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Let $k$ be a field and let $C,D$ be two integral curves in $\mathbb{P}^2_k$. Now let $f:C \to D$ be a birational isomorphism. Can $f$ be extended to $\mathbb{P}^2_k$. To be precise, does there exist a birational isomorphism $F:\mathbb{P}^2_k \to \mathbb{P}^2_k$ such that $F$ and $f$ agree on a (non empty) open subset of $C$?

I am mainly interested in the case when $k=\mathbb Q$.

[Edit] Since it is apparently easy to read over I will state it here explicitly. The map $F$ is allowed to be a birational isomorphism. It is clear that my statement is false when you want $F$ to be an isomorphism since it has to send curves of the same degree to curves of the same degree.

If we take for example $C$ to be the curve $x=0$ and $D$ to be the curve $y^2z=x^3+x^2z$ then $C$ and $D$ are birationally equivalent but clearly no automorphism of $\mathbb{P}^2_k$ sends C to D. However it is possible with a birational isomorphism.

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True if both curves are smooth of the same degree at least four (see Geometry of Algebraic Curves by Arbarello et al). False in general, as Karl points out. – Felipe Voloch May 6 '12 at 23:54
@Felipe: do you know where to find this in ACGH? I don't remember where to find a section discussing birational automorphisms. I certainly suspect this is very false for singular plane curves, which are somehow the "generic" case when viewed from the point of view of the moduli of curves instead of when viewed as zero loci of polynomials--most curves do not embed smoothly in P^2. I don't have a simple argument yet, though--the Cremona group of birational transformations is deceptively complicated. – Jack Huizenga May 7 '12 at 2:25
@Jack. I don't have the book handy, but they prove in an exercise that the $g^2_d$ is unique on a smooth plane curve of degree $d$. – Felipe Voloch May 7 '12 at 12:51
OK, thanks! For anyone interested, the relevant exercise batch is Appendix A, Section 1 (page 56). More precisely, it is an exercise that if $\Gamma \subset \mathbb{P}^2$ is a degree $d$ nodal plane curve with $\delta < d-3$ nodes, then the induced $g_d^2$ on the normalization is unique. So this covers positively the case where the curves have the same degree and do not have too many nodes relative to the degree; in fact in these cases the curves are actually projectively equivalent. – Jack Huizenga May 7 '12 at 17:52
up vote 12 down vote accepted

In general the answer is no.

This kind of question is studied in more generality in the paper by Mella and Polastri "Equivalent birational embeddings", Bull Lond. Math. Soc. 41 (2009) 89-93,

They prove that two birational embeddings of $X$ in $\mathbf{P}^n$ are equivalent up to Cremona transformations of $\mathbf{P}^n$ as long as $n\geq \dim(X)+2$. For instance, any rational variety of codimension at least $2$ in $\mathbb{P}^n$ is Cremona equivalent to a linear space.

The case $n=2$ and $\dim(X)=1$ is outside this range, and indeed there are examples of birational plane curves that are not equivalent up to Cremona transformations, hence not equivalent under the action of $\textrm{Bir}(\mathbb{P}^2)$, since $\textrm{Bir}(\mathbb{P}^2)$ is generated by Cremona transformations.

The following example is given by Mella and Polastri in the last section of their paper. Take a general projection of a curve of bidegree $(1,d)$ on a quadric surface to $\mathbb{P}^2$. This is a plane rational curve $C$ of degree $d$ with only ordinary double points, hence there is a birational isomorphism $C \dashrightarrow L$, where $L$ is a line. However, one proves that if $d \geq 6$ then $C$ is not Cremona equivalent to $L$.

The proof is based on the following

Lemma. Let $X \subset \mathbb{P}^n$ be a rational variety of codimension $1$ and degree $d>1$. If $X$ is Cremona equivalent to a hyperplane, then the singularities of the pair $(\mathbb{P}^n, \frac{n+1}{d} X)$ are not canonical.

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Historical comment: another example of birational plane curves that are not Cremona equivalent comes from the 10-nodal plane sextic. If I understand correctly, this is the lowest-degree rational plane curve that cannot be transformed into a line by Cremona transformations. This was known already to Coble before 1920, and presumably to others much earlier. – user5117 May 7 '12 at 17:52
You are right, $6$ is the lowest possible degree, as it also mentioned in Mella-Polastri's paper. I added this in the answer. – Francesco Polizzi May 7 '12 at 21:45

Another way to see that a nodal rational curve of degree $\ge 6$ cannot be sent to a line by a birational map of the plane is to see that it cannot be contracted by a birational map of the plane. Each birational map of the plane decomposes into blow-ups and contractions of smooth $(-1)$-curves (curves isomorphic to $\mathbb{P}^1$ and of self-intersection $-1$). If you want to contract the curve, you have to blow-up all nodes. The number of nodes is $(d-2)\cdot (d-1)/2$, and the self-intersection of the curve on $\mathbb{P}^2$ is $d^2$. After blowing-up, the self-intersection decreases by $4$ for each double point, so becomes $d^2-2\cdot (d-1)(d-2)=6d-d^2+4$. If $d\ge 6$, this number is $\le -2$, so the curve is not contractible. If $d\le 5$, the curve is contractible and in fact one can check easily that it can be sent onto a line by a birational map of the plane.

