let $\mathbb{C}^{m}$ be the complex $m$-space with the standard complex structure and let $$P:=\left\{p_{1},\ldots,p_{N} \right\}\subset \mathbb{C}^{m}$$

a finite set of points. Now we blow up each of these points and we call $X$ the resulting complex manifold. If $N$ is sufficiently large and points of $P$ are in "sufficiently general position" does $X$ become rigid? (i.e. it has no deformations) If yes how can i see it? I tried to compute $H^{1}(T_{X})$ but i got stuck.

Thank you in advance!

  • $\begingroup$ X will always have deformations corresponding to moving the points. You should be able to compute $H^1(T_X)$ using the Leray spectral sequence associated to the map $X\to \mathbb{C}^m$ to see that those are all of the deformations. $\endgroup$
    – Jim Bryan
    Sep 15, 2015 at 0:16
  • $\begingroup$ All these deformations are not biholomorphic if there are enough points in general position, am i right? $\endgroup$
    – student
    Sep 17, 2015 at 10:50
  • $\begingroup$ I think you are getting confused between deformations and automorphisms. When you deform a variety, you change it's complex structure and you (usually) end up with a variety which is not biholomorphic to the original variety. It sounds like you want to prove that X has no automorphisms. Rigidity usually mean that there are no deformations, but I suppose that not having any automorphisms is a kind of rigidity too. (Infitesimal) deformations are classified by $H^1(T_X)$, whereas (infinitesimal) automorphisms are classified by $H^0(T_X)$. $\endgroup$
    – Jim Bryan
    Sep 18, 2015 at 17:37
  • $\begingroup$ If you compute $H^0(T_X)$ using the Leray spectral sequence associated to the map $X\to \mathbb{C}^n$ it should tell you that infinitesimal automorphisms of $X$ are given by infinitesimal automorphisms of $\mathbb{C}^n$ which vanish at $P$, i.e. vector fields on $\mathbb{C}^n$ which are 0 at $p_1,…,p_N$. $\endgroup$
    – Jim Bryan
    Sep 18, 2015 at 17:51


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