# why are motives more serious than “naive” motives?

I know my question is a bit vague, sorry for this.

Let $k$ be a field of characteristic zero. Consider the Grothendieck ring of varieties over $k$, usually denoted by $K_0(Var_k)$. This is generated by isomorphism classes of varieties over $k$ modulo the relations [X]=[Y]+[X-Y] whenever $Y$ is a closed subvariety of $X$. People usually refer to [X] as the "naive" motive of $X$.

On the other hand, one has Voevodsky's "true" motives $DM_{gm}(k)$ (not as true as we would like to, I know !) and to any variety $X$ we can attach an object $M(X)$in $DM_{gm}(k)$.

Why is this $M(X)$ more serious than the naive one? That is, can you give some examples of properties that cannot be read at the level of $K_0(Var_k)$ but that one sees when working in $DM_{gm}(k)$?

• For the same reason that Betti numbers are more "serious" than the Euler characteristic. – Donu Arapura Nov 5 '14 at 18:18
• Could you elaborate a bit more? Why the E-polynomial, which is a motivic measure on $K_0(Var_k)$ is not enough to study the Hodge numbers? – birk Nov 5 '14 at 18:26
• Voevodsky's motives are a category, and you can consider morphisms between motives. This is relevant if you want to define motivic cohomology and study regulators defined on those (e.g. for arithmetic applications). – Matthias Wendt Nov 5 '14 at 18:47
• Of course, I agree with you that this is extremely important. Still, if you only want the motivic cohomology of X you can use higher Chow groups or K-theory as your definition... – birk Nov 5 '14 at 19:02
• @DonuArapura. I'd rather say: for the same reason that cohomology groups are more "serious" than the Betti numbers. – ACL Nov 5 '14 at 20:58

## 3 Answers

I think the presence of Voevodsky's category of (mixed) motives is a red herring here. Let me briefly explain why I think that, and then say why any "real" notion of motives (say, pure Chow motives, as in Vivek Shende's comment) or, I guess, Voevodsky's motives, are much more serious than the so-called naive motives.

The (partially conjectural) motivic philosophy says we should functorially associate to each variety $X$ a motive, which is an object in some Abelian category with a "weight" (which is just a number) associated to any simple object; any reasonable cohomology theory (e.g. Betti cohomology) is supposed to factor through this functor. If $X$ is a smooth projective variety, this motive is naturally supposed to be a direct sum of (semi-simple) pieces in each weight. A cohomological realization functor applied to the weight $i$ piece should be the degree $i$ cohomology of the original variety. For a general variety, one does not get a semi-simple object, but rather some iterated extension of pieces of large weight by pieces of smaller weight. The associated filtration gives the weight filtration on passing to some cohomological realization. Proving that something like this is true is pretty far out of reach.

But anyway, the picture is supposed to be something like $$\text{Varieties}\to \text{Mixed Motives}\overset{ss}{\to} \text{Pure Motives}$$ where the second map is semi-simplification w.r.t. the weight filtration. We don't know what the middle term in this diagram is, but there is a perfectly good term on the right hand side, given by Grothendieck's Chow Motives; Voevodsky's category is a candidate for the derived category of the middle object. There are several other candidates as well, I guess.

Now, we can also send any variety $X$ to its class $[X]$ in the Grothendieck ring of varieties, $K_0(\text{Var})$. This remembers some of the "motivic" information of $X$; in particular, there is a map $K_0(\text{Var})\to K_0(\text{Chow Motives}_\mathbb{Q})$ so that if $X$ is smooth and projective, the image of $[X]$ under this map will be the class of its associated Chow motive in the K-group of Chow motives. If $X$ is not smooth and projective, it will be (conjecturally) the class of the semi-simplification of its (mixed) motive with respect to the weight filtration (because $K_0$, by definition, forgets all extension information). So in particular, this is the "motivic" manifestation of Dan Petersen's answer: the Grothendieck ring of varieties forgets the weight filtration.

