### Kim on Fundamental Groups in Number Theory

#### Posted by John Baez

My friend Minhyong recently wrote up a talk he gave at Leeds:

- Minhyong Kim, Fundamental groups and Diophantine geometry.

It starts with some pleasant observations of an elementary nature and works its way up to some ideas I find rather terrifying. Maybe we can ask him some questions and get him to explain what’s going on.

Apart from the rather intimidating concepts involved, I think part of my problem is that I’d need to look at some examples to get a feeling for what’s going on. And, I can’t tell if these examples will be comprehensible in full detail by mere mortals… or whether they’ll be big, unfathomable things.

For example, Minhyong talks about how the absolute Galois group of $\mathbb{Q}$,

$G = Gal(\overline{\mathbb{Q}}|\mathbb{Q}),$

acts on the ‘pro-finite étale fundamental group’

$\pi_1(X(\mathbb{C}))\widehat{} \cong \pi_1^{et}(\overline{X})$

where $X$ is an algebraic variety over $\mathbb{Q}$, $X(\mathbb{C})$ is the corresponding variety over $\mathbb{C}$, and $\overline{X}$ is the corresponding variety over the algebraic closure of $\mathbb{Q}$ — I think.

Now, I know the group $G$ is famously hard to understand in detail. But, to what extent can we compute this ‘pro-finite étale fundamental group’ in some simple but nontrivial examples, and to what extent can we understand the action of $G$?

For example, are there any examples where only the abelianized version of $G$ matters? That’s a group I know and love.

## Re: Kim on Fundamental Groups in Number Theory

The description of the etale fundamental groups is rather explicit for smooth algebraic curves over $\bar{\mathbb{Q}}$ because of the isomorphism$\pi_1(X(\mathbb{C}),b)^\simeq \pi_1^{et}(\bar{X},b)$that you’ve referred to. For example, if X is a smooth compact curve of genus g minus one point, then it will be the free pro-finite group on g generators. It is notable that there is no known way to prove even that $\pi_1^{et}(\bar{X})$ is finitely-generated using just algebraic techniques. In any case, the Galois action is very hard to access in most cases.

The best way to understand the Galois action was described in the lecture, namely, via the isomorphism $\pi_1^{et}(\bar{X},b)\simeq \tilde{X}_b,$ where the right hand side refers to the fiber over $b$ of the system of finite covers that come together to form the pro-finite universal object $\tilde{X}$. So what you would try to do in practice is to guess the system over $\mathbb{C}$ geometrically, then try to find a model over $\mathbb{Q}$ for the system. This model, which can be inaccessible in general, is what allows us to compute the Galois action. If $X=\mathbb{A}^1\setminus \{0\}$ and the base-point is $b=1$, things are quite simple because then $X(\mathbb{C})=\mathbb{C}^*$, so you can guess that the pro-finite universal cover is the system $\tilde{X}:=\{X_n \rightarrow X\}_n$where $X_n\to X$ just refers to the $n$-th power map from $X$ to itself. To prove this fact, i.e., that this system has the right universal property, still requires a nice bit of elementary algebraic geometry. Anyways, an element of $\tilde{X}_b$ is a compatible collection of roots of unity. This is often denoted $\hat{\mathbb{Z}}(1)$, group-theoretically isomorphic to $\hat{\mathbb{Z}}$, the pro-finite completion of $\mathbb{Z}$. The action in this case exactly factors through $G^{ab}\simeq \mathbb{Z}^*$ in the natural way.

The annoying thing is that even here, when we view $\pi_1^{et}(\bar{X},b)$ as a sheaf over $Spec(\mathbb{Q})$, the classification of *torsors* still involves the whole group $G$, or at least a non-abelian quotient of it. I can explain this further as needed, but to get a sense of this, consider a usual manifold $B$ and the constant sheaf given by some group $L$. The principal $L$-bundles on $B$ are classified by $H^1(\pi_1, L)$, which for the given trivial action is just $Hom(\pi_1,L)$ modulo the conjugation action. That is to say, the action of $\pi_1$ on $L$ even factors through the *trivial* group. But the classification of bundles still reflects the structure of the fundamental group in a complicated way. Of course the structure of the Galois group $G$ is mysterious, and hence, the difficulty of classifying bundles. This captures pretty well the main issues I’m struggling with over $\Spec(\mathbb{Q})$.

In general, the geometric picture to keep in mind is a fiber bundle $X$ over some base space $B$, and the action of $\pi_1(B,y)$ on $\pi_1(X_y)$. So the ‘analogy,’ in this case is $Spec(\mathbb{Q}) \leftrightarrow B$ $X over Spec(\mathbb{Q}) \leftrightarrow the total space X$ $\bar{X} \leftrightarrow X_y$

Exercise: What is $y$?