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It is easy to show using birational factorization that the only curves on a surface which can be contracted to get an algebraic, smooth surface are smooth $(-1)$-curves. Furthermore, I know of examples of smooth curves of some other negative self-intersection which can be contracted in the algebraic category to result in a singular surface (where the order of the singularity is equal to the negative self-intersection?)

Question 1: Given a curve of negative self-intersection on a complex surface, what is the construction (in the analytic category) of its contraction?

Question 2: Given the curve, what conditions on it that determine whether the contraction is algebraic or not?

Good, clear, elementary references would be fine, as an alternative to an answer!

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  • $\begingroup$ Added the tag ag.algebraic-geometry $\endgroup$
    – Francesco Polizzi
    Commented Mar 2, 2013 at 13:04
  • $\begingroup$ Hi Philip, this article (of mine) gives a necessary and sufficient criterion for algebraicity in a special case: arxiv.org/abs/1301.0126 PS: I myself am interested in your Question 1, and I don't know of any other reference other than Grauert's original article, which is in German and therefore I can't read :( $\endgroup$
    – pinaki
    Commented Apr 27, 2013 at 1:07

1 Answer 1

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The best numerical criterion of contrability for curves on surfaces is perhaps the following result, due to Michael Artin.

Proposition. Let $V$ be a surface and $X=\bigcup X_i \subset X$ be a connected curve. Then the following are equivalent:

$\boldsymbol{(i)}$ $X$ is contractible and if $\pi \colon V \to \bar{V}$ is the contraction map, then $\chi(\mathcal{O}_V) = \chi(\mathcal{O}_{\bar{V}})$;

$\boldsymbol{(ii)}$ the intersection matrix $|(X_i \cdot X_j)|$ is negative definite and for any cycle $Z$ supported in $X$ one has $p_a(Z) \leq 0$.

Moreover, under these conditions, if $V$ is a normal projective surface then $\bar{V}$ is also projective.

In other words, if $V$ is normal, projective and $X$ is "sufficiently rational and negative", then $X$ is contractible and the contraction is algebraic. For further detais, see

M. Artin: Some Numerical Criteria for Contractability of Curves on Algebraic Surfaces,American Journal of Mathematics Vol. 84, No. 3 (Jul., 1962), pp. 485-496,

in particular Theorem 2.3 p. 491.

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  • $\begingroup$ Thanks for your answer, it partially resolves the case of my second question, by giving criteria for contractibility in the algebraic category in certain cases. I mainly wanted an explicit construction {\it in the analytic category} of the contraction. (The conditions would of course be weaker if we allow the contraction not to be an algebraic surface). Since the proposition above seems to be the best result about contractibility, I assume it is hard then to determine whether the resulting surface is algebraic... $\endgroup$
    – Philip Engel
    Commented Mar 6, 2013 at 22:25

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