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I know the definition of $K_X$ on a normal, singular variety, but I don't have a good set of examples in my mind. What's an example of a variety where $K_X$ is $\mathbb Q$-Cartier but not Cartier? Are there any conditions under which an adjunction formula lets me compute the canonical class of a singular divisor?

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3 Answers 3

up vote 18 down vote accepted

An easy way to produce a $\mathbb Q$-Cartier but not Cartier canonical divisor is by a quotient. For instance for the quotient $$X=\mathbb A^3/(x,y,z)\sim (-x,-y,-z)$$ $2K_X$ is Cartier, but $K_X$ is not.

I leave it for you to prove that $2K_X$ is Cartier. Here is how to see that $K_X$ is not: Clearly $X={\rm Spec}k[x^2,y^2,z^2,xy,yz,xz]$ in other words, $X$ is the affine cone over the Veronese surface $\mathbb P^2\simeq V\subset \mathbb P^5$. Blowing up the cone point gives a resolution of singularities $\pi: Y\to X$ with exceptional divisor $E\simeq V$. In fact $E^2\sim -2L$ where $L$ is the class of a line. This follows by considering the blow up as a blow up of the ambient $\mathbb A^6$ (the cone over $\mathbb P^5$) and noticing that $\deg V=2$ in $\mathbb P^5$ so the square of the exceptional divisor of the blow up of $\mathbb A^6$, which is $-1$-times the hyperplane in $\mathbb P^5$ restricts to $-2L$ on $Y$. Now write $K_Y\sim_{\mathbb Q} \pi^*K_X + aE$ and use the adjunction formula ($Y$ is smooth!) to get $$ (a+1)E^2\sim K_E=K_{\mathbb P^2} \sim -3L. $$ Solving for $a$ shows that $a=\dfrac 12$ which shows that $K_X$ cannot be Cartier.

Interesting to note that the same construction does not give a desired example in dimension $2$: The quotient $\mathbb A^2/(x,y)\sim(-x,-y)$ is a cone over a conic which is a surface in $\mathbb P^3$. In particular it is Gorenstein and hence $K_X$ is Cartier.

As for the adjunction formula, it definitely works as long as $K_X+D$ is Cartier and it works up to torsion if it is $\mathbb Q$-Cartier. If it is not $\mathbb Q$-Cartier, it is not clear what the adjunction formula should mean, but even then one can have a sort of adjunction formula involving $\mathscr Ext$'s but this is almost Grotherndieck Duality then.

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Sandor, I'm confused, you still have some sort of adjunction formula (in my answer below). What you might have trouble with is saying things like $n(K_Y + K_X)|_X = n K_X$ (one doesn't have equality in general), but that's a slightly different issue in my mind. –  Karl Schwede Feb 15 '11 at 16:36
    
Karl, I think we agree in the essence, it's a matter of terminology. My main point was that if it is not $\mathbb Q$-Cartier, then one has to be careful what the restriction means. –  Sándor Kovács Feb 15 '11 at 16:42
    
Ah, or perhaps you intended, $K_X + D$. It's not quite clear what the original questioner meant. –  Karl Schwede Feb 15 '11 at 16:55
    
Yes and yes. I need a personal error-catcher... –  Sándor Kovács Feb 15 '11 at 16:57
    
Great, thank you for the answer! I think the -2 in the displayed adjunction should read -3, but indeed a=1/2. –  Anonymous Feb 15 '11 at 17:21

Probably the easiest example is $X = \text{Spec} \; k[x^3, x^2y, xy^2, y^3]$. In this case, $K_X$ is not Cariter, but $3K_X$ is Cartier.

In what way do you have in mind computing the canonical class of a singular divisor? The adjunction formula basically always works assuming things are sufficiently normal.

In particular, one always has the following sequence if you assume that $X$ is a divisor in a normal ambient variety $Y$. $$0 \to \omega_Y \to \omega_Y(X) \to \omega_X \to h^1(\omega_Y^{\bullet}) \to \dots .$$

If $Y$ is Cohen-Macaulay, then the $h^1$ vanishes, if $X$ is normal, then $\omega_X$ can be viewed as a divisor class, and we've obtained some sort of adjunction formula. However, we don't really need those hypotheses.

