What does it mean that $[X]+[Y]=0$ in the Grothendieck ring of varieties?

Question

This is a really basic question. If I have two non-isomorphic varieties $X$ and $Y$, is it possible that $[X]+[Y]=0$ in the Grothendieck ring?

If so, what does this mean geometrically? Obviously $[\emptyset]-[X]-[Y]$ is not one of the standard relations modded into the ring, so $[X]+[Y]$ has to be some non-trivial combination of such relations. I'm having trouble seeing how this could happen though. An example would be especially nice.

Answer 1

The element $[X]+[Y]$ of $K_0({\rm Var}_k)$ is the class of the disjoint union $X\cup Y$. So your question amounts to understand whether the class of a variety, say $X$, can vanish in $K_0({\rm Var}_k)$.

As is implicit in the questions, and in the other answers, the ring $K_0({\rm Var}_k)$ is hard to understand, and is essentially only understood thanks to a few geometrically motivated ``motivic measures'':

• If $k$ is finite, the counting measure $N\colon K_0({\rm Var}_k)\to\mathbf Z$;
• The Euler characteristic using any cohomology theory with compact supports you can imagine;
• If $k=\mathbf C$, the Hodge-Deligne polynomial, valued in $\mathbf Z[u,v]$;
• Using Deligne's theory of weights on étale cohomology, one can define a Poincaré polynomial, valued in $\mathbf Z[t]$;
• If $k$ is a field of characteristic $0$, Larsen and Lunts have defined an exotic — and useful — motivic measure with valued in the free abelian group of stable birational class of varieties.

In any case, any of these motivic measures shows that the class of a non-empty variety does not vanish. They can also be useful to show that the classes of two varieties are not equal.

Answer 2

If $k$ is an algebraically closed field of characteristic $0$, then the map

$$X \mapsto N(X) := \#(\text{connected components of }X)$$

defined for smooth projective varieties extends to a ring homomorphism $K_0(Var/k) \to \mathbf{Z}$. So if both $X$ and $Y$ are smooth projective, then $[X] + [Y] = 0$ implies that $N(X) = N(Y) = 0$, hence $X$ and $Y$ are both empty.

If $X$ or $Y$ is not necessarily smooth projective, then the above invariant is not strong enough and we have to use the Poincaré polynomials (or the Hodge polynomials as SashaP suggested in his comment in Artur Jackson's answer). Here is the outline.

Again, $k$ is an algebraically closed field of characteristic $0$. Fix a Weil cohomology theory $H^\bullet$, then the map defined for smooth projective varieties

$$X \mapsto P(X;t) := \sum_{i}(-1)^ib_{i}(X)t^i $$ where $b_i(X) := \dim H^i(X)$, extends to a ring homomorphism $K_0(Var/k) \to \mathbf{Z}[t]$. One checks that for any variety $X$ (not necessarily smooth nor projective), the associated Poincaré polynomial $P(X;t)$ is of degree $2 \dim X$ (whose leading coeeficient is the number of irreducible components of $X$). Accordingly if $[X] = 0$ for some variety $X$, then $X$ is empty. In particular if $[X \sqcup Y] = [X] + [Y] = 0$, then both $X$ and $Y$ are empty.

Answer 3

Point counting $K_0(Var/\mathbb{F}_q) \to \mathbb{Z}$ induced by $[X] \to \#X(\mathbb{F}_{q^e})$ is a ring homomorphism, so $$[X] + [Y] = 0$$ would imply $\#X(\mathbb{F}_{q^e}) +\#Y(\mathbb{F}_{q^e}) = 0$. And this can only happen if $X$ and $Y$ are the empty varieties.

EDIT: The general idea in such a situation is to use the characterization of $K_0(Var/k)$ as the ring of universal Euler characteristics. In all of the beautiful comments mentioned here what is going on is on finds a function $Var/k \to R$ ($R$ a ring) which behaves as an abstract Euler characteristic. Then this factors through the Grothendieck ring of varieties.