# If Spec Z is like a Riemann surface, what's the analogue of integration along a contour?

Rings of functions on a nonsingular algebraic curve (which, over $\mathbb{C}$, are holomorphic functions on a compact Riemann surface) and rings of integers in number fields are both examples of Dedekind domains, and I've been trying to understand the classical analogy between the two. As I understand it, $\operatorname{Spec} \mathcal{O}_k$ should be thought of as the curve / Riemann surface itself. Is there a good notion of integration in this setting? Is there any hope of recovering an analogue of the Cauchy integral formula?

(If I am misunderstanding the point of the analogy or stretching it too far, please let me know.)

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For a variety $X$ over a finite field, I guess one can take $\ell$-adic sheaves to replace differential forms. Then the local integral around a closed point $x$ (like integral over a little loop around that point) is the trace of the local Frobenius $\operatorname{Frob}_x$ on the stalk of sheaf, the so-called naive local term. Note that $\operatorname{Frob}_x$ can be regarded as an element (or conjugacy class) in $\pi_1(X)$, "a loop around $x$". The global integral would be the global trace map $$H^{2d}_c(X,\mathbf{Q}_{\ell})\to\mathbf{Q}_{\ell}(-d),$$ and the Tate twist is responsible for the Hodge structure in Betti cohomology (or the $(2\pi i)^d$ one has to divide by). The Lefschetz trace formula might be the analog of the residue theorem in complex analysis on Riemann surfaces.

For the case of number fields, each closed point $v$ in $\operatorname{Spec} O_k$ still defines a "loop" $\operatorname{Frob}_v$ in $\pi_1(\operatorname{Spec} k)$ (let's allow ramified covers. One can take the image of $\operatorname{Frob}_v$ under $\pi_1(\operatorname{Spec} k)\to\pi_1(\operatorname{Spec} O_k)$, but the target group doesn't seem to be big enough). For global integral, there's the Artin-Verdier trace map $H^3(Spec\ O_k,\mathbb G_m)\to\mathbb{Q/Z}$ and a "Poincaré duality" in this setting, but I don't know if there is a trace formula. The fact that 3 is odd always makes me excited and confused.

So basically I think of trace maps (both local and global) as counterpart of integrals. Correct me if I was wrong.

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This is interesting, but a little over my head. If you don't mind, could you explain how one thinks about the contour integral in this abstract setting, i.e. in terms of the fundamental group? –  Qiaochu Yuan Nov 4 '09 at 19:58
A loop in this setting is a map from a scheme with a "cyclic" fundamental group. Finite fields have this property, so their spectra can be viewed as circles. For Spec Z, the only interesting loops we see are the canonical maps from spectra of finite fields. One has an analogy between integration of differentials and parallel transport along connections, so we are determining how a vector bundle (our l-adic sheaf) is transformed as we follow flat sections around a "circle". –  S. Carnahan Nov 5 '09 at 2:37
"The fact that 3 is odd always makes me excited and confused." Fantastic. –  Cam McLeman May 8 '11 at 18:56

I think that you have understood the analogy correctly, and you have pinpointed one of its weaknesses. Although number fields are like one dimensional functional fields in many ways, one of the differences is that the vector space of Kahler differentials for a number field has dimension 0, not 1. Here Kahler differentials are the vector space generated by symbols $dx$, subject to the relations $d(x+y) = dx + dy$ and $d(xy) = x dy + y dx$.

Therefore, there is nothing like differentials, and nothing we can integrate.

But it is possible that there is some more sophisticated way to solve this problem. (Maybe using Arakelov geometry?) I'm looking forward to reading the other answers.

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This certainly does have a bit of the Arakelov smell about it. Maybe if you include the infinite places there's some different notion of Kahler differential? –  Ben Webster Nov 4 '09 at 18:40
That makes sense; it's related to the fact that the abc conjecture is much harder than the Mason-Stothers theorem. I think a lot of people would want to see a notion of differential in the number field setting for this reason. –  Qiaochu Yuan Nov 4 '09 at 18:45
I think David's module of differentials on R should include the rule that dc = 0 when c is a constant: here the ring of constants is some specified subring R_0 of R. (So e.g. when R = C[t] one takes R_0 = C; otherwise, the differentials are way bigger than you want.) So we see the problem: David's module is Omega_{Z/Z}, and the relative dimension is 0, while for Omega_{C[t]/C} the relative dimension is 1. In order to have a good notion of differentials you'd need a "field of constants" "inside" Spec Z, which would have... oh, I dunno.... one element? –  JSE Nov 5 '09 at 1:31
Yup! JSE's comment is exactly right, and gives a great way of seeing why we want a field with one element. –  David Speyer Nov 5 '09 at 1:38
Lichtenbaum talked about that and interesting applications last December: institut.math.jussieu.fr/projets/tn/STN/11/lichtenbaum-11.pdf –  Thomas Riepe May 8 '11 at 15:05

For a number $f$ the local-global formula written as

$$f_2\cdot f_3\cdot \dotsb \cdot f_{\mathbb Q} = 1$$

where $f_p$ is the inverse power of $p$ that is equal to $f$ in the local field of $p$-adic numbers and $f_{\mathbb{Q}}$ is $f$ should provide a reasonable analogue (in other words, integration over all holes gives $0$ on a closed surface).

(An example for $f = 75: 1 \times 1/3 \times 1/25 \times 1 \times \dotsb \times 75 = 1$)

A better formulation would involve adeles: the adele ring $\mathbb{A}_{\mathbb Q}$ is a semi-restricted product of $p$-adic rationals and rationals themselves. There is a map $\mathbb Q\to\mathbb A_{\mathbb Q}$ and every element in the image has the property that the product over all places gives $1$.

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I think we have to be careful about the two senses of the word "residue" here. Either way my hope was that one could recover the residue as part of a more general computation. –  Qiaochu Yuan Nov 4 '09 at 18:43
I'm rewriting the answer and making it better. I remember things one by one, but there should definitely be a good answer! –  Ilya Nikokoshev Nov 4 '09 at 18:54
I take the product formula as an analog of the fact that a principal divisor on a curve has degree 0, so it is a (at least special case of the) residue theorem: the local integral at all points add up to the integral over the boundary, and there's no boundary if we also include \infty! –  shenghao Nov 4 '09 at 19:59
That's absolutely what I mean, yes. –  Ilya Nikokoshev Nov 4 '09 at 20:56