1

Let $X\subset \mathbb P^n$ be a smooth projective curve and $p\in X$. Let $\pi:X-p\to \mathbb P^{n-1}$ be the projection from $p$ to a general hyperplane $\mathbb P^{n-1}\subset \mathbb P^n$. How do we define this map to make the image a projective curve? When is the image $\pi(X)$ smooth?

I know that if $p\notin X$, this is the case when $p$ is not on the secant variety of $X$.

This question comes from this MathOverflow answer, which says:

The intersection of the two quadrics in $\mathbb P^3$ is a complete intersection and defines an elliptic curve, so the genus is 1. A way to see this is to pick a point $p$ on $C$ and project from $p$ onto a general hyperplane. The image curve $C'$ is of degree one less than the original curve, hence $C'$ is a plane curve of degree 3. Since cubics have genus 1, we are done.

We need that the image of the projection is smooth curve of degree 3 in order for it to be genus 1.

KReiser
  • 74,746
Chris Z
  • 922
  • 7
  • 10

1 Answers1

1

The map $\pi:X\setminus\{p\}\to\Bbb P^{n-1}$ is a rational map from a smooth curve to a projective space, so by the curve-to-projective extension theorem, we obtain a unique extension of this map $\overline{\pi}:X\to\Bbb P^{n-1}$.

Unless $X$ is a line, the image of $\overline{\pi}$ is a projective curve: assuming $X$ is not a line, the map $\pi$, hence $\overline{\pi}$, is nonconstant. Combining this with the facts that $\overline{\pi}$ has image of dimension at most one, the image of $\overline{\pi}$ is irreducible, and the map $\overline{\pi}$ is projective (so it has closed image), we conclude that the image of $\overline{\pi}$ is a one-dimensional projective variety, aka a curve (if you're working with schemes, you can also see that the image is reduced since $X$ is).

It's possible for this projection to be many-to-one and have the image be a smooth curve - consider the projection of a plane curve to a line in the plane. (I'm not sure what the correct criteria for $\overline{\pi}(X)$ to be a smooth curve is in this case.) But if the projection is injective on points and tangent spaces, the map is a closed immersion, which forces $\overline{\pi}(X)$ to be a smooth curve. The geometric criteria for this is that $P$ is on no multisecant, where we define a multisecant as a line which intersects our curve with multiplicity greater than two.

In the specific case at hand in the MO answer you're linking to, this is automatic: since $X$ is an intersection of quadric hypersurfaces and not a line, the intersection multiplicity of $X$ and a line through a point on $X$ is at most two, so $X$ has no multisecants.

KReiser
  • 74,746