# How many different knot types can have the same shadow (projection)

If we take a knot diagram and ignore the over/under-crossing information, we obtain a shadow (or projection). Thus a shadow is simply a plane embedding of a 4-regular graph. Clearly, two non-equivalent knots can give rise to the same shadow: in the figure below we have a knot diagram of the trefoil knot (left) and of the unknot (center), which yield the same shadow (right). Going in the opposite direction, a shadow with $n$ vertices induces $2^n$ knot diagrams, since we can assign to each vertex an overcrossing or an undercrossing.

In this same forum, Joe O'Rourke asked if at least one of these $2^n$ knot diagrams corresponds to the unknot (the answer is yes).

My question is:

Start with a shadow of $n$ vertices. How many different knot types can be obtained from the $2^n$ knot diagrams induced by the shadow?

For instance, for the shadow on the right hand side of the figure above, the answer is $3$: by assigning overcrossings or undercrossings to each vertex, one can only obtain the trefoil knot, its mirror image, and the unknot.

The answer surely depends on the shadow. Andy Putman's answer highlights an important point (which was also observed by Lou Kauffman in an email message) : one should only consider shadows with no "cut vertices", since such vertices will become nugatory crossings in a knot diagram. If we don't impose this restriction, there are examples such as the one below (Andy's example). No matter what assignment of over/under-crossings this shadow receives, each crossing will be nugatory, and the corresponding knot will always be the unknot: So one should restrict the question to shadows with no cut vertices.

Lou Kauffman has observed that in this case, the two alternating diagrams associated to the shadow correspond to non-trivial knots. This follows since that the number of crossings in an alternating, reduced diagram is an invariant of the alternating knot (for a proof, see http://homepages.math.uic.edu/~kauffman/Bracket.pdf). Thus a partial answer for a shadow with no cut vertices is: at least two of the diagrams it induces correspond to a non-trivial knot.

My guess is that the answer is "exponentially many'' for "most" shadows. Or at least for some? More specifically: is it true that there exists a $c>1$ such that for all sufficiently large $n$, there exists a shadow whose $2^n$ associated diagrams yield at least $c^n$ different knot types?

I have spent quite a bit of time searching for pretty much any result in this direction, without any success.

• There must be a way to explore this computationally, perhaps with snappy? – Neal Sep 24 '16 at 4:32
• I count three, as the left and right trfoils aren't equivalent. – Gerry Myerson Sep 24 '16 at 6:15
• @GerryMyerson: You're right, of course. I was thinking in terms of knot tables, where usually only the trefoil appears; what I stated was wrong. I've edited the question to correct this. Thanks. – Gelasio Salazar Sep 24 '16 at 12:52
• I don't expect much to come out of, say, the standard 2-bridge diagram of (most) 2-bridge knots. On the other hand, using connected sums it is fairly easy to come up a shadow with $n$ crossings and at least $2^{n/\log_3 n}$ knot types above it. There is probably a better count than the naive one I've done, though. – Marco Golla Sep 25 '16 at 20:20
• Actually, I think 2-bridge knots will work to give $c^n$ knots among $2^n$ crossing changes. Starting with a 2-bridge knot with continued fraction expansion $[4,4,4,4,4,...,4]$, one may switch crossings to change any of the $4$s to $2$s, giving mostly distinct knots (2-bridge knots are essentially determined by a rational number, which is uniquely determined by a continued fraction expansion). – Ian Agol Sep 26 '16 at 1:49

I am writing on my phone and cannot draw a picture, but here is a description of one of them. Put a vertex at the points $(k,0)$ of $\mathbb{R}^2$ for $k=1,\ldots,n$. The crossings at each vertex will look locally like the lines with slope $1$ and $-1$. Connect $(1,0)$ to itself entirely to the left of the vertices and $(n,0)$ to itself entirely to the right of the vertices. Finally, connect $(k,0)$ to $(k+1,0)$ for $k=1,\ldots,n-1$ by two arcs, one above the x axis and one below.