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this is a question that came up in my grad knot theory class. my grad class professor raised the question of circles that can be embedded in S3, and he gave us an example by saying the subset {(x, y) : x^2 + y^5 = 0} is an embedded circle in S3 (when viewed as a unit sphere in C2). Then he told us this makes much more sense from knot theory rather than basic algebraic topology. i was curious, how can someone figure out the kind of knot this is in S3 in a simple way using knot theory? is there a simple way at all? thank you.

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closed as off-topic by Ryan Budney, Andrey Rekalo, Daniel Moskovich, Dmitri Pavlov, Chris Godsil Nov 6 '13 at 12:54

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@ Robert: "this makes much more sense from knot theory rather than basic algebraic topology." I'm curious, how can you solve your problem with just basic algebraic topology and no knot theory? – chris May 13 '10 at 1:16
i just assumed we do. i have not actually tried it. i guess you use exponential representations of complex numbers? i honestly don't know – Robert May 13 '10 at 1:28
I mean, quite honestly, I'm not in the grad student phase yet, so I simply don't even see why that subset should be a circle, much less why it should be the cinquefoil knot. – chris May 13 '10 at 1:32
This is a homework problem. – Charlie Frohman May 13 '10 at 2:22
An interesting homework problem, but yes this sounds like you are asking us to do your homework... – Sam Nead May 13 '10 at 11:49

Instead of the unit sphere, consider the sphere of radius $\sqrt2$, containing the unit torus $\{ (z,w) : |z| = |w| = 1\}$. For each fixed value of $z$ in the torus, there are 5 solutions to the equation $z^2 + w^5 = 0$, arranged symmetrically about the circle in the $w$-coordinate. These are the fifth roots of $-z^2$. Similarly, for each value of $w$, there are two solutions to the equation (given by the square roots of $-w^5$) that are diametrically opposite in the unit $z$-circle. As you transport $z$ around its unit circle, the solutions to the equation will complete $2/5$ of a rotation in the $w$ coordinate. This yields a (2,5) torus knot. It is not hard to show that the equation has no other solutions in $S^3$.

I have no idea what was meant by the professor's remark. I usually think of knot theory as a subject defined by the objects one studies (namely knots), while algebraic topology is defined by the tools one uses. However, I don't hold this view with particularly strong conviction.

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how do you get that? my diagram doesn't work out – Robert May 13 '10 at 1:27
@Robert, how do you get what? – Mariano Suárez-Alvarez May 13 '10 at 23:09

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