# Can any smooth hyperelliptic curve be embedded in a quadric surface?

Let's consider algebraic curves over a fixed algebraically closed field $K$.

It's well known, that every smooth elliptic curve (genus $g = 1$) can be embedded in a quadric surface in $\mathbb{P}^3$. This fact follows simply from the Riemann–Roch theorem.

More generally, for smooth hyperelliptic curves of higher genus ($g \ge 2$) it's known that such curves can be embedded in a quadric in weighted projective space $\mathbb{P}(1,1,g)$, see, for example, work of D. Eisenbud. (So, in the case $g=2$ we have embedding in $\mathbb{P}^4$).

But, can any smooth hyperelliptic curve $H$ be embedded in a quadric surface in $\mathbb{P}^3$?

It's natural question, because, by the definition we have mophfism $\phi:H \to \mathbb{P}^1$ of degree 2.

I think, this problem is connected with topics "Families of hyperelliptic curves and double covers of quadric surface" and Quotient Surface of A Hyperelliptic Involution, but I don't get it. (I'm interested in the case of any algebraicaly closed field and in the case of finite field).

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Yes.

Let $C$ be a hyperelliptic curve of genus $g$, and let $L$ be a general line bundle of degree $g+1$. By Riemann-Roch, $\dim|L| = 1$ and $|L|$ is base-point free, so the complete series $|L|$ gives a degree $g+1$ map to $\mathbb{P}^1$. Then the product of this map and the degree $2$ map $C\to \mathbb{P}^1$ gives a map $f:C\to \mathbb{P}^1 \times \mathbb{P}^1$, whose image is a curve of type $(2,g+1)$. But $\mathbb{P}^1\times \mathbb{P}^1$ is just a quadric in $\mathbb{P}^3$.

To see the map is an embedding, it will suffice to show that it is birational. Indeed, the image has arithmetic genus $g$ by adjunction on a quadric surface. But it also has geometric genus $g$ since $C$ is its normalization. Thus the image is smooth if it is reduced.

Finally we must see that the map is birational. The only way the map fails to be injective is if some divisor of $|L|$ contains a pair of points conjugate under the hyperelliptic involution. But in this case the assumption that $\dim |L|=1$ implies that $|L| = g_2^1 + p_1+\cdots+ p_{g-1}$, where $g_2^1$ is the hyperelliptic series and $p_1,\ldots,p_{g-1}$ are base points. Since $L$ was general, this isn't true, and we're done.

Here's a cute (although trivial) kind of partial converse: If $g$ is prime, then any smooth curve $C$ of genus $g$ which embeds in a smooth quadric is hyperelliptic.

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What is $r(L)$? –  rita Oct 31 '11 at 9:06
Just the dimension of $|L|$; I've removed this notation. –  Jack Huizenga Oct 31 '11 at 11:21
One can take L to be the sum of g+1 distinct Weierstrass points; in which case you have two nice generating sections: the g+1 chosen points, or their complimentary points. –  David Lehavi Oct 31 '11 at 13:24
Sorry, but if You have time, can You write the proof above without using the definition of (general?) linear bundle, but only the definition of divisors? I'm beginner in algebraic geometry, so I don't understand formally every step in the proof yet. My main book is Shafarevich's "Basic algebraic geometry", the book of Hartshorne is too comprehensive to me at this moment. Also all links to the used propositions, lemmas etc. will be very useful. Thanks. –  One_math_boy Nov 3 '11 at 21:09
You can replace "line bundle" with divisor everywhere, and change $L$'s to $D$'s. If you're not comfortable with this correspondence yet, now would be a great time to learn it. I'd suggest Rick Miranda's book on algebraic curves as a good place to get familiar with it, as it is made very explicit there. I'm also implicitly using several facts that are covered in the first chapter of Arborello, Cornalba, Griffiths, and Harris "Geometry of Algebraic Curves," particularly the fact that a general degree $d$ divisor $D$ has $\dim |D| = \max(0,d-g)$. –  Jack Huizenga Nov 3 '11 at 23:34

