Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Could you give us an variety X which is Q-Gorenstein variety, but this variety is not Gorenstein variety? What is the canonical divisor K_X on X? How can we compute its canonical divisor? Could you give us an variety which is not Gorenstein variety? What is the canonical divisor K_X on X? How can we compute its canonical divisor?

share|improve this question
1  
Just to quickly answer the other question. The canonical divisor $K_X$ on a normal variety $X$ is any divisor $D$ such that if $U \subseteq X$ is the regular locus, then $$O_U(D|_U) \cong \Omega_{U/k}^{\dim X}.$$ One common way to compute a canonical divisor is by adjunction. For example, if $X \subseteq Y$ is a normal hypersurface in a normal variety $Y$, and we know the canonical divisor on $Y$, then $$(K_Y + X)|_X = K_X.$$ Alternately, for any $X$ of dimension $d$ in $\mathbb{P^n}$, then $$O_X(K_X) \cong \mathcal{E}xt^{n-d}(O_X, O_{\mathbb{P}^n}(-n-1)).$$ –  Karl Schwede Apr 28 '12 at 17:15
add comment

2 Answers 2

Cones are usually a good source of examples when it comes to questions like these. In your case, let $Y\subset \mathbb{P}^5$ be the Veronese embedding of $\mathbb{P}^2$ and let $X\subset \mathbb{P}^6$ be the projective cone of $Y$. Then $X$ is a normal non-Gorenstein variety which is not Gorenstein. Indeed, Let $L\subset Y$ be the image of a line in $\mathbb{P}^2$ and let $D$ be the Weil divisor given by the cone over $L$. Let $H$ be the generator of $Pic(X)$, that is, the hyperplane section from the embedding in $\mathbb{P}^5$. Then $-K_X=3D$ is not Cartier (as you can see by blowing up the vertex of the node and use the adjunction formula). But $2K_X$ is certainly Cartier as $2K_X\sim -6D\sim -2H$.

share|improve this answer
add comment

Let me give a couple more examples to complement what J.C.Ottem already said.

First I'll point out a couple discrepancies in terminology for a normal variety:

Gorenstein: This means that both $K_X$ is Cartier and $X$ is Cohen-Macaulay.

$Q$-Gorenstein: This means that $nK_X$ is Cartier for some integer $n > 0$.

Quasi-Gorenstein (or $1$-Gorenstein): This means that $K_X$ is Cartier but that $X$ is not necessarily Cohen-Macaulay.

In particular, $Q$-Gorenstein does not usually imply Cohen-Macaulay (because of this a lot of people don't like the terminology, notably János Kollár). On the other hand, some people like to require the Cohen-Macaulay condition in their $Q$-Gorenstein definition (adding to the confusion). It would also be "reasonable" to require that some canonical cover of $X$ is Cohen-Macaulay!

Ok, let me at least state the general rule for when a cone (or rather a section ring) is $Q$-Gorenstein or Gorenstein.

First note that because $S$ is graded, a module is free of rank one if and only if it is isomorphic to $S$ itself (up to a shift in grading).

Suppose that $X$ is a smooth (or even Gorenstein) projective algebraic variety over a field $k$ and suppose that $L$ is an ample line divisor. Form the ring: $$ S = \bigoplus_{k \in \mathbb{Z}} O_X(kL) $$ and let $m = S_+$ denote the irrelevant ideal. Also form the $S$-module $$ \omega_S = \bigoplus_{k \in \mathbb{Z}} (\omega_X \otimes O_X(kL)) $$ I claim that $\omega_S$ is the canonical module of $S$. This is actually pretty easy to see, $S$ is always normal so $\omega_S$ is determined on $U = \text{Spec} S \setminus m$. Therefore since $f : U \to X$ is a $\mathbb{A}^1$-bundle it follows that $\omega_U$ is just $f^* \omega_X$ which it is easy to see coincides with $\omega_X|_U$.

Then

Theorem (quasi-Gorenstein): The ring $S$ is quasi-Gorenstein if and only if $K_X \sim nL$ for some integer $n$ (possibly $n = 0$).

