Computations of Cannonical Divisors and Sheaves

There is an intrinsic classification of (smooth) curves by cannonical divisor. In particular, if the anticannonical class is ample, the curve is $\mathbb P^1$; if neither the anticannonical class nor the cannonical class is ample the curve is an elliptic curve; and if the cannonical class is ample the curve is “of general type”. Note that these conditions correspond to the genus of the curve being $0$, $1$, and $>1$, respectively; furthermore, they are equal to the possible values for the Kodaira dimension. This classification for curves is in direct analogy with the classification of surfaces, which I hope to write my thesis on. Here we compute some examples of the cannonical divisor to get a feel for it.

The cannonical divisor of $\mathbb P^1$.

I’ll type this up eventually.

The cannonical divisor of the Fermat cubic.

Let $X = V(x^3 + y^3 + z^3)\subset \mathbb P^2$, and consider the two open affines $U_0 = U(z)$ and $U_1 = U(y)$. We work first in $U_0$, letting $u = \frac xz$ and $v = \frac yz$. Then $U_0 \cong \spec k[u, v] / (u^3+v^3 + 1)$, so that

We consider the section of $(\omega_X\otimes K(X))(U)$ (rational sections of $\omega_X$, where $K(X)$ is the function field of $X$) given by $\alpha = \frac{du}{v^2} = -\frac{dv}{u^2}$. Restricting to $U(u)$, we have that $-\frac{dv}{u^2}$ is regular and nonvanishing; if it vanished at a point $p$ (that is, if it lived in the image of the maximal ideal under the isomorphism $\mathcal O_p\cong \omega_{X,p}$), then we would have $dv=0$ which implies (by $u^2du=-v^2dv=0$) that $du$ is also in the image of the maximal ideal. Since $du$ and $dv$ generate the sheaf of differentials, this would mean that there is a point $p$ where the entire sheaf of differentials lives in the image of the maximal ideal of $\mathcal O_{X, p}$, a contradiction. Restricting to $U(v)$ we can write $\alpha = \frac{du}{v^2}$ and apply a symmetric argument, giving that $\alpha$ is a regular differential - a section of $\omega_X$ - on $U_0$.

Now on $U_1$ let $t=\frac xy$ and $s=\frac zy$, and look at $\alpha' = -\frac{dt}{s^2} = \frac{ds}{t^2}$. By the same argument (thanks to the symmetry of the Fermat cubic!) we have that $\alpha'$ is regular on this open set. Furthermore, if we restrict further to $U(s)$ we get isomorphisms $U_0\setminus V(s)\to U_1\setminus V(v)$ given by the map on rings

This induces a map $\tilde f$ on differentials over these open sets, given by

Note that this map (extended to a map on rational sections) takes $\alpha'= \frac{ds}{t^2}$ to $f\left(\frac 1{t^2}\right)\tilde f(ds) = -\frac{v^2}{u^2}\frac 1{v^2}dv = -\frac{dv}{u^2} = \alpha$. Thus $\alpha$ and $\alpha'$ glue on the intersection $U_1\cap U_0 = U_1\setminus V(v)$ to obtain a global differential form (note that $U_1$ and $U_0$ cover $X$, because if $X$ and $Y$ vanish then $Z$ must likewise vanish).

This differential form vanishes nowhere, and thus gives an isomorphism $\omega_X\cong \mathcal O_X$ globally.