An explicit realization of degree $2$ and degree $g+1$ maps that separate points can be provided. Suppose the equation of a hyperelliptic curve is $$C:y^2=f(x)$$ with $\deg(f)=2g+2$. "Complete the square" by writing $$f(x)=r(x)^2+q(x)$$ with $\deg(r)=g+1$ and $\deg(q)\leq g$. Then, the maps $$A:(x,y) \mapsto x$$ $$B:(x,y)\mapsto y-r(x)$$ are degree $2$ and degree $g+1$ respectively. The map $B$ is degree $g+1$ because if we assume that $B(x,y)=c$, then we get the equation $c(c+2r(x))=q(x)$, which generically has $g+1$ solutions. Furthermore $(A,B)$ is clearly injective. So, the image $$(A,B):C\mapsto C'\subset\mathbb{P}^1\times\mathbb{P}^1$$ is a degree $(2,g+1)$ curve that $C$ normalizes and the logic of Jack's answer applies.

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The answer is yes in the case $g=2$.

In other words, any curve $C$ of genus $2$ can be embedded in a quadric $Q \subset \mathbb{P}^3$. In fact, let us consider a divisor $D$ of degree $5$ on $C$. Then $D$ is very ample [Hartshorne, Corollary 3.2 page 308] and $h^0(D)=4$, so it defines an embedding $\phi \colon C \to \mathbb{P}^3$ such that its image (that we call again $C$) is a curve of degree $5$.

Now we use Castelnuovo theory: if $C$ is any non-degenerate curve of genus $g$ and of odd degree $d$ in $\mathbb{P}^3$, then $$g \leq \frac{1}{4}(d^2-1)-d+1,$$ and if equality holds then $C$ lies on a quadric surface [Hartshorne, Thm. 6.4 page 351].

Since equality does hold for $(g, d)=(2,5)$, we are done.

ADDENDUM. Just for completeness, let me add a deformation theory argument showing that the general smooth hyperelliptic curve of genus $g$ lies on a quadric $Q$. This will replace my previous imprecise naive count of parameters. Notice that Jack Huizenga showed in his answer that this is actually true for any such a curve.

Let us start with a curve $C \subset Q$ of bidegree $(2, g+1)$. This is clearly hyperelliptic and we have a short exact sequence $$0 \to T_C \to T_Q \otimes \mathcal{O}_C \to N_{C/Q} \to 0$$ and since $H^0(T_C)=0$ for $g \geq 2$ this gives in turn a sequence in cohomology $$0 \to H^0(T_Q \otimes \mathcal{O}_C) \to H^0(N_{C/Q}) \stackrel{\delta}{\to} H^1(T_C).$$

Now standard computations yield $$h^0(T_Q \otimes \mathcal{O}_C)=g+6, \quad h^0(N_{C/Q})=3g+5$$ hence the image of the map $\delta \colon H^0(N_{C/Q}) \to H^1(T_C)$ has dimension $3g+5-g-6=2g-1.$

This means that the embedded deformations of $C$ in $Q$, which are all hyperelliptic, form a family of dimension $2g-1$. But this is exactly the dimension of the hyperelliptic locus in $\mathcal{M}_g$, since any hyperelliptic curve of genus $g$ is a double cover of $\mathbb{P}^1$ branched in $2g+2$ points.

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Ah, ok. I misread the question and I worked out only the case $g=2$. I will leave the answer here, maybe it can be useful for the general case. –  Francesco Polizzi Oct 30 '11 at 22:46
Thanks. It's very nice proof for the case $g=2$. –  One_math_boy Oct 30 '11 at 22:51
The naive count of parameters I gave before was not correct. So I deleted it. If I have time, tomorrow I will add the correct computation. –  Francesco Polizzi Oct 31 '11 at 0:12