Why $X$ is a del Pezzo surface? I'm reading the notes for a birational geometry course. There is a statement I do not understand:
Suppose $X$ is a smooth projective rational surface $G\subset \text{Aut}(X)$ is a finite subgroup such that $\text{rk}(\text{Pic}(X)^G)=1$. Then $X$ is a dell Pezzo surface.
It says that to see this the reader should "intersect the orbit of the curve with the canonical divisor and find a positive number". I tried that and got:
$$g(\sum C_i)= \sum C_i \Rightarrow \sum C_i =a K_X-$$
since the fixed part of the Picard group is generated by the canonical divisor (here $\sum C_i$ is the orbit of some curve). So after intersecting we have:
$$\sum C_i\cdot K_X=a K_X \cdot K_X=a\cdot K_X^2$$
I do not see how is gives that $-K_X$ is ample (i.e. $X$ a dell Pezzo surface). I will appreciate any help!
 A: I  think this follows from  the Nakai Moishezon ampleness criteria. The NS criteria says that prove $X$ is del Pezzo it is sufficient to prove that $-K_X.C>0$ for every effective curve $C$.
Firstly, automorphisms preserve tangent bundle so in particular preserve the anticanonical divisor, so $g(-K_X) = -K_X$, so indeed $-K_{X}$ generates $Pic(X)^{G}$ (note that $-K_{X} \neq 0$ since $X$ is assumed to be rational, hence is not weak Calabi-Yau).
The next claim is that it is sufficient to prove that the orbit of any curve intersects $-K_X$ positively. This is because, for any effective curve $C$,  $$\sum_{G} gC. (-K_X) = \sum_{G} gC.g(- K_{X}) = |G|( C.-K_{X}), $$ so I guess this is how to proceed. They claim that the orbit of any curve intersects $(-K_X)$ positively so that would be enough. Actually it is not so clear to my why that is the case but it sounds like you were happy with that part?, I would be interested to know which notes this was in.
Ok, I think I have it, by a slightly messy argument.
Claim 1. There is an effective, rational irreducible curve $C_{+} \subset X$ such that $C_{+}^2 \geq 0$.
This follows from MMP, every rational surfaces is a blow-up of $\mathbb{CP}^2$ or a Hirzebruch surface and so the claim is standerd (*) (I can give more details if you want).
Next, note that $\sum_{G} gC_{+} = a(-K_{X})$ where $a > 0$, and by the fact that $g(C)=0$, the genus formula gives $-K_{X}.C_{+}>0$. So
$0 <\sum_{G} gC_{+}. (-K_{X}) = a(-K_{X})^2$.
In particular $(-K_{X})^2>0$.
Now, for any curve in $C \subset X$, we write $$\sum_{G} gC = a' (-K_{X}) $$ where $a'>0$. Then $$\sum_{G} gC (-K_{X}) = a' (-K_{X})^2 >0$$, hence we are done by the Nakai Moaishezon ampleness criterion.
(*) if the surface is a blow up of $\mathbb{CP}^2$ then you can take a line which misses all of the blow-up points. If the surface is the blow up of a Hirzebruch surface i.e. a $\mathbb{P}^1$-bundle over $\mathbb{P}^1$ you can take a $\mathbb{P}^1$-fibre which misses all of the blow up points. Lines have self intersection $+1$ fibres have self-intersection $0$, since they miss the blow-up points their self-intersection doens't change, you just need to note you have enough of them to miss the blow-up points for example because they cover the surface.
