course for $k=2$ we find $s=2$... again, much less than your equations would dictate. – David G. Stork Jan 25 '18 at 6:15
• @David Stork: Right, that figure is not minimal. I meant that I don't know whether $6k$ is asyptotically best. It's definitely not the best possible for some small values of $k$. – Douglas Zare Jan 25 '18 at 9:00

There are Brunnian links with $2, 6, 14, 26$ crossings with $2,3,4, 5$ components respectively:    On the other hand, Doug Zare shows that there are $k$-component Brunnian links with crossing number $6k$. Here is another picture for $k=3$: This seems plausibly optimal (ie linear growth of crossings, with coefficient $6$).

There is a trivial lower bound on the crossing number of $2k$ for a $k$-component Brunnian link, $k>2$. Define the number of crossings of a component as the number of crossings that it goes through, counted with multiplicity (so self-crossings count twice). If the number of crossings were $< 2k$, when we add up the number of crossings of all of the components, we get $< 4k$. So one component must have at most 2 crossings. Since each component must cross any other component an even number of times, we see that this component must cross a single other component twice. Then the two components cannot be linked, so we can slide the two crossing one off of the other (possibly over other components in the diagram) to see that it has a trivial component, and hence the link isn't Brunnian.

• Thanks. Yes... I had found your first three examples as well. Apparently, though, nobody has solved the general $k$ case exactly. I suspect there is an approach based on roots of knot polynomials of Brunnian links that would be fully general: a great research project. – David G. Stork Jan 26 '18 at 16:58