The Larsen–Lunts theorem

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’

Leave a Reply Cancel reply