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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.

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.

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.

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

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.

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 simple answer yes.

A degenerate isosceles trapezoids on a knot (when two of the points come together) is a secant together with a parallel tangent vector. 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.

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, such isosceles trapezoids always exist on the knot.

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.

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