Let $X$ be a smooth quasi-projective variety and $Y$ be a positive dimensional subvariety. Let $Z$ be a variety obtained from $X$ by contracting $Y$. My question is what is the relation between $K_X$ and $K_Z$, where $K_X$ and $K_Z$ are canonical divisors of $X$ and $Z$ respectively ?
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1$\begingroup$ The normal thing to do is to compare $K_X$ and the pullback of $K_Z$ (their difference being the discrepancy divisor), but if the codimension of $Y$ is greater than 1, then $K_Z$ fails to be ($\mathbb Q$)-Cartier, so you can't pull back $K_Z$ (at least not in the usual sense). If you're interested in the case where $Z$ is not a divisor, then, it's maybe unclear how one should compare the two canonical divisors. $\endgroup$– Devlin MalloryCommented Jun 25, 2020 at 15:52
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$\begingroup$ is it true that if $Y$ is of codimension at least $2$ then $Z$ is not Cohen-Macualay ? $\endgroup$– user130022Commented Jun 26, 2020 at 5:28
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1$\begingroup$ One has to be a little careful with terminology. It is possible to contract a subvariety of codimension 2 to get $Z$ with $\mathbf Q$-Cartier, even Cartier, canonical divisor. The basic example is the Atiyah flop: here the base of the contraction is an ordinary double point, which is Gorenstein (in particular CM). However, sometimes people use contraction to mean exclusively K-negative contraction: for a $K$-negative contraction (or any $K$-nontrival contraction), if the contracted locus has codimension 2 or more, it's true that the canonical divisor of the base is not $\mathbf Q$-Cartier. $\endgroup$– PopCommented Jun 26, 2020 at 9:40
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$\begingroup$ By K-negative did you mean that anti canonical is ample ? How to prove that for K-negative contraction with co-dim at least $2$ contracted locus, the canonical divisor is not Q-Cartier ? $\endgroup$– user130022Commented Jun 26, 2020 at 13:39
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1$\begingroup$ Related to the comment of @Pop: If $\mathrm{codim}(Y \subset X) > 1$ then $Z$ can't be $\mathbb{Q}$-factorial, but $K_Z$ could still be $\mathbb{Q}$-Cartier, for instance if $Z = C(\mathbb{P}^1 \times \mathbb{P}^1)$ (this is a hypersurface so $K_Z$ is Cartier) and $X = \mathrm{Bl}_{C(0 \times \mathbb{P}^1)} Z$ ($X$ is smooth, since the preimage of $C(0 \times \mathbb{P}^1)$ is a smooth Cartier divisor on X, and the exceptional set is a $\mathbb{P}^1$). $\endgroup$– cgodfreyCommented Jun 26, 2020 at 18:57
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1 Answer
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As mentioned in the comments, at least when $Y \subset X$ is a divisor and $K_Z$ is ($\mathbb{Q}$-)Cartier, $K_X, K_Z$ and $Y$ are related through the discrepancy $a(Y, X)$, defined by $$ K_X = \pi^* K_Z + a(Y, Z)Y $$ where $\pi: X \to Z$ is the contraction. Section 2.3 of Kollár-Mori's Birational geometry of algebraic varieties covers this in detail.
