most recent 30 from http://mathoverflow.net 2013-05-20T03:53:59Z http://mathoverflow.net/feeds/question/43936 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43936/relative-canonical-divisors Karl Schwede 2010-10-28T04:50:54Z 2010-10-28T15:47:52Z

<p>Suppose that $X$ is a Gorenstein variety and that $\pi : Y \to X$ is a birational map of varieties with normal $Y$. </p> <p>In this case the relative canonical divisor is defined to be $K_Y - \pi^*K_X$ (if you choose $K_Y$ and $K_X$ that agree where $\pi$ is an isomorphism, $K_Y - \pi^*K_X$ is independent of the particular representatives of $K_X$ and $K_Y$ you pick). $\mathcal{O}_Y(K_Y - \pi^* K_X)$ can also be identified with $\pi^! \mathcal{O}_X$ in this case.</p> <p>Suppose that that $X$ is $\mathbb{Q}$-Gorenstein (which means that $nK_X$ is Cartier for some $n > 0$, also assume Cohen-Macaulay if it helps). In birational geometry, the relative canonical divisor (a $\mathbb{Q}$-divisor) is defined by the formula $$K_{Y/X} = K_Y - {1 \over n} \pi^* (nK_X).$$ Again, if you choose $K_X$ and $K_Y$ that agree where $\pi$ is an isomorphism, then this $\mathbb{Q}$-divisor $K_{Y/X}$ is independent of the choices involved.</p> <p><strong>Question:</strong> Is it known whether $\mathcal{O}_Y(\lceil K_Y - {1 \over n} \pi^* (nK_X) \rceil)$ (or maybe with $\lfloor \cdot \rfloor$) is at all related to the zeroth cohomology of $\pi^! \mathcal{O}_X$? </p>