5
$\begingroup$

So I'm trying to understand a proof of Belyi's theorem from http://eprints.soton.ac.uk/29785/1/b45h1koe.pdf

specifically lemma 3.4.

The setup is as follows: Let $X/\mathbb{C}$ be a curve, and let $t : X\rightarrow\mathbb{P}^1_\mathbb{C}$ be a meromorphic function on $X$ thought of as a covering map of degree $n$. Further, let $\text{Crit}(t)$ denote the critical points of the cover $t$ - ie, the points in $\mathbb{P}^1_\mathbb{C}$ which have fewer than $n$ pre-images under $t$.

Then, he claims that $\text{Crit}(t) = t(\text{supp}(\Omega^1_{X/\mathbb{P}^1_\mathbb{C}}))$.

Now, it's my understanding that the critical points of $t$ should be the images of the ramification points of $t$ under $t$, so I've been trying to understand why it should be the case that the sheaf of relative differentials of $X/\mathbb{P}^1_\mathbb{C}$ should be nonzero only on the ramification points (or at least only above the critical points).

To this end, I'm trying to understand the definition given in Hartshorne (III.8), namely: $\Omega^1_{X/\mathbb{P}^1_\mathbb{C}} = \Delta^*(\mathcal{I}/\mathcal{I}^2)$, where $\Delta : X\rightarrow X\times_{\mathbb{P}^1_\mathbb{C}} X$ is the diagonal map, and $\mathcal{I}$ is the sheaf of ideals of the image $\Delta(X)$ in some open subset $W\subset X\times_{\mathbb{P}^1_\mathbb{C}} X$.

I kind of understand sheaves of ideals (they're essentially functions on the ambient space that vanish on the closed subscheme), but I'm still not very comfortable with the notion of $\Delta^*(\mathcal{I}/\mathcal{I}^2)$ (in this case defined to be $\Delta^{-1}\mathcal{I}/\mathcal{I}^2\otimes_{\Delta^{-1}\mathcal{O}_{X\times X}}\mathcal{O}_X$, where the fibred product is taken over $\mathbb{P}^1$).

Any comments on how I should think of $\Delta^*(...)$ and why the sheaf of relative differentials only have nonzero stalks at ramification points would be awesome!

thanks.

$\endgroup$
3
  • 1
    $\begingroup$ Try looking at the local picture. Pick local coordinates on $X$ and $\mathbb P^1$, centered on a ramification point on $X$, such that the morphism is given by $z \mapsto z^k$. Write down the short exact sequence associated of tangent sheaves associated to the morphism, and note that the differential of $f$ is $k z^{k-1} dz$. For a fixed $z \not= 0$ this morphism $T_X \to f^*T_{\mathbb P^1}$ is surjective, because it is injective and both spaces are of dimension 1, so the sheaf of relative differentials is zero. For $z = 0$, the morphism is zero, whence the support of the relative sheaf. $\endgroup$ Apr 29, 2012 at 7:58
  • 1
    $\begingroup$ In my understanding, $\Delta^*(\mathcal{I}/\mathcal{I}^2)$ is technically convenient because it is obviously well-defined, but not a good way to reason about differentials in practice. First, you need to understand $\Omega_{X/\Bbbk}$, where $X$ is a smooth variety over a field $\Bbbk$. (This is essentially the sheaf of differential forms.) Once you understand this, $\Omega_{X/Y}$ is obtained (for a morphism $X \to Y$) by taking $\Omega_{X/\Bbbk}$ and modding out by pullbacks of differential forms on $Y$. $\endgroup$ Apr 29, 2012 at 21:48
  • $\begingroup$ Qualification: $\Delta^*(\mathcal{I}/\mathcal{I}^2)$ is often not the best way to reason about differentials in practice, particularly when you are learning. $\endgroup$ Apr 29, 2012 at 21:50

2 Answers 2

4
$\begingroup$

Ravi Vakil has a good explaination for the definition $\Delta^*(I/I^2)$ in his notes. See his AG notes here or here (chapter 23). In particular, I guess thinking about this locally makes it a little clearer what's going on, in terms of derivations etc. Also, when $X$ is smooth, it is instructive to see that this really gives the cotangent bundle on $X$.

As for your question about ramification points: Let $f:X\to Y$ be a finite morphism of curves (I will assume that these are smooth in the following). It is useful to have in mind the exact sequence $$0\to f^*\Omega_{Y}\to \Omega_X \to \Omega_{X|Y}\to 0.$$(This is exact at the right in the smooth case, but not in general). Note that $\Omega_{X|Y}$ is a torsion sheaf since the two other sheaves are locally free of the same rank (they are line bundles on $X$). At a point $q\in Y$ and $p\in X$ in the preimage of $q$, let $dx$ denote a generator for $\Omega_{Y,q}$ as a $O_{Y,q}$-module. Now, $(\Omega_{X|Y})_P=0$ if and only if $f^*dx$ is a generator of $\Omega_{X,p}$, which happens if and only if $f$ pulls back a local parameter to a local parameter, that is $p$ is unramified. Moreover, the exact sequence above shows that the ramification index is exactly the length of the sheaf $\Omega_{X|Y}$. Finally, note that this sequence gives the Riemann hurwitz formula, relating the canonical divisors of $X$ and $Y$ and the ramification divisor of $f$.

$\endgroup$
2
$\begingroup$

Let $\pi:X\to Y$ be a finite morphism of curves.

Then, for any point $x$ in $X$ lying over $y$ in $Y$, the coefficient $v_x(\pi)$ of $\Omega_{\pi}$ is the valuation of the different of the extension of dvr's $\mathcal{O}_{y}\subset \mathcal{O}_x$. If you are working in characteristic zero, then $$v_y(\pi) = e_x-1,$$ where $e_x$ is the ramification index. So you see that $\Omega_{\pi}$ is supported on the ramification points.

Also, you have a short exact sequence (it's on page 2 of Chapter IV.2 in Hartshorne) which relates $\Omega_\pi$ with $\Omega_X$ and $\Omega_Y$. The above actually shows the important Riemann-Hurwitz formula: $$K_X = \pi^{\ast} K_Y + R.$$ Here $R$ is the ramification divisor. This equals $\Omega_{\pi}$ in this case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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