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Let $C$ be a non hyperelliptic complex algebraic curve of genus $g$, then the vector space $I_2(C)$ of quadrics containing the canonical image of $C$ is $\binom{g+1}{2}-h^0(2\omega_C) = (g-2)(g-3)/2$ dimensional. Moreover, if $g > 4$ then $C$ is the intersection of the nulls of all these quadrics (see ACGH VI.4.1 for a proof, I'm not sure how "classical" this is, or who originally proved it).

Question (edited following a comment from David Speyer) what is the least $d$ so that if $V\subset I_2(C)$ is any $d$ dimensional vector space, and $X$ is the intersection of the nulls of the quadrics in $V$, then the only irreducible component of $X$ which linearly spans $|\omega_C|^*$ is the canonical image of $C$ ?

I don't even know the generic bound, or indeed what is the bound for hyperelliptic curves (in which case the canonical curve is a rational normal curve).

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Do you want to cut out the curve set-theoretically or scheme-theoretically? Set-theoretically, the rational normal curve is the intersection of the $g-2$ quadrics $x_{i+1} x_{i-1} = x_i^2$. I don't know about the more interesting questions here. –  David Speyer Nov 21 '10 at 13:57
@David: to begin with, set theoretically would be fine. Regarding the RNC, you gave a very special V, not a bound on the dimension on all possible V's. –  David Lehavi Nov 21 '10 at 18:00
Oh, you want a $d$ such that any $V$ of dimension $d$ will work? But then you have to use the full (g−2)(g−3)/2 dimensional vector space. Proof: Choose any point $x$ not on the curve. Consider the vector space $W$ of quadrics that vanish on the curve and on $x$. Since vanishing at $x$ is a codimension $1$ condition, $W$ has dimension $(g-2)(g-3)/2 - 1$. So, if our $V$ is contained in $W$, then $V$ will not cut out the curve. –  David Speyer Nov 21 '10 at 21:36
@David: you are right of course, which means that my original question was morally wrong: let me fix it then :) –  David Lehavi Nov 21 '10 at 22:17
Historically, Enriques proved that, with the noted exceptions, the canonical curve can be cut out set theoretically by quadrics, with the possible exception of a finite set of points common to all the quadrics. Then Charles Babbage argued that there are indeed no such extraneous points. Finally Petri proved the ideal is generated by quadrics. –  roy smith Nov 23 '10 at 19:22

3 Answers 3

Petri's theorem (see ACGH III.3) states that if $C\subset P^{g-1}$ is a canonical curve (i.e. the canonical image of a non hyperelliptic curve) of genus $g\ge 4$ then the ideal of $C$ is generated by quadrics iff $C$ is not trigonal or a plane quintic (for trigonal curves or plane quintics the intersection of all quadrics through $C$ is a surface).

If $C$ is hyperelliptic, the dimension of the space of quadrics through the canonical image $\Gamma$ is $(g+1)g/2-2g+1=(g-1)(g-2)/2$. The ideal of $\Gamma$ is also generated by quadrics.

If $g-2$ is not a power of 2, a trivial lower bound is $d\ge g-1$. Indeed by Bezout's theorem (see Fulton, Intersection theory, 8.4) if the intersection of $g-2$ quadrics of $P^{g-1}$ is a curve of degree $d$, then $d$ divides $2^{g-2}$.

The following $2g-5$ quadrics cut the rational normal curve in $P^{g-1}$ set theoretically: $x_0x_2-x_1^2,x_0x_3-x_1x_2, \dots, x_0x_{g-1}-x_{g-2}x_1, x_2^2-x_1x_3, x_3^2-x_2x_4,\dots,x_{g-2}^2-x_{g-3}x_{g-1}$.

I don't know if one can do better.

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@rita: per your paragraphs: 1) you can do much better, see the reference I gave in the question - ACGH VI.4.1. 2) sure. 3) Obviously d > g-2, but can you give a sane upper bound on it ? e.g. d subquadratic in g would be nice - I have no idea if this is true. 4) see David Speyers comment to my question for g-2 quadrics that give the RNC set theoretically. –  David Lehavi Nov 21 '10 at 21:26
@David Lehavi: 1) maybe I misunderstood. I thought that in your question you were claiming that the intersection of the quadrics through a canonical curve is the canonical curve itself, without excluding trigonal and plane quintic. 2) In your question you seemed to imply that the number of quadrics is the same for hyperelliptic, which is not true. 4)David Speyer equations don't seem to work: if you set all the variables equal to $0$ except the first and the last you get a line on which all the given quadrics vanish. –  rita Nov 21 '10 at 23:24
@David: 1. I don't see how the statement can be correct. a) plane quintics: by adjunction, the bicanonical divisors are cut by quadrics of the plane. The canonical image $D$ of $C$ lies on the Veronese surface in $\pp^5$, which is the intersection of all the quadrics containing $D$. b) trigonal not hyperelliptic: if $A+B+C$ is in the $g^1_3$, then by geometric RR the points $A$, $B$,$C$ lie on a line in the canonical immage. Letting the divisor $A+B+C\in g^1_3$ vary one gets a ruled surface containing $D$ which is the intersection of all the quadrics containing $D$. –  rita Nov 22 '10 at 8:03
@David Speyer: I think that if one adds the equation $x_0x_{g-1}-x_1x_{g-2}=0$ to the ones you wrote, then you get the rational normal curve (set theoretically). –  rita Nov 23 '10 at 20:45
@David Lehavi: a canonical curve is the image via the canonical map of a curve of genus $g\ge 2$. So I can consider the canonical image $D$ of a smooth plane quintic $C$. Then $g(C)=6$ and $|K_C|$ is the restriction of $|O_{P^2}(2)|$. This means that the canonical map of $C$ is the restriction of the Veronese map $P^2\to P^5$ and $D$ lies on the Veronese surface. Let $Q(y_0,\dots y_5)$ be a quadric that vanishes on $D$: pulling $Q$ back to $P^2$ I get a quartic $F(x_0,x_1,x_2)$ that vanishes on $C$. By degree reasons, $F=0$, namely $Q$ vanishes on the Veronese surface. –  rita Nov 23 '10 at 21:28

