The n-Category Café

I can tell you that the answer to the first question is “in general, no”. Let

$$A = \sum_{\mathbb{N}} \mathbb{Z},$$

a coproduct of countably many copies of $\mathbb{Z}$ with the discrete topology. Then

$$\hom(\sum_{\mathbb{N}} \mathbb{Z}, \mathbb{C}^*) \cong \prod_{\mathbb{N}}\hom(\mathbb{Z}, \mathbb{R} \times S^1) \cong \mathbb{R}^{\mathbb{N}} \times (S^1)^{\mathbb{N}}$$

but $\mathbb{R}^{\mathbb{N}}$ is not locally compact Hausdorff. (Notice also that $A$ in this case is countable as a set and therefore second countable.)

Unless I’m making a stupid mistake, I think the answer to question (1) is “yes” in the case where $A$ is connected. Since $\mathbb{C}^* \cong \mathbb{R} \times S^1$, we really just have to worry about the factor $\mathbb{R}$. If $A$ is connected, then we should have an injection

$$\hom(A, p): \hom(A, \mathbb{R}) \to \hom(A, S^1)$$

where $p: \mathbb{R} \to S^1$ is the usual covering projection, realizing $\hom(A, \mathbb{R})$ as a closed subgroup of its Pontryagin dual.

I’m beginning to want to compile a table of theorems of this form:

If the locally compact Hausdorff topological group $A$ has property $P$, then its Pontryagin dual $A^*$ has property $P^*$.

I mentioned the most famous ones:

If $A$ is compact, $A^*$ is discrete.

If $A$ is discrete, $A^*$ is compact.

and of course there’s this:

If $A$ is finite, $A^*$ is finite.

I’ve bumped into a few more:

If $A$ is torsion-free and discrete, $A^*$ is connected and compact.

If $A$ is connected and compact, $A^*$ is torsion-free and discrete.

If $A$ is a Lie group, $A^*$ has finite rank.

If $A$ has finite rank, $A^*$ is a Lie group.

If $A$ is second countable, $A^*$ is second countable.

If $A$ is separable, $A^*$ is metrizable.

To add to my collection, I should get ahold of these books:

• Lev Semenovich Pontryagin, Topological Groups, Gordon & Breach, 1966.
• Sidney A. Morris, Pontryagin Duality and the Structure of Locally Compact Abelian Groups, London Math. Soc. Lecture Notes 29, Cambridge U. Press, 1977.

By the way, as I mentioned here Ulrich Bunke has the notion and applications for Pontryagin duality of higher categorical groups. He says that as far as the gerbes involved are concerned, T-duality is just Pontryagin duality of higher groups.

See

Bunke, Schick, Spitzweck, Thom, Duality for topological abelian group stacks and T-duality

Abstract: We extend Pontrjagin duality from topological abelian groups to certain locally compact group stacks. […] As an application of the theory we interpret topological T-duality of principal Tⁿ-bundles in terms of Pontrjagin duality of abelian group stacks.

Todd showed that even if the abelian group $A$ is locally compact, the abelian group

$$A' = hom(A,\mathbb{C}^*)$$

may not be. His example was an infinite coproduct of copies of $\mathbb{Z}$:

$$A = \bigoplus_{n \in \mathbb{N}} \mathbb{Z}$$

Yves de Cornulier has allowed me to share an email in which he argues that in a certain nice sense, this group $\bigoplus_{\mathbb{N}} \mathbb{Z}$ is the ‘only obstacle’ to $A'$ being locally compact. There are aspects of his argument that I don’t understand, but it seems promising.

Here is his email after a bit of LaTeXing:

$A'$ is not necessarily locally compact. E.g, this fails with $\bigoplus_{\mathbb{N}} \mathbb{Z}$, an infinite direct sum of copies of $\mathbb{Z}$ (see below). I think that this is essentially the only counterexample: if a locally compact abelian group does not have a discrete quotient of infinite rank (i.e. containing $\bigoplus_{\mathbb{N}} \mathbb{Z}$ as a subgroup), then $A'$ is locally compact.

Here is how I see this.

As a topological group, $\mathbb{C}^*$ is nothing else than $\mathbb{R} \times \mathrm{U}(1)$, $\mathbb{R}$ being the reals. So

$$hom(A,\mathbb{C}^*) = hom(A,U(1)) \times hom(A,\mathbb{R}),$$

and the discussion boils down to the study of $hom(A,\mathbb{R})$.

Now $A$ has a compact subgroup $K$ such that $A/K$ is a Lie group (maybe with infinitely many components). We can pick $K$ so that the unit component of $A/K$ is $\mathbb{R}^d$. As

$$hom(A,\mathbb{R})=hom(A/K,\mathbb{R}),$$

we can reduce to the study of $A/K$, so I suppose $A=A/K$.

