Condensed Pontryagin duality

Question

Has Pontryagin duality been extended to condensed abelian groups? The obvious approach being to define $\hat M$ as the internal hom to the circle group. Is it true that $\hat{\hat M}=M$ with this definition?

Answer 1

This cannot be true for all condensed abelian groups. Indeed, in this answer to Are there (enough) injectives in condensed abelian groups?, Scholze explains that there are no nonzero injective condensed abelian groups. So it suffices to prove the following:

Lemma. Let $\mathscr A$ be an abelian category and let $\mathbf D \colon \mathscr A^{\text{op}} \to \mathscr A$ be a left exact functor such that $\mathbf D^2$ is faithful and preserves monomorphisms (e.g. $\mathbf D^2 \cong \mathbf 1_{\mathscr A}$). Then $\mathbf D$ is exact.

Indeed, this would prove that $\mathbf T = \mathbf R/\mathbf Z$ is injective if $\mathbf D = \mathbf{Hom}(-,\mathbf T)$ is an auto-duality.

Proof of Lemma. If $0 \to A \to B \to C \to 0$ is an exact sequence in $\mathscr A$, then $$0 \to \mathbf D(C) \to \mathbf D(B) \to \mathbf D(A)$$ is exact by hypothesis. Let $M$ and $N$ be the image and cokernel of $\mathbf D(B) \to \mathbf D(A)$ respectively, so that we get exact sequences $$\begin{array}{ccccccccc} 0 & \to & \mathbf D(C) & \to & \mathbf D(B) & \to & M & \to & 0,\\ 0 & \to & M & \to & \mathbf D(A) & \to & N & \to & 0. \end{array}$$ Dualising again gives exact sequences $$\begin{array}{ccccccc} 0 & \to & \mathbf D(M) & \to & \mathbf D^2(B) & \to & \mathbf D^2(C),\\ 0 & \to & \mathbf D(N) & \to & \mathbf D^2(A) & \to & \mathbf D(M). \end{array}$$ The composition $\mathbf D^2(A) \to \mathbf D(M) \to \mathbf D^2(B)$ is injective by hypothesis, so the second sequence gives $\mathbf D(N) = 0$. Faithfulness of $\mathbf D^2$ gives $N = 0$, so $\mathbf D(B) \to \mathbf D(A)$ was surjective to begin with. $\square$

Presumably there is a direct example showing that $\mathbf T$ is not injective. If you want to work with $\mathbf{Cond}_\kappa$ instead of $\mathbf{Cond}$, then the same argument doesn't work (because $\mathbf{Cond}_\kappa$ does have enough injectives), and I don't know if $\mathbf T$ is injective in $\mathbf{Cond}_\kappa$ for any strong limit cardinal $\kappa$. (Surely not, right?)

Answer 2

Here is an attempt to give a more concrete example.

Let $V$ be a condensed $\mathbb R$-vector space. Then we have \[ \DeclareMathOperator\RHom{RHom} \newcommand\iRHom{\underline{\RHom}} \iRHom_{\mathbb Z}(V,\mathbb R/\mathbb Z)\simeq\iRHom_{\mathbb R}(V,\iRHom_{\mathbb Z}(\mathbb R,\mathbb R/\mathbb Z))\simeq\iRHom_{\mathbb R}(V,\mathbb R), \] where we used the fact that the Pontryagin dual of $\mathbb R$ is itself. It follows that, if the underlying condensed abelian group $V$ satisfies the Pontryagin duality, then the map from the condensed $\mathbb R$-vector space $V$ to its double $\mathbb R$-dual is an isomorphism. Such $V$ seems to be closely related to stereotype spaces in functional analysis. In particular, it seems that, for non-locally-convex spaces, this map is far from being an isomorphism.

Answer 3

In fact, the evaluation map $M \to M^{\vee\vee}$ is not even an isomorphism for all topological groups $M$. If we take $M$ such that the topologies on $M$ and $M^\vee$ are compactly generated¹⁾, then Proposition 4.2 in Lectures on Condensed Mathematics shows that the formation of $M^\vee$ and $M^{\vee\vee}$ is preserved by the passage from $\mathbf{TopAb}$ to $\mathbf{Cond}(\mathbf{Ab})$.

Here is a simple and concrete example.

Example. Let $M = \mathbf Q$ with the subspace topology from $\mathbf R$. This is not locally compact, but it is compactly generated: suppose $S \subseteq \mathbf Q$ is a subset such that $S \cap K$ is closed for any compact subset $K$. If $x_1,x_2,\ldots$ is a sequence of elements in $S$ with limit $x \in \mathbf Q$, then $K = \{x\} \cup \{x_1,x_2,\ldots\}$ is compact as it is closed and bounded in $\mathbf R$. Since $S \cap K$ is closed and contains $\{x_1,x_2,\ldots\}$, we conclude that $x \in S$.

But the inclusion $M \hookrightarrow \mathbf R$ induces an isomorphism $\operatorname{Hom}_{\text{cts}}(\mathbf R,\mathbf T) \stackrel\sim\to \operatorname{Hom}_{\text{cts}}(M,\mathbf T)$, for instance because continuous group homomorphisms are automatically uniformly continuous (for the left or right uniformity ― they agree in the abelian case), so they factor uniquely via the completion. Thus, $M^{\vee\vee} = \mathbf R^{\vee\vee} = \mathbf R$ as well. $\square$

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¹⁾ Beware that some authors use the separate and unrelated notion that a Hausdorff abelian group $A$ is compactly generated if there exists a compact subset $S \subseteq A$ that generates $A$ as a group. I stress that I am using the topological notion here.