# When do blow-ups at $N$ points give isomorphic varieties?

Let $$X$$ be a smooth projective variety over an algebraically closed field $$k$$. For any length $$n$$ ideal sheaf $$P$$ of $$X$$ (e.g $$N$$ different points $$P=(P_1)(P_2)..(P_N)$$), we can consider $$Bl_{P}(X)$$, the blow up of $$X$$ at $$P$$. Assume $$Bl_{P}(X) \cong Bl_{Q}(X)$$ for a length $$n$$ ideal sheaf $$P$$ and a length $$m$$ ideal sheaf $$Q$$ , what can we say about $$P$$ and $$Q$$ ? If neccessary, one can assume $$P,Q$$ are radical.

Note blow up at $$I$$ and $$I^2$$ are the same. So there is a trivial case i.e when there exists an automorphism $$\phi:X \rightarrow X$$ s.t $$\phi^*(P^k)=Q^l$$.

I am interested in some examples e.g projective spaces, abelian varieties, surfaces.

• The question has more content and a different emphasis over non-algebraically closed fields. Let $P$ and $Q$ be reduced zero-dimensional subschemes of $X$ such that the blow ups are isomorphic. Are $P$ and $Q$ isomorphic schemes? – Evgeny Shinder Jan 6 '20 at 23:15
• Note that in spite of blow ups of $I$ and $I^2$ being the same, ideals with the same radical as $I$ do not in general lead to isomorphic blow up. Indeed, if $I = (x,y)$ in the plane, then $I^2 \subset (x,y^2) \subset I$ but blowing up the intermediate ideal leads to a singular point (a node). – Evgeny Shinder Jan 6 '20 at 23:22

## 1 Answer

The answer in general is a mess, because isomorphisms between the blow-ups do not necessarily descend to automorphisms of $$X$$. If you take $$X = \mathbb P^2$$ and fix a general configuration $$P$$ of $$n \geq 9$$ points, then the set of $$Q$$ for which $$Bl_P(X) \cong Bl_Q(X)$$ is a countable union of subvarieties of $$(\mathbb P^2)^n$$, which is probably Zariski dense (though I'm not sure whether anybody has actually tried to prove this).

In some sense the point is that such blow-ups have infinitely many $$(-1)$$-curves on them, and so you can blow down to $$\mathbb P^2$$ in infinitely many different ways by choosing which curves to contract. The resulting configurations in $$\mathbb P^2$$ that you get as the images of the contracted curves do not simply differ by elements of $$PGL(3)$$, as you might hope.

If $$X$$ is not uniruled, things are probably easier. For example, if $$X$$ is abelian, any isomorphism $$f : Bl_P(X) \to Bl_Q(X)$$ descends in the obvious way to an isomorphism $$g : X \to X$$, and so the answer is that the blow-ups are isomorphic if and only if $$P$$ and $$Q$$ differ by some element of $$Aut(X)$$.

• Just a comment on your last paragraph for those interested: If $X$ is smooth projective and has no rational curves, then every isomorphism $Bl_P(X) \to Bl_Q(X)$ descends to an isomorphism $X\to X$. – Ariyan Javanpeykar Jan 7 '20 at 20:00