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Let $y^2 = x^2(x+1)$ be the nodal cubic, where $x,y \in \mathbb{R}$. Parametrize the curve by $x(t) = t^2 - 1$ and $y(t) = t^3 - t$. Then the curve fails to be regular if $\gamma'(t) = \langle x'(t), y'(t) \rangle = 0$, i.e., if $x'(t) = y'(t) =0$. It should be that this occurs at $(0,0)$ since the curve fails to smooth at this point, but we find the following:

$x'(t) = 2t =0 \iff t=0$ and $y'(t) = 3t^2 - 1 = 0 \iff t = \pm \frac{1}{\sqrt{3}}$.

The point $(0,0)$ corresponds to $t=1$, so why is this point not arising from the computation? Is it because regular points are only defined for smooth points of the curve?

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The parametrization is regular everywhere, but the curve itself is not smoothly embedded because the parametrization is not injective. The curve has two separate "branches" at the node $(0,0)$, and the parametrization passes through them at different times (once at $t=1$ and once at $t=-1$). Near each of $t=1$ and $t=-1$ you have a perfectly good regular curve, but when you look at the curve globally it has a singularity at $(0,0)$ since the parametrization goes through it twice.

Eric Wofsey
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  • This is the proper for cubics of genus $0$ which does not occur for cubics of genus $1$ or elliptics. – Piquito Mar 01 '20 at 02:28
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The definition of "regular" in the differential geometry of curves just means that for each value of the parameter, the parameterization gives a nonzero tangent vector at the corresponding point on the curve. This doesn't mean that the tangent line containing that tangent vector is unique, and the nodal cubic illustrates this: when $t=-1$ we get one tangent vector at $(0,0),$ and when $t=1$ we get another tangent vector which is not parallel to the first one.

With that in mind, we might strengthen the definition of regular to something like "a very regular parametrised curve is one which whose parametrization is injective and whose velocity is never zero". Then we see that the injectivity condition implies that the curve fails to be regular at $(0,0)$, as expected.

Finally, note that your computation actually does tell us that there are no values of $t$ for which the velocity vector is zero. So away from the point where injectivity fails, the curve is regular (which is expected because it is smooth everywhere else).

By the way, when I first read the question, I thought that regularity assumed injectivity anyway (and this is my own personal preference, if that counts for anything); but then I did some googling and it seems that a number of courses don't assume the parametrization is injective!

Will R
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