What is the intuition for \$\mathbb{Q}^{ab}\$ having cohomological dimension \$1\$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:36:44Z http://mathoverflow.net/feeds/question/97051 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97051/what-is-the-intuition-for-mathbbqab-having-cohomological-dimension-1 What is the intuition for \$\mathbb{Q}^{ab}\$ having cohomological dimension \$1\$? James D. Taylor 2012-05-15T20:57:46Z 2012-05-15T20:57:46Z <p>I frequently talk to people who think of finite fields as arithmetic analogs of punctured discs. This makes some sense: the absolute Galois group of a finite field is the profinite completion of \$\mathbb{Z}\$. Since the absolute Galois group of a field is its algebraic fundamental group, this gives the feel of punctured disc. </p> <p>The cohomological dimension of a field is the cohomological dimension of its absolute Galois group. Therefore the cohomological dimension of a finite field is \$1\$. This agrees with the intuition that finite fields are "like" punctured discs. (Given a projective curve \$C\$ over \$\mathbb{C}\$ and a point \$P\$ on \$C\$, a small punctured analytic neighborhood of \$P\$ has dimension \$1\$ in the sense that it is a neighborhood of a curve.)</p> <p>The picture gets murky when we get to \$\mathbb{Q}^{ab}\$. The absolute Galois group of \$\mathbb{Q}^{ab}\$ is not known, but is conjectured to be profinite free. It is known that the cohomological dimension of \$\mathbb{Q}^{ab}\$ is \$1\$. Is there some geometric intuition associated with \$\mathbb{Q}^{ab}\$? It is surely much more complex than a punctured disc, because its algebraic fundamental group (absolute Galois group) is more complicated. </p>