The n-Category Café

Well, firstly, if G is a compact ring, then the connected component of the identity is also a compact ring, so the basic question is why there are no non-trivial compact connected rings.

The Peter-Weyl theorem implies that every compact group is the inverse limit of Lie groups (or in fact linear groups). In the case of connected abelian groups, they are the inverse limit of tori. So just from the additive structure, a compact connected ring is an inverse limit of tori.

Now, tori are rigid; they don’t have any endomorphisms near the identity. (By Pontryagin duality, this is equivalent to the discreteness of the Pontryagin dual.) But multiplication on a ring by an element close to 1 is an endomorphism. So there are no non-trivial elements close to 1, and so the ring is trivial.

So I think the main ingredients here are (a) compact connected abelian groups are basically tori (in particular, they have much less variability than arbitrary compact groups), and (b) tori are basically rigid.

Not to belabor this. But looking over the past comments, might this be a satisfactory way of summarizing the argument?

Proposition: Every compact Hausdorff ring $R$ is totally disconnected.

Proof: Let $\hat{R}$ be the Pontryagin dual of (the additive group of) $R$; this $\hat {R}$ is discrete. As $R$ and $\hat{R}$ are locally compact Hausdorff, they are exponentiable as spaces, and it follows quickly that the functorial maps

$$Hom(R, R) \to Hom(\hat{R}, \hat{R}), \qquad Hom(\hat{R}, \hat{R}) \to Hom(\hat{\hat{R}}, \hat{\hat{R}}) \cong Hom(R, R)$$

are continuous homomorphisms of topological groups. They are mutually inverse, and so they give an isomorphism of topological groups.

The multiplication map $R \times R \to R$ transforms to a continuous injection $i: R \to Hom(R, R) \cong Hom(\hat{R}, \hat{R})$. As $R$ is compact and $Hom(\hat{R}, \hat{R})$ is Hausdorff, the injection $i$ maps $R$ homeomorphically onto its image in $Hom(\hat{R}, \hat{R})$, i.e., is a subspace embedding. But $Hom(\hat{R}, \hat{R})$ is manifestly a subspace of a product of discrete spaces and hence totally disconnected, so $R$ is totally disconnected as well. (In more detail: if $C \subseteq R$ is connected, then the image of $C$ under the composite

$$C \subseteq R \stackrel{i}{\to} Hom(\hat{R}, \hat{R}) \hookrightarrow \prod_{x \in \hat{R}} \hat{R} \stackrel{proj_x}{\;\;\to\;\;} \hat{R}$$

is also connected and hence a one-point space for each $x \in \hat{R}$, so that $i(C)$ and therefore $C$ are one-point spaces.)

Considering all the aforementioned contributions and meaning-fying previous remarks , we may comment correspondence mainstreams in the article by Jan Dobrowolski and Krupiński in J Algebra 401 , 161-178 ( 2014 ) about the denotation of small compact G-rings , allowing the description of possible interactions of G on the underlying ring.

Very nice proofs! I had the opposite reaction to Tom and thought that this should be true.. I’d like to go a little further in the classification. Suppose that we start with a compact (commutative) Hausdorff ring. Then we know that it is totally disconnected. There are numerous examples of this. Take any non-trivial commutative ring $R$ and form the $p$-adic topology using a prime ideal $p$. My question is: are these the only possible examples? I mean, if we start with a compact commutative Hausdorff ring $R$, does there exist a ideal $I$ such that $R$ can be identified as a subring of the completion of $R$ by $I$?

I had never thought about this before, but it seems completely natural to me. (And this was just when I had only read the part above the fold, visible from this blog's main page.)

The main examples of nontrivial rings that I know are finite (hence compact but also totally disconnected) or locally compact (and connected but not compact)[^1]. I quickly realized that the finite ones generalize to profinite ones (still totally disconnected), and I was a little disappointed that I couldn't come up with any other compact ones; so having read here that the only compact ones are profinite, I'm actually feeling pretty good about my familiarity with rings.

Well, you can get profinite spaces from finite ones, but you can also get compact spaces from locally compact ones. So with so many locally compact rings around, why can't we complete them to compact rings? But this is the old problem of learning to compute with infinity, and this doesn't give us a ring, because we can't handle $\infty - \infty$. So we can't turn locally compact rings into compact ones, and in the end I'm not at all surprised that the only compact rings are totally disconnected.

On the other hand, you can easily calculate with infinity if you give up subtraction. The nontrivial semiring $[0,\infty]$ (with the lower semicontinuous topology) is compact and connected, and there are many others in this vein. (Actually, maybe infinity is not the point here. Even ${[0,\infty[}$ is compact and connected, as long as you use the lower semicontinuous topology. That's because the only neighbourhood of $0$ in that topology is the entire space. Infinity merely serves to motivate that topology, since it is needed to make multiplication by infinity continuous.)

[^1]: I'm just taking it for granted that everything's Hausdorff.

For no particularly compelling reason, I decided to write up an article on the result cited in this post at the nLab, here.