# How does f_* O_X measure ramification and Grothendieck-Riemann-Roch

Let $f:X\longrightarrow Y$ be a finite morphism of smooth projective varieties over a field $k$ of characteristic zero, where $\dim X=\dim Y$. Then $f$ is flat. Hence $f_\ast \mathcal{O}_X$ is a coherent locally free sheaf on $Y$.

Now, my question is based on the following example. (For simplicity, take $k=\overline{k}$.)

Example. Suppose that $\dim X =\dim Y = 1$ (i.e., curves). Apply Grothendieck-Riemann-Roch to $f$ and $\mathcal{O}_X$. In degree 0, we get the fact that $\textrm{deg} \ f = \textrm{rk} \ f$. In degree 1 we get a "Hurwitz theorem". In fact, with little effort the formula reads $$2c_1(f_\ast \mathcal{O}_X) =\deg f \cdot K_Y - f_\ast(K_X) = f_\ast(-R),$$ where $R$ is the ramification divisor on $X$.

Now for my two questions that are based on this formula.

Q1. The divisor $R$ is not called the ramification divisor for nothing. Its support is the set of ramification points and the multiplicity of $R$ at a point $P$ is precisely $e_P-1$. So in my opinion, it "measures" the ramification. What about $c_1(f_\ast \mathcal{O}_X) = c_1(\det f_\ast \mathcal{O}_X)$? How does he "measure" the ramification? (I'm probably missing something really elementary here.)

Q2. In higher-dimensions, if I understand correctly, one should get a "higher-dimensional" Hurwitz formula: $$2c_1(f_\ast \mathcal{O}_X) =f_\ast(\textrm{td}(X/Y)).$$ I doubt that this "measures" all the ramification. And, to be frank, I don't really know what it "measures". Can anyone provide some insight?

-

The following does not exactly answer your question, but you may find it interesting. It is the Riemann-Hurwitz formula for surfaces.

Let $\phi:S_1\to S_2$ be a finite morphism between smooth, projective surfaces (over an algebraically closed field of characteristic zero) of degree $n$, and let $B\subseteq S_2$ be the set of $y\in S_2$ such that $\phi^{-1}(y)$ does not contain $n$ points (i.e. $B$ is the ramification locus). Zariski's purity theorem states that $B$ is pure of dimension one; let $B_1,\dots,B_r$ be its irreducible components, and let $n_i$ be the degree of the morphism $\phi|_{\phi^{-1}(B_i)}:\phi^{-1}(B_i)\to B_i$. Then

$$\chi(S_1)=\chi(S_2)\deg \phi-\sum_{i=1}^r(n-n_i)\chi(B_i)+\sum_{y\in B}\left(|\phi^{-1}(y)|-n+\sum_{i=1}^r(n-n_i)m_i(y)\right)$$

where $m_i(y)$ denotes the number of local branches of $B_i$ at $y$. Here $\chi$ is the $\ell$-adic Euler characteristic of the surface ( topological Euler characteristic if $k=\mathbb{C}$), which can be translated into a Chern class if you prefer.

The proof is B. Iversen, 'Numerical invariants and multiple planes', Amer. J. Math. 92 (1970), 968-996. When $k=\mathbb{C}$, you can prove it by thinking of the topological Euler characteristic as a measure on constructible sets (e.g. O. Ya. Viro, Some integral calculus based on Euler characteristic); then the formula is equivalent to Fubini's theorem ($\int\int dxdy=\int\int dydx$) for the graph of $\phi$.

-
That's beautiful. If I remember correctly, here it's the second Chern class that comes into play. I would guess, for an n-dimensional Riemann-Hurwitz formula, the n-th Chern class would come into play. Anyway, I like this alot. Thnx. –  Ariyan Javanpeykar May 8 '10 at 20:35
Yes, second Chern class sounds right. I did once ask an algebraic topologist about higher dimensional versions of the formula, and the concensus seems to be that there is no problem in theory, but the formula will get very very ugly as the dimension increases. –  Matthew Morrow May 8 '10 at 20:43
Alright. But I'm still fuzzy about my first question. Do you happen to know how one can get the ramification indices from f_* O_X ? –  Ariyan Javanpeykar May 9 '10 at 10:39
If your varieties are curves then $f_*O_X$, as an $O_Y$-module, locally looks like a finite extension of Dedekind domains $A/B$. And for any prime ideal $p$ in $A$ you can define the ramification index $e_p$ of $A/B$ at $p$. So surely $f_*O_X$ encodes all the ramification data? But maybe this isn't what you are after. In the higher dimensional case, you need to tell me what you mean by ramification (I usually study things over finite characteristic, where higher dimensional ramification theory is mysterious). –  Matthew Morrow May 9 '10 at 17:54