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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.

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.

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A quadrisecant of a smooth knot generically means that as you move along the knot and project the rest to the celestial $P^2$, you see a Reidemeister move of type III (and will from the other $3$ points, too). Is there an analogous meaning for an isoceles trapezoid? –  Douglas Zare May 30 '11 at 3:58
    
Also, I don't think the codimension is right, $3$ while $K^4$ has dimension $4$. –  Douglas Zare May 30 '11 at 4:02
    
@Douglas Zare: There is a 1-dimensional family of cocircular 4-tuples of points (cyclic quadrilaterals) on the knot. Each trisecant on a long knot gives one of these when you 1-point compactify. Requiring the cyclic quadrilaterals to be isosceles picks a 0-dimensional family (a set of points) out of this 1-dimensional family, doesn't it? I'm confused... –  Daniel Moskovich May 30 '11 at 4:20
    
You get a generically $1$-dimensional family of trisecants for each way to send a point of the knot to infinity, so the cyclic quadrilaterals should be $2$-dimensional. –  Douglas Zare May 30 '11 at 4:33
    
@Douglas: I see. Sorry about that. I'm still interested in the question- if it's an easy yes, then what about four vertices of a rectangle? –  Daniel Moskovich May 30 '11 at 8:38
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2 Answers 2

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).

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Very nice references, thanks. –  Douglas Zare May 30 '11 at 13:47
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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.

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.

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I would say an isosceles trapezoid would correspond to a projection with two crossings such that the midpoints of the crossing chords are at the same heights. If the tangent is only in the same direction but not necessarily at the median height, it seems to be a degenerate trapezoid but not necessarily a degenerate isosceles trapezoid. –  Douglas Zare May 30 '11 at 4:22
    
Hi Ryan! This is really nice- but I'm still a bit confused. Every trisecant on a long knot gives rise to a cocircular 4-tuple of points (a cyclic quadrilateral) by 1-point compactifying. This is a 1-dimensional family. But now you have to get the angles right, so there are only a 0-dimensional family of isosceles quadrilaterals, right? Or am I missing something obvious? Anyway, if it's a 1-parameter family, then might every knot even contain four points of a rectangle? –  Daniel Moskovich May 30 '11 at 4:27
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