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S Jan 31, 2020 at 23:09 history suggested cgodfrey CC BY-SA 4.0
Since this part is about \pi_* K_Y = K_X
Jan 28, 2020 at 19:11 review Suggested edits
S Jan 31, 2020 at 23:09
Jun 10, 2013 at 5:40 comment added Karl Schwede It doesn't come from there exactly. The right way to obtain it is from Hom'ing the exact sequence $$0 \to O_X(-D) \to O_X \to O_D \to 0$$ into $\omega_X$, and noting $Ext^1(O_D, \omega_X) = \omega_D$. The next Ext vanishes which gives you surjectivity on the right. There are no other Ext's for obvious reasons (at least as long as $D$ is Cartier). This isn't birational geometry, it's Grothendieck/Serre-duality theory. Perhaps you could look at Residues and Duality by Hartshorne?
Jun 10, 2013 at 4:04 comment added Li Yutong Thank you for the hint, is this short exact sequence valid for all the scheme? It seems that the short exact sequence comes from tensoring $$0 \to O_X \to O_X(D) \to O_D(D) \to 0$$ by $\omega_X$, and using the adjuction formula for the third term. Here comes two problem:(1)Why it is exact after tensoring $\omega_X$(2)Why adjunction formula can be applied to this case. Could you please point out the reference of this sort to me, because I am not good at birational geometry.
Jun 10, 2013 at 3:54 vote accept Li Yutong
Jun 10, 2013 at 3:42 comment added Karl Schwede To me, $K_X$ is a divisor, $\omega_X$ is a sheaf, $\omega_X \cong O_X(K_X)$. This is probably the most common convention in birational geometry but not the only one. For your problem, let me give you a hint: do it one $D_i$ at a time (I'm assuming the $D_i$ are Cartier divisors?). Note that we have a short exact sequence: $$0 \to \omega_X \to \omega_X(D_1) \to \omega_{D_1} \to 0.$$ Then do the short exact sequence $$0 \to \omega_{D_1} \to \omega_{D_1}(D_2|_{D_1}) \to \omega_{D_2|_{D_1}} \to 0.$$ Repeat, and keep track of everything, you'll get the answer you want.
Jun 10, 2013 at 3:36 comment added Li Yutong con't, and the complete intersection preserve the CM-condition, the result can be proved. But the first thing confused me is the meaning of $K_X,K_Y$.
Jun 10, 2013 at 3:36 comment added Li Yutong (1)In your definition of $K_X$, do you require $X$ to be projective? Moreover, I do not understand the difference between $K_X$ and $\omega_X$. (2)My adjunction problem is: $X$ is a normal, Cohen-Macaulay variety with anticanonical sheaf $−K_X=D_1+\dots +D_n$, where $D_i$ are effective Cartier divisors. Suppose $Y$ is a complete intersection of $D_i′s$. I want to use adjunction formula to show $K_Y=0$. By doing adjunction formula as in Birational geometry of algebraic varieties (Page182, Prop5.73) for each complete intersection ,
Jun 10, 2013 at 3:27 comment added Karl Schwede Oh, let me explain this perhaps in a better way. Every integral scheme is S1, and hence Cohen-Macaulay outside a set of codimension 2. Thus for some purposes you can just work on the Cohen-Macaulay locus.
Jun 10, 2013 at 3:05 comment added Karl Schwede Indeed, if $X$ is Cohen-Macaulay, then $\omega_X$ is the dualizing sheaf. But the canonical sheaf is still defined by the same Ext formula even for non-Cohen-Macaulay schemes. If your singularity is regular in codimension-1, then you can still use the pushforward from the regular locus. What exactly is your adjunction problem? Maybe it would be better to state that.
Jun 10, 2013 at 2:40 comment added Li Yutong Thank you for your answer, when $X$ is normal, one can defined the canonical divisor on its smooth locus, and then pullback to $X$. This is explained in the book Toric variety by Cox etc. But in my case, the singularity is worse than that, I hope something can be said when the scheme is Cohen-Macaulay. My original problem is to do adjunction formula on such scheme, and it is mention in the book Birational geometry of algebraic varieties (Page182, Prop5.73) one has adjuntion formula on CM-scheme, but I do not know how could one define $K$ in this setting. Beside, is $\omega_X$ dualizing sheaf?
Jun 10, 2013 at 2:32 history edited Karl Schwede CC BY-SA 3.0
added 2668 characters in body
Jun 10, 2013 at 2:11 history answered Karl Schwede CC BY-SA 3.0