Let us remenber that we have the following proposition of Artin and Mumford (in "Some elementary examples unirational varieties which are not rational" proposition 1.):

"The torsion subgroup $T_2\subset H^3(V,\mathbb{Z})$ is a birational invariant of a complete non-singular complex variety $V$ of any dimension $n$."

Question : Are there other any other type of singular cohomological information (even homotopical) which is known to be a birational invariant of complete non-singular complex varieties?

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    $\begingroup$ the fundamental group is another example $\endgroup$ – Piotr Achinger Nov 5 '16 at 11:52
  • $\begingroup$ The spaces $H^0(X,\Omega^i_X)$ and $H^i(X,\mathcal O_X)$ are birational invariants in arbitrary characteristic! (The former is an exercise in Hartshorne, and the latter follows in characteristic $0$ from Hodge symmetry, and in characteristic $p$ from a paper by Chatzistamatiou and Rülling.) $\endgroup$ – R. van Dobben de Bruyn Nov 5 '16 at 13:57
  • $\begingroup$ Also, the Brauer group is a birational invariant of smooth projective varieties, which Artin and Mumford use in the second half of their article to reinterpret and generalise the results of the first half. $\endgroup$ – R. van Dobben de Bruyn Nov 5 '16 at 14:02

Basically, any invariant $T$ which satisfies the following purity property will turn into a birational invariant for proper smooth varieties (aka complex compact manifolds): let $X$ be a smooth variety and $U$ be an open subspace of $X$ such that $\mathop{\rm codim}_X(X\setminus U)\geq 2$; then the injection $U\hookrightarrow$ induces an isomorphism $T(X)\simeq T(U)$.

The proof is as follows. Let $X,Y$ be proper smooth varieties and let $f\colon X\dashrightarrow Y$ be a birational isomorphism. Because $X$ is smooth and $Y$ is proper, the valuative criterion applies and shows that $f$ is defined on an open subset $U\subset X$ such that $\mathop{\rm codim}_X(X\setminus U)\geq 2$. Similarly, the inverse isomorphism $f^{-1}$ is defined on an open subset $V\subset Y$ such that $\mathop{\rm codim}_Y(Y\setminus V)\geq 2$. Say $T$ is covariant, for definiteness. One gets maps $f_*\colon T(X)\simeq T(U)\to T(Y)$ and $g_*\colon T(Y)\simeq T(V)\to T(X)$. One checks that their composite is the identity.

A loot of interesting examples are given in the answer of R. van Dobben de Bruyn.

Another class of examples is given by the unramified cohomology. They are important in the study of rationality of varieties.

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    $\begingroup$ This is very nice! I guess I was thinking of invariants which are not necessarily easy to define on open varieties (e.g. crystalline cohomology), or where it's not clear they are still finite-dimensional (for some of the examples, it only follows a posteriori from the purity property). $\endgroup$ – R. van Dobben de Bruyn Nov 5 '16 at 23:40
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    $\begingroup$ For a concrete example where you really need a more elaborate argument, consider the Chow group of $0$-cycles. It is not true that the map $\operatorname{CH}_0(X) \to \operatorname{CH}_0(U)$ is an isomorphism if $\operatorname{codim}_X(X\setminus U) \geq 2$. For example, take $U$ to be the complement of a point $p \in \mathbb P^n$; then $\operatorname{CH}_0(U) = 0$ since we can move any $0$-cycle to the point $p$. This gives a counterexample (if $n \geq 2$). $\endgroup$ – R. van Dobben de Bruyn Apr 29 '17 at 0:50

Here is some of the general philosophy of birational invariants, at least those coming from (co)homology (I don't think this approach quite works for homotopical invariants.)

Philosophy. If $f \colon X \dashrightarrow Y$ is a birational map, consider the cycles $\bar \Gamma_f \subseteq X \times Y$ and $\bar \Gamma_{f^{-1}} \subseteq Y \times X$. Then $\bar\Gamma_{f^{-1}} \circ \bar \Gamma_f$ and $\bar\Gamma_f \circ \bar\Gamma_{f^{-1}}$ differ from the identity by a cycle supported on $D \times D$ and $E \times E$ respectively, where $D \subseteq X$ and $E \subseteq Y$ are divisors. (For a proof, see e.g. this answer.)

Since cycles "act on any cohomology theory", this has implications for the parts of cohomology that are acted upon trivially by cycles supported on $D \times D$. Let me give some examples.

