Does every knot contain all four vertices of an isosceles trapezoid?

Question

I ask this question with some trepidation, because it may be trivial and/or of entirely recreational interest.

Erika Pannwitz proved in 1933 that every non-trivial knot contains a quadrisecant (four points which lie along a common line). Generically, a knot has finitely many quadrisecants. But what is so special about the condition of collinearity? In a dream I had a few nights ago (I apologise that this question arose in such a context), somebody asked me whether every knot contains all four vertices of an isosceles trapezoid (not in those words). I could answer the question neither in my sleep nor after I woke up. The claim sounded plausible to me (the codimension seems about right), and if it's true, then one might dream that a signed count of such trapezoids might give rise to a knot invariant along the lines of Budney-Conant-Scannell-Sinha's New perspectives on self-linking.

Does every knot contain all four vertices of an isosceles trapezoid? More generally, is there a nice description of the subclass $\mathcal{C}$ of quadrilaterals such that every knot contains all four vertices of at least one, and generically finitely many, quadrilaterals in $\mathcal{C}$?
EDIT: Does every knot contain all four vertices of a rectangle?

I have thought about this problem a bit (trying to put it in the framework people use for dealing with colinearity problems), and it feels like it should be easy (and well-known to experts), but I'm a bit stuck.

Answer 1

In fact, there is a large number of inscribed quadrilateral results (rhombi, rectangles, etc.), mostly inspired by the square peg problem of Toeplitz and Schnirelmann. Here are some references (in alphabetical order): by Griffiths, Makeev, Stromquist, and Wu. There are other related papers, but these are all on quadrilaterals inscribed into space curves. A small warning: not all of these are correct and precise everywhere (please forgive me for being cryptic - this is not the place to elaborate). Finally, if you would like to see more context, see Nielsen's site, Jordan Ellenberg's blog post, my short preprint, and my book, Chapter 5 (sorry for the self-promotion).

Answer 2

In the configuration space of 4 points in $\mathbb R^3$, the subspace of isosceles trapezoids has codimension three. So on a generic knot, you'd expect a 1-parameter (possibly empty) family of isosceles trapezoids whose vertices sit on the knot.

edit:

Regarding your question about inscribed rectangles, I doubt there's a way to extract an invariant of knots from this. Generically a knot has only finitely many inscribed rectangles, but 1-parameter families of inscribed rectangles can degenerate, allowing two edges to come together.

At this degeneracy you have a configuration of two tangent vectors on the knot, where the vectors are parallel and the base-points are separated by an orthogonal vector. You can check that in the configuration space of two points along the knot, such configurations are co-dimension 3. So in general, a 1-parameter family of knots can have such degeneracies, and this would allow for a 1-parameter family of inscribed rectangles to collapse, allowing for an individual inscribed rectangle to slide-off the knot via a 1-parameter family of knots.

It's possible you can find a correction-term -- when the rectangle slides off the knot, perhaps there's another corresponding thing to count. But it's not clear to me what that should be.

As a concrete example -- consider a type-2 Reidemeister move, when you have two parallel strands but before you cross them. If you were to apply a type-1 Reidemeister move to one of the strands, you would create an inscribed rectangle, one for every twist. So inscribed rectangles might be a non-diagrammatic analogue to writhe of a knot diagram.