Is the Euler characteristic a birational invariant - MathOverflow

Is the Euler characteristic a birational invariant

2010-05-25T20:10:35Z

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?

Answer by Emerton for Is the Euler characteristic a birational invariant

2010-05-25T20:28:14Z

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.

Answer by Mike Roth for Is the Euler characteristic a birational invariant

2010-05-25T21:03:43Z

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).

Answer by Dan Petersen for Is the Euler characteristic a birational invariant

2010-05-25T21:25:54Z

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.

Answer by VA for Is the Euler characteristic a birational invariant

2010-05-26T04:19:02Z

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.