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Is there a smooth compact complex curve that does not admit an injective holomorphic map to $\mathbb CP^2$ ? Let me stress, that the image of the curve in $\mathbb CP^2$ can have singularities.

I should probably add, that I see this question as a baby version of the following question of Fancesco Polizzi: Surfaces in $\mathbb{P}^3$ with isolated singularities

Remark. Any curve $x^n+y^n+z^n=0$ admits injective morphisms to $\mathbb CP^2$ of arbitrary high degree (the proof is given by Jeremy Blanc here: Injective morphism from an elliptic curve to $\mathbb CP^2$. )

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If the map is an closed embedding, then the curve is smooth in the plane. Writing $d$ its degree, the adjunction formula says that it has genus $(d-1)(d-2)/2$, so of course not all curves are possible. If the map is not an embedding, the image is singular, with only cuspidal singularities. There are thus more possibilities for the curve. In this case, I don't know if each curve is possible. –  Jérémy Blanc Feb 22 '13 at 18:28
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@Jérémy: I assume that aglearner is allowing unibranch singularities. –  Daniel Litt Feb 22 '13 at 19:08
    
@Jérémy: why must the singularities be cuspidal? –  Jesus Martinez Garcia Feb 22 '13 at 19:11
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@Daniel: for me a cusp is exactly a unibranch singularity, as for most people studying cuspidal curves. @Jesus: the injectivity implies that when you make a resolution of singularities, there should be only one point above each singular pt. This is equivalent to ask for a cusp (or unibranch singularity). –  Jérémy Blanc Feb 23 '13 at 9:56
    
Oh, OK. I thought cusp meant $y^2-x^3$ (never read it anywhere, just from conversations), but I realise that analytically that is also $y^m-x^n$ for $m,n>1$ , which I think is equivalent to what you say. Thanks a lot :) –  Jesus Martinez Garcia Feb 26 '13 at 13:16
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3 Answers

Update.

Estimates coming from the Bogomolov-Miyaoka-Yau inequality yield:

Corollary. (to Theorem A in this paper of Sakai) There is no $M$ such that every curve admits an injective morphism to a curve in $\mathbb{P}^2$ whose points all have multiplicity at most $M$.

Proof. The geometric genus $g$ curves of degree $d$ move in a family of dimension $3d + g - 1$, so if these are to fill the moduli space we must have $2g \le 3d + 2$. On the other hand, Theorem A of this paper of Sakai gives after rearranging that for any genus $g$, degree $d$, unibranch plane curve, $$ 2g \ge \frac{d^2}{2m+1} + O(md) $$ where $m$ is the maximal multiplicity of a point on the curve. For any bounded $m$, these inequalities cannot both be satisfied for large $g$. $\square$

Remark: the argument also shows that the same statement holds restricted to hyperelliptic curves. This is not incompatible with Jeremy's answer below, which requires a cusp of multiplicity $2g$ to inject a hyperelliptic curve of genus $g$.


I leave the older argument below because it is a different and possibly still useful approach.

Theorem. There is no finite set $\Sigma$ of topological types of unibranch singularities such that every smooth curve admits an injective map to a curve in $\mathbb{P}^2$ with only singularities from $\Sigma$.

Proof. I will argue that curves with unibranch singularities of total cogenus $\delta$ occur in the linear system $|\mathcal{O}(d)|$ with codimension at least $\mathrm{min}(\kappa, 3d - 1 + \delta)$, where $\kappa$ is the codimension of the equiclassical locus in the versal deformation of the singularities; it is $2\delta$ in the case of curves with only cusps, and for any finite set of singularities is bounded below by $(1 + \epsilon)\delta$ for $0 < \epsilon \le 1$ some constant depending on the set of singularities. Let us first see why this implies the result.

Saying a geometric genus $g$ curve appears in degree $d$ in codimension $3d - 1 + \delta = 3d + (d-1)(d-2)/2 - g - 1$ means it comes in a $g + 1 + (d+1)(d+2)/2 - (d-1)(d-2)/2 -3d = g + 3$ dimensional family. On the other hand $\kappa$ is given by $\delta + r$, where $r$ is the total ramification divisor on the normalization of the curve (for the normalization map), i.e. the sum over the branches of the singularities of (multiplicity - 1). So in this case the curves come in a family of dimension $g + 4 + 3d - r$. I don't know what to do with this, but under the further assumption that the set of singularities from which we choose must be finite and so we get $\kappa = (1 + \epsilon) \delta$ as above (e.g., for only cusps, $\epsilon = 1$), then the argument in Will's answer above shows this can fill the moduli space for only finitely many $g$.

To get the advertised codimension estimate, let $C$ be such a curve of degree $d$. Let $\Lambda$ be the tangent space to the versal deformation of the singularities of $C$. Then there exists a linear subspace $L \subset \Lambda$ such that

  1. equigeneric ideal $\ge$ L $\ge$ equiclassical ideal
  2. aside from the above restriction, L may be taken so that the dimension of the equigeneric ideal mod L is up to $3d - 1$.
  3. the image of $T_C|\mathcal{O}(d)| \to \Lambda$ is transverse to $L$.

This is essentially classical, for the sort of argument to prove it you can see e.g. the statement of Lemma 8 and the proof of Lemma 10 in my paper with Steve. $\square$

It's not yet clear to me how to improve this to answer the OP, maybe the better sort of bounds of Greuel, Lossen, and Shustin will help? (Or maybe there's just some easy argument not involving deformation theory...)

