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My question is about Corollary 6.1(ii) in Lectures on Condensed Mathematics by Scholze (page 41). Here is the claim:

The derived category $D(\mathrm{Solid})$ is compactly generated, and the full subcategory $D(\mathrm{Solid})^{\omega}$ of compact objects consists of the bounded complexes all of whose terms are of the form $\Pi_{I} \mathbb{Z}$. The category $D(\mathrm{Solid})^{\omega}$ is contravariantly equivalent to the category $D^{b}(\mathbb{Z})$ via the functor $$ C \mapsto R \underline{\operatorname{Hom}}(C, \mathbb{Z}) : D^{b}(\mathbb{Z})^{\mathrm{op}} \rightarrow D(\mathrm{Solid}) $$

I'm not quite understanding how to interpret this internal hom. I know how to view $R\underline{\operatorname{Hom}}(C, \mathbb{Z})$ as an object in $D^b(\mathbb{Z})$, but I'm expecting to get an object in $D(\mathrm{Solid})$. One way to get from $D^b(\mathbb{Z})$ to $D(\mathrm{Solid})$ is to first move to $D(\mathrm{Cond}(Ab))$, then apply the derived solidification functor $(-)^{L\blacksquare}$. So should I think of this as $$C \mapsto \left(\underline{R \underline{\vphantom{\_}\operatorname{Hom}}(C, \mathbb{Z})}\right)^{L\blacksquare}?$$ Alternatively, is there some reason $\underline{R \underline{\vphantom{\_}\operatorname{Hom}}(C, \mathbb{Z})}$ is solid? Or am I completely off base?

Edit: As mentioned in the lectures and by Dan below, the fact that every bounded complex of abelian groups can be resolved by a complex of free abelian groups means that the objects in the internal hom are direct products of $\mathbb{Z}$. This roughly answered my question about $\underline{R \underline{\vphantom{\_}\operatorname{Hom}}(C, \mathbb{Z})}$ being solid, but made me realize I don't know exactly what I mean by $\underline{R \underline{\vphantom{\_}\operatorname{Hom}}(C, \mathbb{Z})}$.

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    $\begingroup$ The 1-minute answer is that every object of $D^b(\mathbb Z)$ is given by a bounded complex all of whose terms are of the form $\bigoplus_I \mathbb Z$, and that applying $\mathrm{Hom}(-,\mathbb Z)$ turns those sums into products. $\endgroup$
    – Dan Petersen
    Commented Feb 22, 2021 at 6:35
  • $\begingroup$ Thank you! This helps somewhat, but I still have some silly questions. I think this produces just a chain complex of abelian groups. How do I turn this into a chain complex of condensed abelian groups? Do we give everything the discrete topology? $\endgroup$
    – diracdeltafunk
    Commented Feb 22, 2021 at 7:03
  • $\begingroup$ Also, perhaps I'm misunderstanding notation: does $R\underline{\operatorname{Hom}}$ here refer to the internal hom of $D(\mathbb{Z})$? $\endgroup$
    – diracdeltafunk
    Commented Feb 22, 2021 at 7:08
  • $\begingroup$ You're suppose to think of $\prod_I \mathbb Z$ as the condensed cartesian product, not the abelian group $\prod_I \mathbb Z$ with the discrete topology. And the homs are the internal derived Hom, but if you decide to represent your objects of $D^b(\mathbb Z)$ by bounded complexes of free abelian groups then this coincides with the internal Hom of chain complexes. $\endgroup$
    – Dan Petersen
    Commented Feb 22, 2021 at 8:17

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The notation is actually amibiguous which I guess causes the confusion. In fact this $\underline{RHom}(C,\mathbb{Z})$ stands for the internal hom from $C$ to $\mathbb{Z}$ in derived $\textit{condensed}$ abelian groups. Here $C$ and $\mathbb{Z}$ start in $D(\mathbb{Z})$ but we can view that as the ``discrete" full subcategory of $D(Cond(\mathbb{Z}))$. As Dan indicates in the comments, this internal RHom is calculated by representing $C$ by a bounded complex of free $\mathbb{Z}$-modules, and then noting that $\underline{RHom}(\oplus_I \mathbb{Z},\mathbb{Z})=\prod_I\mathbb{Z}$.

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  • $\begingroup$ Thank you very much! Can I ask for some clarification about what ``discrete'' means here? Is the embedding $D(\mathbb{Z}) \to D(\mathrm{Cond}(\mathbb{Z}))$ given by the derived functor of $A \mapsto \underline{A_{\text{discrete}}} : \mathrm{Ab} \to \mathrm{Cond}(\mathbb{Z})$? $\endgroup$
    – diracdeltafunk
    Commented Feb 22, 2021 at 19:17
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    $\begingroup$ In a word, yes! Note also that $A\mapsto \underline{A_{discrete}}$ is an exact functor so this derived functor is not very subtle. $\endgroup$
    – Dustin Clausen
    Commented Feb 22, 2021 at 20:14
  • $\begingroup$ Thanks! Just wanted to make sure I'm understanding everything precisely. $\endgroup$
    – diracdeltafunk
    Commented Feb 22, 2021 at 20:20

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