As $\mathbb{R}^d$ is a divisible abelian group, it has a direct factor $B$ in $A$ (viewed as a discrete group), but as $B\cap \mathbb{R}^d$ is trivial $B$ is actually discrete in the topological group $A$. Again, since morphisms to $\mathbb{R}$ vanish on the torsion, we can suppose that $B$ is torsion-free.

Now suppose that $B$ has finite rank: it contains a subgroup isomorphic to $\mathbb{Z}^k$ such that the natural restriction morphism

$$hom(B,\mathbb{R}) \to hom(\mathbb{Z}^k,\mathbb{R})$$

is an isomorphism (actually of topological groups). Now $hom(\mathbb{Z}^k,\mathbb{R})$ is locally compact and second countable, so $hom(A,R)$ is so as well.

On the other hand, suppose that $B$ has infinite rank. Let $V$ be a neighbourhood of 0 in $hom(B,\mathbb{R})$: it contains, for some compact(=finite) subset $F$ of $B$ and some $r\gt 0$, the set $V(K,B(r))$ of homomorphisms $B \to \mathbb{R}$ mapping $F$ into the $r$-ball $B(r)$. I can find, in this $V(K,B(r))$, a sequence with no cluster point. Namely, since $B$ has infinite rank, there exists $y$ in $B$ such that the cyclic subgroup generated by $y$ has trivial intersection with the subgroup generated by $K$. So there exists a homomorphism $u_n:B \to \mathbb{Z}$ mapping $K$ to zero and $y$ to $n$. Clearly, this has no cluster point, but it lies in $V(K,B(r))$.

So we get the equivalences:

$hom(A,\mathbb{C}^*)$ is locally compact $\iff$

$hom(A,\mathbb{R})$ is locally compact $\iff$

$hom(A,\mathbb{Z})$ is locally compact $\iff$

$\bigoplus_{\mathbb{N}} \mathbb{Z}$ is not a subgroup of a discrete quotient of A $\iff$

$\bigoplus_{\mathbb{N}} \mathbb{Z}$ is not a discrete subgroup of A.

In these cases, $hom(A,\mathbb{R})$ and $hom(A,\mathbb{Z})$ are also 2nd-countable.

The other question should reduce to the study of whether $hom(\bigoplus_{\mathbb{N}} \mathbb{Z},R)$ is 2nd-countable, which shouldn’t be a big deal.

[EDIT: note the correction here.]

In response to a question, he added:

Note that by Lie group I don’t assume connectedness. So any discrete group is a Lie group, for instance.

It’s well known (and not trivial) that any locally compact abelian group $A$ has a compact subgroup $K$ such that $A/K$ is a Lie group. Clearly, any homomorphism from $A$ to $\mathbb{R}$ has to vanish on $K$, so factors thru $A/K$.

Concerning your article in preparation on infinite dimensional 2-representations:

we talked about this a little bit before somewhere, but since you don’t mention it: it looks to me like the 2-category $\mathbf{Meas}$ you describe should pretty closely be related to the 2-category $Bimod_{vN}$ whose

- objects are (commutative) von Neumann algebras

- morphisms are von Neumannn bimodules

- 2-morphisms are homomorphisms of these,

or rather the image in $Vect-Mod$ under $$Bimod_{vN} \to Vect-Mod$$ which sends

- algebras to their categories of modules;

- bimdodules to functors induced by tensoring with these

- bimodule homomorphisms to natural transformations of these.

Commutative vonNeumann algebras are precisely the algebras of measurable functions, I think. Indeed, as you mention on p. 24, with reference to Dixmier and Arveson, the “fields of Hilbert spaces” which you discuss are precisely the modules for commutative vonNeumann algebras.

Given that, a bimodule of commutative von Neumann algebras should be interpretable equivalently as a “field of Hilbert spaces” on a product space $X \times Y$, as it indeed appears in the definition of morphisms on $\mathbf{Meas}$ on p. 29.

I understand that you have a different angle here, but wouldn’t it be fun to relate this a bit for instance to the interesting 2-representation of the strict String-2-group on – in this language – “fields of Hilbert spaces” over non-commutative measurable spaces (i.e. using a non-commutative von Neumann algebra, namely that of a highest weight rep of the centrally extended loop group sitting in $String(n)$, the way it is described for instance in the context of String 2-bundles in my article with Konrad?

Everyone interested in Pontryagin duality should read this:

• Michael Barr, On duality of topological abelian groups.

He explains:

Did you know that there is a *-autonomous category of topological abelian groups that includes all the LCA groups and whose duality extends that of Pontrjagin? The groups are characterized by the property that among all topological groups on the same underlying abelian group and with the same set of continuous homomorphisms to the circle, these have the finest topology. It is not obvious that such a finest exists, but it does and that is the key.