Given a cubic and a point S not on the cubic, how many tangent lines to F can we draw from S?

Question

Let $F\in \mathbb{C}[x,y,z]$ be an irreducible homogeneous polynomial with total degree 3, defining a cubic in $\mathbb{CP}^2$. Given a point $S$ on the projective plane but not on $\mathbb{V}(F)$, how many tangent lines to $F$ can we draw from $S$ ? By saying "a tangent line" I mean a line intersect with $F$ at some point with intersection multiplicity $\geqslant 2$.

For example, given a cubic with cusp: $y^3-x^2z=0$. If the point $S=[0,1,0]\notin \mathbb{V}(F)$, one can show that there are exactly 2 tangent lines through $S$: $x=0$ and $z=0$, both with intersection multiplicity 3.

So I guess although the number of tangent lines may not be a constant, we can conclude that: $$ \sum I_P(F,l)=6 $$ where $l$ is the tangent line to F through S, P is the tangent point of $F$ and $l$, $I_P(F,l)$ stands for the intersection multiplicity. Does anyone know something about this question and proof? Any references are welcomed. Thanks a lot!

Answer 1

The really cool trick here is to introduce the correct definition: for a projective hypersurface $X = V(F) \subset \mathbf{P}_\mathbf{C}^n$ of degree at least $2$ we can define the dual variety as the projective variety lying in the dual projective space $$ X^\ast :=\{ H \in \mathbf{P}^{n \ast}_\mathbf{C} \mid H\text{ intersects }X \text{ tangentially} \} \subset\mathbf{P}^{n\ast}_\mathbf{C}. $$ $X^\ast$ can be seen to be a projective variety since it's quite explicitly given by the image of the Gauss map $$ g_X : X \to \mathbf{P}^{n \ast}_\mathbf{C} $$ sending the point $p \in X$ to its tangent hyperplane $$ g_X(p) = \{\partial_{0}(F)(p)\cdot Y_0 + \ldots + \partial_n(F)(p)\cdot Y_n = 0\} \in \mathbf{P}^{n \ast}_\mathbf{C} $$ where $Y_0,\ldots,Y_n$ are projective coordinates for $\mathbf{P}^{n \ast}_\mathbf{C}$ and $\partial_i(F) = \frac{\partial F}{\partial Y_i}$ denotes the partial derivative of the homogeneous polynomial $F$ with respect to the variable $Y_i$.

The study of the dual hypersurface $X^\ast$ is rather subtle and requires some machinery to properly understand - for instance, it turns out that $g_X$ is actually a birational map between $X$ and $X^\ast$; this follows fairly directly once you know $X^\ast \subset \mathbf{P}^{n\ast}_\mathbf{C}$ is also a hypersurface, by identifying $g_{X^\ast}$ as an inverse for $g_X$ on some dense open subset. Showing $X^\ast$ and $X$ have the same dimension requires some work though and is a statement that falls under the umbrella term "reflexivity of projective varieties".

Given all these black-boxes, if $X = V(F) \subset \mathbf{P}^2_\mathbf{C}$ is the vanishing locus of a homogeneous polynomial $F \in \mathbf{C}[x,y,z]$ of degree $d > 1$ and $p \in \mathbf{P}^2_\mathbf{C}\setminus X$, as in your situation, we see that $$ \{H \in \mathbf{P}^{n \ast}_\mathbf{C} \mid p \in H\} \cap X^\ast \subset \mathbf{P}^{n \ast}_\mathbf{C} $$ is the intersection of $X^\ast$ with the projective line $$ L = \{H \in \mathbf{P}^{n \ast}_\mathbf{C} \mid p \in H\} \subset \mathbf{P}^{n \ast}_\mathbf{C} $$ of lines passing through $p \in \mathbf{P}^2_\mathbf{C}$. Since $g_X$ is a birational map, we see that (if $L$ is chosen so that $L \cap X^\ast$ lies in an open subset on which $g_X$ is an isomorphism - i.e. $L$ is chosen generically) $$ \# X^\ast \cap L = \# X \cap g^{-1}_X L $$ and since $g_X$ is quite visibly a morphism of degree $d-1$, $g^{-1}_X L \subset \mathbf{P}^2_\mathbf{C}$ is a hypersurface of degree $d-1$. By Bézout's theorem, this means the number of points you're interested in is (with multiplicities)$$ \deg X \cdot \deg g_X^{-1}L = d \cdot (d-1) $$ which equals $6$ if $d=3$ :D

Where I learned this from is Eisenbud's and Harris' textbook 3264 and all that and I think it's a great read in case you're interested in learning more about intersection theory :)

Hope this helps!

Answer 2

As $F$ is irreducible, no information is lost in considering the likewise cubic polynomial $f(x,y)=F(x,y,1)$. The points you are looking for satisfy the equations $f(p)=0$ and $f'(p)[S-p]=0$.

Now the remainder $f'(p)[p]-3f(p)$ is a quadratic polynomial of degree $2$, so that the second equation can be reduced to degree $2$. Now return to the homogeneous version to find that the number of solutions counting multiplicities is indeed $6$.

This generalizes, if $F$ is homogeneous of degree $d$, then $f'(p)[S]$ and $f'(p)[p]-d\cdot f(p)$ both have degree $d-1$, so the tangency equation can be reduced to degree $d-1$ and the Bezout number is $d\cdot(d-1)$.

Not considered in this short argument is that the reduced tangency condition could have a degree smaller than $d-1$ (but I suspect that the irreducibility of $F$ prevents that).