Problem in Rick Miranda: finding genus of a Projective curve - MathOverflow

Problem in Rick Miranda: finding genus of a Projective curve

2011-02-13T12:56:36Z

I asked the following question in stack exchange (http://math.stackexchange.com/questions/21164/problem-in-rick-miranda-finding-genus-of-a-projective-curve) a few days ago, but didnt get any solution. Somebody please help me with it.

I have just started learining Reimann Surfaces and I am using the book by Rick Miranda: Algebraic curves and Reimannn Surfaces. #F in section 1.3 asks to determine the genus of the curve in $\mathbb{P}^3$ defined by the two equations $x_0x_3=2x_1x_2$ and $x_0^2 + x_1^2 +x_2^2 +x_3^2 = 0$. #G also has a similar question in which he asks to determine the genus of the twisted cubic. Please explain how to approch this type of question.

Answer by J.C. Ottem for Problem in Rick Miranda: finding genus of a Projective curve

2011-02-13T13:37:05Z

The intersection of the two quadrics in $\mathbb{P}^3$ is a complete intersection and defines an elliptic curve, so the genus is 1. A way to see this is to pick a point $p$ on $C$ and project from $p$ onto a general hyperplane. The image curve $C'$ is of degree one less than the original curve, hence $C'$ is a plane curve of degree 3. Since cubics have genus 1, we are done.

Another way to see that $g(C)=1$ is by computing cohomology of the sequence $$ 0 \to O_{P^3}(-4) \to O_{P^3}(-2)\oplus O_{P^3}(-2) \to O_{P^3}\to O_C \to 0 $$(This is the resolution of $O_C$ as an $O_{P^3}$ module, which is easy to write down for complete intersections). Using this and the standard formulae for cohomology on $P^n$, we get $g=h^1(O_C)=1$.

Yet another way to see it is by looking at the curve as a divisor of type $(2,2)$ on the quartic surface $X_0X_3-2X_1X_2$. In general, by the adjunction formula, divisors of type $(a,b)$ have arithmetic genus $(a-1)(b-1)$, so again we get g=1.

The twised cubic $C$ is the (isomorphic) image of $P^1$ under the 3-uple embedding $f_3:(u,v)\to (u^3,u^2v,uv^2,v^3)$, so since $P^1$ has genus 0, C has genus 0. Of course, this computation could be carried out using a projection, and $C'$ would be a plane curve of degree 2.

Answer by David Speyer for Problem in Rick Miranda: finding genus of a Projective curve

2011-02-13T18:58:46Z

Here is the most algebraic way I can see to compute this. Let $Q_1$ and $Q_2$ be two quadratic polynomials in four variables. Let $R$ be the graded ring $k[x_1, x_2, x_3, x_4]/(Q_1, Q_2)$. Let $V_d$ be the vector space of degree $d$ polynomials in $(x_1, x_2, x_3, x_4)$, and let $R$ be the degree $d$ part of $R$. We have an exact sequence: $$0\to V_{d-4} \to V_{d-2}^{\oplus 2} \to V_d \to R_d \to 0.$$ The first (nontrivial) map is $f \mapsto (f Q_2, - f Q_1)$, the second is $(g,h) \mapsto g Q_1 + h Q_2$, the third is the degree $d$ part of the quotient map $k[x_1, x_2, x_3, x_4] \to R$. So $$\dim R_d = \dim V_d - 2 \dim V_{d-2} + \dim V_{d-4} =$$ $$\frac{(d+3)(d+2)(d+1)}{6} - 2 \frac{(d+1)(d)(d-1)}{6} + \frac{(d-1)(d-2)(d-3)}{6} = 4d$$ for $d>0$. (For $d=0$, this computation fails because $\dim V_{-4}$ is $0$, not $(-4+3)(-4+2)(-4+1)/6=-1$.)

Let $X$ be the curve $Q_1 = Q_2 = 0$, and let $L$ be the line bundle on $X$ gotten by restricting the line bundle $\mathcal{O}(1)$ on $\mathbb{P}^3$. For sufficiently large $d$, we have $R_d = H^0(X, L^{\otimes d})$. So, for large $d$, we have $\dim H^0(X, L^{\otimes d}) = 4d$. By Riemman-Roch, this dimension should be $(\deg L)d - (\mathrm{genus}(X)-1)$. So $\deg L=4$, and $X$ has genus $1$.

Answer by Jack Huizenga for Problem in Rick Miranda: finding genus of a Projective curve

2011-02-13T21:04:43Z

While I think it is very difficult to solve the question using only the machinery presented in Miranda at this point, the Hurwitz formula comes up in the next chapter and is probably the most elementary tool to use. In case of the first curve, you can consider restricting a projection $\mathbb{P}^3 \setminus \{x_1=x_2=0\} \to \mathbb{P^1}$ given by $[x_0,x_1,x_2,x_3]\mapsto [x_1,x_2]$ to the curve C.

