## Plücker Formulas

Let $C$ be a curve and $C^*$ be the dual curve. We say that $C$ has traditional singularities if every point of $C$ and $C^*$ is smooth, a node or a cusp. That is, if we take affine coordinates around $p$ and get equation $f$ for $C$, and look in the ring of formal power series $\mathbb{C}[[x,y]]/(f)$, then we have a ring isomorphic to one of $\mathbb{C}[[x,y]]$, $\mathbb{C}[[x,y]]/(xy)$ or $\mathbb{C}[[x,y]]/(y^2-x^3)$, the first being a regular point, the second a node, and the third a cusp.

Geometrically, a node on the dual curve corresponds to a bitangent on the original curve, that is, a tangent line which is tangent to two points on $C$ simultaneously, and a cusp on the dual curve corresponds to a flex, that is, a smooth point of $C$ where the tangent line has contact of order three. As $C^{**}=C$, we have the converse statements as well.

Now, let $g$ be the geometric genus of $C$, and let $d=\deg C$, $b$ be the number of bitangents of $C$, $f$ the number of flexes of $C$, $\kappa$ the number of cusps and $\delta$ the number of nodes, and let $d^*,b^*,f^*,\kappa^*,\delta^*$ be the corresponding numbers for $C^*$.

Immediately, we note that $b=\delta^*$, $b^*=\delta$, $f=\kappa^*$ and $f^*=\kappa$. Now, the degree of $C^*$ is the number of points where $C^*$ intersects a generic line in the dual projective space. That is, the number of tangent lines to $C$ passing through a generic point in $\mathbb{P}^2$. Assume that $p$ is a point in $\mathbb{P}^2$ which doesn’t lie on the tangents to any singular points of $C$. We can choose coordinates such that $p=(0:0:1)$, and then if $C$ is given by a polynomial $g(x_0,x_1,x_2)=0$, the tangent lines to $C$ through $p$ are given by the smooth points of $C$ with $\partial g/\partial x_2=0$. Now, this is an equation of degree $d-1$, and it passes through each node with multiplicity 2 and each cusp with multiplicity 3. So then we have that the number of solutions, which is the number of tangents through $p$, is $d(d-1)-2\delta-3\kappa=d^*$, by applying Bezout and using these observations.

Now we’ll look at the projection map from $p$ to the line and apply Hurwitz’s Theorem. It gives us that $2g-2=-2d+B$ where $B$ is the number of branch points of the map. So a smooth point is a branch point if and only if its tangent passes through the point we’re projecting from, so we have $d(d-1)-2\delta-3\kappa$ branch points among the smooth points, and it happens that though nodes aren’t branch point, cusps are, so Hurwitz tells us that $2-2g=2d-d(d-1)+2\delta+2\kappa$. This is then $g=\frac{(d-1)(d-2)}{2}-\delta-\kappa$, a more refined genus formula, which gets the geometric genus of mildly singular curves.

Now, we combine all this to get the classical Plücker formulas, which are the above genus formula, as well as $g=\frac{(d^*-1)(d^*-2)}{2}-b-f$, $d^*=d(d-1)-2\delta-3\kappa$ and $d=d^*(d^*-1)-2b-3f$. We can use these to say quite a bit about the geometry of specific curves, now.

For instance, take a smooth cubic curve in the plane. Bezout’s Theorem tells us that there aren’t any bitangents, because then a line would have to intersect the cubic in four points. Similarly, a tangent with multiplicity higher than two can have at most three, so the cubic has traditional singularities and the Plücker formulas hold. So then, we plug $d=3$ and $\kappa=\delta=0$ into the formulas to get $d^*=3(3-1)=6$ and $3=6(6-1)-3f$ which tells us that $f=9$, so an elliptic curve will have nine flex points. In fact, these points are rather special for a second reason: any line connecting two flex points on an elliptic curve will intersect it in a third one.

Now look at a smooth plane quartic with traditional singularities. then $d^*=4(4-1)=12$, and $2b+3f=12(12-1)-4=128$. But the genus formula tells us that $3=11(10)/2-b-f=55-b-f$, and so we have a system of linear equations. We can solve this, and get that there are exactly 24 flexes and 28 bitangents on a quartic curve.

More generally, if we have a smooth plane curve of degree $d$ with traditional singularities, we can solve the Plücker formulas to tell us that there are $3d(d-2)$ flexes and $\frac{1}{2}d(d+1)(d-1)(d-2)-4d(d-2)$ bitangents.