Algebraic Curves Lecture 7:

The seventh lecture of algebraic curves by Karl Christ.
  1. Connection to last lecture
  2. Curves in P2\mathbb P^2
  3. Curves in P1×P1\mathbb P^1\times \mathbb P^1

Connection to last lecture

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

Theorem: (Adjunction formula). Let XX be smooth and Y⊆XY\subseteq X smooth of codimension kk. Then we can express the canonical sheaf of YY in terms of the canonical sheaf of XX in the following way:
KY=KX⊗⋀kNY/X⊗OY,\begin{aligned} K_Y = K_X\otimes \bigwedge^k N_{Y/X} \otimes \mathcal O_Y, \end{aligned}
where we include the last tensor just to ensure that this equality happens as sheaves on YY. I believe the tensor products are taken over C\mathbb C.

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

KY=KX⊗OX(Y)⊗OY\begin{aligned} K_Y = K_X\otimes \mathcal O_X(Y) \otimes \mathcal O_Y \end{aligned}

or in the divisor notation

KY=(KX+Y)∣Y0.\begin{aligned} K_Y = (K_X + Y)|_{Y^0}. \end{aligned}

Here's a sketch of the proof for k=1k=1:

Proof.   (Sketch). We have an exact sequence

0→TY→TX∣Y→NY/X→0.\begin{aligned} 0 \to T_Y \to T_X|_{Y} \to N_{Y/X} \to 0. \end{aligned}
We have a picture on the board at this point which shows a curve YY together with a tangent line drawn at a point on YY and a normal space (a plane) at the same point.

Dualize this to get

0→NY/X∗→TX∣Y∗→TY∗→0\begin{aligned} 0 \to N_{Y/X}^* \to T_{X|_{Y}}^*\to T^*_{Y}\to 0 \end{aligned}
noting that the rank of these bundles are 11, nn and n−1n-1 respectively. It's a general fact that if you have an exact sequence
0→F′→F→F′′→0,\begin{aligned} 0 \to \mathcal F'\to \mathcal F\to \mathcal F'' \to 0, \end{aligned}
then
⋀rF=⋀r′F′ ⊗ ⋀r′′F′′,\begin{aligned} \bigwedge^r \mathcal F = \bigwedge^{r'} \mathcal F' ~\otimes~\bigwedge^{r''}\mathcal F'', \end{aligned}
so
⋀nTX∣Y∗=NY/X∗⊗⋀n−1TY∗.\begin{aligned} \bigwedge^n T_{X|_Y}^* = N_{Y/X}^* \otimes \bigwedge^{n-1}T_Y^*. \end{aligned}
The leftmost term is KX∣YK_{X|_Y} and the rightmost term is KYK_Y, so rearranging, we get
KY=NY/X⊗KX⊗OY.\begin{aligned} K_Y = N_{Y/X}\otimes K_X\otimes \mathcal O_Y. \end{aligned}

□\square
 

Curves in P2\mathbb P^2

Let's see Bezout's theorem.

Theorem: (Bezout's Theorem). If C1C_1, C2⊆P2C_2\subseteq \mathbb P^2 are projective curves of degree d1d_1 and d2d_2 respectively, then C1⋅C2=d1⋅d2C_1\cdot C_2 = d_1\cdot d_2.
Proof.   C1C_1 is linearly equivalent to a union of d1d_1 general lines; likewise, C2C_2 is linearly equivalent to a union of d2d_2 general lines. Any two linear intersect in exactly one point, and C1⋅C2C_1\cdot C_2 is well-defined on linear equivalence classes. Thus, counting intersection of lines gives C1⋅C2=d1⋅d2C_1\cdot C_2 = d_1\cdot d_2.
□\square
 

Remember: C1C_1 and C2C_2 are divisors of P2\mathbb P^2 and two divisors CC and DD are linearly equivalent (C∼DC\sim D) if and only if C−D=div⁡(f)C - D = \operatorname{div}(f). The statement that every curve of degree dd is linearly equivalent to a union of lines is something like saying that every polynomial of degree dd factors as a product of dd linear terms.

Proposition: (genus-degree formula). If C⊆P2C\subseteq \mathbb P^2 is a smooth plane curve of degree dd and genus gg, then
g=(d−1)(d−2)2.\begin{aligned} g = \frac{(d-1)(d-2)}{2}. \end{aligned}
Proof.   Recall that we saw, as a corollary of Riemann Roch, that deg⁡(KC)=2g−2\deg(K_C) = 2g - 2. By adjunction we get
KC=(KP2+C)∣C=(OP2(−3)⊗OP2(d))∣C=(OP2(d−3))∣C\begin{aligned} 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{aligned}
which implies
deg⁡(KC)=(d−3)⋅d=d2−3d=2g−2.\begin{aligned} \deg(K_C) = (d-3)\cdot d = d^2 - 3d = 2g - 2. \end{aligned}
Solving for gg gives the result.
□\square
 