Similarly, one sees that a nodal rational curve of degree $d\ge 6$ cannot be sent onto a nodal rational curve of degree $d'\ge 6$ when $d\not=d'$.

Similar arguments work with curves of genus $1$. See for example Proposition 3.3.3 of "On birational transformations of pairs in the complex plane", J. Blanc, I. Pan, T. Vust, Geom. Dedicata 139 (2009), 57-73.

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EDIT: As René Pannekoek points out in a comment, I misread the question. This question was asking for birational automorphisms of $\mathbb{P}^2$, not just automorphisms.

I'll leave this answer (which doesn't answer the question) since it might be useful.

A negative result for elliptic curves: Not in general. For example, suppose that $C$ and $D$ are the same elliptic curve inside $\mathbb{P}^2$. Set $P \in C$ to be an inflection point, ie a point such that $L \cap C = 3P$ for some line $L \in \mathbb{P}^2$. Fix $Q$ to be another point which is not of that form (ie, such that the tangent line intersects $Q$ at another point of $C = D$). In particular, let's say that $Q$ is an element of infinite order.

Suppose that $f : C \to D$ to be a map which sends $P$ to $Q$ (note that such maps always exist since $C$ is an elliptic curve and thus an Abelian variety). No automorphism of $\mathbb{P}^2$ (ie, an element of $PGL(2)$) will restrict to $f$, since such an $F$ will send the line $L$ going through $P$ to a line going through $Q$ (and only $Q$).

EDIT: As François Brunault points out, this sort of construction can't work to avoid birational automorphisms.

A positive result for elliptic curves: On the other hand, if your elliptic curve is $C = V(y^2 - x(x-1)(x-\lambda) )$, with inflection point at infinity $P$, then Exercise 4.3 in Chapter IV of Hartshorne says that every isomorphism of $C$ that leaves $P$ fixed comes from a linear automorphism of $\mathbb{P}^2$.

A positive result for genus 3 curves: The result is true for genus 3 curves, see Hartshorne, Chapter IV, Exercise 5.7(a).

A positive result vacuously: Of course, ``most'' curves of higher genus have no automorphisms at all (ie, most elements of the moduli space have no automorphisms except the identity). See for example Baily, On the automorphism group of a generic curve of genus >2.

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Re your counter-example: Maarten's $F$ was allowed to be a birational automorphism. Hence $L$ is not necessarily mapped to a line, I guess. – René May 6 '12 at 23:52
Isn't it the case that any point on a cubic curve can be made into an inflection point using a birational transformation ? There is an algorithm due to Nagell to transform any cubic curve into Weierstrass form, but I don't know whether this applies here. – François Brunault May 7 '12 at 6:11
Thanks for the answer. Altough it was not the answer to my question it stil was instructive. – Maarten Derickx May 7 '12 at 9:38
Yesterday during lunch at Univeriteit Leiden I also heard another argument (by bas Edixhoven) why the "negative result for elliptic curves" argument cannot work in the birational case. The argument is by considering the Hesse pencil $u(x3+y3+z3)+vxyz$ . Now by blowing up in the 9 points where rational map $\mathbb{P}^2 \to \mathbb{P}^1$ given by $(x:y:z)\mapsto(x^3+y^3+z^3:xyz)$ is not defined we get $E(\Gamma(3))\mapsto \mathbb{P}^2$. $E(\Gamma(3))$ together with the map to $\mathbb{P}^1$ (via $\mathbb{P}^2$) will be the universal elliptic curve with full level 3 structure. – Maarten Derickx May 11 '12 at 21:02
Now $E(\Gamma(3))/\mathbb{P}^1$ doesn't have a section of infinite order. But by choosing a suitable line $L \in \P^2$ to parameterize the elliptic elliptic curves in the Hesse pensil by one can find a different elliptic curve $E \to \P^1$ wich has a section of infite order, and this $E$ will also be birational equivalent to $\P^2$. Now the translation by this point of infinite order is an isomorphism from $E$ to itself and hence gives a birational map from $\P^2$ to itself. This birational automorphism will be the translation by a point of infinite order in most of its fibers over $L$. – Maarten Derickx May 11 '12 at 21:15

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