Timeline for Can real algebraic knots be recovered from their projections?
Current License: CC BY-SA 3.0
7 events
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| Sep 20, 2017 at 20:35 | comment | added | j.c. | 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 | |
| Sep 20, 2017 at 20:09 | comment | added | David G. Stork | 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. | |
| Sep 20, 2017 at 19:25 | comment | added | j.c. | @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? | |
| Sep 20, 2017 at 18:51 | comment | added | David G. Stork | Yes... Simply assign the different crossings (as described) to the projections to see that the resulting knots (or un-knot) preserve real-algebraic curveness. | |
| Sep 20, 2017 at 18:47 | comment | added | Dustin G. Mixon | Is it clear that your joining operation preserves real-algebraic-curveness? | |
| Sep 20, 2017 at 18:44 | history | edited | David G. Stork | CC BY-SA 3.0 |
added 129 characters in body
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| Sep 20, 2017 at 18:35 | history | answered | David G. Stork | CC BY-SA 3.0 |