I was describing Manish Kumar's work a few weeks ago to a fellow graduate student, and she stumped me with a big-picture question I couldn't answer.
Manish Kumar proved that the commutator subgroup of $\pi_1(\mathbb{A}^1_K)$, where $K$ is a characteristic $p$, algebraically closed field, is pro-finite free. He proved this, in fact, for any smooth affine curve over $K$.
(He proved this for algebraically closed fields of char $p$ which are uncountable in his thesis: http://www.math.msu.edu/~mkumar/Publication/thesis.pdf; and without the cardinality restriction in: http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.4472v2.pdf)
As I explained to my colleague, this is a geometric analogue of Shafarevich's conjecture, that $Gal(\mathbb{Q}^{ab})$ is pro-finite free. Indeed $Gal(\mathbb{Q}^{ab})=\pi_1^c(Spec(\mathbb{Q}))$ ($c$ stands for the commutator subgroup). But why is $\mathbb{A}^1_K$, for $K$ an algebraically closed characteristic $p$ field (or indeed, any smooth affine curve over $K$), an analogue of $Spec(\mathbb{Q})$? Usually $\mathbb{A}^1_K$ (for $K$ an algebraically closed field) is an analogue of $Spec(\mathbb{Z})$. I came up with some partial explanations, but no full heuristic. Can you think of one?
Best Answer
One can see that the commutator subgroup of the topological fundamental group of a complex curve is free for elementary reasons, but this is a pretty weak analogy. In fact I don't have a good heuristic of why it should be true, other than the fact that is. I was Manish's adviser, and I was pretty surprised by the result when he proved it.