This map has degree 4, and there are 4 fibers of the map with cardinality 2 instead of 4. We can calculate this explicitly from the defining equations. For if $[1,c]$ is a point in $\mathbb{P}^1$, we must solve the system of equations

$x_0x_3 = 2c$

$x_0^2+x_3^2+1+c^2 = 0.$

Setting $x_3= 2c/x_0$ yields

$x_0^2+\frac{4c^2}{x_0^2} + 1 + c^2 = 0.$

This equation has four solutions $x_0$ unless $c$ is one of the four roots of $c^4-14c^2+1=0$, and in those cases there are two solutions.

It follows from the Hurwitz formula that

$2g(C)-2 = 4(2g(\mathbb{P}^1)-2)+4\cdot 2,$

and thus $g(C)=1$.

Answer by Qiaochu Yuan for Problem in Rick Miranda: finding genus of a Projective curve

2011-02-13T23:38:13Z

If you know that genus is a birational invariant, you can explicitly write down some maps: $x_0 x_3 - 2x_1 x_2 = 0$ is birational to $\mathbb{P}^2$ via the substitutions $x_0 = RS, x_3 = 2T^2, x_1 = RT, x_2 = ST$. Substituting these into the second quadric gives $R^2 S^2 + 4T^4 + R^2 T^2 + S^2 T^2 = 0$, which is more or less an elliptic curve in Edwards normal form $x^2 + y^2 = a^2 + a^2 x^2 y^2$.

This argument generalizes to the intersection of two quadrics in $\mathbb{P}^3$: if $A, B$ are $4 \times 4$ matrices such that your quadrics are given by $x^T A x = x^T B x = 0$, then their intersection is birational to the curve $y^2 = P(t)$ where $P(t) = \det(A - tB)$ (at least over an algebraically closed field). I have no idea how well-known this is; I imagine it is an exercise somewhere, but (embarrassingly enough) I wrote an entire paper about this result in high school.

Answer by Sándor Kovács for Problem in Rick Miranda: finding genus of a Projective curve

2011-02-13T23:56:00Z

Here is a solution in the spirit of Miranda's book. Given the way the question was asked I think the point is to give a proof/computation that does not use much algebraic geometry if anything at all.

First consider the intersection of the quadrics $x_0x_3=x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$.

This is easy to deal with because one can solve the equation system: Take $x_3=\dfrac{x_1x_2}{x_0}$ and substitute it in the second equation. It easily leads to $$(x_0^2+x_1^2)(x_0^2+x_2^2)=0$$ This is the equation of two pairs of skew lines forming a $4$-gon. In other words $4$ spheres, each intersecting two others forming a cycle.

Now observe that the intersection of the quadrics $x_0x_3=x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$ is a continuous degeneration of the intersection of the quadrics $x_0x_3=2x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$. Therefore the later intersection is a compact Riemann surface $T$ (I leave it to the reader to verify that this intersection is smooth) degenerating to the above cycle of $4$ spheres. It is easy to see that then $T$ is a torus and hence its genus is $1$.

Remark The algebraic geometer's way to think about this solution is the following: The quadric $x_0x_3=\lambda x_1x_2$ is the Segre embedding of $\mathbb P^1\times \mathbb P^1\to \mathbb P^3$ given by $[a:b]\times [c:d]\mapsto [\lambda ac:ad:bc:bd]$ and then the intersection of $x_0x_1=\lambda x_1x_2$ and $x_0^2+x_1^2+x_3^2+x_4^2=0$ pulls back to $\mathbb P^1\times \mathbb P^1$ as the curve defined by the equation $$\lambda^2a^2c^2+a^2d^2+b^2c^2+b^2d^2=0.$$ Now this defines a divisor of degree $(2,2)$ on $\mathbb P^1\times \mathbb P^1$ which can be represented (choosing $\lambda=1$ for instance) by two pairs of lines as described above. If one knows about the behaviour of the (arithmetic) genus in flat families, then everything claimed above is clear.

Answer by Steven Gubkin for Problem in Rick Miranda: finding genus of a Projective curve

2012-04-05T23:06:34Z

This was meant to be a comment on the ending remark of Sándor Kovács' answer, but it got too long to fit:

In a student seminar today, some people had the old edition of Miranda, and some had the new edition, so we had both the original problem and your degenerate version. (The old edition has the degenerate version).

The way we ended up seeing that solution set $X$ of $x_0x_1 = x_2x_3$ is $\mathbb{P}^1\times\mathbb{P}^1$ was to observe that we could rewrite it as $det\begin{pmatrix} x_0 & x_2 \\ x_3 & x_1 \end{pmatrix} = 0$ or $det\begin{pmatrix} x_0 & x_3 \\ x_2 & x_1 \end{pmatrix} = 0$. So both matrices must be rank 1. So we have two maps from $X$ to $\mathbb{P}^1$, namely the maps which send a point of $X$ to the corresponding element of the nullspace of one or the other matrix. This gives a map $X \to \mathbb{P}^1\times\mathbb{P}^1$, which is not too hard to compute explicitly (in fact $(\langle a,b\rangle, \langle c,d\rangle) \mapsto \langle ac,ad, bc, bd \rangle$ just as you say). So we are really looking at the zero set of $a^2c^2+a^2d^2+b^2c^2+b^2d^2=0$ in $\mathbb{P}^1\times\mathbb{P}^1$. At this point we basically followed the rest of your post. I just thought someone might like the observation about determinants!