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Hi. I have a question about automorphisms of smooth ratonal surfaces. (I am not an algebraic geometer, so my question could be stupid. I am sorry about that.)

Let $X_k$ be a blow-up of $\mathbb{P}^2$ at $k$ generic points. (Especially for $k \leq 3$.) Fix a homology class $A \in H_2(X_k, \mathbb{Z})$, and let $C_1, C_2$ be two algebraic curves on $X_k$ such that $[C_1] = [C_2] = A$. Then my question is,

Q1 : Can we always find an automorphism of $X_k$ which sends $C_1$ to $C_2$ ?

Q2 : If Q1 is not true, the same question Q1 when $C_1$ and $C_2$ are rational.

For a given symplectic manifold $(M,\omega)$ and two symplectic surfaces $C_1$ and $C_2$ with $[C_1]=[C_2] \in H_2(M,\mathbb{Z})$, I wanted to know whether there exists a symplectomorphism of $(M,\omega)$ which sends $C_1$ to $C_2$ or not.

On $\mathbb{P}^2$, B. Seibert and G. Tian proved that any symplectic curve of degree less than 18 is symplectically isotopic to an algebraic curve (along the Hamiltonian isotopy on $\mathbb{P}^2$). And they also prove that for the Hirzebruch surface or $\mathbb{P}^1 \times \mathbb{P}^1$ when a degree of curve is less than 8. (ON THE HOLOMORPHICITY OF GENUS TWO LEFSCHETZ FIBRATIONS)

And V. Shevcishin proved that any two curves with the same degree less than 7 on $\mathbb{P}^2$ is symplectically isotopic each other.

My question is much weaker problem than the symplectic isotopy problem, so I hope that there would be an answer for the Question 1 and 2.

Thank you for reading and I appriciate for any comment.

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This is not true even for $k=0$. In fact, $H_2(\mathbb{P}^2)=\mathbb{Z}$, so the homology class of a curve is determined uniquely by its degree. Now take a smooth cubic $C_1$ and a nodal cubic $C_2$. They belong to the same cohomology class but there is no element in $\textrm{Aut}(\mathbb{P}^2)$ exchanging them, symply because they are not isomorphic: $C_1$ is an elliptic curve, whereas $C_2$ is rational. You can find counterexamples to Q2 in the same way: take a quartic curve with three nodes and one with a triple point: they are cohomologous and both rational, but they are not isomorphic – Francesco Polizzi Aug 14 '11 at 19:19
If $C_1,C_2$ are two smooth cubics in $\mathbb{P}^2$ with different $j$-invariants (such pairs exist). Then no automorphism will take one to the other. – Donu Arapura Aug 14 '11 at 19:19
Note that you can also find counterexamples to Q1 where both curves are smooth, if the degree is at most $3$. In fact, plane curves of degree $\geq 3$ have moduli, so two of them are not isomorphic in general. For instance, it sufficies to take as $C_1$ and $C_2$ two smooth cubic curves whose $j$-invariant is different – Francesco Polizzi Aug 14 '11 at 19:22
@Donu: It seems that we wrote the same example at the same time :) – Francesco Polizzi Aug 14 '11 at 19:23
Perhaps the OP meant the curves in Q2 to be smooth and rational. But this is still false, e.g., conics on a generic cubic surface ($\infty^1$ conics but a finite automorphism group). – Jason Starr Aug 15 '11 at 12:55

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