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Posts Tagged ‘profinite groups’

Previously we looked at several examples of n-ary operations on concrete categories (C, U). In every example except two, U was a representable functor and C had finite coproducts, which made determining the n-ary operations straightforward using the Yoneda lemma. The two examples where U was not representable were commutative Banach algebras and commutative C*-algebras, and it is possible to construct many others. Without representability we can’t apply the Yoneda lemma, so it’s unclear how to determine the operations in these cases.

However, for both commutative Banach algebras and commutative C*-algebras, and in many other cases, there is a sense in which a sequence of objects approximates what the representing object of U “ought” to be, except that it does not quite exist in the category C itself. These objects will turn out to define a pro-object in C, and when U is pro-representable in the sense that it’s described by a pro-object, we’ll attempt to describe n-ary operations U^n \to U in terms of the pro-representing object.

The machinery developed here is relevant to understanding Grothendieck’s version of Galois theory, which among other things leads to the notion of ├ętale fundamental group; we will briefly discuss this.

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An interesting result that demonstrates, among other things, the ubiquity of \pi in mathematics is that the probability that two random positive integers are relatively prime is \frac{6}{\pi^2}. A more revealing way to write this number is \frac{1}{\zeta(2)}, where

\displaystyle \zeta(s) = \sum_{n \ge 1} \frac{1}{n^s}

is the Riemann zeta function. A few weeks ago this result came up on math.SE in the following form: if you are standing at the origin in \mathbb{R}^2 and there is an infinitely thin tree placed at every integer lattice point, then \frac{6}{\pi^2} is the proportion of the lattice points that you can see. In this post I’d like to explain why this “should” be true. This will give me a chance to blog about some material from another math.SE answer of mine which I’ve been meaning to get to, and along the way we’ll reach several other interesting destinations.

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