The canonical line bundle of a normal variety - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:27:25Z http://mathoverflow.net/feeds/question/35736 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35736/the-canonical-line-bundle-of-a-normal-variety The canonical line bundle of a normal variety Jesus Martinez Garcia 2010-08-16T09:06:18Z 2012-10-13T01:36:01Z <p>I have heard that the canonical divisor can be defined on a normal variety X since the smooth locus has codimension 2. Then, I have heard as well that for ANY algebraic variety such that the canonical bundle is defined:</p> <p>$$\mathcal{K}=\mathcal{O}_{X,-\sum D_i}$$</p> <p>where the $D_i$ are representatives of all divisors in the Class Group.</p> <p>I want to prove that formula or I want to find a reference for that formula, or I want someone to rephrase it in a similar way if they heard about it.</p> <p>Why do I want to prove it? Well, I use the definition that something is Calabi Yau if its canonical bundle is 0. In the case of toric varieties, $\sum D_i$~0 if all the primitive generators for the divisors lie on a hyperplane. Then the sum is 0 and therefore the toric variety is Calabi-Yau.</p> <p>Can someone confirm or fix the above formula? I do not ask for a debate on when something is Calabi-Yau, I handle that OK, I just ask whether the above formula is correct. A reference would be enough. I have little access to references at the moment.</p> http://mathoverflow.net/questions/35736/the-canonical-line-bundle-of-a-normal-variety/40226#40226 Answer by Josh Guffin for The canonical line bundle of a normal variety Josh Guffin 2010-09-27T22:26:07Z 2010-09-29T18:31:20Z <p>Your formula is not quite right for toric varieties. In particular, the sum is not over "representatives of the class group", but over a set of minimal generators for the free group on torus-invariant divisors. Such a set is furnished by the 1-cones in the fan. More precisely,</p> <blockquote> <p>Let $X_\Sigma$ be the toric variety associated to a fan $\Sigma$, and assume that $X_\Sigma$ has no torus factors. Then for each $\rho \in \Sigma(1)$, there is a torus invariant divisor $D_\rho$, and $$\mathcal O_{X_\Sigma}\Big(-\sum_{\rho} D_\rho\Big)\cong \omega_{X_\Sigma}$$</p> </blockquote> <p>This is Proposition 8.2.7 in Cox Little Schenck.</p> <p>An easy toric proof that no projective toric variety is Calabi-Yau is that, as you said, the minimal generators of the one-cones must lie in a hyperplane. The positive hull over such a set of generators is strongly convex, so that the support of the fan cannot be all of $N_{\mathbb R}$, and thus $X_\Sigma$ is not complete and thus not projective.</p> <p>I believe that your question really concerns the existence of a natural set of generators for the "total coordinate ring" of an Calabi-Yau variety and if they obey a linear relation. The total coordinate ring is defined to be $$R=\bigoplus_{D\in Cl(X)} \Gamma(X,\mathcal O_X(D)).$$</p> <p>Here $Cl(X)$ is the class group of $X$. See <a href="http://arxiv.org/abs/0801.3995" rel="nofollow">0801.3995</a> for more details.</p> <p>For toric varieties $X_\Sigma$, this is simply $\mathbb C[x_\rho | \rho \in \Sigma(1)]$. If this is indeed your question, it is probably too much to hope for, as $R$ is not known to be finitely generated! It is known to be so for toric varieties (<a href="http://arxiv.org/abs/alg-geom/9210008" rel="nofollow">Cox</a>), and for varieties of Fano type (<a href="http://www.ams.org/journals/jams/2010-23-02/S0894-0347-09-00649-3/home.html" rel="nofollow">Birkar–Cascini–Hacon–McKernan</a>). Some other specifically constructed examples (<a href="http://www.iag.uni-hannover.de/~prendergast/Papers/Example.pdf" rel="nofollow">Prendergast-Smith</a>) are known, but a general characterization is not.</p> <p><strong>Edit:</strong> Updated the links and fixed the unclear notation - thanks Artie!</p> http://mathoverflow.net/questions/35736/the-canonical-line-bundle-of-a-normal-variety/46663#46663 Answer by Sándor Kovács for The canonical line bundle of a normal variety Sándor Kovács 2010-11-19T18:23:13Z 2012-10-13T01:36:01Z <blockquote> <p><strong>Edit</strong> (11/12/12): I added an explanation of the phrase "this is essentially equivalent to $X$ being $S_2$" at the end to answer aglearner's question in the comments. [See also <a href="http://mathoverflow.net/questions/45347/why-does-the-s2-property-of-a-ring-correspond-to-the-hartogs-phenomenon/45354#45354" rel="nofollow">here</a> and <a href="http://mathoverflow.net/questions/45347/why-does-the-s2-property-of-a-ring-correspond-to-the-hartogs-phenomenon/45616#45616" rel="nofollow">here</a>]</p> </blockquote> <p>Dear Jesus,</p> <p>I think there are several problems with your question/desire to define a canonical divisor on <em>any</em> algebraic variety. </p> <p>First of all, what is <em>any</em> algebraic variety? Perhaps you mean a quasi-projective variety (=reduced and of finite type) defined over some (algebraically closed) field.