Motives | Lovely little lemmas

This post is about a (very weak) shadow in characteristic $p$ of the Larsen–Lunts theorem. See my previous post for the statement and sketch of the proof of Larsen–Lunts.

Remark. In characteristic $p$ , we do not even know the weakest form of resolution of singularities (e.g. find a smooth proper model for any function field). Thus, we certainly do not know the Larsen–Lunts theorem. However, we can still try to prove corollaries (and if they fail, we know that resolution must fail).

Today, I want to talk about the following statement:

Theorem. (Ekedahl) Let $k = \mathbb F_q$ . Let $X$ and $Y$ be smooth proper varieties, and assume $X$ and $Y$ are stably birational. Then $|X(k)| \equiv |Y(k)| \pmod{q}$ .

Remark. This would follow immediately from Larsen–Lunts if we knew a sufficiently strong form of resolution of singularities. Indeed, the map

$K_0(\operatorname{Var}_k) \to \Z/q\Z$

given by counting $\F_q$ -points modulo $q$ factors through $K_0(\operatorname{Var}_k)/(\mathbb L)$ since $|\mathbb A^1(\F_q)| = q$ . Hence, by Larsen–Lunts, it factors through $\mathbb Z[\operatorname{SB}]$ .

It turns out that the theorem is true without assuming resolution of singularities, and the proof is due to Ekedahl (although in his paper he never explicitly states it in this form). The reader should definitely check out Ekedahl’s article (see references below), because his proof is more beautiful than the one I present here, and actually proves a bit more.

We will need one fairly deep theorem:

Theorem. Let $X$ be a variety of dimension $n$ over $k = \F_q$ . Let $\alpha$ be an eigenvalue of Frobenius on $H^i_c(\bar X\et, \Q_\ell)$ . Then $\alpha$ and $q^n\alpha^{-1}$ are both algebraic integers.

The first part (integrality of $\alpha$ ) is fairly well-known. For the second part (integrality of $q^n\alpha^{-1}$ ), see SGA 7 $_{\text{II}}$ , Exp. XXI, Corollary 5.5.3(iii).

The statement that appears in Ekedahl’s article is the following:

Theorem. (Ekedahl’s version) Let $k = \mathbb F_q$ . Let $X$ and $Y$ be smooth connected varieties (not necessarily proper!), and assume $X$ and $Y$ are birational. If $\alpha$ is an eigenvalue of Frobenius on $H^i(\bar X, \Q_\ell)$ which is not an eigenvalue on $H^i(\bar Y, \Q_\ell)$ , then $\alpha$ is divisible by $q$ .

This statement should be taken to include multiplicities; e.g. a double eigenvalue for $X$ which is a simple eigenvalue for $Y$ is also divisible by $q$ . By symmetry, we also get the opposite statement (with $X$ and $Y$ swapped). Thus, the eigenvalues (with multiplicities) that are not divisible by $q$ are the same for $X$ and $Y$ .

Proof. We immediately reduce to the case where $X \sbq Y$ is an open immersion, with complement $Z$ . We have a long exact sequence for étale cohomology with compact support:

$\cdots \to H^{i-1}_c(Z) \to H^i_c(X) \to H^i_c(Y) \to H^i_c(Z) \to H^{i+1}(X) \to \cdots.$

If $\alpha$ is an eigenvalue on some $H^i_c(Z)$ , then $q^{n-1}\alpha^{-1}$ is an algebraic integer (see above). Hence, for any valuation $v$ on $\bar \Q$ with $v(q) = 1$ , we have $v(\alpha) \leq n-1$ . We conclude that the eigenvalues for which some valuation is $> n-1$ on $H^i_c(X)$ and $H^i_c(Y)$ agree. Hence, by Poincaré duality, the eigenvalues of $H^{2n-i}(X)$ and $H^{2n-i}(Y)$ for which some valuation is $< 1$ agree. These are exactly the ones that are not divisible by $q$ . $\qedsymbol$

The theorem I stated above immediately follows from this one:

Proof. Since $|X\times\P^n(\F_q)| = (q^n + \ldots + 1) |X(\F_q)|$ , we may replace $X$ by $X \times \P^n$ . Thus, we can assume $X$ and $Y$ are birational; both of dimension $n$ .

