February 19th
Today I learned about the topology of profinite groups. A profinite group is the inverse limit of a family of finite groups; that is, for a directed set (proset where finite subsets are upper-bounded) $\mathcal I$ and a contravariant functor $F$ from $\mathcal I$ to $\texttt{Grp}$ with image only finite groups, we take\[G:=\varprojlim_{\mathcal I}F(\bullet)\]as our profinite group. In practice, we realize $G$ by writing $G_k:=F_k$ for $k\in\mathcal I$ and $\varphi_{\ell k}:G_\ell\to G_k$ for the morphism when $k\le\ell.$ Then we can define $G$ by coherent sequences of the $G_\bullet$ by writing\[G:=\left\{(g_\bullet)\in\prod_\mathcal IG_\bullet:k\le\ell\implies\varphi_{\ell k}(g_k)=g_\ell\right\}.\]Note the morphisms $\varphi_{\ell k}$ are best read from right to left. Our prototypical example of this construction is $\ZZ_p.$ We remark that it is a fact that this $G$ is non-empty when the $G_\bullet$ is non-empty, but this is nontrivial to prove.
Now we provide a topology to this. The $G_\bullet$ are finite (nonempty) groups, so we give them the discrete topology. Then we note our coherent sequences lets us think\[G\subseteq\prod_\mathcal IG_\bullet,\]so we first give $\prod_\mathcal IG_\bullet$ the product ("finite gate'') topology, and second give $G$ the subspace topology from this.
From this presentation, the topology on $G$ feels somewhat contrived, but it has some nice properties. Doing fact collection, we quickly remark that $G$ is Hausdorff because it is the subspace of a Hausdorff space, where $\prod_\mathcal IG_\bullet$ is Hausdorff because it is the product of Hausdorff spaces. Additionally, $G$ is closed in $\prod_\mathcal IG_\bullet$ because its complement is\[G^c=\left\{(g_\bullet)\in\prod_\mathcal IG_\bullet:\exists k,\ell,k\le\ell,\varphi_{\ell k}(g_k)\ne g_\ell\right\},\]which rearranges to\[G^c=\bigcup_k\bigcup_{k\le\ell}\left\{(g_\bullet)\in\prod_\mathcal IG_\bullet:\varphi_{\ell k}(g_k)\ne g_\ell\right\}.\]This last set is open because it is the union of various $\{(g_\bullet):\varphi_{\ell k}(g_k)\ne g_\ell\},$ a condition which only focuses on $G_k$ and $G_\ell.$ In particular, the $G_k$ and $G_\ell$ components of this set will be ugly—but open because their topology is discrete—while the other $G_\bullet$ components have all $G_\bullet.$ This means our sets are open in the product topology, so $G^c$ is open, so $G$ is closed.
Finally, our final piece of fact collection is that $G$ is compact, for it is a closed set in a compact space, where $\prod_\mathcal IG_\bullet$ is compact by Tychonoff's theorem. I don't actually know a proof of either of these facts (being closed means compact or Tychonoff's), but it won't affect my sleep schedule.
Anyways, what's really nice about profinite groups is that we have an equivalent topological definition: a topological group is profinite if and only if it is compact and totally disconnected. As usual, think $\mathbb Z_p.$ We recall that a space is connected if and only if it cannot be decomposed into two disjoint open subsets, and totally disconnected means that points are the largest connected subspaces.
Currently I know only one direction of this proof currently, where $G$ being profinite implies totally disconnected. Our definition of totally disconnected focuses on open sets, so it suffices to look only at the connected component of the identity $e\in G.$ Name this component $G^\circ,$ and we would like to show that $G^\circ=\{e\}.$
Quickly, we note it is generally true that $G^\circ$ is a subgroup of $G,$ in any topological group. Indeed, $g\in G^\circ$ gives $g^{-1}G^\circ$ connected and containing $e,$ so\[g^{-1}\in g^{-1}G^\circ\subseteq G^\circ,\]so $G^\circ$ is closed under inverses. Doing the same for $gG^\circ$ tells us $G^\circ$ is closed under products as well, so $G^\circ$ is indeed a subgroup.
Returning to the proof, the key trick here is to look at open subgroups. In particular, open subgroups of $G$ we expect to have "substance,'' in that because multiplication is continuous, we expect to be able to "wiggle'' $e$ around inside of an open subgroup. So intuitively, we expect $G^\circ$ to be inside every open subgroup. To show this, we use the fact $G^\circ$ is connected: fix any open subgroup $U,$ and look at\[V=\bigcup_{x\in G^\circ\setminus U}x(G^\circ\cap U).\]Note $V$ is open in $G^\circ$ because $G^\circ\cap U$ is open in $G^\circ$; in fact $V\subseteq G^\circ$ because $G^\circ$ is a subgroup. Because $e\in G^\circ\cap U,$ we see $G^\circ\setminus U\subseteq V,$ so $G^\circ$ splits into open sets $G^\circ\cap U$ and $V.$ And finally, $G^\circ\cap U$ and $V$ are disjoint because if $gu\in V$ with $u\in G^\circ\cap U$ also has $gu\in G^\circ\cap U,$ then $g\in G^\circ\cap U.$
Finishing up here, $G^\circ\cap U$ and $V$ decompose $G^\circ$ into disjoint open sets, so one of them must be empty. Certainly $e\in U,$ so $V$ must in fact be empty. But this requires $G^\circ\setminus U$ to be empty, so indeed $G^\circ\subseteq U.$
Now we know $G^\circ$ to be quite small, inside of every open subgroup of $G.$ This is not quite good enough to deduce that $G^\circ=\{e\}$—we haven't even used the fact $G$ is profinite yet! But we can finish now. Fix any $g\in G\setminus\{e\},$ which we label $(g_\bullet)$ to get us thinking profinitely. We want to know $g\notin G^\circ,$ which we can show $x$ not in some open subgroup of $G.$
Well, $(g_\bullet)\ne e$ cannot be the identity on all coordinates, so fix $g_k\ne e_k\in G_k.$ Then we look at\[U=\{(x_\bullet)\in G:x_k=e_k\}.\]Note $U$ is open, for all but the $k$th coordinate is $G_\bullet,$ and $\{e_k\}$ is an open set in $G_k.$ Additionally, being the identity on one coordinate is a property closed under inversion and multiplication, so $U$ is also a subgroup. Thus, $G^\circ\subseteq U,$ but $U$ doe not contain $g,$ so $G^\circ$ doesn't either. This completes the proof.