# Is the Euler characteristic a birational invariant

Suppose that $X$ and $Y$ are smooth projective varieties which are birationally equivalent. I would like to have that $$\textrm{deg} \ \textrm{td}(X) = \textrm{deg} \ \textrm{td}(Y).$$ Invoking the Hirzebruch-Riemann-Roch theorem, this boils down to showing that $$\chi(X,\mathcal{O}_X) = \chi(Y,\mathcal{O}_Y).$$

This is probably a basic fact. A stronger statement is apparently shown in Birationale Transformation von linearen Scharen auf algebraischen Mannigfaltigkeiten by van der Waerden. The only problem is that I can't seem to find it in that article (probably because I don't read that well German).

For $\dim X =2$ one can prove this as Hartshorne does as follows.

Any birational transformation of nonsingular projective curves can be factored into a sequence of monoidal transformations and their inverses. For such a monoidal transformation, the result follows from Proposition 3.4 in Chapter V of Hartshorne.

Does this work in the general case?

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In characteristic 0 there is a result of Wlodarczyk known as Weak Factorization Theorem which claims that every birational map between smooth varieties can be decomposed as a sequence of smooth blowups and smooth blowdowns. – Sasha May 26 '10 at 3:29

If you are willing to stick to characteristic zero, then you can assume that there is actually a morphism $f\colon X\longrightarrow Y$ realizing the birational equivalence (reason: look at the graph $\Gamma\subset X\times Y$ realizing the birational equivalence and take its closure, use resolution of singularities to resolve $\Gamma$, and then replace $X$ by $\Gamma$). In this case, $f_{*}\mathcal{O}_{X}=\mathcal{O}_{Y}$, and all higher direct images are zero, the Leray spectral sequence then implies that the Euler characteristics are equal.

More generally, if $Y$ has rational singularities and $f\colon X\longrightarrow Y$ is a proper birational map, with $X$ smooth, then $f_{*}\mathcal{O}_{X}=\mathcal{O}_Y$ and all higher direct images are zero (this is the definition of rational singularities) and so the same conclusion follows. Smooth varieties have rational singularities! (The computation for smooth varieties is necessary to show that the definition makes sense, i.e., that checking that this property holds for one resolution $X$ implies that it holds for all resolutions).

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The dimensions $h^i(\mathcal O_X)$ of the cohomology groups of $\mathcal O_X$, and thus the Euler characteristic, are birational invariants of smooth proper varieties in positive characteristic as well, by a recent work of Andre Chatzistamatiou and Kay Rülling. It is not published yet but a preprint is available.

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If you can settle for complex numbers, then there is the following simple argument. By the symmetry of the Hodge numbers, you have that $\dim H^p(X,\mathcal{O}_X) = \dim H^0(X,\Omega^p_X)$. So it suffices to show that the latter numbers are birational invariants. But whenever $X\to Y$ is birational, then its locus of indeterminacy has codimension at least two. On the other hand, Hartog's extension lemma says that functions (hence also p-forms) can be extended to the whole space if they are defined off a codimension two subset. One concludes that any birational map induces an isomorphism of global sections of the sheaf of p-forms, so we are done.

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This argument works for any subfield of the complex numbers as well of course. And it extends to any field of characteristic 0 by a Lefschetz principle argument (although as always there is a lot hidden in that statement). – Ravi Vakil May 25 '10 at 21:45
Hodge symmetry also follows from the strong Lefschetz theorem for Hodge cohomology and Serre duality. The strong Lefschetz theorem for Hodge cohomology follows, I believe, from the strong Lefschetz theorem for étale cohomology and $p$-adic Hodge theory. Hence there should be an algebraic, though very involved, proof. (Though using a $p$-adic Lefschetz principle, instead of a complex one.) – Torsten Ekedahl May 26 '10 at 5:53

The quantity that you are asking about (i.e. $\chi(X,\mathcal O_X)$) is closely related to the arithmetic genus (namely, the arithmetic genus is equal to $(-1)^{\dim X}(\chi(X,\mathcal O_X)-1)$), and so you will probably have better luck searching for "birational invariance of arithmetic genus". Doing so, one finds for example the following link, which states that over a field of char. $0$, the arithmetic genus is birationally invariant. I'm not sure whether the situation in char. $p$ has improved since that article was written.

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Yes, I read that article too. Since I'm only interested in the characteristic zero case this does confirm that it's true. It's just that I would like to see "why" it is true. I looked through the Hirzebruch's book referred to in Dolgachev's article and couldn't find it. – Ariyan Javanpeykar May 25 '10 at 20:42
False alarm. It's written in Hirzebruch's book on page 176. – Ariyan Javanpeykar May 25 '10 at 20:46
@Emerton: Dear Matt, I'm confused. If we normalize a nodal cubic, we get a smooth curve of geometric and arithmetic genus 0. But in the original curve, the arithmetic and the geometric genus are not the same: en.wikipedia.org/wiki/Genus%E2%80%93degree_formula. In fact, it's mentioned here en.wikipedia.org/wiki/Geometric_genus that the geometric genus is a birational invariant. So the arithmetic one cannot be. I don't know how much I can trust the wikipedia article but the fact that geometric genus is a birational invariant makes sense to me (e.g. a pinched sphere has genus 0). – adrido Sep 28 '15 at 19:01