More generally if $Y$ is normal but not necessarily Cohen-Macaulay, it is Cohen-Macaulay outside a set of codimension at least 3 (because $Y$ is S2). Thus $h^1(\omega_Y^{\bullet})$ (the first cohomology of the dualizing complex of $Y$) is supported at a codimension 3 (or more) subset. In particular, the map $\omega_Y(X) \to \omega_X$ is surjective in codimension 2 on $Y$ and so surjective in codimension 1 on $X$.

This is enough to compute the canonical class of $X$ because $\omega_X$ is always S2 (if $X$ is S1, which a reduced scheme is). Thus $\omega_X$ is determined by its codimension 1 behavior, which we can completely determine by the above short exact sequence.

EDIT: This sort of restriction stuff is certainly contained in chapter 16 of Kollár et al's "Flips and abundance for algebraic threefolds." book.

EDIT2: One should be slightly careful about saying $(K_Y + X)|_X = K_X$. It can be a little hard to make sense of the restriction of $K_Y + X$ when $K_Y + X$ isn't Cartier or $\mathbb{Q}$-Cartier (at least at the codimension 1 points of $X \subseteq Y$). This is what Sándor is talking about. However, you always have the exact sequence above and so one can always in some sense still compute $\omega_X$.

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The dot on your dualizing complex is barely visible, but I am not sure what one can do about it. :( I have a macro named \kdot to produce a better looking dot, but it might be too elaborate for the online LaTeX engine. (\kdot="Karl's dot" :) –  Sándor Kovács Feb 15 '11 at 16:47
    
Sandor, thanks. I made it a \bullet, I suspect the rendering here would probably die on the circle command I/we usually use for those dots. –  Karl Schwede Feb 15 '11 at 16:52
    
Thanks for the latest edit! –  Sándor Kovács Feb 15 '11 at 16:58
    
Ah, thanks for the help! That clears up my adjunction confusion. –  Anonymous Feb 15 '11 at 17:22

Here is a more algebraic perspective on your question. If $X=\text{Spec} (R)$ is affine and $R$ is a Cohen-Macaulay algebra over some field (the following is true in more general setting), then $K_X$ is Cartier is equivalent to $R$ is Gorenstein. On the other hand, $K_X$ is $\mathbb Q$-Cartier is the same as the class of $K_X$ is torsion in the divisor class group (assuming $R$ is normal).

So to find a class of examples, you need normal Cohen-Macaulay rings with torsion class group but not Gorenstein. If $R$ is a Veronese $S^{(d)}$ of $S=\mathbb C[x_1,\cdots, x_n]$ then $Cl(R)$ is always torsion, but $R$ is Gorenstein if and only if $d|n$ (Sandor's example is indeed the simplest one in this class, with $d=2, n=3$).

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Karl, you are right, I am just about to add Cohen-Macaulay –  Hailong Dao Feb 15 '11 at 16:53
    
Long, by the way this is just a question of terminology (and I know different people ahve differing opinions on this), when you say $\mathbb{Q}$-Gorenstein, do you always assume Cohen-Macaulay? –  Karl Schwede Feb 15 '11 at 17:04
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Karl, I think generally $\mathbb Q$-Gorenstein does not mean Cohen-Macaulay, but this is confusing for many. I know that when I first saw it as a graduate student, I did not understand the difference between $1$-Gorenstein and Gorenstein. The main problem is that many people do not realize that this is even an issue. I was once asked after a talk why I kept saying that $K_X$ was Cartier instead "simply" saying that $X$ was Gorenstein... As far as "quasi-Gorenstein" is concerned I always thought it was only known by commutative algebraists, but apparently it is not even widely known among them. –  Sándor Kovács Feb 15 '11 at 20:07
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One more thing about this: Kollár very much dislikes the accepted usage of $\mathbb Q$-Gorenstein. He thinks it should mean $K_X$ is $\mathbb Q$-Cartier and $X$ is Cohen-Macaulay. I agree with that, but I think it is too late for that. –  Sándor Kovács Feb 15 '11 at 20:08
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@Sandor: +1 for "most people do not realize that this is even an issue". As for quasi-Gorenstein, I think it is a bit unfair to deduce from my ignorance that it is not well-known among commutative algebraists (-: May be they all know it, just didn't tel me! –  Hailong Dao Feb 15 '11 at 20:14

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