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I asked the following question in stack exchange ( a few days ago, but didn't get any solution. Somebody please help me with it.

I have just started learning Riemann Surfaces and I am using the book by Rick Miranda: Algebraic curves and Riemannn 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 approach this type of question.

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I don't think this should be closed. Remember that lots of people who aren't algebraic geometers still want to learn something about algebraic curves. If someone is considering voting to close this, please consider leaving a comment to cancel mine instead, as discussed at… – David Speyer Feb 13 '11 at 18:44
I voted up David's comment. As one posible approach: it should be possible to show the genus of the intersection of two quadrics is genus one by projecting onto something simple and counting the branches and ramification points (the way that one would show a plane cubic is genus one by projecting to the $x$-axis). This is hinted at in Roy Smith's comments to J.C. Ottem's answer, but it would be nice if someone wrote this (or at least the underlying method) out in detail as an answer, since it would probably be quite helpful to the OP and to others. – Emerton Feb 13 '11 at 19:15
With all respect, it seems helpful to see how researchers answer even elementary questions like this. It is all too common for the younger generation to believe that one needs cohomology and adjunction formulas as in Harthorne to answer these questions. When I was young I used to wonder how then was it possible for the ancients to know the answer to all these questions? – roy smith Feb 14 '11 at 4:22
roy, you are right that cohomology was probably asked and very well answered.I don't know how is called line + cubic curve. And you remind me very nice experience: when I learned english, all Americans were named Smith. I am very happy to meet real one: how do you do Mr. Smith? – evgeniamerkulova Feb 14 '11 at 18:19
thank you, i am indeed fine. evgenia, are you perhaps russian? when i learned russian all men were named ivan. ivan mixanyek. ivan rabotayet na fabrikye. I do not remember the generic feminine names, but it is a pleasure also to meet you! – roy smith Feb 15 '11 at 3:55

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 $g(C)=1$.

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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.

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Of course this is a correct answer, but if the OP is has just started learning Riemann surfaces, they may not be in a very good position to apply "standard formulae for cohomology on $P^n$". – Emerton Feb 13 '11 at 16:54
Yes, that's true. I modified the answer above. – J.C. Ottem Feb 13 '11 at 18:20
Degenerate one quadric to 2 planes. Then the intersection is a union of 2 plane conics meeting at the 2 points where the line common to the two planes meets the other quadric. A plane conic is ≈ P^1, so the union is a degeneration of a torus. (2 "spheres" meeting at two points.) Or project from a point of one quadric, collapsing 2 ruling lines, secants of the curve, mapping the curve to a plane quartic with 2 nodes, so g = 3-2 = 1. A plane curve of degree d can be degenerated to a union of d general lines. Hence the genus is the number of holes they form: "(d-1) choose 2". – roy smith Feb 13 '11 at 18:28
In the first version of the book there was a typo in problem F, omitting the 2 in the equation for the first quadric. What is the intersection then? In what sense does it still have genus one? – roy smith Feb 13 '11 at 18:37
Oh, I hadn't seen Roy's comment until now. This is pretty much the answer I gave below. (Although I hadn't known about the typo. In the remark I explained how I arrived to doing it this way.) – Sándor Kovács Feb 14 '11 at 0:16

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.

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Dear Jack, Sandor's answer is (of course) correct when understood in the appropriate sense. Rather than simply criticizing it, why don't you work on understanding what that sense is? Regards, Matthew – Emerton Feb 14 '11 at 1:05
I am sorry if I came off as overly critical. I was in particular remarking on his comment that "the topological genus is clearly 1. If in doubt, triangulate it." Triangulating the space, I get an Euler characteristic of 4, which does not suggest "genus 1" to me. What is the sense in which I should interpret this? Clearly the arithmetic genus solves all the problems, but then there's the work of relating arithmetic genus to topological genus. – Jack Huizenga Feb 14 '11 at 1:11
Great answer! Makes me wonder why the question chose $x_0x_3=2x_1x_2$ instead of $x_0x_3=x_1x_2$. Anyways, there are some typos above (some $x_1$s should be $x_3$s). – J.C. Ottem Feb 14 '11 at 1:15
@Jack: Your comment made me realize that there is a better way to present this solution. I edited my answer accordingly. Thanks for the contribution! – Sándor Kovács Feb 14 '11 at 1:42
@J.C.: Thanks for the compliment and the typo catching. I think I got all the renegades. Did I? – Sándor Kovács Feb 14 '11 at 1:46

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 $1$.

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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.

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This is classical. It is, e.g. in Weil's book "Number Theory: an approach to history" pg 136. Still, that's some high school you went to! – Felipe Voloch Feb 14 '11 at 0:22
@Felipe: actually it was at RSI ( Thank you for the reference! I have always wondered just how trivial my paper was... – Qiaochu Yuan Feb 14 '11 at 0:41
There is nothing trivial about it. And if you do the same thing in P^5, you get a hyperelliptic curve whose jacobian is isomorphic to the intermediate jacobian of the threefold base locus of that pencil of quadrics. Now you are already at the level of a PhD thesis of the 1970's. – roy smith Feb 14 '11 at 23:47
I.e. the odd dimensional non singular base locus of a pencil of quadrics of form [A-tB], has always an intermediate Jacobian which is also the Jacobian of a hyperelliptic curve, defined as the double cover of the P^1 parametrizing the pencil, and branched over the singular quadrics in the pencil, i.e. the zeroes of det[A-tB]. as I recall. Perhaps this is due to Weil as well. Going to the base locus of three quadrics in P^2g, we get a base locus whose intermediate Jacobian is the Prym variety of the associated double cover of the curve of singular quadrics in the net. as memory serves. – roy smith Feb 15 '11 at 3:44

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!

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