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

Question

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.

Answer 1

There exist projections with any number of crossings such that any choices of over/under crossings gives the unknot.

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.

Proving that this only gives the unknot is a fun exercise.

Answer 2

For questions of this type Jean Lannes' result on the Arf invariant is helpful: if the canonical genus of the Surface surface constructed from the shadow is $g$, then the fraction of knots with Arf invariant 1 (with respect to the total number of knots, $2^n$) is $\frac{1}{2}(1-\frac{1}{2^g})$.

This shows convincingly a dependency of the number of non-trivial knots induced by the shadow on its complexity, measured by the shadows's genus. (Of course, some of the knots with Arf invariant 0 might also be knotted.)

In the trefoil shadow case there are 8 possibilities and $g=1$ yields a fraction of $\frac{1}{2}(1-\frac{1}{2}) = \frac{1}{4}$ for knots with Arf invariant 1. These 2 possibilities are the two trefoils and the other 6 possibilities are trivial knots.

Reference: Jean Lannes, Sur l’invariant de Kervaire des noeuds classiques, Proposition 6.

Answer 3

Here is a way to get the desired exponential growth, at least for some shadows. (This is morally similar to Ian's comment, but I thought of this method before I read that (ha!), and my proof is different.)

Fix $\Gamma$ a four-valent planar graph without cut vertices, cut pairs (of edges), or "Conway spheres". For example, a large-ish, square-ish region of the square tiling of $\mathbb{R}^2$, plus some extra edges along the boundary, will work. I also want that $\Gamma$ has no symmetries. (I'll add an additional hypothesis below).

Fix $k < \ell - 2$ both "large enough" and independent of $\Gamma$. We now replace all of the vertices of $\Gamma$ with twist regions having somewhere between $k$ and $\ell - 2$ twists. (Some fussing about is needed to ensure that the result is the diagram of a knot, not a link.)

We choose $\Gamma$, $k$, $\ell$, and the sizes of the twist regions so that the resulting knot is hyperbolic and so that each clasp (about each twist region) is a very short geodesic in the hyperbolic metric. Thus, hyperbolic symmetries of the knot complement preserve the set of clasps, so preserve the diagram $\Gamma$, and thus are trivial.

We can now add (or not) one more full twist to each twist region. We count - the number of crossings is (at most) $\ell \cdot |V(\Gamma)|$ while the number of knots produced is more than $2^{|V(\Gamma)|}$. Since $\ell$ is independent of $\Gamma$, we are done.