12
$\begingroup$

Let $X$ be a variety which might be singular, how to defined the canonical sheaf $K_X$ on $X$?

When $X$ is a proper, irreducible variety over $\mathbb{C}$, Ueno defined $K_X$ as the pushforward of the canonical sheaf of its nonsigular model (see Chapter2 in his book " Classification Theory of Algebraic Varieties and Compact Complex Spaces"). However, I was wondering if there is a well-defined canonical sheaf on singular variety which is not proper (over $\mathbb{C}$ is OK for me). I will be very appreciated if someone can point out referenced of this kind.

$\endgroup$
1
  • $\begingroup$ You can find it in page 8 of the book singularities of minimal model program of Kollar $\endgroup$
    – user21574
    Sep 29, 2016 at 19:59

1 Answer 1

21
$\begingroup$

For $X$ normal, saying that canonical divisor is the pushforward of the canonical divisor of a resolution of singularities is totally fine (it even works in characteristic $p > 0$ if you happen to have a resolution). In particular, if $\pi : Y \to X$ is a resolution of singularities, then $\pi_* K_Y$ is $K_X$ (here we define $\pi_* K_Y$ by simply throwing away any components of $K_Y$ that get contracted to non-divisorial varieties).

However: The pushforward of the canonical sheaf $\omega_Y$ is not the canonical sheaf $\omega_X$ in general. Indeed, $\pi_* \omega_Y = \omega_X$ is very close to requiring that $X$ has rational singularities. (Actually one definition of rational singularities, typically attributed to Kempf, is that $X$ is Cohen-Macaulay and $\pi_* \omega_Y = \omega_X$.

A good exercise is to show that if $X = \text{Spec} k[x,y,z]/(x^3+y^3+z^3)$ and $\pi : Y \to X$ is the blowup of the cone point, then $\pi_* \omega_Y = \mathfrak{m} \cdot \omega_X$ where $\mathfrak{m}$ is the maximal ideal of the origin. (Hint: use the adjunction formula and the formula for the canonical divisor when blowing up a point on $\mathbb{A}^3$).

The canonical sheaf

Ok, so what is the right definition of the sheaf $\omega_X$ in general?

Well, for a projective variety of dimension $d$, $i : X \hookrightarrow \mathbb{P}^N$, define $$i_* \omega_X := \text{Ext}^{N-d}\big(i_* O_X, O_{\mathbb{P}^N}(-N-1)\big).$$ Since $i$ is a closed embedding, this uniquely determines $\omega_X$. For $X$ quasi-projective, you can define this by localizing. There are generalizations which apply to other schemes of finite type over $k$ defined using the $f^!$ functor for $f : X \to k$ the structural map, but I won't get into that here.

The canonical sheaf is S2

It turns out that for any variety, $\omega_X$ is an S2 sheaf, this means it satisfies Hartog's phenomena (search math overflow). In particular, the sheaf is determined by its codimension-1 behavior. There's an easy way to see this, it turns out that if $h : X \to Z = \mathbb{P^d}$ is a generic projection to a hyperplane of the same dimension, then $$h_* \omega_X = \text{Hom}(h_* O_X, O_Z(-d-1)).$$ Now, it easily follows that this sheaf is S2 since it is reflexive on $Z$ and reflexive sheaves on $Z$ are always S2 (see for example Hartshorne's Generalized divisors on Gorenstein schemes).

Why does this matter? Well it means that if $U$ is the regular locus of $X$, and furthermore $X \setminus U$ has codimension 2 on $X$ (which happens for example if $X$ is normal), then if $j : U \hookrightarrow X$ is the inclusion, then $j_* \omega_U = \omega_X$ since both sheaves are S2 and they agree outside a codimension-2 set.

Back to divisors

This also explains our first statement about divisors. Indeed, any divisor, like $\pi_* K_Y$ is determined outside a codimension 2 set, it is determined on $U$ in fact. And so if $X$ is normal, $\pi : Y \to X$ is an isomorphism outside of a codimension-2 set of $X$, and so the canonical divisor on that set works fine as a canonical divisor everywhere. In particular, it can be computed on $Y$ as claimed.

For non-normal $Y$, something can be done, but the formula isn't quite so simple (you also have to describe by what exactly you mean by a divisor on a non-normal variety).

$\endgroup$
8
  • $\begingroup$ 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? $\endgroup$
    – Li Yutong
    Jun 10, 2013 at 2:40
  • $\begingroup$ 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. $\endgroup$ Jun 10, 2013 at 3:05
  • $\begingroup$ 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. $\endgroup$ Jun 10, 2013 at 3:27
  • $\begingroup$ (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 , $\endgroup$
    – Li Yutong
    Jun 10, 2013 at 3:36
  • $\begingroup$ 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 $K_X,K_Y$. $\endgroup$
    – Li Yutong
    Jun 10, 2013 at 3:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.