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Motivation: First I present my motivation for this question but this motivational part is not my main question.

I participated in a talk on knot theory. Then I presented the following question:

Prequestion: In the usual consideration of knots we project a knot from $\mathbb{R}^3$ to $\mathbb{R}^2$. So if this projection make a homeomorphism between the knot and its projection then the knot is an unkonted curve. What about the same situation if we replace the standard projection $\mathbb{R}^3\to \mathbb{R}^2$ with the Hopf fibration $q: S^3\to S^2$? Namely assume that we have a knot $K$ in $S^3$ for which the Hopf projection $q$ gives a homeomorphism between $K$ and $q(K)$. Does this implies that $K$ is unknoted?

I realized that perhaps the question received some interest or perhaps it is not trivial for participants. But after the talk I continue to think about this question. I did not have any idea but it was a motivation for my following main question:

My main Question:

Is there a horizontal knot in $\mathbb{R}^3$? By horizontal I mean horizontality with respect to the standard structure of $S^3$: Orthogonal connection associated to the Hopf fibration? If the answer is negative is there another connection on $S^3$ associated to the hopf fibration which admit a horizontal knot? Can horizontal knots be very complicated: namely are there horizontal knots with sufficiently large unknoting number?

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  • $\begingroup$ The answer to your prequestion is no, such a curve must be unknotted. If the Hopf fibration maps the curve homeomorphically to a circle in $S^2$, then either of the discs that circle bounds lifts to an embedded disc in $S^3$ (lifting property of fibrations), so your knot is bounded by a disc, i.e. trivial. For your main question I'm not sure I know what connection you are referring to. $\endgroup$
    – Ryan Budney
    Commented Aug 24 at 0:46
  • $\begingroup$ @RyanBudney The connection is the 2 dimensional distribution orthogonal to the fibers $\endgroup$
    – Ali Taghavi
    Commented Aug 24 at 1:10
  • $\begingroup$ @RyanBudney I am aware of lifting but why embeded lifting. Any way my main question is some thing else $\endgroup$
    – Ali Taghavi
    Commented Aug 24 at 1:16
  • $\begingroup$ A lift of an embedded disc has to be embedded. $\endgroup$
    – Ryan Budney
    Commented Aug 24 at 2:28
  • $\begingroup$ @RyanBudney Yes BTW is my connection explanation clear now? $\endgroup$
    – Ali Taghavi
    Commented Aug 24 at 2:33

1 Answer 1

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The plane field orthogonal to the Hopf fibration is a “standard contact structure” on the three-sphere. Knots whose tangent plane lies in this contact structure are called Legendrian knots. These have a rich theory.

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  • $\begingroup$ Thank you very much for this answer. $\endgroup$
    – Ali Taghavi
    Commented Aug 27 at 11:50
  • $\begingroup$ But the relation to horizontal curves in 3 sphere? $\endgroup$
    – Ali Taghavi
    Commented Aug 27 at 11:58
  • $\begingroup$ Do you mean the corresponding concept in sphere $\endgroup$
    – Ali Taghavi
    Commented Aug 27 at 12:00
  • $\begingroup$ The standard contact structures $\xi_S$ on the three-sphere and $\xi_E$ on the three-space are related by (a) restricting to the southern hemisphere and (b) orthogonal projection. So the theory of Legendrian knots "is the same" in the three-space and the southern hemisphere of the three-space. $\endgroup$
    – Sam Nead
    Commented Aug 27 at 14:09
  • $\begingroup$ For a discussion of (a) and (b) (just above) see the first two pages of Contact structures induced by skew fibrations of $\mathbb{R}^3$ by Michael Harrison - arxiv.org/abs/1904.00405. $\endgroup$
    – Sam Nead
    Commented Aug 27 at 14:10

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