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When I read the paper "A survey on Cox rings" with the link below:

http://math.berkeley.edu/~velasco/Survey.pdf

In Section 4.2 it is mentioned that the Del Pezzo surface given by blowup of $\mathbb{P}^2$ at $s>3$ general points is not toric. Is there an easy way to see this?

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Dear minimax, I think the title question is much more interesting than the particular case you want! I suggest adding "what are the general techniques for proving that a given variety is not toric"? – Piotr Achinger Jan 17 at 6:57
Related to my comment above: mathoverflow.net/questions/93224/… – Piotr Achinger Jan 17 at 6:58
@Piotr: Changed! – minimax Jan 17 at 7:05
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Dear minimax, the inequality in the question should read $s>3$, not $s\geq 3$ (unless the 3 points are collinear). The blow-up of $\mathbb{P}^2$ at three non-collinear points is toric. – Piotr Achinger Jan 17 at 7:11
@Piotr: Thanks, it has been changed! – minimax Jan 19 at 0:55

3 Answers

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I'm thinking about the following argument. It's not entirely precise, I will be grateful for comments how to improve it.

Suppose such an $X$ is toric. Because the exceptional curves $E_i$ do not move inside $X$, they are fixed by the torus action. Therefore after blowing them down, the torus still acts and their images are fixed points. But there are only 3 fixed points of the torus action on $\mathbb{P}^2$!

Different idea (using fans): since $Pic (X) = \mathbb{Z}^{1+s}$, the fan defining $X$ has $3+s$ rays, and $s$ of them correspond to the exceptional curves $E_1, \ldots, E_s$. Let us denote the torus invariant divisors corresponding to the other 3 rays $D_1, D_2, D_3$. Since $\dim X=2$, these $3+s$ rays/divisors are arranged in a circular order, so that only neighbors on this circle intersect. Because the curves $E_i$ and $E_j$ do not intersect for $i\neq j$, no two $E_i$ are neighbors on the circle, which shows that there at most as many $E$'s are $D's$, that is, $s\leq 3$.

Note that the first argument works more generally for the blow-up of $\mathbb{P}^n$ in $s > n+1$ points.

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Here is another way to see why this is true (though as already noted you need $s >3$).

Note that if $X$ is a toric variety with respect to an algebraic torus $T$, then by definition $T$ acts faithfully on $X$. In particular there is an injective homomorphism $T \to \mathrm{Aut}(X)$.

Now if $X$ the blow-up of $\mathbb{P}^2$ in $s >3$ general points, then it is "well-known" that the automorphism group of $X$ is finite (see e.g. The main theorem of Koitabashi - Automorphism groups of generic rational surfaces). In particular such surfaces are not toric varieties. Moreover Koitabashi even shows that the automorphism group is trivial for $s \geq 9$.

Note that this method also shows that such $X$ do not admit a faithful action of any algebraic group of positive dimension, so they are very far from being homogeneous.

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It is definitely very easy to show that the automorphism group of the blowup of $\mathbb{P}^2$ in $s>3$ general points is discrete, since its Lie algebra is the space of holomorphic vector fields; on $\mathbb{P}^2$ this space has dimension $3$, and every time you blow up a point the dimension drops by one. – YangMills Jan 19 at 2:13
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Nobody has mentioned that you can just classify the complete smooth toric surfaces using some rather simple combinatorics (a fun exercise or chapter 10 of the book by Cox, Little and Schenck). The upshot is that the only complete smooth toric surfaces are obtained from toric blowups of $P^2$, $P^1 \times P^1$, or the Hirzebruch surfaces. Now we are just left to observe that the ones in the original question are not on the list. Of course, I understand people are interested in the more general techniques.

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