The n-Category Café

Are we secretly saying that there’s exactly one way to make a finite-dimensional real vector space into a compact Hausdorff topological $\mathbb{R}$-module, where $\mathbb{R}$ is given its discrete topology?

This is weird but fun stuff!

I don’t think you can get a finite-dimensional vector space over $\mathbb{R}$ to be compact Hausdorff (except for $\{0\}$)! The compact Hausdorff ones are of the form

$$Hom(V, S^1)$$

by Pontryagin duality, where $V$ is a discrete vector space. The smallest nonzero example would be where $V = \mathbb{R}$, and already $Hom(\mathbb{R}, S^1)$ has uncountable dimension as a vector space over $\mathbb{R}$.

It’s hard or impossible to visualize this compact space, but it sits as a closed subspace in a product of continuum many copies of $S^1$:

$$Hom(\mathbb{R}, S^1) \hookrightarrow \prod_{r \in \mathbb{R}} S^1$$

It’s as if in order to get a vector space to be compact, the vector space structure has to be “smeared out” across many (continuum many) dimensions.

Incidentally, it was Mike Barr that I thought I had heard say that thing about Pontryagin duality you mentioned – but he was talking to someone else at the time, not me, and my memory may be playing tricks. We should look this stuff up.

Todd wrote:

I don’t think you can get a finite-dimensional vector space over $\mathbb{R}$ to be compact Hausdorff (except for $\{0\}$)! The compact Hausdorff ones are of the form

$$Hom(V, S^1)$$

by Pontryagin duality, where $V$ is a discrete vector space. The smallest nonzero example would be where $V = \mathbb{R}$, and already $Hom(\mathbb{R}, S^1)$ has uncountable dimension as a vector space over $\mathbb{R}$.

It’s hard or impossible to visualize this compact space, but it sits as a closed subspace in a product of continuum many copies of $S^1$:

$$Hom(\mathbb{R}, S^1) \hookrightarrow \prod_{r \in \mathbb{R}} S^1$$

Oh, duh — I get it.

I guess it suffices to take a product over $r$ lying in a Hamel basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$. That simplifies things enormously, and makes this gadget vastly easier to visualize.

How is this gadget related to Bohr compactification of $\mathbb{R}$? It smells vaguely similar… but the Bohr compactification depends on the topology of $\mathbb{R}$, which is usually taken not to be discrete.

How about the Bohr compactification of the discrete topological group $\mathbb{R}$?

I guess it suffices to take a product over r lying in a Hamel basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$. That simplifies things enormously, and makes this gadget vastly easier to visualize.

Yes indeed, for as we know from that interesting letter from Kevin Buzzard (reproduced at the end of Week 273), the Pontryagin dual of $\mathbb{Q}$ is the puny little group $\mathbb{A}/\mathbb{Q}$, the adele group of $\mathbb{Q}$ divided by its diagonally embedded subgroup $\mathbb{Q}$. So the Pontryagin dual of the discrete group $\mathbb{R}$ is merely a cartesian product of continuum many copies of $\mathbb{A}/\mathbb{Q}$.

I’m afraid I don’t know much of anything about Bohr compactifications. But yes, this dual of discrete $\mathbb{R}$ is the Bohr compactification of the usual $\mathbb{R}$. For if $G$ is any (let’s say locally compact Hausdorff) topological abelian group, and $G'$ is its Pontryagin dual, then you get $Bohr(G)$ as $(G', dis)'$ where the thing inside parentheses is $G'$ retopologized by the discrete topology.

I’d just like to write this out. Let’s define $Bohr(G)$ as we did in the preceding paragraph. First, we get the map

$$i: G \to Bohr(G)$$

just by applying the contravariant functor $\hom(-, S^1)$ to the continuous homomorphism

$$id: (G', dis) \to G'$$

(if $G$ is locally compact Hausdorff, we have a canonical isomorphism $G \cong \hom(G', S^1)$). This map $i: G \to Bohr(G)$ is an epi, because $id: (G', dis) \to G'$ is a mono and the functor

$$\hom(-, S^1): LCHAb^{op} \to LCHAb$$

is an equivalence.

