Consider a knot represented by a real algebraic curve $C$ in $\mathbb{R}^3$. For example, this answer on SE gives a representation of the trefoil knot. Given a projection of $C$ onto the plane (sans crossing information), can one recover the knot up to chirality?
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asked Sep 20, 2017 at 17:55
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2 Answers
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No, this is not possible. For example, the MSE answer of user porst17 that you linked works out a set of algebraic equations for the following parametrization of the trefoil: $$ x = \sin t + 2 \sin 2t, \quad y=\cos t - 2 \cos 2t, \quad z=-\sin 3t. $$
I plotted this in SageMath using:
parametric_plot3d((sin(t)+2*sin(2*t), cos(t)-2*cos(2*t), -sin(3*t)), (t,0,2*pi))
Here are screenshots of the four views of this curve the SageMath notebook JMol viewer, first the positive octant view, then the XY, XZ, and YZ, projections, from left to right.
However, we can carry out exactly the same procedure to get a set of algebraic equations for the following parametrization of the unknot (pay close attention to $z$): $$ x = \sin t + 2 \sin 2t, \quad y=\cos t - 2 \cos 2t, \quad z=-\sin t, $$ and this has the same projection to the xy plane as the trefoil above.
Here's the SageMath code to plot this unknot:
parametric_plot3d((sin(t)+2*sin(2*t), cos(t)-2*cos(2*t), -sin(t)), (t,0,2*pi))
And here are the resulting screenshots, in the same order as above.
To compute a set of implicit equations for this unknot, I followed almost exactly the steps of porst17 in the abovementioned answer, and this culminated in the following SINGULAR code:
ring R = 0,(x,y,z,s,c),dp;
ideal I = 4*s*c-x+s, 4*s^2-y+c-2, -z-s, s^2+c^2-1;
ideal J = eliminate(I,sc);
The unknot is therefore the variety corresponding to the ideal $J=\langle P_1,P_2,P_3,P_4\rangle$ with $$ \begin{aligned} P_1&=x^2+y^2+4xz-12z^2+4y+3\\ P_2&=16z^3-4yz-x-9z\\ P_3&=4yz^2-y^2-xz+6z^2-4y-3\\ P_4&=8xz^2-2xy+2yz-3x-3z. \end{aligned} $$
I don't know how to check if there's a simpler generating set, or e.g. if this curve might be a complete intersection. If anyone has suggestions I can try to work it out since I'm curious.
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$\begingroup$ Do your two knots differ merely in chirality though? $\endgroup$ Commented Sep 20, 2017 at 18:41
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$\begingroup$ The first knot is the trefoil, and the second knot is the unknot. $\endgroup$– j.c.Commented Sep 20, 2017 at 18:44
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$\begingroup$ Ah... OK. Thanks. Easier to see in a diagram than in the equations, but OK. $\endgroup$ Commented Sep 20, 2017 at 18:45
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1$\begingroup$ Good! But what if we know all three projections? $\endgroup$ Commented Sep 20, 2017 at 20:28
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$\begingroup$ Great question! I have no idea yet, but perhaps ask it separately? Actually, I just found an article which seems to address that. arxiv.org/abs/math/0208099 $\endgroup$– j.c.Commented Sep 20, 2017 at 20:37
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No. The top figures show projections of two trefoil knots, each of which can be assigned a different (or same) crossings. Join the different assignments in the different crossings, as shown at the bottom to get two different knots, not related by a mirror reflection.
After all, one can be the un-knot (loop) and another a true knot, and these cannot be related by a mirror symmetry (of course).
answered Sep 20, 2017 at 18:35
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2$\begingroup$ Is it clear that your joining operation preserves real-algebraic-curveness? $\endgroup$ Commented Sep 20, 2017 at 18:47
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$\begingroup$ Yes... Simply assign the different crossings (as described) to the projections to see that the resulting knots (or un-knot) preserve real-algebraic curveness. $\endgroup$ Commented Sep 20, 2017 at 18:51
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$\begingroup$ @DavidG.Stork Sorry, I don't understand your comment. To be more explicit, let us take as given that the two trefoils that you show can be represented by (say, [parametrized by] and/or [the solution set of]) real algebraic (polynomial) equations (though even this is not obvious to me). How do you ensure that the connected sum that you show (your joining operation) is also represented by real algebraic equations? $\endgroup$– j.c.Commented Sep 20, 2017 at 19:25
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$\begingroup$ Dustin and j.c.: I can parameterize the component trefoil portions in any number of ways. I can also use parameterizations of basis sets, such as Bezier curves (which obey real algebraic equations) to continuously splice the components together, thereby preserving algebraic curveness. $\endgroup$ Commented Sep 20, 2017 at 20:09
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1$\begingroup$ I may be missing something but the "splicing" one typically does with Bézier curves does not preserve "algebraic curveness" as I understand it. Splicing gives you a curve that is piecewise parametrized by polynomials but the condition of being an algebraic curve is that the curve is globally the solution set to a system of polynomials. See the definitions here en.wikipedia.org/wiki/Algebraic_curve $\endgroup$– j.c.Commented Sep 20, 2017 at 20:35