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I am sorry, my question is very naive.

2nd Edit: Let us suppose that $V$ is a smooth complex projective variety, and $Y\subset V$ is a smooth divisor and has an ample conormal line bundle. We would like to contract $Y$ down to a point in $V$, such contractions do not always exist in the category of schemes (thanks to Donu Arapura and abx, see V Example 5.7.3 in, Hartshorne's "algebraic geometry"). However such contractions exist in the category of analytic spaces (Grauert) and algebraic spaces both under positivity conditions on the conormal line bundle (Artin), see for example: Conditions for the Existence of Contractions in the Category of Algebraic Spaces, Joseph Mazur, Transactions of the American Mathematical Society, Vol. 209, (Aug., 1975), pp. 259-265.

$\textbf{Question 1:}$ if a contraction of $V$ is a projective variety $V'$ what can be said about the singular points of $V'$? Are these singularities classical do they have a special name?

$\textbf{Question 2:}$ let us start with a singular projective variety $V'$ of dimension $n$ with at worst isolated singularities, and we suppose that the tangent cone of $V'$ at each of these singular points is an affine cone over a smooth projective hypersurface $Q$ of dimension $n-1$. Am I correct if I say that there exists a resolution $V$ of $V'$ whose exceptional divisors are isomorphic to the $Q$'s and that the conormal bundles are very ample?

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  • $\begingroup$ I think you have the wrong sign, very ample divisors are not (birationally) contractible. $\endgroup$
    – Donu Arapura
    Commented Mar 3, 2014 at 17:10
  • $\begingroup$ Donu is right, you can contract only divisors which are negative in some sense. For instance you can contract an irreducible curve $C$ on a smooth projective surface iff $C^2<0$. $\endgroup$
    – abx
    Commented Mar 3, 2014 at 17:13
  • $\begingroup$ I am sorry, my knowledge of algebraic geometry is very very elementary. I hope that the question sounds better now. $\endgroup$
    – David C
    Commented Mar 3, 2014 at 17:28
  • $\begingroup$ Your new Q2 should be fine. What sort of answer do you want for Q1? $\endgroup$
    – Donu Arapura
    Commented Mar 4, 2014 at 11:54
  • $\begingroup$ For Q1, I think the point under the contraction should be normal by Stein factorization. $\endgroup$
    – Fei YE
    Commented Dec 13, 2020 at 4:21

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