The proof idea is as follows. Since $K_X \sim nL$, it follows that $\omega_S$ is just $S$ shifted over by $n$. But that's a free module. Conversely, if $\omega_S$ is a free module, since $S$ is graded, it must be a shift of $S$.

Theorem ($Q$-Gorenstein): $S$ is $Q$-Gorenstein if and only if $mK_X \sim nL$ for some integers $n$ and $m \neq 0$.

The proof idea is similar, it follows from the same sorts of arguments as above that $$O_{\text{Spec} S}(mK_S) = \omega_S^{(m)} = \bigoplus_{k \in \mathbb{Z}} O_{X}(kL + mK_S).$$ Therefore for this to be free, we need exactly $mK_S = nL$ for some integer $n$.

Theorem (Cohen-Macaulay): Finally $S$ is Cohen-Macaulay if and only if $H^i(X, O_X(kL)) = 0$ for all $\dim X > i > 0$ and all $k \in \mathbb{Z}$.

Note that the points away from the origin are Cohen-Macaulay since $X$ was smooth (or at least Gorenstein $\Rightarrow$ Cohen-Macaulay). The point is then that $[H^{i+1}_m(S)]_k = H^i(X, O_X(kL))$ where $[\bullet]_k$ is the $k$th graded piece of the module. This follows from computations in Cech cohomology.

The upshot:

This gives us lots of examples of varieties that are $Q$-Gorenstein or not.

Fano: If $X$ is Fano (meaning $-K_X$ is ample) and we take $L = -K_X$, then $S$ is certainly quasi-Gorenstein. On the other hand, if we take $L = -mK_X$, then $mK_X = (-1)L$, $S$ is $Q$-Gorenstein.

Now let's suppose we are in characteristic zero and that $X$ is smooth. If $k > 0$, we have $H^i(X, O_X(kL)) = H^i(X, O_X(K_X + (m-1)K_X + (k-1)L) = 0$ by Kodaira vanishing. If $k < 0$, by Serre-duality, $H^i(X, O_X(kL)) = H^{\dim X - i}(X, O_X(K_X - kmL)) = 0$ by Serre vanishing. On the other hand for $k = 0$, $H^{i}(X, O_X) = H^i(X, O_X(K_X - K_X)) = 0$ by Kodaira vanishing. Thus $S$ is Cohen-Macaulay.

Calabi-Yau: If $X$ is Calabi-Yau (meaning $K_X = 0$) and we choose any ample $L$, then $S$ is quasi-Gorenstein. It may not be that $X$ is NOT Gorenstein though, it depends on the vanishing of $H^i(X, O_X(kL))$. For example, an Abelian surface has $H^1(X, O_X) \neq 0$.

Others: If $K_X$ is neither ample nor anti-ample, nor $Q$-trivial, then you are out of luck. The cone will never be $Q$-Gorenstein.

share|improve this answer
    
is there a relation between gorenstein and having quotient singularities? –  Yosemite Sam Apr 28 '12 at 17:49
    
No, there are non-quotient singularities that are Gorenstein (any hypersurface singularity is Gorenstein). There are quotient singularities that are not Gorenstien, for example $k[x^3, x^2y, xy^2, y^3]$. Normal finite quotient singularities are always Cohen-Macaulay and $Q$-Gorenstein though. –  Karl Schwede Apr 28 '12 at 18:15
    
thanks for the quick reply! where would I go about learning the normal finite case for example? –  Yosemite Sam Apr 28 '12 at 21:19
    
For the Cohen-Macaulay bit, this follows since the ring is a direct summand of the ring it is the invariants of. There is a paper of Hochster and J.~Roberts that does this. In fact, Boutot's theorem actually says that the singularity is rational, which amoung other things is Cohen-Macaulay. This stuff works for reductive groups. $$\text{ }$$ For the $Q$-Gorenstein bit, see for example Lemma 5.16 in Kollár-Mori, Birational geometry of algebraic varieties. –  Karl Schwede Apr 29 '12 at 1:40
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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