A rational singularity is a singularity of a complex variety $X$ such that for any resolution $\pi:\; \tilde X\rightarrow X$ the higher direct images $R^i\pi_*(O_{\tilde X})$ vanish for all $i>0$. Suppose that $X$ has isolated rational singularity, and $\tilde X\rightarrow X$ its resolution. I expect that the fiber of $\pi$ over the singular point is rationally connected; I would be very grateful for any reference to this. I need to apply this to the local situation, so it would be especially nice if the argument does not use projectivity.
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asked Sep 23, 2021 at 13:16
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3$\begingroup$ Do you want to assume that the singularities are also Gorenstein? Then it should follow from Elkik together with Hacon-McKernan's proof of Shokurov's conjecture. $\endgroup$– Jason StarrCommented Sep 24, 2021 at 1:02
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2$\begingroup$ Many thanks, yes, the examples I meant have crepant resolutions $\endgroup$– Misha VerbitskyCommented Sep 24, 2021 at 12:28
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No. For instance the cone over an Enriques surface (with respect to any projective embedding) has rational singularity, but Enriques surface is not rationally connected.
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$\begingroup$ Does this also work for the cone over a K3 surface or a surface of general type? $\endgroup$ Commented Sep 23, 2021 at 18:27
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2$\begingroup$ The rationality of singularity condition is $H^{>0}(S, \oplus_{k \ge 0} L^k) = 0$, where $L$ is the ample line bundle defining the embedding. In particular, $H^{>0}(S,\mathcal{O}_S) = 0$ is necessary, so K3 surface won't work. But some surface of general type, such as fake projective planes, do work. $\endgroup$– SashaCommented Sep 23, 2021 at 18:32