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:

$$\mathcal{K}=\mathcal{O}_{X,-\sum D_i}$$

where the $D_i$ are representatives of all divisors in the Class Group.

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.

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.

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.

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    $\begingroup$ Talking off the top of my head: a toric variety has a dense open subset isomorphic to a torus, so (over an algebraically closed field say) is rational, and in particular never Calabi--Yau. Also a minor point "since the smooth locus has codimension 1" is a confusing way to say what you mean. "since the singular locus has codimension 2" would be better. $\endgroup$
    – user5117
    Commented Aug 16, 2010 at 10:09
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    $\begingroup$ To clarify the comments above about toric varieties not being Calabi-Yau: by Calabi-Yau, sometimes people colloquially mean that the canonical bundle is trivial and sometimes they mean more specifically that the variety should be projective, have trivial canonical, and that the intermediate cohomology groups of the structure sheaf be trivial (so that the total cohomology looks like that of a sphere). There are non-proper toric varieties with trivial canonical bundle (for instance a torus), but I don't think it is possible to have a proper toric variety with trivial canonical bundle. $\endgroup$
    – Chris Brav
    Commented Aug 16, 2010 at 12:01
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    $\begingroup$ Dear Chris, Of course you're right. I was assuming say projective (maybe proper is enough). In that case I think my argument above suffices to prove your assertion. $\endgroup$
    – user5117
    Commented Aug 16, 2010 at 12:11
  • $\begingroup$ Hi Artie. I just wanted to clarify for the question-asker. Also, yes, you are right: since toric varieties are rational, proper ones do not admit any non-trivial forms of top degree, so they can't be Calabi-Yau, even in the weak sense. $\endgroup$
    – Chris Brav
    Commented Aug 16, 2010 at 12:47
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    $\begingroup$ Dear Jesus, Just to clarify, the point of my comment was that no projective (or more generally as Chris says, proper) toric variety is Calabi--Yau. Also, as John Christian says in the first comment, the formula for toric varieties is K_X = O(-\sum D_i) where the sum is over the torus invariant divisors --- not representatives of all divisors in the class group. Indeed, the class group will almost always be infinite, in which casw such a sum will not be defined. $\endgroup$
    – user5117
    Commented Aug 16, 2010 at 14:02

2 Answers 2


Edit (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 here and here]

Dear Jesus,

I think there are several problems with your question/desire to define a canonical divisor on any algebraic variety.

First of all, what is any algebraic variety? Perhaps you mean a quasi-projective variety (=reduced and of finite type) defined over some (algebraically closed) field.

OK, let's assume that $X$ is such a variety. Then what is a divisor on $X$? Of course, you could just say it is a formal linear combination of prime divisors, where a prime divisor is just a codimension 1 irreducible subvariety.

OK, but what if $X$ is not equidimensional? Well, let's assume it is, or even that it is irreducible.

Still, if you want to talk about divisors, you would surely want to say when two divisors are linearly equivalent. OK, we know what that is, $D_1$ and $D_2$ are linearly equivalent iff $D_1-D_2$ is a principal divisor.

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 order of vanishing 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.

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 $S_2$.

So we get that in order to define divisors as we are used to them, we would need that $X$ be $R_1$ and $S_2$, that is, normal.

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.

As far as I can tell, the best way is to work with Weil divisorial sheaves 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.

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 sheaf of $X$ is $$ \omega_X=h^{-n}(\omega_{\overline X}^\bullet)|_X=\mathscr Ext^{N-n}_{\mathbb P^N}(\mathscr O_{\overline X},\omega_{\mathbb P^N})|_X $$ 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 torsion-free $S_2$ sheaf of rank $1$. (And torsion-free and $S_2$ is equivalent to reflexive on a normal scheme).

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.

(Added on 11/12/12):

Lemma 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.

Proof 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 $2$.

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 $x\in X. \square$

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    $\begingroup$ This was a pleasure to read, Sandor! I love your top-down reasoning. It makes the abstruce accessible. Thanks. $\endgroup$ Commented Nov 19, 2010 at 22:00
  • $\begingroup$ Dear Sandor, when you say "...this is essentially equivalent..." is this really equivalent? Is it true that if for every regular function $f$ its zero set has codimension $1$ then the variety has property $S_2$? Can property $S_2$ be formulated in such a language, or this is just implied by $S_2$? (maybe I miss-understand what you wrote). $\endgroup$
    – aglearner
    Commented Oct 11, 2012 at 21:04
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    $\begingroup$ @aglearner: I added a statement to explain that. I believe it tells you that in fact you can define property S2 this way. See also mathoverflow.net/questions/45347/… and mathoverflow.net/questions/45347/… $\endgroup$ Commented Oct 13, 2012 at 1:38
  • $\begingroup$ Dear Sandor, thank you for adding this reasoning, this is very helpful. I realised though that I have one more question about the same paragraph in your answer. Do you know an example of a variety $X$ that satisfies $R1$ but not $S_2$ and has a function that vanishes on a set $Z$ of co-dimension $2$ or larger in $X$? I have a problem to imagine such an example (I don't think that such example exists among affine varieties over $\mathbb C$...) $\endgroup$
    – aglearner
    Commented Oct 16, 2012 at 21:52
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    $\begingroup$ @aglearner: OK, it only took me almost 10 years to notice this question... Sorry! Anyway, how about this. Let $X=Z(xz,xt,yz,yt)\subseteq \mathbb A^4$. In other words, $X=X_1\cup X_2$, where $X_1=Z(x,y),X_2=Z(z,t)$. Let $f:X\to k$ be defined as $k(P)=z$ if $P\in X_1$ and $f(P)=x$ if $P\in X_2$. (Notice that $z=x=0$ at the point $X_1\cap X_2$). Then $f$ is $0$ only at the point $(0,0,0,0)=X_1\cap X_2$. $\endgroup$ Commented Jul 13, 2022 at 18:15

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,

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 $$\mathcal O_{X_\Sigma}\Big(-\sum_{\rho} D_\rho\Big)\cong \omega_{X_\Sigma}$$

This is Proposition 8.2.7 in Cox Little Schenck.

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.

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 $$R=\bigoplus_{D\in Cl(X)} \Gamma(X,\mathcal O_X(D)).$$

Here $Cl(X)$ is the class group of $X$. See 0801.3995 for more details.

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 (Cox), and for varieties of Fano type (Birkar–Cascini–Hacon–McKernan). Some other specifically constructed examples (Prendergast-Smith) are known, but a general characterization is not.

Edit: Updated the links and fixed the unclear notation - thanks Artie!

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    $\begingroup$ Dear Josh, a couple of minor comments: 1. In your definition of the total coordinate ring, I guess you should say what K is (otherwise it's a little mysterious). 2. A more up-to-date reference for Birkar--Cascini--Hacon--McKernan is the published version, available online at ams.org/journals/jams/2010-23-02/S0894-0347-09-00649-3/… $\endgroup$
    – user5117
    Commented Sep 29, 2010 at 13:04

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