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Suppose I have a morphism $f:\overline{\mathcal{M}}_{0,n} \to \mathbb{P}^N$ birational onto its image, and I know exactly what $F$-curves are contracted (or "dually", what divisors are contracted). Suppose furthermore that the image is a normal variety. Is this information in some way sufficient to determine what type of singularities will the image have? How do I detect the kind of singularities just by knowing the exceptional curves?

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    $\begingroup$ In general the knowledge of the dual graph of the singularity does not identify its analytic type. When this happens, the singularity is called "taut". Taut two-dimensional singularities were classified by Laufer, see Math. Ann.205 (for instance, quotient singularities are taut). I do not know whether there are similar results in higher dimension. $\endgroup$
    – Francesco Polizzi
    Commented Jan 31, 2015 at 18:04
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    $\begingroup$ At the very least, you will need to specify that the image of $f$ is normal. Otherwise, you cannot specify the image uniquely just by knowing the pullback of the ample cone (which, I suspect, is what you are getting at via the contracted $F$-curves). $\endgroup$
    – Jason Starr
    Commented Jan 31, 2015 at 21:38
  • $\begingroup$ @Jason: and would normality be enough, in your opinion? $\endgroup$
    – IMeasy
    Commented Feb 2, 2015 at 14:56
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    $\begingroup$ Suppose that $f:X\to Y$ is a birational morphism of normal projective varieties, then doesn't $X$ and the set of exceptional curves determine $Y$? (i.e. given $f':X\to Y'$ birational of normal varieties with the same set of exceptional curves, then $f'=f$. Maybe I misunderstood the question?) $\endgroup$
    – Hacon
    Commented Feb 3, 2015 at 0:24
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    $\begingroup$ I see. The only results of this kind that I know are from the MMP. If you contract a codim \geq 2 subset you get something not Q-factorial; if you contract a K_X+D negative extremal ray via f:X->Y (where (X,D) is klt) then (Y,f_*(D+H)) is klt where H is an appropriate ample divisor on X. Unluckily, I suspect that this is not what you are after.... $\endgroup$
    – Hacon
    Commented Feb 4, 2015 at 14:10

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