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.

  • $\begingroup$ 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


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).

  • $\begingroup$ 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

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