Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For reasons which are hard to articulate (due to they not being very clear in my mind), but having to do with the eprint From Matrix Models and quantum fields to Hurwitz space and the absolute Galois group by Robert de Mello Koch and Sanjaye Ramgoolam, I have been wondering whether there is a notion of integration over the algebraic numbers $\overline{\mathbb{Q}}$, as there is over the p-adic completions of the rationals.

share|improve this question
1  
Reading the comments below, it seems that people would like the algebraic numbers to have a nice topology first, and do measure theory compatibly with this topology then. Why don't we just look for a way to assign to each algebraic number a certain "weight" (perhaps somehow related to some Galois theoretic property of that algebraic number) and then just sum the series? –  Qfwfq Apr 19 '10 at 14:23
    
This is essentially what the Weil height (and other similar "height functions") on algebraic numbers does. –  Keenan Kidwell Apr 19 '10 at 15:19
add comment

2 Answers

up vote 18 down vote accepted

You can definitely talk about integration on $\overline{\mathbb{Q}}$. The question is "with respect to what measure?" The reason integration theory works so well over a completion of $\mathbb{Q}$ is that such a field is, as an additive group, locally compact, and so possesses a Haar measure (a non-zero, translation invariant Radon measure) which is in fact regular because any $p$-adic field is second countable (think of Lebesgue measure on $\mathbb{R}$). In less technical language, the completion has a canonical "nice" measure against which to integrate. The (additive) group of algebraic numbers is definitely not complete in any of its (many) valuations (obtained by embedding it into the algebraic closure of some $\mathbb{Q}_p$ or into $\mathbb{C}$), and therefore is not locally compact. The main tool for constructing interesting measures on topological spaces (that interact nicely with the topology) is the Riesz representation theorem, but this is specifically for locally compact Hausdorff spaces, so doesn't apply to the algebraic numbers. Really the reason people care about completions of $\mathbb{Q}$ and of number fields in general, is because the completions are amenable to lots of tools for analysis, whereas $\mathbb{Q}$ (and I guess also $\overline{\mathbb{Q}}$) just isn't. Completing fields allows one to bring analysis to the study of, say, number-theoretic problems.

Now, in another direction, integration definitely works on the Galois group of $\mathbb{Q}$, $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$. This is a profinite group (meaning it is compact and totally disconnected) in its natural topology, and so, like the completions of number fields, admits a Haar measure.

In yet another direction, one can view the multiplicative group of algebraic numbers, $\overline{\mathbb{Q}}^\times$, modulo torsion (roots of unity) as a (multiplicative) vector space over $\mathbb{Q}$. J. Vaaler and D. Allcock sort of initiated the study of this group (as such) and worked out its completion as a certain subspace of $L^1(Y,\mu)$, where $Y$ is a totally disconnected, locally compact Hausdorff space whose underlying set is related to the places of $\mathbb{Q}$ and where $\mu$ is a Radon measure (got using the theorem I mentioned above) that satisfies some kind of relative invariance with respect to the absolute Galois group of $\mathbb{Q}$.

I guess maybe I went way off topic with this. My point is that interesting measures on $\overline{\mathbb{Q}}$ are maybe difficult to come by, and without an interesting measure, there's no reason to do integration.

share|improve this answer
2  
Here's a link with the paper of Vaaler and Allcock: ma.utexas.edu/users/allcock I should have said that one gets a norm on $\overline{\mathbb{Q}}^\times/\text{torsion}$ from the Weil height...then it makes sense to ask about its completion. –  Keenan Kidwell Apr 19 '10 at 1:47
    
Interesting. Thanks for the detailed answer the link to the paper. Perhaps when I've a clearer idea of what it is that I'd like to do with this, I will post a more detailed and hopefully better motivated question. –  José Figueroa-O'Farrill Apr 19 '10 at 1:51
add comment

Pick an algebraic number at random. What is its expected degree? As noted, since the set is countable, we cannot expect to do conventional integrals on it. But we still have things like this: pick a natural number at random---then it is square-free with probability $6/\pi^2$.

So, maybe the question is: What are "natural" Folner sets for the algebraic numbers? The usual Folner sets for $\{1,2,3,\dots\}$ are the sets $\{1,2,\dots,n\}$, which define the "density" (not an actual measure) $$ \delta(A) = \lim_{n\to\infty}\frac{\\#(A \cap \{1,\dots,n\})}{n} $$ for sets $A \subseteq \{1,2,\cdots\}$. See, for example, http://en.wikipedia.org/wiki/F%F8lner%5Fsequence for information on Folner sequences.

share|improve this answer
2  
The fact that the set of algebraic numbers is countable is not what precludes applying the usual integration machinery to it. The additive group of integers is a perfectly good (countable) locally compact group and people do Fourier analysis on it all the time. The problem with $\overline{\mathbb{Q}}$ is that it fails to be locally compact in any of its usual topologies. –  Keenan Kidwell Apr 19 '10 at 11:59
1  
@KK: As with the integers, you could put the discrete norm on $\overline{\mathbb{Q}}$ and then it becomes locally compact and complete, with natural (Haar) measure being the counting measure. This feels like a rather trivial answer, but I can't tell from the question exactly what Jose has in mind. –  Pete L. Clark Apr 19 '10 at 14:32
    
@Pete Yeah, I thought about that as soon as I posted my response. I'm also not really sure what Jose has in mind, but it's good to point this out, that there is one obvious locally compact topology on $\overline{\mathbb{Q}}$. –  Keenan Kidwell Apr 19 '10 at 14:47
add comment

Your Answer

 
discard

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.