# symplectic classes on rational surfaces.

Hi. I have a stupid question. Let $M$ be a blow-up of the complex projective plane at $k$ generic points. Then we can choose an orthoginal basis (with respect to the cup product) $H, E_1, \cdots, E_k$ of $H^2(M;\mathbb{Z})$ such that $H^2 = 1, E_i^2 = -1$ for each $i=1,\cdots,k$. Then my question is,

For a given class $C = aH + b_1E_1 + \cdots b_kE_k \in H^2(M;\mathbb{Z})$, how can we check whether $C$ is a symplectic class or not? (I mean, how can we know there exists a symplectic form which represents $C$?). For example, if $k=2$, is there a symplectic form which represents a class $2H - E_1 - E_2$?

I'd really appriciate for your any comment. Thank you in advance.

## 1 Answer

This answer is rewritten and include more details

First of all I highly recommend you the article of Paul Biran From Symplectic Packing to Algebraic Geometry and Back available on the page http://www.math.tau.ac.il/~biranp/Publications/Pubications.html , especially theorem 3.2.

Your question basically asks "what is the symplectic cone of $\mathbb CP^2$ blown up in a finite number of points?" This question was answered by Paul Biran (check theorem 3.2. from the above article), though the answer is not 100% explicit. Also, it is known that the symplectic cone of $\mathbb CP^2$ blown up in up to $9$ points coincides with the Kahler cone if the points are chosen so that the resulting surface has only $-1$ curves (in particular it is Fano if the number of points is at most $8$). This permits one to answer your last question (that is done below). In fact Kahler cones of Fano surfaces rather classical objects and all basic questions about them can be answered.

I would like to add that from a certain conjecture from algebraic geometry - Harbourne-Hirschowitz conjecture, it follows that the Kahler cone of $\mathbb CP^2$ blown up in a very generic collection of points coincides with its symplectic cone (Literately this conjecture says the following : any integral curve with negative self-intersection on the blow-up of $\mathbb CP^2$ at a set of points in very general position is a smooth rational curve with self-intersection $−1$. In order to deduce the statement that the symplectic cone coincides with the Kahler one you have to use SW theory). Habourine-Hirschwitz conjecture is open even for $\mathbb CP^2$ blown up in $10$ points, and the famous Nagata conjecture is a partial case of it.

Now let us answer the last bit of the question. The class $2H-E_1-E_2$ is not symplectic. The symplectic cone of $\mathbb CP^2$ blown up in $2$ points coincides with the Kahler cone of $\mathbb CP^2$ blown up in two points. And $H-E_1-E_2$ is a rational $-1$-curve on this surface, while $(2H-E_1-E_2)\cdot (H-E_1-E_2)=0$, so $2H-E_1-E_2$ is not ample, and hence not symplectic.

• You are very welcome :) ! – Dmitri Panov May 22 '11 at 13:36
• @Dmitri: I didn't think the question was that involved: Isn't it true that if $C$ is a symplectic class, then $C^2>0$ (the symplectic class is a volume form) and $C\cdot X>0$ for any complex curve $X$ (since every complex curve is a symplectic submanifold). So $C$ has to satisfy $C^2=a^2-\sum b_i^2 >0$ and $aA-\sum b_iB_i>0$ when $X=AH+\sum_i B_iE_i$ is represented by any complex curve. Maybe I'm missing something obvious. – Paul May 22 '11 at 16:44
• OK, I get it, its the converse that is tricky. – Paul May 22 '11 at 20:05
• Paul, yes, indeed $C\cdot X$ must be positive, the point is that if you we blow up $\mathbb CP^2$ in $n\ge 9$ generic points the number of rational $-1$-curves is infinite, and the structure of this set infinite is quite involved. If you blow up $\mathbb CP^2$ in $8$ points the number of $-1$ curves is around 250 (I am lazy now to make the precise calculation) – Dmitri Panov May 22 '11 at 23:32