</p> <p>OK, let's assume that $X$ is such a variety. Then what is a <em>divisor</em> on $X$? Of course, you could just say it is a formal linear combination of <em>prime divisors</em>, where a prime divisor is just a codimension 1 irreducible subvariety. </p> <p>OK, but what if $X$ is not equidimensional? Well, let's assume it is, or even that it is irreducible. </p> <p>Still, if you want to talk about divisors, you would surely want to say when two divisors are <em>linearly equivalent</em>. OK, we know what that is, $D_1$ and $D_2$ are linearly equivalent iff $D_1-D_2$ is a <em>principal divisor</em>.</p> <p>But, what is a principal divisor? Here it starts to become clear why one usually assumes that $X$ is normal even to just talk about divisors, let alone defining the canonical divisor. In order to define principal divisors, one would need to define something like the <em>order of vanishing</em> of a regular function along a prime divisor. It's not obvious how to define this unless the local ring of the general point of any prime divisor is a DVR. Well, then this leads to one to want to assume that $X$ is $R_1$, that is, regular in codimension $1$ which is equivalent to those local rings being DVRs.</p> <p>OK, now once we have this we might also want another property: If $f$ is a regular function, we would expect, that the zero set of $f$ should be 1-codimensional in $X$. In other words, we would expect that if $Z\subset X$ is a closed subset of codimension at least $2$, then if $f$ is nowhere zero on $X\setminus Z$, then it is nowhere zero on $X$. In (yet) other words, if $1/f$ is a regular function on $X\setminus Z$, then we expect that it is a regular function on $X$. This in the language of sheaves means that we expect that the push-forward of $\mathscr O_{X\setminus Z}$ to $X$ is isomorphic to $\mathscr O_X$. Now this is essentially equivalent to $X$ being $S_2$.</p> <p>So we get that in order to define <em>divisors</em> as we are used to them, we would need that $X$ be $R_1$ and $S_2$, that is, <em>normal</em>. </p> <p>Now, actually, one can work with objects that behave very much like divisors even on non-normal varieties/schemes, but one has to be very careful what properties work for them.</p> <p>As far as I can tell, the best way is to work with <em>Weil divisorial sheaves</em> which are really reflexive sheaves of rank $1$. On a normal variety, the sheaf associated to a Weil divisor $D$, usually denoted by $\mathcal O_X(D)$, is indeed a reflexive sheaf of rank $1$, and conversely every reflexive sheaf of rank $1$ on a normal variety is the sheaf associated to a Weil divisor (in particular a reflexive sheaf of rank $1$ on a regular variety is an invertible sheaf) so this is indeed a direct generalization. One word of caution here: $\mathcal O_X(D)$ may be defined for Weil divisors that are not Cartier, but then this is (obviously) not an invertible sheaf.</p> <p>Finally, to answer your original question about canonical divisors. Indeed it is possible to define a canonical divisor (=Weil divisorial sheaf) for all quasi-projective varieties. If $X\subseteq \mathbb P^N$ and $\overline X$ denotes the closure of $X$ in $\mathbb P^N$, then the dualizing complex of $\overline X$ is $$\omega_{\overline X}^\bullet=R{\mathscr H}om_{\mathbb P^N}(\mathscr O_{\overline X}, \omega_{\mathbb P^N}[N])$$ and the canonical <em>sheaf</em> of $X$ is <code>$$\omega_X=h^{-n}(\omega_{\overline X}^\bullet)|_X=\mathscr Ext^{N-n}_{\mathbb P^N}(\mathscr O_{\overline X},\omega_{\mathbb P^N})|_X$$</code> where $n=\dim X$. (Notice that you may disregard the derived category stuff and the dualizing complex, and just make the definition using $\mathscr Ext$.) Notice further, that if $X$ is normal, this is the same as the one you are used to and otherwise it is a reflexive sheaf of rank $1$.</p> <p>As for your formula, I am not entirely sure what you mean by "where the $D_i$ are representatives of all divisors in the Class Group". For toric varieties this can be made sense as in Josh's answer, but otherwise I am not sure what you had in mind.</p> <blockquote> <p>(Added on 11/12/12):</p> </blockquote> <p><strong>Lemma</strong> A scheme $X$ is $S_2$ if and only if for any $\iota:Z\to X$ closed subset of codimension at least $2$, the natural map $\mathscr O_X\to \iota_*\mathscr O_{X\setminus Z}$ is an isomorphism.</p> <p><strong>Proof</strong> Since both statements are local we may assume that $X$ is affine. Let $x\in X$ be a point and $Z\subseteq X$ its closure in $X$. If $x$ is a codimension at most $1$ point, there is nothing to prove, so we may assume that $Z$ is of codimension at least $2$.</p> <p>Considering the exact sequence (recall that $X$ is affine): $$0\to H^0_Z(X,\mathscr O_X) \to H^0(X,\mathscr O_X) \to H^0(X\setminus Z,\mathscr O_X) \to H^1_Z(X,\mathscr O_X) \to 0$$ shows that $\mathscr O_X\to \iota_*\mathscr O_{X\setminus Z}$ is an isomorphism if and only if $H^0_Z(X,\mathscr O_X)=H^1_Z(X,\mathscr O_X)=0$ the latter condition is equivalent to $$\mathrm{depth}\mathscr O_{X,x}\geq 2,$$ which given the assumption on the codimension is exactly the condition that $X$ is $S_2$ at $x\in X$. $\qquad\square$</p>