If we Blow-up $\mathbb P^2_{\mathbb C}$ at $5$ points $T=\{p_1,\ldots,p_5\}$ we obtain a Del Pezzo surface $X$ of degree $4$. Now take another set of $5$ points $T'=\{q_1,\ldots,q_5\}$ ($T'\neq T$), we obtain another Del Pezzo surface $X'$. We might have some $q_i=p_j$ for some indexes $i,j$.

How do we have to choose $T'$ in oder to get a surface $X'$ which is not isomorphic to $X$? Is there any algorithm on the choice of the points that allows to obtain a set of non-isomorphic Del Pezzo surfaces?

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    $\begingroup$ First to obtain a del Pezzo surface, you need to assume that the points are in general position. Next, up to the action of $PGL_3$, you can assume the first four points are $(0,0,1),(0,1,0),(1,0,0),(1,1,1)$. So only the last point is "free". I don't know in general if different last points give non-isomorphic surfaces. But I expect that there is any answer coming from invariant theory. $\endgroup$ Commented Apr 2, 2019 at 19:05
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    $\begingroup$ For a general del Pezzo surface of degree $4$, the symmetry group of the Mordell-Weil lattice (with the canonical bundle marked) is $W(D_5)$. This acts on the hyperplane class, and I believe at least for a general surface that the entire orbit consists of ample classes and thus defines a map to $\mathbb P^2$ with $6$ exceptional fibers. This group also acts on orderings of the $6$ fibers, which combined with what Daniel Loughran says implies it should have an action on $\mathbb P^2$ by birational automorphisms. You want the invariants under this action to be distinct. $\endgroup$
    – Will Sawin
    Commented Apr 2, 2019 at 19:53
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    $\begingroup$ Take any one of the $5$ points and project $\mathbb{P}^2$ from the point to a line; the cross ratio of the remaining $4$ points is an invariant of $X$ marked with the five $(-1)$-curves coming from the blow up points. If you symmetrize that cross ratio over all sets of markings, you get a continuous invariant of Del-Pezzo's. $\endgroup$ Commented Apr 2, 2019 at 20:55

2 Answers 2


Lemma: Let us work over a field ${\mathbf{k}}$. Let $p_1,\ldots,p_5,q_1,\ldots,q_5\in \mathbb{P}^2=\mathbb{P}^2_{\mathbf{k}}$ be ten points such that no $3$ of the $p_i$ are collinear and no $3$ of the $q_i$ are collinear (in particular, $\{p_1,\ldots,p_5\}$ consists of $5$ points and the same holds for $\{q_1,\ldots,q_5\}$). Then, the following are equivalent:

$a)$ There exists $\alpha\in \mathrm{Aut}(\mathbb{P}^2)=\mathrm{PGL}(3,{\mathbf{k}})$ such that $\alpha(\{p_1,\ldots,p_5\})=\{q_1,\ldots,q_5\}$.

$b)$ Denoting by $C\subset \mathbb{P}^2$ the unique irreducible conic through the $p_i$ and by $C'\subset \mathbb{P}^2$ the unique irreducible conic through the $q_i$, there is an isomorphism $C\to C'$ sending $\{p_1,\ldots,p_5\}$ onto $\{q_1,\ldots,q_5\}$.

$c)$ The Del Pezzo surfaces $X$ and $X'$ obtained by the blowing-ups $\pi\colon X\to \mathbb{P}^2$ of the $p_i$ and $\pi'\colon X'\to \mathbb{P}^2$ of the $q_i$ respectively are isomorphic.

Remark: Condition $b)$ is in practice easy to check, as it is only a question on $5$ points of $\mathbb{P}^1$; one can for instance use the cross-ratios of $4$ of the $5$ points. In particular, over $\mathbf{k}=\mathbb{C}$ (or more generally when $\mathbf{k}$ is infinite), not moving $p_1,\ldots,p_4$ and changing $p_5$ gives infinitely many distinct non-isomorphic del Pezzo surfaces.

Proof of the Lemma: The equivalence between $a)$ and $b)$ follows from the uniqueness of the conic through five points (which follows from the hypothesis that no $3$ are collinear).

$a)\Rightarrow c)$ follows from the universal property of blow-ups.

Hence, the only non-trivial (which if our interest here) is $c)\Rightarrow a)$.

Let $\tau\colon X\to X'$ be an isomorphism and consider the birational map $\varphi=\pi'\circ \tau\circ\pi^{-1}\mathbb{P}^2\dashrightarrow \mathbb{P}^2$. It has base-points only at the points $p_1,\ldots,p_n$. The preimage of a line is a curve of degree $d\ge 1$ having multiplicity $m_i$ at each $p_i$. Computing the self-intersection and using the adjunction formula on $X$, we get the two so-called Noether equalities $$\sum m_i^2=d^2-1, \sum m_i=3d-1$$ and the only solutions are then

$(i)$: $d=1$, $(m_1,\ldots,m_5)=(0,0,0,0,0)$.

$(ii)$: $d=2$, $(m_1,\ldots,m_5)=(1,1,1,0,0)$ (up to permutation)

$(iii)$: $d=3$, $(m_1,\ldots,m_5)=(2,1,1,1,1)$ (up to permutation)

In case $(i)$, the map $\varphi$ is an automorphism of $\mathbb{P}^2$ which sends $\{p_1,\ldots,p_5\}$ onto $\{q_1,\ldots,q_5\}$, so we obtain $a)$.

In case $(ii)$, we change the coordinates such that $p_1=[1:0:0]$, $p_2=[0:1:0]$, $p_3=[0:0:1]$, $p_4=[1:1:1]$, $p_5=[a:b:c]$ with $a,b,c\in {\mathbf{k}}^*$. Then, $\varphi$ is equal to $\alpha\circ \sigma$, where $\alpha\in \mathrm{Aut}(\mathbb{P}^2)$ and $\sigma$ is the involution $$[x:y:z]\mapsto [ayz:bxz:cxy]$$ As the points $p_4,p_5$ are exchanged by $\sigma$ and as $p_1,p_2,p_3$ are also base-points of the inverse of $\sigma$, we obtain $\alpha(\{p_1,\ldots,p_5\})=\{q_1,\ldots,q_5\}$.

In case $(iii)$, we again change the coordinates such that $p_1=[1:0:0]$, $p_2=[0:1:0]$, $p_3=[0:0:1]$, $p_4=[1:1:1]$, $p_5=[a:b:c]$ as before, and use the same $\sigma$ as before. Then, $\sigma\pi\colon X\to \mathbb{P}^2$ is again a birational morphism, contracting $5$ $(-1)$-curves onto the five points $p_1,\ldots,p_5$, and the birational map $\varphi\circ \sigma$ has now degree $2$. We again obtain $a)$ by using a second time a quadratic map as in $(ii)$.


Moduli theory of quartic Del Pezzp surfaces goes back to Coble. I hope the following tow refrences are usefull: moduli of quartic del pezzo surfaces (Colombo, van Geemen, Looijenga) on the moduli of degree 4 del pezzo surfaces (Hasset, Kresch, Tschinkel) They are both pretty concrete, (for required background see e.g. Dolgachev's classical algebraic geometry book), and the second one pretty much seals the deal on the problem.


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