By the Weil conjectures, we know that

$|X(\F_q)| = \sum_{i=0}^{2n} \sum_\alpha \alpha^i,$

where the inner sum runs over all eigenvalues of Frobenius. If we reduce mod $q$ , then we only need to consider eigenvalues that are not divisible by $q$ . By Ekedahl’s version of the theorem, the set (with multiplicities) of such $\alpha$ are the same for $X$ and $Y$ . $\qedsymbol$

Historical remark. Although the theorem above was essentially proven in 1983 (but not explicitly stated), a separate proof for threefolds appeared in a paper by Gilles Lachaud and Marc Perret in 2000. It uses Abhyankar’s results on resolution of singularities, and is much closer to the proof of Larsen–Lunts than Ekedahl’s proof was. In 2002, Bruno Kahn provided a different proof for the general case using some (fairly advanced?) motive machinery (‘almost without cohomology’).

References.

Torsten Ekedahl, Sur le groupe fondamental d’une variété unirationelle. Comptes rendus de l’académie des sciences de Paris, Serie I: mathématiques, 297(12), p. 627-629 (1983).

Bruno Kahn, Number of points of function fields over finite fields. arXiv:math/0210202

Gilles Lachaud and Marc Perret, Un invariant birationnel des variétés de dimension 3 sur un corps fini. Journal of Algebraic Geometry 9 (2000), p. 451-458.

The Larsen–Lunts theorem is one of the most beautiful theorems I know. But first, let me recall some definitions.

Definition. The Grothendieck ring of varieties $K_0(\operatorname{Var}_k)$ over a field $k$ is the free abelian generated by (formal) symbols $[X]$ for $X$ a variety over $k$ (which I will take to mean a geometrically reduced, separated scheme of finite type over $k$ ), subject to the relations

$[X] = [U] + [Z]$

whenever $Z \to X$ is a closed immersion and $U = X \setminus Z$ . It becomes a ring by setting $[X] \cdot [Y] = [X \times Y]$ (exercise: show that this is well-defined). The class $\mathbb L = [\A^1]$ is called the Lefschetz motif.

Remark. Recall that a rational map $f \colon X \dashrightarrow Y$ is a morphism $U \to Y$ defined on some dense open $U \subseteq X$ . Varieties with rational morphisms form a category, and $f$ is called a birational map if it is an isomorphism in this category. We say that $X$ and $Y$ are birational if there exists a birational map $f \colon X \stackrel\sim\dashrightarrow Y$ . If $X$ and $Y$ are integral, this is equivalent to the equality of function fields $K(X) = K(Y)$ .

We say that $X$ and $Y$ are stably birational if $X \times \P^n$ is birational to $Y \times \P^m$ for some $m, n \in \Z_{\geq 0}$ . This is equivalent to the existence of an isomorphism

$K(X)(x_1,\ldots,x_n) \cong K(Y)(y_1,\ldots,y_m).$

There are examples of stably birational varieties that are not birational.

Definition. Write $\operatorname{SB}$ for the set of stable birational classes of smooth proper varieties over $k$ . To avoid confusion, I shall denote the class of $X$ by $(X)$ . This set becomes a commutative monoid by setting $(X) \cdot (Y) = (X \times Y)$ (again: show that this is well-defined).

Theorem. (Larsen–Lunts) Let $k = \C$ . There exists a unique ring homomorphism

$\phi \colon K_0(\operatorname{Var}_\C) \to \Z[\operatorname{SB}]$

such that for any smooth proper $X$ , the image of $[X]$ is $(X)$ . Moreover, the kernel of $\phi$ is the ideal generated by $\mathbb L$ .

Proof (sketch). The map $\phi$ is constructed by induction on the dimension. For smooth proper $X$ , it is clear what $\phi(X)$ should be (namely $(X)$ ). If $X$ is smooth, we can find a smooth compactification $X \to \bar X$ (using resolution of singularities). Then we set $\phi(X) := (\bar X) - \phi(\bar X \setminus X)$ , where the right-hand side is defined by the induction hypothesis.