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Vivek, thanks a lot! It will take me time to absorb your proof. –  aglearner Feb 26 '13 at 9:48
    
I'll try and write about the deformation theory when I'm less lazy (& assuming no-one puts a proof of the whole thing...) –  Vivek Shende Feb 26 '13 at 23:10
    
Vivek, huge thanks for this update! Now I am completely convinced in your statement. Your answer is really nice (the only reason I have not yet accepted it is that it seems to me now that the original question is an open problem...) –  aglearner Mar 13 '13 at 21:39
    
Having asked around a bit, I also think it's open. The fact that hyperelliptic curves inject, but not with bounded multiplicity, suggests that giving an answer of 'yes' to your original question requires a subtler analysis that sees the difference between 2g and 3g which sounds discouraging. I haven't given up yet though! –  Vivek Shende Mar 14 '13 at 12:21
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The moduli space of curves of degree $d$ in $\mathbb P^2$ has dimension $\left( \begin{array}{c} d + 2 \\ 2 \end{array}\right)$. The most typical type of singularity is a simple cusp. Having a cusp at a given point is a codimension $4$ condition, so having $k$ cusps anywhere is codimension $2k$. This gives a moduli space of dimension $\left( \begin{array}{c} d + 2 \\ 2 \end{array}\right) -2k$ of objects of geometric genus $\left( \begin{array}{c} d -1 \\ 2 \end{array}\right)-k$, since each cusp reduces the geometric genus, relative to the arithemetic genus, by $1$.

Setting $k=\left( \begin{array}{c} d -1 \\ 2 \end{array}\right)-g$, if we wish these curves to fill up the moduli space of curves of genus $g$, $d$ and $g$ must satisfy the inequality:

$\left( \begin{array}{c} d + 2 \\ 2 \end{array}\right) +2g -2 \left( \begin{array}{c} d -1 \\ 2 \end{array}\right) \geq 3g-3$

or

$g \leq 3 + \left( \begin{array}{c} d + 2 \\ 2 \end{array}\right) - 2\left( \begin{array}{c} d -1 \\ 2 \end{array}\right)$

which is satisfiable for only finitely many values of $g$, because the right side goes to $-\infty$ as $d$ goes to $\infty$. (In particular, $g\leq 12$.)

Thus, one cannot embed all curves of large genus using only simple cusps. More complicated singularities could give larger reductions in the genus, at the price of comparatively larger reductions in the size of the moduli space. I don't think that can save you, but I don't know.

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Dear Will, thank you for this answer. I agree that "one cusp on the curve" gives you codimension $2$. But I don't see why having $k$ cusps, when $k$ is around $(d+1)(d+2)/4$ gives you codimension $2k$. You need some transversality for this. Where would you get it from? –  aglearner Feb 22 '13 at 23:17
    
I can't think of an explanation. You certainly have transversality for $d$ sufficiently large with respect to $k$, since the map from functions to their values, first derivatives, and second derivatives at $k$ fixed points is surjective for large $d$. But I'm not sure how to make this effective or if that would be enough. However, I think it's a pretty good heuristic. You would need to get a lot of non-transversality to prove the reverse of this. –  Will Sawin Feb 23 '13 at 5:43
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One more comment: there exist sextic curves of genus 1 with 9 simple cusps of multiplicity 2 (dual to a smooth cubic curves), but any rational curves of the plane has at most 8 cusps (Tono, K.: On the number of the cusps of cuspidal plane curves. Math. Nachr. 278 (2005), 216–221.) In particular, there is no rational sextic with simple cusps of multiplicity 2 (this is easy check in this case: look at the dual curve). This shows that imposing the conditions at cusps is really a global question. –  Jérémy Blanc Feb 23 '13 at 10:25
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@Will: "The most typical type of singularity is a simple cusp" --- I would say that the most typical is a node. –  Sasha Feb 23 '13 at 18:52
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For $2k \le d+1$ you have enough transversality to show codimension $2k$ (since $\mathcal{O}(d)$ is $(d+1)$-very-ample and therefore induces locally versal deformations of singularities with Tjurina number up to $d+1$), and always having $k$ cusps is at least codimension $k+1$, because the Severi varieties are of codimension $k$ and the nodal locus inside is open and dense. I don't know of better bounds. –  Vivek Shende Feb 23 '13 at 22:51
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The question is probably hard to answer in general. Just an easy observation: let $g>0$ be an integer, take a curve of equation $y^2z^{2g}=p_{2g+2}(x,z)$, where $p_{2g+2}$ is a polynomial of degree $2g+2$ with only simple roots, and such that $z$ does not divides $p_{2g+2}(x,z)$. This is the equation of a curve of genus $g$ in $\mathbb{P}^2$, and the only singularity is the point $[0:1:0]$, which is a cusp. In consequence, you get an injective morphism from the resolution of the curve to $\mathbb{P}^2$. And all hyperelliptic curves are obtained by this way. At least, you see that there is no obstruction on the genus. But of course, a general curve of genus $g$ with $g>2$ is not hyperelliptic, so this is only a partial answer.

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Jeremy, thank you! nice to hear that the question is not completely trivial –  aglearner Feb 23 '13 at 14:18
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