6

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.

Community
  • 1
Li Yutong
  • 4,324
  • 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. – Moishe Kohan Oct 03 '13 at 02:43

2 Answers2

5

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_{\tilde X} \pi^\ast\Bigl(\mathrm{ch}(\mathcal O_X)\cdot\mathrm{td}(\mathcal T_X)\Bigr) \\ &= \int_{\tilde X} \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.

Jesko Hüttenhain
  • 15,116
  • @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 06 '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 06 '13 at 21:39
  • @Jesko Huttenhain, is this true when $\tilde{X}$ is smooth, and $X$ has finitely many singular points? And $\pi$ outside these finitely many points is a an etale cover? – gradstudent Sep 23 '15 at 12:09
  • @poorna: We need Hirzebruch-Riemann Roch only for the Bundle $\mathcal O_{\tilde X}$ on the smooth variety $\tilde X$, which works. The Lemma requires the morphism $\pi:\tilde X\to X$ to be flat and finite, according to Fulton. I would assume that this condition can not be left out. Flatness is implicit if both $X$ and $\tilde X$ are regular. So, I think it works if you can show that $\pi$ is flat. I do not know off the top of my hat if your conditions imply that, though. Flatness is quite hard to grasp for me. If something comes to mind, I will post it. – Jesko Hüttenhain Sep 23 '15 at 13:09
  • @poorna: Ok, I was a bit slow there. But assuming that $\pi$ is etale outside a finite set of points does not even guarantee that the morphism is finite, it could for instance be the blow-up of a singular surface. I suspect one might be able to construct counterexamples this way. – Jesko Hüttenhain Sep 23 '15 at 13:18
  • @Jesko Huttenhain, thanks for the prompt reply! The specific case that I am interested in when $\tilde{X}$ is an abelian surface over $\mathbb{C}$ and $X$ is the kummer surface which is the quotient of $\tilde{X}$ by the action of involution. Then $\pi$ is finite and a flat, etale map outside the finite set of 2-torsion points. Can we say anything then? – gradstudent Sep 23 '15 at 16:28
  • 1
    @poorna: sounds like a tame situation, but I don't know. Euler characteristic is sensitive to removing a finite set of points, so I am unsure. I think it might be best to ask this as a separate question, I am sure someone will know better. – Jesko Hüttenhain Sep 24 '15 at 09:24
1

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

Bruno Joyal
  • 55,975
  • Thank you! The only Riemann-Hurwitz I know is about curves, any reference on arbitrary (generic) finite morphism? – Li Yutong Oct 03 '13 at 02:20
  • @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 03 '13 at 07:49
  • @JeskoHüttenhain I am sorry, but I did not see how did this formula related to Euler character. – Li Yutong Oct 03 '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 04 '13 at 21:43