To check that it is independent of the compactification chosen, we need a strong form of weak factorisation: any two compactifications $X \to \bar X_1, \bar X_2$ differ by a series of blow-ups and blow-downs along smooth centres disjoint from $X$ . Now if $X \to Y$ is the blow-up along a smooth centre $Z \sbq Y$ with exceptional divisor $E \sbq X$ , then $E$ is a $\P^r$ -bundle over $Z$ for some $r$ ; thus $Z$ and $E$ are stably birational. Now well-definedness of the map on lower-dimensional varieties proves independence on the smooth compactification.

Finally if $X$ is singular, we simply set $\phi(X) = \phi(X^{\operatorname{nonsing}}) + \phi(X^{\operatorname{sing}})$ . After some further checks (like additivity and multiplicativity), this finishes the construction of $\phi$ .

Now clearly $\mathbb L \in \ker \phi$ , since $(\P^1) = (\P^0)$ , and $\mathbb L = [\P^1] - [\P^0]$ . Conversely, let $\alpha \in \ker \phi$ . We can write any $\alpha$ as

$\alpha = \sum_{i=1}^n [X_i] - \sum_{j=1}^m [Y_j]$

for certain $X_i, Y_j$ smooth proper (we again use resolution here). Since $\Z[\operatorname{SB}]$ is the free algebra on $\operatorname{SB}$ , we conclude that $n = m$ and $(X_i) = (Y_i)$ after renumbering. Thus it suffices to consider the case $\alpha = [X] - [Y]$ for $X$ and $Y$ smooth proper and stably birational (to each other). We may replace $X$ by $X \times \P^n$ since their difference is $[X] \cdot (\mathbb L + \ldots + \mathbb L^n)$ , which is in the kernel. Thus, we may assume $X$ and $Y$ are birational.

Now by weak factorisation, we reduce to the case of a blow-up $X \to Y$ in a smooth centre $Z \sbq Y$ . Let $E$ be the exceptional divisor, which is a $\P^r$ -bundle over $Z$ . Thus $Z$ and $E$ differ by a multiple of $\mathbb L$ , since $[\P^r] - [\P^0] = \mathbb L + \ldots + \mathbb L^r$ . $\qedsymbol$

Remark. The hard part of the theorem is the definition of the map. In order to define $\phi(X)$ for $X$ not necessarily smooth and proper, we need to assume resolution of singularities (for this, a very mild version of resolution suffices). To check that it is independent of choices, we need the weak factorisation theorem (which in turn uses a very strong version of resolution of singularities). The computation of the kernel again uses resolution of singularities and weak factorisation.

This is why we restrict ourselves to $k = \C$ . I suspect that it is also fine for arbitrary algebraically closed fields of characteristic $0$ .

Corollary. Let $X$ and $Y$ be smooth proper. Then $X$ and $Y$ are stably birational if and only if $[X] \equiv [Y] \pmod{\mathbb L}$ .

Proof. Since $\Z[\operatorname{SB}]$ is the free algebra on $\operatorname{SB}$ , we have $(X) = (Y)$ if and only if $X$ and $Y$ are stably birational. The result is now immediate from the theorem. $\qedsymbol$

Remark. If we knew weak factorisation (without knowing resolution), then one implication would follow immediately: if $X$ and $Y$ are stably birational, then $X \times \P^n \stackrel\sim\dashrightarrow Y \times \P^m$ for some $m, n$ . Clearly $[X \times \P^n] - [X] = (\mathbb L + \ldots + \mathbb L^n) \cdot [X]$ is divisible by $\mathbb L$ , so we may assume $X \stackrel\sim\dashrightarrow Y$ . Now by weak factorisation, a birational map factors as a chain of blow-ups and blow-downs along smooth centres, so we reduce to that case. But if $X = \operatorname{Bl}_Z Y$ has exceptional divisor $E$ , then $E$ is a $\mathbb P^r$ -bundle over $Z$ for some $r$ , hence $[X] - [Y] = [E] - [Z]$ is divisible by $\mathbb L$ .

However, for the other implication there is no direct proof even if we knew weak factorisation.

In my next post, I will address a statement in positive characteristic (where neither resolution of singularities nor weak factorisation are currently known) that is related to the corollary (but much weaker).

Lovely little lemmas

Or maybe ‘amusing observations’

Category Archives: Motives

Number of points modulo q is a stable birational invariant

The Larsen–Lunts theorem