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

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.

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.

• Thank you for this answer. It is true that my main confusion was about try to wrap my head around what it even means for an element to be $0$ in this ring. I understand now that the motivic measures are useful in trying to understand some basic things about this ring – user2520938 Apr 13 '17 at 16:05
• Right. First understanding what is 0 in the ring is a big step. But be careful! It is possible that there are torsion elements! (Although the proofs on this page show that the class of a variety [X] cannot be torsion.) Also the existence of zero divisors is extremely subtle. – Artur Jackson Apr 13 '17 at 16:13

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.

• Nice. Lots of nice proofs. – Artur Jackson Apr 13 '17 at 12:26
• A comment: the fact that the functions $N(-)$ and $P(-,t)$ defined on smooth projective varieties give rise to well defined motivic measures is not obvious; if you want to prove it, you should use the Bittner presentation of the Grothendieck ring. – Dan Petersen Apr 13 '17 at 16:11
• @DanPetersen I figured that much yea. No one here has proven that $[X]=0\Rightarrow X=\emptyset$ "directly". I figured that these approaches that make it seem easy must have some more stuff going on underneath the surface. – user2520938 Apr 13 '17 at 16:14

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.

• Thanks, that's nice and easy. Any ideas for the case $k=\mathbb{C}$? – user2520938 Apr 13 '17 at 12:02
• Perhaps: using semiring tactics? I think there is a more subtle version $K_0(Var/k)^+$ which is a semiring defined essentially the same way but with free semigroup and congruences. And try positivity? – Artur Jackson Apr 13 '17 at 12:12
• or better: $[X \cup Y] = [X] + [Y] = 0$. So then $X \cup Y$ is the empty variety. – Artur Jackson Apr 13 '17 at 12:14
• For $\mathbb{C}$ the Hodge polynomial can be used instead. – SashaP Apr 13 '17 at 12:18
• Edited to include general proof strategy. – Artur Jackson Apr 13 '17 at 12:30