Definition. The coniveau filtration on $H^i(X,\mathbb Z/n \mathbb Z)$ (singular or étale cohomology, assuming $(n,p) = 1$ if $k$ has positive characteristic) is defined by $$N^r H^i(X,\mathbb Z/n\mathbb Z) = \left\{x\ \bigg|\ \begin{array}{cc} x|_{X-Y} = 0 \in H^i(X-Y,\mathbb Z/n\mathbb Z) \\ \text{ for some } Y \subseteq X \text{ with } \operatorname{codim}(Y,X) \geq r\end{array}\right\}.$$

Lemma. Let $Z \in \operatorname{CH}^n(X \times Y)$ by a cycle supported on $X \times E$. Then the image of the induced map $$Z_* \colon H^*(X, \mathbb Z/n\mathbb Z) \to H^*(Y, \mathbb Z/n\mathbb Z)$$ lands in $N^1H^*(Y,\mathbb Z/n\mathbb Z)$.

Proof. If $x \in H^i(X,\mathbb Z/n\mathbb Z)$, then $Z_* x$ is supported on $E$. Thus, $(Z_* x)|_{Y-E} = 0$. $\square$

Examples. Let me sketch how this philosophy manifests itself in certain cohomology theories:

Hodge realisation. In the Hodge realisation, the coniveau $r$ part of the cohomology will land inside $$H^{k-r,r}(X) \oplus \ldots \oplus H^{r,k-r}(X) \subseteq H^k(X,\mathbb C).$$ Indeed, an element of $H^k(X,\mathbb C)$ that vanishes on $X-Y$ with $\operatorname{codim}(Y,X) = r$ comes from the cohomology $H^{k-2r}(Y,\mathbb C)$ under the Gysin map, at least when $Y$ is smooth (in general, replace $Y$ by a resolution $\tilde Y \to Y$). This is a morphism of Hodge structures of bidegree $(r,r)$, which proves the claim.

This shows that $H^0(X,\Omega^k_X)$ and $H^k(X,\mathcal O_X)$ are birational invariants: $(\bar\Gamma_{f^{-1}} \circ \bar\Gamma_f) _* = \operatorname{Id} + Z$, with $Z$ supported on $D \times D$. By the lemma above, $Z_*$ maps everything into the coniveau $1$ part, so it acts as $0$ on $H^k(X,\mathcal O_X) \oplus H^0(X,\Omega^k_X)$. Hence, the composition $$H^k(X,\mathcal O_X) \to H^k(Y,\mathcal O_Y) \to H^k(X,\mathcal O_X)$$ is the identity, and similarly for the opposite composition, as well as for $H^0(-,\Omega^k)$. $\square$

In this example, one can also use the explicit description of the Hodge diamond of a blow-up, and use weak factorisation. Or even use the elementary argument that $H^0(-,\Omega^k)$ is a birational invariant and invoke Hodge symmetry.

Frobenius realisation. Over a finite field $k$, we have the slope filtration on crystalline cohomology (after inverting $p$), which is the closest analogue to the Hodge decomposition on de Rham cohomology. This gives a filtration on $$H^k_{\operatorname{crys}}(X/K) = H^k_{\operatorname{crys}}(X/W(k))[\tfrac{1}{p}]$$ whose successive subquotients are the spaces where Frobenius acts with slope $[i,i+1)$. These spaces can alternatively be described (cf. Illusie) as the de Rham-Witt cohomology $$H^k_{\operatorname{crys}}(X/K)_{[i,i+1)} = H^i(X,W\Omega^{k-i}_X)[\tfrac{1}{p}],$$ which is why it is similar to the Hodge decomposition of de Rham cohomology in characteristic $0$.

It follows from Berthelot's work that the coniveau $r$ part has slopes $\geq r$ (see e.g. Lemma 2.1 of this paper by Hélène Esnault; the argument is identical to the one given above in the Hodge realisation: use purity of the Gysin maps). Thus, the slope $[0,1)$ part is a birational invariant (hence by slope symmetry, so is the slope $(k-1,k]$ part).

Explicitly, this says that the multiset of Frobenius-eigenvalues not divisible by $q = |k|$ is a birational invariant. This was actually proven in an elementary way by Ekedahl in 1983 (Sur le groupe fondamental d'une variété unirationelle).

Cohomology of $\mathcal O_X$ in positive characteristic. In this paper by Chatzistamatiou and Rülling, it is shown that $H^i(X,\mathcal O_X)$ is a birational invariant in positive characteristic. The main idea of the paper is to define a well-defined action of Chow groups on Hodge cohomology [with compact support] and use the philosophy above. (The main difficulty is to define pushforwards, which is where the supports come in.)

