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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.

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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.

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Hmm

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Regarding your question about inscribed rectangles, but do they actually occur -- I think doubt there's a (?maybe?) simple answer yes. A sketch of way to extract an argument appears below but it has some gapsinvariant of knots from this.

A degenerate isosceles trapezoids on Generically a knot (when two has only finitely many inscribed rectangles, but 1-parameter families of the points inscribed rectangles can degenerate, allowing two edges to come together) is a secant together with .

At this degeneracy you have a parallel configuration of two tangent vector which is vectors on plane (!) orthogonal to the mid-point of the secant (!). You get these by finding (non-regular) planar diagrams of the knot, where there is a "cusp" type singularity (corresponding to the intermediate step in a Reidemeister 1 move) together with a regular double-point. Edit: technically that's not an argument that such exist but it seems that since there's generically such a large family of these, one of them should satisfy vectors are parallel and the (!) conditionbase-points are separated by an orthogonal vector.

It's a fairly elementary transversality argument to say You can check that in a neighbourhood (in the sense of the compactified configuration space of 4 two points on along the knot) , such configurations are co-dimension 3. So in general, a 1-parameter family of that "degenerate" isosceles trapezoidknots can have such degeneracies, there is and this would allow for a 1-parameter family of genuine ones that approximate inscribed rectangles to collapse, allowing for an individual inscribed rectangle to slide-off the degenerate oneknot via a 1-parameter family of knots.

Well, I suspect such isoceles trapezoids always exist but

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

show/hide this revision's text 2 Added qualifiers (!) to indicate where the argument isn't complete.

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.

Hmm, but do they actually occur -- I think there's a (?maybe?) simple answer yes. A sketch of an argument appears below but it has some gaps.

A degenerate isosceles trapezoids on a knot (when two of the points come together) is a secant together with a parallel tangent vector . which is on plane (!) orthogonal to the mid-point of the secant (!). You get these by finding (non-regular) planar diagrams of the knot where there is a "cusp" type singularity (corresponding to the intermediate step in a Reidemeister 1 move) together with a regular double-point. Edit: technically that's not an argument that such exist but it seems that since there's generically such a large family of these, one of them should satisfy the (!) condition.

It's a fairly elementary transversality argument to say that in a neighbourhood (in the sense of the compactified configuration space of 4 points on the knot) of that "degenerate" isosceles trapezoid, there is a 1-parameter family of genuine ones that approximate the degenerate one.

So yes, since every non-trivial knot has a diagram with a cusp and a regular double-point

Well, I suspect such isosceles isoceles trapezoids always exist on but the knotabove argument has some holes in it.

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