Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Fix a base ring $R$, and consider pairs $(X,f)$ where $X$ is a scheme of finite type over $R$ and $f:X\to R$ is an $R$-valued algebraic (not constructible!) function on $X$.

I want to consider a Grothendieck $R$-algebra of such pairs, where if $X = Y \coprod Z$, then $[(X,f)] = [(Y,f|_Y)] + [(Z,f|_Z)]$, but also $[(X,f+g)] = [(X,f)] + [(X,g)]$ and $[(X,rf)] = r[(X,f)]$.

Surely this is a standard extension of the usual notion of the Grothendieck ring of varieties (which only has $f=1$, and the first sort of relation)? If so, where can I read about it?

Maybe I'm misreading the motivic integration survey literature (by K. Smith, and E. Looijenga), but it seems like they're insisting on constructible functions, not algebraic. Ordinarily when a construction like this isn't in the literature, I assume it's because it has too many relations and is $0$, but if $R = {\mathbb Z}$ it seems to me that this ring has many functionals, like $[(X,f)] \mapsto \sum_{x \in X_p} (f(x) \bmod p) \in {\mathbb Z}/p.$ (I don't see an analogue of $[X] \mapsto$ the Euler characteristic $\chi(X_{\mathbb C})$.)

EDIT: One problem I see is that $({\mathbb A}^1, f(x)=x)$ is isomorphic under translation to $({\mathbb A}^1, f(x)=x+1)$. So $[({\mathbb A^1}, 1)] = [({\mathbb A}^1, (x+1)-x)] = [({\mathbb A}^1, x+1)] - [({\mathbb A}^1, x)] = 0$. Of course this fits with point-counting $\bmod p$.

share|cite|improve this question
$[(X,f)]*[(Y,g)] := [(X\times Y,fg : (x,y) \mapsto f(x)g(y)]$ –  Allen Knutson Sep 25 '12 at 21:59
When you write $X=Y\coprod Z$, do you assume disjoint union, or may $Y$ be closed and $Z$ the complementary open subset? If it is so, as I would believe, then this extension is fairly standard. Note that in your ring, you will automatically get pairs $(X,f)$, where $f$ is constructible, but such pairs are of course generated by elements of the form $(X,f)$ with $f$ regular. –  ACL Sep 26 '12 at 0:59
Concerning your last question, if $R\to S$ is a morphism of rings, there is a ring morphism from $K_0(Sch/R)$ to $K_0(Sch/S)$. If you take $R=\bf Z$ and $S=\bf C$, you see that $K_0(Sch/{\bf Z})$ has an Euler characteristic. But considering $S=\overline{{\bf F}_p}$ and étale $\ell$-adic cohomology, it has also many other ones... –  ACL Sep 26 '12 at 1:02
I assume disjoint-but-not-topologically-disjoint, as you say. (Do you have a reference in which this extension is defined or used?) I guess what you're saying in your first comment is that I could extend the definition beyond algebraic functions to constructible ones; sure. But I don't understand your last comment, since I'm not dealing with $K_0(Sch/{\mathbb C})$, but $K_0($ schemes with functions $)$; I still don't know what the "Euler characteristic" of ${\mathbb A}^1$ carrying the function $f(x)$ should be. –  Allen Knutson Sep 26 '12 at 18:48
Another problem in addition to your edit is that if $R$ contains $\mathbb{Q}$, then dilation invariance implies $(\mathbb{A}^1, x^n) = 2^n(\mathbb{A}^1, x^n) = 0$ for all $n>0$. By additivity, $(\mathbb{A}^1,f) = 0$ for all polynomials $f$. I imagine you get a similar collapse for non-constant functions on toric varieties. –  S. Carnahan Sep 27 '12 at 5:41

1 Answer 1

To answer your question is a bit difficult because anytime you sum elements of something built out of the grothendieck ring of varieties you are doing (or attempting to do) motivic integration. Pairs $\phi = (X, f)$ certainly already exist in the literature (well actually one should consider the tensor product). But, note that you want $f$ to be a function in $n\in \mathbb{Z}$. Thinking of $f$ being algebraic to your base doesn't add anything because that part can be moved over to the 1st factor in the tensor product via the characteristic function on the graph. In fact, I think it is generally accepted that $f$ should be a function $$f : X(K) \to \mathbb{Z}[\mathbb{L}, \mathbb{L}^{-1}, (\mathbb{L}^{i} -1)_{i\in\mathbb{N}}^{-1}]$$ where $K$ is a field, and $X(K)$ is definable in the three sorted language of Denef-Pas and where $X$ above lives over $S$ in some suitable sense (factors through some projection $S\times \mathbb{A}^{n}\rightarrow S$).
The reason to use the language denef-pas is so that when the summation of a function $\phi(s,n)$ over $n$ exists, then there is in fact a function of the form $I(\phi)$ which can legitimately be called its integral. In other words, the measure exists and generating series will become rational.

Perhaps you already know of this article: which lays the foundation for both geometric and arithmetic integration and connects it partly to motivic integration over formal schemes.

Personally, I think what might be interesting here is trying to develop something non-commutative by taking pairs (X,f) where f is an endomorphism of X. But, to get something non-commutative, you cannot just take an extension of the usual grothendieck ring of varieties -- the multiplicative structure on such pairs needs to correspond with composition of functions.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.