1
$\begingroup$

Question. Let $C$ be a generic smooth curve of degree $d$ in $\mathbb{CP}^2$, and let $P$ be an arbitrary point away from this curve. How many lines are there through point $P$ that are tangent, or have tangency of order $k$ (for any $k$ between 3 and $d$) with $C$? Probably this can be done for small $d$ using the equation for $C$, but I would like to find out if there is a formula for general $d$.

$\endgroup$

2 Answers 2

3
$\begingroup$

You have some polynomial $f(x,y,z)$. A line through the point $(1:0:0)$ can be paramaterized by a map from $\mathbb P^1: (u:v) \to (u:av:bv)$ for some constant $a$ and $b$. $f$ restricts to a degree $d$ polynomial in $u$ and $v$. Since it has no roots where $v=0$, set $v=1$. You now have a univariate polynomial such that the $k$th coefficient is a degree $k$ polynomial in $a$ and $b$. The discriminant of this polynomial determines whether it has a double root, which is the same as if the line is tangent. The discriminant is a degree $d(d-1)$ polynomial in $a$ and $b$.

Thus, the generic number of roots is $d(d-1)$. This is similarly the generic number of tangent lines. The generic number of inflection points or higher is zero, as this would require a multiple root of the discriminant.

To check that there is not some extra constraint on the discriminant polynomial that forces a multiple root, we can just find an example that does not have a multiple root. If $f(x,y,z)=x^d+xy^{d-1}+z^d$, then $f(u)=u^d+ua^{d-1}+b^d$, whose discriminant, which is $a^{d^2-d}+b^{d^2-d}$ up to some constant factors, has no multiple roots.

$\endgroup$
1
  • $\begingroup$ You can also find the tangency points as intersections with the polar curve $\partial f/\partial x=0$. This is an alternative way to see that there are $d(d-1)$ such points. If the base point of the pencil is $[a:b:c]$, you need the polar to be $a \partial f/\partial x + b \partial f/\partial y +c \partial f/\partial z=0$. For non generic curves the amount by which $d(d-1)$ is bigger than the actual number can be controlled (depending on higher order tangencies and singularities). Look for "Plücker formulas". $\endgroup$
    – quim
    May 24, 2012 at 9:34
6
$\begingroup$

Let $\tilde\pi:\mathbb{PC}^2\setminus\{P\}\to\mathbb{PC}$ be the projection from $P$, which restricts to a surjective morphism $\pi:C\to\mathbb{PC}$ of degree $d$. You are asking for the number of points of ramification of this morphism with multiplicity, or more precisely the degree of the ramification divisor $R$. By Hurwitz' theorem, it can be computed as $$\deg(R) = 2g(C)-2 - d(2g(\mathbb{PC}) - 2)=2\binom{d-1}{2} + 2(d-1) = d(d-1).$$

$\endgroup$
1
  • $\begingroup$ A technical note: There is of course no morphism from $\mathbb {CP}^2$ to $\mathbb {CP}$. You mean to say that $\bar{\pi}$ is a morphism from $\mathbb {CP}^2-\{P\}$. $\endgroup$
    – Will Sawin
    May 24, 2012 at 6:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.