But aside from this, I think "mixedness" is not really important for understanding this question--the Grothendieck ring of varieties forgets a lot more than the weight filtration!

Because smooth projective varieties are "cohomologically pure," the Grothendieck ring of varieties does "remember," say, the Hodge structures on their cohomology groups. Namely, one can take a class in $K_0(\text{Var})$ and send it to its Euler characteristic in the $K$-theory of rational Hodge structures; because of purity, one can pick out its individual cohomology groups. The main issue is that the Grothendieck ring of varieties forgets morphisms! Namely, one can recover the map $$H^i: \text{Smooth Varieties} \mapsto H^i(-, \mathbb{Q})$$ sending a smooth projective variety to its $i$-th cohomology group, but not the functor. And if you can get much algebraic geometry done without using any of the functorial properties of cohomology, I'd be very impressed.

Let me just remark that the map $K_0(\text{Var})\to K_0(\text{Chow Motives}_\mathbb{Q})$ is quite mysterious, and does forget some information. For example, it is not injective--given two isogenous elliptic curves $E_1, E_2$ over $\mathbb{C}$, the class $[E_1]-[E_2]$ is not zero in $K_0(\text{Var})$ (by results of Larsen-Lunts, for example), but its image in $K_0(\text{Chow Motives}_\mathbb{Q})$ is zero. On the other hand, I see no reason for the map to be surjective either (for example, I have no idea how to hit a random Chow Motive of pure weight).

• Chow motives are not the "right" candidate for pure motives (in the sense of being in the conjectural abelian category of mixed motives), one should rather use pure motives a la Grothendieck for the numerical equivalence (which are known to form a semi-simple abelian category by a theorem of Jannsen). They should be the subcategory of semi-simple objects of MM. The "only" problem is that H^i only factors through them if you believe a substantial part of Grothendieck's standard conjectures. – Simon Pepin Lehalleur Nov 19 '14 at 16:09

Let me give a rather trivial remark: $K_0(\mathrm {Var}_k)$ doesn't remember any nontrivial extensions, but Voevodsky's category does. For instance, let $C$ be a smooth projective curve of positive genus, and let $x$ and $y$ be two points on $C$. Then $h^1(C \setminus \{x,y\})$ is a mixed motive (whatever that means) which depends nontrivially on the class of $[x]-[y]$ in $J(C)$. But the expression $[C] - [x]-[y] \in K_0(\mathrm{Var}_k)$ is obviously independent of the choice of $x$ and $y$.

Note that the Grothendieck ring of varieties does, at least conjecturally, remember some information about varieties that the category of motives does not.

Under the cut-and-paste conjecture, two varieties are equivalent in the Grothendieck ring if and only if they can both be decomposed into the same set of locally closed pieces. There are many pairs of varieties which have the same motive but cannot be cut and pasted into each other (like a $\mathbb P^2$ and a fake projective plane). So assuming the cut-and-paste conjecture, the Grothendieck group remembers a lot of extra information about the varieties.

In particular, because the fundamental group is a birational invariant for smooth projective varieties, the Grothendieck group class tells you the fundamental group, which is highly non-abelian invariant.

So the category of motives loses some information that is contained the Grothendieck group of varieties - in some sense, the non-abelian information. One reason that the category of motives is good to work with is that it simplifies things by getting rid of this additional structure. For tasks where you only need that structure, motives are more serious.

• One doesn't need the full cut-and-paste conjecture to get examples; by work of Liu and Sebag, cut-and-paste is known for varieties containing finitely many rational curves, and by work of Larsen-Lunts, the Grothendieck ring knows varieties up to "stable birational equivalence." Either of these facts let your fake projective plane example work unconditionally, or more simply let you e.g. distinguish isogenous but non-isomorphic elliptic curves (which $\mathbb{Q}$-motives can't do). – Daniel Litt Nov 14 '14 at 16:45