Algebraic Curves Lecture 7

Last time we stopped with the calculation that \(K_{\mathbb P^r} = \mathcal O_{\mathbb P^r}(-r-1)\). Using this together with the following theorem is quite profitable:

The most common situation in which we use this is where \(k = 1\), in which case the adjunction formula says

\begin{align*} K_Y = K_X\otimes \mathcal O_X(Y) \otimes \mathcal O_Y \end{align*}

or in the divisor notation

\begin{align*} K_Y = (K_X + Y)|_{Y^0}. \end{align*}

Here’s a sketch of the proof for \(k=1\):

Proof

(Sketch). We have an exact sequence

\begin{align*} 0 \to T_Y \to T_X|_{Y} \to N_{Y/X} \to 0. \end{align*}

We have a picture on the board at this point which shows a curve \(Y\) together with a tangent line drawn at a point on \(Y\) and a normal space (a plane) at the same point.

Dualize this to get

\begin{align*} 0 \to N_{Y/X}^* \to T_{X|_{Y}}^*\to T^*_{Y}\to 0 \end{align*}

noting that the rank of these bundles are \(1\), \(n\) and \(n-1\) respectively. It’s a general fact that if you have an exact sequence

\begin{align*} 0 \to \mathcal F'\to \mathcal F\to \mathcal F'' \to 0, \end{align*}

then

\begin{align*} \bigwedge^r \mathcal F = \bigwedge^{r'} \mathcal F' ~\otimes~\bigwedge^{r''}\mathcal F'', \end{align*}

so

\begin{align*} \bigwedge^n T_{X|_Y}^* = N_{Y/X}^* \otimes \bigwedge^{n-1}T_Y^*. \end{align*}

The leftmost term is \(K_{X|_Y}\) and the rightmost term is \(K_Y\), so rearranging, we get

\begin{align*} K_Y = N_{Y/X}\otimes K_X\otimes \mathcal O_Y. \end{align*}

Let’s see Bezout’s theorem.

Proof

\(C_1\) is linearly equivalent to a union of \(d_1\) general lines; likewise, \(C_2\) is linearly equivalent to a union of \(d_2\) general lines. Any two linear intersect in exactly one point, and \(C_1\cdot C_2\) is well-defined on linear equivalence classes. Thus, counting intersection of lines gives \(C_1\cdot C_2 = d_1\cdot d_2\).

Remember: \(C_1\) and \(C_2\) are divisors of \(\mathbb P^2\) and two divisors \(C\) and \(D\) are linearly equivalent (\(C\sim D\)) if and only if \(C - D = \div(f)\). The statement that every curve of degree \(d\) is linearly equivalent to a union of lines is something like saying that every polynomial of degree \(d\) factors as a product of \(d\) linear terms.

Proof

Recall that we saw, as a corollary of Riemann Roch, that \(\deg(K_C) = 2g - 2\). By adjunction we get

\begin{align*} K_C = (K_{\mathbb P^2} + C)_{|_C} &= \left(\mathcal O_{\mathbb P^2}(-3) \otimes \mathcal O_{\mathbb P^2}(d)\right)_{|_C} \\ &= \left(\mathcal O_{\mathbb P^2}(d - 3)\right)_{|_C} \end{align*}

which implies

\begin{align*} \deg(K_C) = (d-3)\cdot d = d^2 - 3d = 2g - 2. \end{align*}

Solving for \(g\) gives the result.

A curve on \(\mathbb P^1\times \mathbb P^1\) (take coordinates \([u_0:u_1]\times [v_1\times v_2]\)) is given by a bi-homogeneous polynomial of bi-degree \((d_1, d_2)\); that is, a polynomial in \(u_0,u_1,v_0,v_1\) which is homogeneous of degree \(d_1\) in \(u_0,u_1\) and homogeneous of degree \(d_2\) in \(v_0, v_1\). For instance, \(u_0^2 \cdot u_1 \cdot v_0 \cdot v_1 + u_1^3 \cdot v_0^2\) has bidegree \((3, 2)\).

Fact: \(C\sim D\subseteq \mathbb P^1\times \mathbb P^1\) if and only if \(C\) and \(D\) have the same bi-degree.

The proof is, again, done by counting the intersections of lines. Omitted.

Now consider an arbitrary curve \(C\) in \(\mathbb P^1\times \mathbb P^1\) of bi-degree \((d_1,d_2)\). It has canonical divisor

\begin{align*} K_C = (K_{\mathbb P^1\times \mathbb P^1} + C)_{|_C} \end{align*}

by the adjunction formula. Once you know that \(K_{\mathbb P^1\times \mathbb P^1} = \mathcal O_{\mathbb P^1\times \mathbb P^1}(-2)\) (see below for this computation) the above equality implies that \(\textrm{bideg}(K_Q) = (-2, -2)\).

Here are two ways of seeing that \(K_{\mathbb P^1\times \mathbb P^1} = \mathcal O_{\mathbb P^1\times \mathbb P^1}(-2)\).

Method 1: We realize \(Q = \mathbb P^1\times \mathbb P^1\) as a smooth quadric in \(\mathbb P^3\) (in fact, every smooth quadric in \(\mathbb P^3\) is isomorphic to \(\mathbb P^1\times \mathbb P^1\)). To see this, take the Segre embedding \([u_0:u_1]\times [v_0:v_1]\mapsto [u_0v_0:u_0v_1:u_1v_0:u_1v_1]\) and notice that this is exactly the surface \(xy = zw\) in \(\mathbb P^3\) after choosing homogeneous coordinates \([x:y:z:w]\).

Now apply adjunction:

\begin{align*} K_Q = (K_{\mathbb P^3} + Q)_{|_Q} = (\mathcal O_{\mathbb P^3}(-4) + \mathcal O_{\mathbb P^3}(2))_{|_Q} = (\mathcal O_{\mathbb P^3}(-2))_{|_Q}. \end{align*}

Method 2: (Thanks Abhishek!) We can instead calculate \(K_{\mathbb P^1\times \mathbb P^1}\) without considering an embedding into \(\mathbb P^3\) by applying the adjunction formula to one copy of \(\mathbb P^1\) inside of \(\mathbb P^1\times \mathbb P^1\). Adjunction says

\begin{align*} K_{\mathbb P^1} = (K_{\mathbb P^1\times \mathbb P^1} + \mathbb P^1). \end{align*}

We know \(K_{\mathbb P^1} = \mathcal O_{\mathbb P^1}(-2)\) since \(K_{\mathbb P^r} = \mathcal O_{\mathbb P^r}(-r - 1)\).

Applying the Corollary about \(\deg(K_C)\) to our curve \(C\) of bi-degree \((d_1,d_2)\), we have

\begin{align*} 2g - 2 &= \deg(K_C) = \deg((K_{\mathbb P^1\times \mathbb P^1} + C)_{|_C}) \\ &= (d_1 - 2)\cdot d_2 ~+~ (d_2 - 2)\cdot d_1 \\ &= 2d_1d_2 - 2d_2-2d_1, \end{align*}

which implies \(g = (d_1-1)(d_2 - 1)\).

This leads to the following proposition.

Comparing the genus/degree formulas for curves in \(\mathbb P^2\) and \(\mathbb P^1\times \mathbb P^1\) warrants the following remark.