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Jan 15, 2017 at 19:24 history edited Qfwfq CC BY-SA 3.0
I'm really sorry to spoil the "pun" in the title, but it occurred to me more than once that -on stumbling upon this question- I got quite confused. So I though a parenthetic remark in the title may avoid confusion also for other readers....
Jan 14, 2017 at 20:04 comment added Libli Why didn't you consider in the first place the curve defined by $y^3 +x^2y-x^2y^2-x^4 = 0$? Doesn't it satisfy the same hyptohesis as your original curve? But the explanation why it is $\mathcal{C}^{\infty}$ manifold is so much easier... ($y^3 +x^2y-x^2y^2-x^4 = (y-x^2)(y^2+x^2)$).
Jan 14, 2017 at 1:05 answer added Scot Adams timeline score: 3
Jan 5, 2017 at 23:17 history edited Gro-Tsen CC BY-SA 3.0
fix slight computational error / typo (as per comment by John Sidles)
Jan 5, 2017 at 22:28 answer added user73454 timeline score: 7
Dec 29, 2016 at 1:43 comment added John Sidles Thank you for this lovely question-and-answer. There is (as it appears to me) one minor typographic error. Namely, in regard to the second edit, the curve $X$ is in fact the image of $\mathbb{R}$ under $\phi(t)=(t(t^2+2),t^2(t^2+2))$, rather than the expression given for $\phi(t)$ (as can be verified by direct substitution).
Jun 29, 2012 at 12:38 history edited Georges Elencwajg CC BY-SA 3.0
Added Second Edit
Jun 1, 2012 at 11:18 vote accept Georges Elencwajg
Jun 1, 2012 at 6:05 comment added Olivier This question and the answers are fantastic. MO at its very best.
May 30, 2012 at 17:46 answer added Robert Bryant timeline score: 53
May 30, 2012 at 17:16 comment added Martin Brandenburg +1 for your example: Pictures cannot always be trusted ..
May 30, 2012 at 13:26 history edited Georges Elencwajg CC BY-SA 3.0
Added Edit
May 30, 2012 at 12:31 answer added Francesco Polizzi timeline score: 46
May 30, 2012 at 11:52 history asked Georges Elencwajg CC BY-SA 3.0