Suppose given a continuous homomorphism $f: G \to K$ to a compact abelian group $K$. Then $K'$ is discrete, so

$$f': K' \to G'$$

factors uniquely through $id: (G', dis) \to G'$, i.e., $f'$ equals a composite of the form

$$K' \stackrel{\phi}{\to} (G', dis) \stackrel{id}{\to} G'$$

and now by Pontryagin duality, $f = f''$ equals the dual of this composite, which is

$$G \stackrel{i}{\to} Bohr(G) \stackrel{\phi'}{\to} K$$

so indeed $\phi'$ is the extension we want for the desired universal property of Bohr compactification mentioned in the Wikipedia article you linked to. It’s the unique such extension because $i: G \to Bohr(G)$ is epi.

I think all this goes through without much change even if we drop the assumption that $G$ is locally compact Hausdorff, but I’m a little tired right now and don’t feel like checking it out at the moment.

John wrote:

I guess it suffices to take a product over $r$ lying in a Hamel basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$.

I no longer believe this. Lately I tend to wake up in the middle of the night and think mostly pointless thoughts about the events of the day; sometimes my thoughts turn to math, and that’s when I notice mistakes I made in over-hasty guesses on this blog.

I had been thinking a group homomorphism $$f : \mathbb{R} \to S^1$$ was determined by its values on a Hamel basis — that is, a basis of $\mathbb{R}$ as an uncountable-dimensional vector space over $\mathbb{Q}$. I still think that would be true for a group homomorphism $$f: \mathbb{R} \to \mathbb{R}$$ But the problem with $$f : \mathbb{R} \to S^1$$ is that $S^1$ has torsion: if we know $f(x)$ for some $x \in \mathbb{R}$, we know $f(2x)$ but not, say, $f(x/2)$, since there are two consistent choices of what that could be.

So, to determine $f$ I think we need to know its values not on a Hamel basis of $\mathbb{R}$ a vector space over $\mathbb{Q}$, but on a set of generators of $\mathbb{R}$ as a $\mathbb{Z}$-module. (Duh.) And I think we can get this by taking a Hamel basis $x_\alpha$ and then using the numbers $x_\alpha/n$ for $n = 1,2,...$ Or, if we wish to be ‘more efficient’, we can use $x_\alpha/p^n$ for $p$ prime and $n = 0,1,2,....$

Here is a relevant paper by Michael Barr:

• Michael Barr, On duality of topological abelian groups.

He constructs a category of topological abelian groups containing the category of LCA groups, which is *-autonomous with the circle as the dualizing object.

Here is a comment he emailed to me. He allowed me to post it here:

I looked at the web site you mentioned. It seems to have left coalgebras behind. There is no mystery as to $Vect^op$ (assuming that is the opposite of vector spaces of a field). If the field is finite, then it is just the compact vector spaces. More generally, it consists of “linearly compact” vector spaces. Lefschetz defined them maybe 80 years ago in order to try to make cohomology theory of spaces dual to homology. The definition is the finite intersection property on closed linear subspaces, but turns out to be characterized as cartesian powers of the field. This assumes the field is discrete. Compactness cannot work as there cannot be a compact topology even on a one dimensional space over an infinite field.

You can go on and construct a *-autonomous category following the same path as I trod for abelian groups. You could define a locally linearly compact space as one with a linearly compact nbd of 0, but vector spaces have so much structure that it just amounts to being a product of a discrete and a linearly compact space. (You must know that a locally compact abelian group is a product of a compact group, a discrete group, and a finite dimensional real vector space.) Taking the same tack, that is starting with a space that can be embedded into a product of locally linearly compact spaces and then the largest topology with the given set of linear functionals, you get a *-autonomous category that extends the duality of the locally linearly compact ones.

Also someone should explain that a quasivariety is a subcategory of a variety closed under products and subobjects (but not homomorphic images). In the case of $Top^op$, it is actually a subvariety characterized by a single Horn clause. The whole variety, not surprisingly, is frames.

The last remark addresses this comment by David Corfield.