Is there a notion of integration over the algebraic numbers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T08:27:20Z http://mathoverflow.net/feeds/question/21800 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21800/is-there-a-notion-of-integration-over-the-algebraic-numbers Is there a notion of integration over the algebraic numbers? José Figueroa-O'Farrill 2010-04-19T01:12:40Z 2010-04-19T11:57:59Z <p>For reasons which are hard to articulate (due to they not being very clear in my mind), but having to do with the eprint <a href="http://arxiv.org/abs/1002.1634" rel="nofollow"><em>From Matrix Models and quantum fields to Hurwitz space and the absolute Galois group</em></a> 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.</p> http://mathoverflow.net/questions/21800/is-there-a-notion-of-integration-over-the-algebraic-numbers/21803#21803 Answer by Keenan Kidwell for Is there a notion of integration over the algebraic numbers? Keenan Kidwell 2010-04-19T01:45:36Z 2010-04-19T01:45:36Z <p>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. </p> <p>Now, in another direction, integration definitely works on <em>the Galois group</em> 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. </p> <p>In yet another direction, one can view the <em>multiplicative group</em> 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 $\mathbb{Q}$. </p> <p>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. </p> http://mathoverflow.net/questions/21800/is-there-a-notion-of-integration-over-the-algebraic-numbers/21834#21834 Answer by Gerald Edgar for Is there a notion of integration over the algebraic numbers? Gerald Edgar 2010-04-19T11:52:49Z 2010-04-19T11:57:59Z <p>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 $6/\pi^2$.</p> <p>So, maybe the question is: <strong>What are "natural" Folner sets for the algebraic numbers?</strong> 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, <a href="http://en.wikipedia.org/wiki/F%F8lner%5Fsequence" rel="nofollow">http://en.wikipedia.org/wiki/F%F8lner%5Fsequence</a> for information on Folner sequences.</p>