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Suppose that $X$ is a smooth projective surface over $\mathbb C$ and $f\colon X\to\mathbb P^2$ is a finite morphism branched over a curve $S\subset\mathbb P^2$. Assume in addition that all the singularities of the curve $S$ are nodes and cusps and that the covering $f$ is ``simple'' (i.e., for a general line $L\subset \mathbb P^2$, the branched covering $f^{-1}L\to L$ has only one ramification point over each $p\in S\cap L$, and the ramification index at this point is $2$). In this setting, when can one say that, for a given curve $S$, there exists at most one morphism $f$?

I am aware of the following two results.

(a) (Kulikov, Nemirovski, 2001) The assertion is true whenever $\deg f\ge 12$;

(b) (Kulikov, 2008) The assertion is true whenever $f^*\mathcal O_{\mathbb P^2}(1)$ is very ample and $f$ is not a projection of the Veronese surface $v_2(\mathbb P^2)\subset\mathbb P^5$.

Could somebody tell me what else is known in this direction?

Thank you in advance,
Serge

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  • $\begingroup$ Just to clarify: you are interested in positive results, not in negative results, correct? It seems to me that one can produce negative examples as a composition $f=(g_1\times g_2)\circ h$ of a degree $2$ morphism $h:\mathbb{P}^1\times \mathbb{P}^1\to \mathbb{P}^2$ with a product morphism $g_1\times g_2:C_1\times C_2 \to \mathbb{P}^1\times \mathbb{P}^1$, where $g_i:C_i\to \mathbb{P}^1$ is a simple branched cover. Did you want to add a hypothesis that $\pi_1(\mathbb{P}^2\setminus S, p) \to \text{Aut}(f^{-1}\{p\})$ is surjective? $\endgroup$
    – Jason Starr
    Dec 23, 2013 at 15:48
  • $\begingroup$ @Jason: if the surface $X$ is irreducuble, then the homomorphism $\pi_1(\mathbb P^2\setminus S,p)\to \mathrm{Aut}(f^{-1}(p))$ is automatically surjective. $\endgroup$
    – Serge Lvovski
    Dec 23, 2013 at 19:04
  • $\begingroup$ And yes, I'm interested in positive results, and I am not sure the suggested construction will yield counterexamples. $\endgroup$
    – Serge Lvovski
    Dec 23, 2013 at 19:06
  • $\begingroup$ You are right. My covering is not simple: the branch divisor of $h$ shows up multiply in the branch divisor of $f$. $\endgroup$
    – Jason Starr
    Dec 23, 2013 at 21:32

1 Answer 1

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I believe this is also true if $\text{deg}\ f$ equals $2$; presumably this is well-known. To prove that $f$ is unique up to isomorphism (over $\mathbb{P}^2$), it is equivalent to prove that the $\mathcal{O}_{\mathbb{P}^2}$-algebra $f_*\mathcal{O}_X$ is unique up to isomorphism. Of course, by Zariski's Main Theorem (or easier arguments), it suffices to prove uniqueness over the open complement $U$ of finitely many points of $\mathbb{P}^2$. Then there is the standard short exact sequence, $$ 0 \to \mathcal{O}_{U} \xrightarrow{f^\#} f_*\mathcal{O}_X \to \mathcal{L} \to 0,$$ where $\mathcal{L}$ is an invertible sheaf on $U$ (after deleting finitely many points). The algebra structure on $f_*\mathcal{O}_{X}$ determines an injective homomorphism of coherent sheaves, $$u:\mathcal{L}^{\otimes 2} \to \mathcal{O}_{U},$$ whose image equals the ideal sheaf $\mathcal{O}_{U}(-\underline{S})$ (again after deleting finitely many points). Thus uniqueness of $f$ is equivalent to uniqueness of the pair $(\mathcal{L},u)$ of an invertible sheaf $\mathcal{L}$ and an isomorphism $u:\mathcal{L}^{\otimes 2}\to \mathcal{O}_{U}(-\underline{S})$ up to "equivalence", i.e., $(\mathcal{L},u)$ is equivalent to $(\mathcal{M},v)$ if there exists an isomorphism of invertible sheaves $w:\mathcal{M}\to \mathcal{L}$ such that $v\circ w^{\otimes 2}$ equals $u$.

Of course for two such pairs, $(\mathcal{L},u)$ and $(\mathcal{M},v)$, then $\mathcal{T} := \mathcal{M}\otimes \mathcal{L}^\vee$ is an invertible sheaf and $v\otimes u^\dagger$ is an isomorphism $\mathcal{T}^{\otimes 2} \to \mathcal{O}_{U}$. In other words, $\mathcal{T}$ is a $2$-torsion element in $\text{Pic}(U)$. However, for every open complement $U$ of finitely many points of $\mathbb{P}^2$, $\text{Pic}(U)$ is just $\mathbb{Z}$. Thus $\mathcal{T}$ is isomorphic to $\mathcal{O}_U$, so that the pair $(\mathcal{L},u)$ is unique up to equivalence.

I am sure the case that $\text{deg}\ f$ equals $2$ is well-known. Just to make one observation: with the one exception where $X$ is a quadric surface, for $f$ as above, $f^*\mathcal{O}_{\mathbb{P}^2}(1)$ is not very ample. Indeed, the pullback map $$H^0(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}(1)) \to H^0(X,f^*\mathcal{O}_{\mathbb{P}^2}(1))$$ is an isomorphism. Of course, also $\text{deg}\ f \not\geq 12$.

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  • $\begingroup$ Dear Jason, is it correct to say that your argument works for projective spaces (or even smooth quadrics) of any dimension, if one assumes that singularities of the ramification are good enough (say, double points)? Of course $U$ would just be an open set whose complement is at least codimension two. $\endgroup$
    – IMeasy
    Jan 13, 2017 at 10:05
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    $\begingroup$ @IMeasy. Yes, I think the argument works very generally. I learned after I wrote this answer that Pardini has a lovely article on Abelian covers. I am certain the answer above follows from her article. Probably what you ask also follows from Pardini's article. $\endgroup$
    – Jason Starr
    Jan 14, 2017 at 0:11

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