I guess I am not getting something, as it seems elementary the maximum number of quadrics needed is usually $g$. I.e. given a canonical curve $C$ in $P^{g-1}$ which can be cut out by quadrics, it seems any general choice of $g-2$ quadrics containing it cuts out a union of curves including $C$. Then any general quadric containing $C$ cuts out $C$ and a finite set of points on the other curves. Then another general quadric through $C$ omits those points. Is this nonsense? I see now the question in the title is no longer the same as the edited question.

edit: David, do you really want the property in your question or do you just want to know when every $d$ dimensional subspace of $I_2(C)$ determines the canonical curve somehow? i.e. a Torelli result.

Here is an example suggesting d may be large, a plane sextic, re embedded canonically in $P^9$ via plane cubics. The image is a del Pezzo surface $S$ of degree $9$, on which any one quadric cuts out the canonical curve, unless the quadric contains the del Pezzo. But the $55$ dimensional space of quadrics in $P^9$ cuts out the $28$ diml space of plane sextics, hence a $27$ diml space of quadrics contains the del Pezzo. Since the ideal $I_2$ has dimension $28$, we actually need the whole space $I_2$ to get the curve, or to get any set with the curve as a component.

Is this right? If so, plane curves of other degrees may be problematic as well....The situation seems to improve as the degree goes up. A plane septic seems to lie canonically on an embedded copy of $P^2$ that is contained in only $75$ independent quadrics among the $78$ containing the curve itself, so d seems to equal at least $76$, maybe $77$ since it seems to need two more quadrics this time. For a plane octic $d$ seems to be at least $166$, out of a space $I_2$ of dim = $171$. well we're gaining on it, but somewhat slowly.

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the issue is that I want it to be correct for any V not for a generic one. –  David Lehavi Nov 23 '10 at 21:06
@Roy Smith: I think what you observe in the case of plane curves is even more general. If the curve $C$ lies on a surface $Y$ with $h^0(K_Y)=h^1(K_Y)=0$ (for instance $Y$ is rational), then the restriction $|K_Y+C|\to|K_C|$ is an isomorphism, so the canonical map of $C$ is induced by the map $Y\to P^{g−1}$ given by $|K_Y+C|$ and you have a similar situation. In fact I think this is what happens for trigonal curves. –  rita Nov 24 '10 at 7:20
Thanks rita. So the usual counterexamples to cutting out the canonical curve by quadrics, i.e. plane quintics and trigonals, seem just the extreme case of a sequence of examples where canonical curves lie on surfaces which themselves lie on lots of quadrics. It may be too early to say this, but it suggests to me that the d sought here may not be very small in general. –  roy smith Nov 24 '10 at 18:56
Thanks Roy, Rita. I agree that your examples seem to suggest that for very special curves, this $d$ is very large. Do you have any ideas regarding the second part of the question (where the curve, but not the vector space, is generic) ? –  David Lehavi Nov 24 '10 at 21:52
@Roy, I'm not loking for a Torelli result - see my reply your remark to aginesky's question below –  David Lehavi Nov 26 '10 at 11:19

Amplifying on of Speyer's comments, if p is a point on a secant line of C, then the quadrics vanishing on C and p are of codimension one in the space of all quadrics vanishing on C. Such a quadric vanishes at 3 points of the secant line ( p and the two points of C defining the line as a secant) and hence vanished on L. Am I doing something silly?

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This looks right. But this causes no problem with David's question, since the line is not spanning. I guess you can generalize your example and compute the codimension of quadrics vanishing on a rational spanning curve. I.e. I suppose a rational normal curve of degree g-1, lies on any quadric that contains 2g-1 points of it. So this gives a subspace of that codimension in I(2) that contains the two curves. –  roy smith Nov 25 '10 at 20:49
so you seem to be giving a lower bound for d of (1/2)(g-2)(g-3) -2g+1, for all embedded canonical curves? –  roy smith Nov 25 '10 at 20:53
indeed if you make the rational normal curve go through k points on the canonical curve, your bound goes up by k. I forget what k is. –  roy smith Nov 25 '10 at 20:58
I guess k=g, so you get a lower bound for d of (1/2)(g-2)(g-3)-g+1? –  roy smith Nov 25 '10 at 21:14
@Roy: no, I'm not looking for a Torreli result, which at least generically (for the curve, not for the vector space), is probably not that complicated. The original question indeed came from a Torelli problem, but then I started wondering.... –  David Lehavi Nov 26 '10 at 11:18

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