Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\pi: \tilde{X} \to X$ be an etale finite cover, then why the Euler character has relation: $$\chi(\tilde{X},\mathcal{O}_{\tilde{X}})=\deg(\pi)\chi({X},\mathcal{O}_{{X}}).$$

I try to use Riemann-Roch, but do not know how to relate Chern characters and Todd class of them.

Besides, I found a similar question on topological setting.

share|improve this question
    
For general dimension, I think you can repeat Mariano's argument (from the link) using spectral sequences, at least if you are working cover the complex numbers. –  studiosus Oct 3 '13 at 2:43

2 Answers 2

Because $\pi$ is étale, you have $\pi^\ast(\mathcal T_X)=\mathcal T_{\tilde X}$ and $\pi^\ast(\mathcal O_X)=\mathcal O_{\tilde X}$ anyway. Using Hirzebruch-Riemann-Roch and the below Lemma, we get \begin{align*} \deg(\pi)\cdot \chi(X,\mathcal O_X) &=\deg(\pi)\cdot\int_X \Bigl(\mathrm{ch}(\mathcal O_X)\cdot\mathrm{td}(\mathcal T_X)\Bigr) = \int_Y \pi^\ast\Bigl(\mathrm{ch}(\mathcal O_X)\cdot\mathrm{td}(\mathcal T_X)\Bigr) \\ &= \int_Y \mathrm{ch}(\mathcal O_{\tilde X})\cdot\mathrm{td}(\mathcal T_{\tilde X}) = \chi(\tilde X,\mathcal O_{\tilde X}). \end{align*}

Lemma Let $\pi:\tilde X\to X$ be a finite surjective morphism of nonsingular varieties of dimension $n$. Denote by $A(X)$ the Chow ring of $X$. The composite $$A(X)\xrightarrow{\quad\textstyle\pi^\ast\quad} A(\tilde X)\xrightarrow{\quad\textstyle\pi_\ast\quad} A(X)$$ is multiplication by $N=\deg(\pi)$. In particular, for all $\alpha\in A^n(X)$, $$\int_{\tilde X} \pi^\ast(\alpha) = N\cdot\int_X \alpha$$ Proof This is Example 1.7.4 from Fulton's book Intersection Theory, but I can give a really short proof using Formula 12.6.2 from Görz-Wedhorn if we are dealing with varieties and the morphism is étale. By said formula, for any point $P\in X$, we have $N=\sum_{\pi(Q)=P} [\Bbbk(Q):\Bbbk(P)]$ because the ramification indices are all $1$. Hence, $$N\cdot\int_X [P] = N\cdot[\Bbbk(P):\Bbbk]=\sum_{\pi(Q)=P} [\Bbbk(Q):\Bbbk(P)][\Bbbk(P):\Bbbk] = \sum_{\pi(Q)=P} [\Bbbk(Q):\Bbbk] = \int_{\tilde X} \pi^\ast[P].$$ We are done because this extends to formal sums of points.

Comment 1: The Lemma does not require the étale condition because points can effortlessly be moved out of the ramification locus. I left this out because, well, we don't have ramification.

Comment 2: The more I look at this, the more I think that this can probably be generalized to quasi-projective schemes over fields, using Grothendieck-Riemann-Roch. I am not sure how general you would like your objects $X$ and $\tilde X$ to be.

share|improve this answer
    
@Cantlog: I am not familliar with the term: Does "ouvert" refer to the quasi in quasi-projective? If so, do you think projective schemes over fields is reasonable? Or do we only need to require finite Euler characteristic? Do we need to be over a field, anyways? I don't know how much of this works at what levels of abstraction. –  Jesko Hüttenhain Oct 6 '13 at 19:26
    
"Ouvert" probably meant open. This denotes in general quasi-projective but not projective varieties. For example, affine varieties of positive dimension have infinite (coherent) Euler characteristic. So for open varieties, if $\chi(X)$ is infinite, is $\chi(\bar{X})$ infinite too ? –  Cantlog Oct 6 '13 at 21:39

This follows from Riemann-Hurwitz, and from the fact that an étale cover is unramified.

share|improve this answer
    
Thank you! The only Riemann-Hurwitz I know is about curves, any reference on arbitrary (generic) finite morphism? –  Li Yutong Oct 3 '13 at 2:20
    
@LiYutong: Yes, there is. Check out formula (12.6.2), Page 329 of the book Algebraic Geometry I by Ulrich Görtz and Torsten Wedhorn. –  Jesko Hüttenhain Oct 3 '13 at 7:49
    
@JeskoHüttenhain I am sorry, but I did not see how did this formula related to Euler character. –  Li Yutong Oct 3 '13 at 13:35
    
@LiYutong: I realize now that I can probably only prove this to you for varieties. You can use the referenced formula to deduce that the fiber cardinality is constant, then use this to prove the statement which is also Example 1.7.4 in Fulton's book Intersection Theory. That statement should yield your result simply applying it to $\mathrm{ch}(\mathcal O_X)\cdot\mathrm{td}(\mathcal T_X)$ and using Hirzebruch-Riemann-Roch. If you want, I will write something more detailed for varieties in an answer. –  Jesko Hüttenhain Oct 4 '13 at 21:43

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.