Curves in P1×P1\mathbb P^1\times \mathbb P^1

A curve on P1×P1\mathbb P^1\times \mathbb P^1 (take coordinates [u0:u1]×[v1×v2][u_0:u_1]\times [v_1\times v_2]) is given by a bi-homogeneous polynomial of bi-degree (d1,d2)(d_1, d_2); that is, a polynomial in u0,u1,v0,v1u_0,u_1,v_0,v_1 which is homogeneous of degree d1d_1 in u0,u1u_0,u_1 and homogeneous of degree d2d_2 in v0,v1v_0, v_1. For instance, u02⋅u1⋅v0⋅v1+u13⋅v02u_0^2 \cdot u_1 \cdot v_0 \cdot v_1 + u_1^3 \cdot v_0^2 has bidegree (3,2)(3, 2).

Fact: C∼D⊆P1×P1C\sim D\subseteq \mathbb P^1\times \mathbb P^1 if and only if CC and DD have the same bi-degree.

Proposition: Let C1C_1 and C2C_2 be curves of bidegrees (d1,e1)(d_1,e_1) and (d2,e2(d_2,e_2). Then
C1⋅C2=d1e2+d2e1\begin{aligned} C_1\cdot C_2 = d_1e_2 + d_2 e_1 \end{aligned}

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

Now consider an arbitrary curve CC in P1×P1\mathbb P^1\times \mathbb P^1 of bi-degree (d1,d2)(d_1,d_2). It has canonical divisor

KC=(KP1×P1+C)∣C\begin{aligned} K_C = (K_{\mathbb P^1\times \mathbb P^1} + C)_{|_C} \end{aligned}

by the adjunction formula. Once you know that KP1×P1=OP1×P1(−2)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 bideg(KQ)=(−2,−2)\textrm{bideg}(K_Q) = (-2, -2).

Here are two ways of seeing that KP1×P1=OP1×P1(−2)K_{\mathbb P^1\times \mathbb P^1} = \mathcal O_{\mathbb P^1\times \mathbb P^1}(-2).

Method 1: We realize Q=P1×P1Q = \mathbb P^1\times \mathbb P^1 as a smooth quadric in P3\mathbb P^3 (in fact, every smooth quadric in P3\mathbb P^3 is isomorphic to P1×P1\mathbb P^1\times \mathbb P^1). To see this, take the Segre embedding [u0:u1]×[v0:v1]↦[u0v0:u0v1:u1v0:u1v1][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=zwxy = zw in P3\mathbb P^3 after choosing homogeneous coordinates [x:y:z:w][x:y:z:w].

Now apply adjunction:

KQ=(KP3+Q)∣Q=(OP3(−4)+OP3(2))∣Q=(OP3(−2))∣Q.\begin{aligned} 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{aligned}

Method 2: (Thanks Abhishek!) We can instead calculate KP1×P1K_{\mathbb P^1\times \mathbb P^1} without considering an embedding into P3\mathbb P^3 by applying the adjunction formula to one copy of P1\mathbb P^1 inside of P1×P1\mathbb P^1\times \mathbb P^1. Adjunction says

KP1=(KP1×P1+P1).\begin{aligned} K_{\mathbb P^1} = (K_{\mathbb P^1\times \mathbb P^1} + \mathbb P^1). \end{aligned}

We know KP1=OP1(−2)K_{\mathbb P^1} = \mathcal O_{\mathbb P^1}(-2) since KPr=OPr(−r−1)K_{\mathbb P^r} = \mathcal O_{\mathbb P^r}(-r - 1).

Applying the Corollary about deg⁡(KC)\deg(K_C) to our curve CC of bi-degree (d1,d2)(d_1,d_2), we have

2g−2=deg⁡(KC)=deg⁡((KP1×P1+C)∣C)=(d1−2)⋅d2 + (d2−2)⋅d1=2d1d2−2d2−2d1,\begin{aligned} 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{aligned}

which implies g=(d1−1)(d2−1)g = (d_1-1)(d_2 - 1).

This leads to the following proposition.

Proposition: If CC is a smooth curve of bi-degree (d1,d2)(d_1,d_2) on P1×P1\mathbb P^1\times \mathbb P^1, then
g(C)=(d1−1)(d2−1).\begin{aligned} g(C) = (d_1 - 1)(d_2 - 1). \end{aligned}

Comparing the genus/degree formulas for curves in P2\mathbb P^2 and P1×P1\mathbb P^1\times \mathbb P^1 warrants the following remark.

Remark:

  1. For some gg (e.g. g = 2) there exists no smooth plane curve of that genus, simply because d↦(d−1)(d−2)2d\mapsto \frac{(d-1)(d-2)}{2} is not a surjective map on N\mathbb N.

  2. In contrast, for any gg there exists a smooth curve of genus gg on P1×P1\mathbb P^1\times\mathbb P^1. For example, a smooth curve of bi-degree (g−1,2)(g-1, 2).

©Isaac Martin. Last modified: February 07, 2024.