(Partial) crepant resolutions

Question

Consider de orbifold $\mathbb{C}^2$/$\mathbb{Z}_n$. In this case a full crepant resolution exists and it is unique. However, this orbifold admits partial resolutions. So my question is: Are all those partial resolutions crepant? or on the contrary, there is a unique partial crepant resolution too.

Answer 1

You should really make this question more clear. What do you mean by a "partial resolution"? For instance, is it normal, or at least $S_2$? Is it $\mathbb Q$-factorial? Or is at least the canonical divisor $\mathbb Q$-Cartier? You need some assumption for the question to make sense.

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Consider the crepant resolution $\pi:Y\to X$. (By the way if a surface has a crepant resolution it must be the minimal resolution and hence it is unique). Next, let's blow up a point on a $\pi$-exceptional curve $E\subset Y$ (say one that's not contained in any other exceptional curves) $\sigma: Z\to Y$. Now $Z$ is smooth and $F=\sigma^{-1}_*E$ has self-intersection $F^2=-3$ and $F\simeq \mathbb P^1$, so it can be contracted to a normal surface $Z\to W$. By construction we get a proper birational morphism $W\to X$. This is a partial resolution, but it is not crepant.

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So, perhaps you mean a partial resolution that's partial to the crepant resolution. I mean that it is an intermediate partial resolution. That kind of partial resolutions are also crepant by the following Lemma (but let's do a definition first).

Definition Let $\alpha:X\to Y$ be a proper birational morphism between normal varieties. Assume that $K_Y$ is $\mathbb Q$-Cartier, then $\alpha$ will be called crepant if $K_X\sim \alpha^* K_Y$.

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Lemma Let $\alpha:X\to Y$ and $\beta:Y\to Z$ be two proper birational morphisms and assume that $X,Y,Z$ are normal algebraic surfaces such that $K_Y$ and $K_Z$ are $\mathbb Q$-Cartier. then if $\beta\circ\alpha$ is crepant, then so are $\alpha$ and $\beta$.

Proof: Write $$K_X=\alpha^*K_Y+\sum a_iE_i$$ where $E_i$ are $\alpha$-exceptional irreducible divisors and $$K_Y=\beta^*K_Z+\sum b_jF_j$$ where $F_j$ are $\beta$-exceptional irreducible divisors. It follows that then $$K_X=\alpha^*\beta^*K_Z+\sum a_iE_i+\sum b_j\alpha^* F_j.$$ If $\beta\circ\alpha$ is crepant, then $$\sum a_iE_i+\sum b_j\alpha^* F_j\sim 0.$$ However, for each $j$, we have that $\alpha^*F_j= \widetilde F_j +\sum a_{ij}E_i$ where $\widetilde F_j$ is the strict transform of $F_j$ on $X$, so we get that for some other set of coefficients we have $$\sum c_iE_i+\sum b_j\widetilde F_j\sim 0.$$ The $E_i$ and $\widetilde F_j$ together are all $\beta\circ\alpha$ exceptional and hence linearly independent. Therefore $b_j=0$ for all $j$ and then $$\sum a_iE_i\sim 0$$ with the original $a_i$'s. The same reasoning says that then the $a_i=0$ which proves the statement.$\square$

Remark
It seems to me that the same proof works in arbitrary dimension, so one does not need to limit to surfaces for the Lemma.