4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ – Evgeny Shinder Jan 6 '20 at 23:15
  • 1
    $\begingroup$ 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). $\endgroup$ – Evgeny Shinder Jan 6 '20 at 23:22
6
$\begingroup$

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 $Aut(X)$.

$\endgroup$
1
  • $\begingroup$ 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 $X\to X$. $\endgroup$ – Ariyan Javanpeykar Jan 7 '20 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.