Higher direct images and singular varieties - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:01:57Z http://mathoverflow.net/feeds/question/29623 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29623/higher-direct-images-and-singular-varieties Higher direct images and singular varieties unknown (google) 2010-06-26T17:25:46Z 2010-06-28T00:45:33Z <p>The specific question: I've got a projective variety Y and a subvariety X cut out of Y as the zero scheme of a regular section of a vector bundle E on Y. In the end, I'd like to prove that X has rational singularities... and I was hoping to try to use a theorem I found in a paper of Kovacs: </p> <p>Let $\phi: Y \rightarrow X$ a surjective morphism of projective varieties over $\mathbb{C}$, with dim Y = N and dim X = n. Assume Y has rational singularities. Then X has rational singularities iff X is Cohen-Macaulay and $R^{N-n}\phi_* \omega_Y \cong \omega_X$.</p> <p>I think I've got Cohen-Macaulay from the fact that X is cut out of Y as a zero scheme of a regular section. The silly part: how do I compute $R^{N-n}\phi_* \omega_Y$ or any other $R^i\phi_* \omega_Y$? I just realized everything I know from Hartshorne assumes nonsingularity.</p> <p>(A more general question is whether there might be a better way to prove my variety X has rational singularities!)</p> http://mathoverflow.net/questions/29623/higher-direct-images-and-singular-varieties/29649#29649 Answer by Karl Schwede for Higher direct images and singular varieties Karl Schwede 2010-06-26T22:41:29Z 2010-06-27T12:21:23Z <p>There might be many ways to prove a variety has rational singularities. </p> <p>I certainly agree with Zsolt's comment above that you should be careful with your notation, Kov\'acs theorem refers to a $Y \to X$ and above you mention $X \subset Y$, I'm assuming you are simply abusing notation. </p> <p>With regards to your initial question, I would try to put $R^i \phi_* \omega_Y$ into a long exact sequence (so it depends on what $Y$ is mapping to $X$), see #3 below.</p> <p>Anyway, here are some things I would try for a subvariety $X$ in $Y$ (obtained in some way), some of which don't use the subvariety structure.</p> <ol> <li><p>Is your variety some quotient by a group action (then more info is likely needed). </p></li> <li><p>Is $O_X$ (locally) a summand of something with rational singularities? (Apply Boutot's theorem)</p></li> <li><p>Sandor's result, but you may as well try with a resolution $\pi : X' \to X$ and try to show that $\pi_* \omega_{X'} = \omega_X$ (at which point, this is due to Kempf, not Kovacs) (I assume you have already shown that $X$ is CM), this is easier to compute. If you have some other $Y$ with rational singularities mapping to it in some natural way, then you might be in business via Sandor's theorem. Of course, the quickest hope for computing some higher cohomology like this is sticking it in some long exact sequence.</p></li> <li><p>If $X$ is a divisor, you could try to show that the pair $(Y, X)$ is purely log terminal (see also Lazarsfelds's book, and adjoint ideals). In this same direction, if $X$ is NOT a divisor, you could try to show that $X$ is a minimal log canonical center of some log canonical pair and then apply Kawamata's subadjunction theorem. Of course, you could also just try to show that $X$ is log terminal directly. </p></li> <li><p>If you have specific equations, you could also try some reduction to characteristic p techniques (like things related to F-splitting and F-rationality, some of these are very effective if you have explicit equations). Even without specific equations, some of these techniques still might be useful.</p></li> <li><p>I suppose you could also do some Bertini type tricks if somehow this subvariety is sufficiently general (for example, a general section of a base point free linear system of something with rational singularities still has rational singularities).</p></li> <li><p>You can also see this question: <a href="http://mathoverflow.net/questions/23091/is-there-an-obvious-way-for-showing-singularities-are-quotient/23137#23137" rel="nofollow">http://mathoverflow.net/questions/23091/is-there-an-obvious-way-for-showing-singularities-are-quotient/23137#23137</a></p></li> <li><p>Does your variety have a small resoluation ($Y \to X$)? If it is also Cohen-Macaulay and normal, then it has rational singularities.</p></li> <li><p>Does your variety have a Cartier divisor $D$ on it with log canonical (or maybe Du Bois) singularities such that $X \setminus D$ is has log terminal singularities (or maybe smooth). Then $X$ can be show to have rational singularities. Some things like this appeared in a paper of Koll\'ar and Shepherd-Barron (also see the related work of Karu as well as a paper of mine on Du Bois singularities).</p></li> </ol> <p>That's all I can think of right now.</p> http://mathoverflow.net/questions/29623/higher-direct-images-and-singular-varieties/29739#29739 Answer by Zhiyu for Higher direct images and singular varieties Zhiyu 2010-06-28T00:45:33Z 2010-06-28T00:45:33Z <p>For a normal variety $X$, if $K_X$ is Cartier, then $X$ has rational singularity if and only if $X$ is canonical. In your case, suppose the ambient variety is smooth, then $X$ is l.c.i. In particular, $K_X$ is Cartier. But $X$ can certainaly fail to be canonical, e.g. the cone over elliptic curve in ${P}^3$. </p> <p>Anyway, my point is that you cannot prove $X$ has rational singularity with only these assumptions. You probably need to use some properties of the ambient variety and the locally free sheaf.</p>