Timeline for Dualizing sheaf on a Cohen-Macaulay variety
Current License: CC BY-SA 2.5
15 events
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| Jul 19, 2010 at 22:40 | comment | added | BCnrd | The pushforward formula is only correct when the open set has complement with codimension $\ge 2$, also used in the vanishing aspect of the local duality argument in my initial comment. In general I don't like to say the formulas (pushforward, Ext on an affine space, etc.) are the "definition" of a dualizing sheaf, but rather a way to compute one (coupled with a suitable trace map, which can be quite hard to make "explicit" away from simple examples). A dualizing object yields a "codimension function", and demanding it match usual codimension (can always be done) eliminates shift ambiguity. | |
| Jul 19, 2010 at 22:06 | comment | added | user717 | @Donu, Karl: Okay, as I don't know exactly what I want (the reason for asking this question), I am not qualified to comment on that. At least in Liu Quing's book the canonical sheaf on a smooth variety $X$ is defined as $\mathrm{det}\Omega_X^1$ (6.4.2). But I'll try to put things together... | |
| Jul 19, 2010 at 21:36 | comment | added | Karl Schwede | Donu is right, you probably want normality if you want to do that. Another characterization in the affine case that you might find useful is the following: If $X$ is an affine CM equidimensional subscheme of $\mathbb{A}^n$, then you can define a dualizing sheaf to be $Ext^{n-dim X}(O_X, O_{\mathbb{A}^n})$. Something like this is stated in Hartshorne's algebraic geometry and it's certainly is Residues and duality (as well as many other sources -- Brian Conrad's book). | |
| Jul 19, 2010 at 21:27 | comment | added | Donu Arapura | Arminius: Not convinced. If $X$ is the $y^2=x^3$, $j_*\Omega_X^1$ isn't even coherent. | |
| Jul 19, 2010 at 21:23 | comment | added | user717 | @t3suji: 'semi-separated' was perhaps what I was looking for. But except for separated schemes I don't know of any semi-separated ones. Is there some condition that for a morphism $f:X \rightarrow Y$ of varieties into a semi-separated/separated/affine variety ensures that $X$ is also semi-separated? | |
| Jul 19, 2010 at 21:19 | history | edited | user717 | CC BY-SA 2.5 |
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| Jul 19, 2010 at 21:19 | comment | added | user717 | @t3suji: Of course! I was somewhere else... @Donu: I think it is $j_*\Omega_X^{\dim X}$ as long as $X$ is irreducible!? @all others: I will see if I can understand your comments... | |
| Jul 19, 2010 at 21:03 | comment | added | Donu Arapura | OK, I see, thanks. I guess I was using a narrower definition. Hopefully this hasn't made things worse for Arminius: People in birational geometry often call $j_*\Omega_X^{\dim X}$ the canonical sheaf, which perhaps is less ambiguous. | |
| Jul 19, 2010 at 20:55 | comment | added | t3suji | By the way, in response to last question of Arminius: `separated' does include affine varieties. I believe semi-separatedness actually works just as well. | |
| Jul 19, 2010 at 20:52 | comment | added | t3suji | @Donu Arapura (and BCnrd can correct me): in general, dualizing sheaf/complex (if it exists) is unique up to twist by a line bundle and a shift. If you deal with a scheme of finite type over a field, there is a natural choice (among all possible dualizing sheaves) that you suggest: $f^!O_{spec(k)}$. | |
| Jul 19, 2010 at 20:49 | comment | added | BCnrd | @Donu: the formula you give is a dualizing object, but not the only one. The property that defines a dualizing object $R^{\bullet}$ in $D^b_c(X)$ is finite injective dimension and on $D^b_c(X)$ the functor $\mathbf{D} = {\mathbf{R}}\mathcal{H}om^{\bullet}(\cdot,R^{\bullet})$ satisfies the "double duality" map ${\rm{id}} \rightarrow \mathbf{D} \circ \mathbf{D}$ (which makes sense due to the finite injective dimension hypothesis) is an isom. By inspection, applying shift and line bundle twist has no effect! Remarkable fact is that this is the only ambiguity, for connected $X$. | |
| Jul 19, 2010 at 20:28 | comment | added | Donu Arapura | I was hoping you'd jump in, but now I'm confused. $\omega=f^!\mathcal{O}_{spec k}$ is only unique up to shift & twist? | |
| Jul 19, 2010 at 20:16 | comment | added | BCnrd | Uniqueness of $\omega$ up to shift & twist by line bundle when $X$ conn'd. For smooth $X$, any line bundle is dualizing; top-diff'tls makes duality "explicit" via residues. CM equiv. to $\omega$ being sheaf (in some degree). For CM $X$ and $j:U \rightarrow X$ open with complement of codim $\ge 2$, want $\omega \rightarrow j_{\ast}(\omega|_U)$ an isom. By SGA2, III, Cor. 3.5, need depth$(\omega_x) > 1$ for $x \in X-U$. Crux: $\omega_x$ satisfies local duality, so by Ext depth criterion same as $H^i_x(k(x))=0$ for $i \ge {\rm{dim}}O_x-1$ ($> 0$!). Verify via excision (cf. SGA2, II, Cor. 4)! | |
| Jul 19, 2010 at 19:18 | comment | added | Donu Arapura | I think $X$ would have to be normal in addition to Cohen-Macaulay for this formula to work. There are many experts who can explain this better than I could. In one sentence that the dualizing sheaf (or complex) is the thing which makes the Grothendieck-Serre duality theorem work; it would be unique up to isomorphism if it exists. | |
| Jul 19, 2010 at 19:10 | history | asked | user717 | CC BY-SA 2.5 |