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?
3 Answers
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 autoduality.
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?)

$\begingroup$ Thanks for the nice answer! I think the noninjectivity of the torus $\mathbb T$ can already be seen in the category of locally compact abelian groups. You map $\mathbb Z$ injectively to $\mathbb T$ and you also map $\mathbb Z$ injectively to $\mathbb{Z}_p$ for some prime $p$. Since any continuous homomorphism from $\mathbb{Z}_p$ to the torus has finite image, the map from $\mathbb Z$ to $\mathbb T$ cannot be lifted. $\endgroup$– EchoFeb 6 at 18:16

1$\begingroup$ @Echo be careful, the category $\mathbf{LCA}$ of locally compact abelian groups is not abelian, so you have to think about what you mean by injective and so on. The inner RHom is actually not inner: it goes $D^b(\mathbf{LCA})^{\text{op}}\times D^b(\mathbf{LCA}) \to D^b(\mathbf{TopAb})$, and is computed in a cumbersome way (see the Hoffman–Spitzweck reference in the Clausen–Scholze notes). For this RHom, the torus is of course injective, because Pontryagin duality holds! The point is that only strict short exact sequences in $\mathbf{LCA}$ are sent to exact sequences in $\mathbf{Cond}$. $\endgroup$ Feb 6 at 18:44

1$\begingroup$ I used to think that the notion of injective object exists for any category. And as LCA embeds into Cond, where the notion of monomorphism is preserved, the torus being not injective in LCA implies that it is not injective in Cond, or did I miss anything? $\endgroup$– EchoFeb 7 at 6:17

$\begingroup$ Oh right, I was using the "Hom is exact" definition which exactly doesn't work in this setting. The usual categorical definition indeed immediately shows $\mathbf T$ is not injective. If $A\to B$ is the map $\mathbf Z\hookrightarrow\mathbf Z_p$, then $\mathbf D(B)\to\mathbf D(A)$ is $\mathbf Q_p/\mathbf Z_p\hookrightarrow\mathbf T$. If $N$ is the cokernel, running the argument above shows $\mathbf D(N)=0$ but $N \neq 0$. This makes sense as the image of $\mathbf Q_p/\mathbf Z_p\to\mathbf T$ is dense, somewhat similar to my second answer. $\endgroup$ Feb 7 at 13:07

$\begingroup$ So the point is that $\mathbf{Hom}(,\mathbf T)$ only turns strict monomorphisms into surjections, not arbitrary injections. $\endgroup$ Feb 7 at 13:14
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 nonlocallyconvex spaces, this map is far from being an isomorphism.

$\begingroup$ Ah, you're absolutely right, it's not even an isomorphism for topological groups! I have worked out another example in my second answer. $\endgroup$ Feb 6 at 18:35
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. $\square$
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.

$\begingroup$ It seems worthwhile to note that $\underline{\operatorname{RHom}}_{\mathbb Z}(\mathbb Q,)$ is easier to compute a priori since $\mathbb Q$ is a filtered colimit of $\mathbb Z$, while $\underline{\operatorname{RHom}}_{\mathbb Z}(\mathbb R,)$ seems nontrivial (using Breen–Deligne resolution, etc.). $\endgroup$– Z. MFeb 7 at 13:19

$\begingroup$ @Z.M that is for the discrete topology on $\mathbf Q$, which is usually what the topological abelian group $\mathbf Q$ means. But I'm using the subspace topology from $\mathbf R$. Also, I'm just computing in $\mathbf{LCA}$ rather than $\mathbf{Cond}$ (which for underived inner Hom doesn't matter). Even for derived inner hom, I would probably use Hoffman–Spitzweck's weird construction of the inner hom in $D^b(\mathbf{LCA})$ to compute it. $\endgroup$ Feb 7 at 16:24