Chow group of zero-cycles. The group $\operatorname{CH}_0(X)$ is a birational invariant over any field. Indeed, it suffices to show that a cycle supported on $D \times D$ induces the zero map on $\operatorname{CH}_0(X)$. This is because we can move any zero-cycle to a cycle not meeting $D$. See this answer for more details.

Other examples. Low degree homology $H_2(X,\mathbb Z)$ "should only pick up things coming from $H_0(D, \mathbb Z)$" if we act by a cycle supported on $X \times D$. This is torsion-free, hence the torsion in $H_2$ is a birational invariant. Under universal coefficients, this gives the torsion in $H^3$.

Similarly, the $n$-torsion of the Brauer group is given by $H^2(X,\mathbb Z/n \mathbb Z)$ via the exact sequence $$0 \to \mu_n \to \mathbb G_m \to \mathbb G_m \to 0.$$ Splitting this up into $n$-torsion in $H^3(-,\mathbb Z)$ and the part coming from $H^2(-,\mathbb Z)/n$, we see that it should be a birational invariant. (In positive characteristic, use $\mathbb Z_\ell$ instead of $\mathbb Z$, at least when $n = \ell$. There is also a characteristic-independent proof using Brauer-theoretic arguments.) (In characteristic $0$ we can bypass $\mathbb Z/n \mathbb Z$ by using the exponential sequence instead of the argument I gave here.)

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    $\begingroup$ Ekedahl's proof was exactly analogous to the arguments given by Esnault later one. The apparent difference lied in the way they got grip on the $p$-adic absolute values of eigenvalues of Frobenius acting on cohomology. Esnault (at least in the paper about Fano varieties) used the $p$-adic cohomology of Berthelot — which is made for that purpose. Ekedahl used $\ell$-adic cohomology and had thus to rely on a theorem of Deligne asserting that the eigenvalues are algebraic integers. $\endgroup$ – ACL Nov 5 '16 at 21:20
  • $\begingroup$ @ACL: thanks, you're right. The argument still boils down to a Gysin map. But somehow not mentioning coniveau at any point makes it seem more elementary... $\endgroup$ – R. van Dobben de Bruyn Nov 5 '16 at 23:26
  • $\begingroup$ Perhaps I am being silly, but your discussion of the crystalline case seems to imply the following, which I find surprising. Namely, if $X/k$ is smooth proper then $H^1_\text{cris}(X/K_0)$ always has slopes bounded by $1$. In fact, it's the Dieudonne isocrystals associated with $A[p^\infty]$ where $A=\text{Alb}(X)$. So from what you said above it seems to imply, unless I've misread something, that if $X$ and $Y$ are birational that the Dieudonne isocrystals of their Albanese varieties are isomorphic and thus, consequently, that the $p$-divisible groups of their Albanese varieties are $\endgroup$ – Alex Youcis Nov 6 '16 at 7:52
  • $\begingroup$ isomorphic. This is surprising since, of course, the Albanese variety itself is far from being a birational invariant. Have I made a mistake somewhere and, if not, can you say anything about this? $\endgroup$ – Alex Youcis Nov 6 '16 at 7:53
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    $\begingroup$ Oh, wait, I'm being dumb. The Albanese variety IS a birational invariant. I was stupid and thinking the fact that Picard group was not a birational invariant meant that the Albanese variety is not, but this is obviously false--it's the component group of the Picard scheme which is not a birational invariant, the Albanese is fine. I'll leave my silly comment above there in case anyone else makes the same mistake. I guess it also explains a fractional, fractional part of the result you cited. In fact, I guess that the birational invariance of the Albanese is why $H^1$ is usually OK. Cheers! $\endgroup$ – Alex Youcis Nov 6 '16 at 8:10

Here a few quick observations. A necessary condition for birational invariance of something is invariance under blow up along a smooth centre. This is also sufficient, over $\mathbb{C}$, by the weak factorization theorem. Most invariants cooked up from singular cohomology will fail this test. But here a few that do:

  1. Torsion in $H^i(X,\mathbb{Z})$ for $i=2,3$. Of course you already knew $i=3$; the remaining case is easy, because by universal coefficients, $H^2(X)_{tors}=H_1(X)_{tors}= H_1(\pi_1(X))_{tors}$.
  2. $H^1(X,\mathbb{Z})$ and in fact the sub algebra of $H^*(X,\mathbb{Z})$ generated by $H^1(X,\mathbb{Z})$. The quickest way to see binational invariance is to observe that this is the image of $H^*(Alb(X),\mathbb{Z})$ under the Albanese mapping.

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