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Let $X=\mathbb{C}^3$ and take the weighted blow up $f:Y\to X$ where the weights $\omega(x,y,z)=(1,1,2)$. Then by checking affine charts of $Y$, $K_Y=f^*K_X+(1+1+2-1)E$ where $E$ is the exceptional divisor of $f$ and $E\cong \mathbb{P}(1,1,2)$. Blow up the singular point of $Y$, we get $g: Z\to Y$ and $K_Z\sim g^*K_Y+\frac{1}{2}E'$ since $Y$ is locally a cone over Veronese surface, which is explained well in this question:

Cone over the Veronese surface

Also $g^*E=\tilde{E}+E'$, which can be computed by blowing up $\mathbb{P}(1,1,1,2)$ and consider $E$ as a cone over a conic. The details can be found in:

strict transform under resolution of singularity along a singular $\mathbb{Q}$-Cartier divisor

Then I am confused. We can get $K_{Z/X}=K_Z-(g\circ f)^*K_X=3\tilde{E}+\frac{7}{2}E'$ and this relative canonical divisor seems not Cartier, but only $\mathbb{Q}$-Cartier.

My question is : How can $K_{Z/X}$ be not Cartier when $Z, X$ are all smooth varieties and $h=g\circ f$ is a birational morphism, isomorphic out of $0\in X$?

I have known $K_Y$ is not Cartier, but $2K_Y$ is Cartier. (Here is an answer: Example of a variety with $K_X$ $\mathbb Q$-Cartier but not Cartier) $E$ passes through the singular point of $Y$ and $E$ itself is singular at that point. $K_Z$ should be Cartier since $Z$ is smooth, and $(g\circ f)^*K_X$ should be Cartier. So I don't know where it goes wrong.

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  • $\begingroup$ I think I have found where it goes wrong. Anyway, I will leave this question here for further discussions. $\endgroup$ – Mingyi Zhang Oct 29 '17 at 4:08

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