I started reading the Lectures on Condensed Mathematics. I am looking at the material at page 32-34. I have three fundamental computation questions:

  1. At the last line of pg 32 - it seems to imply that for finite sets $S$, $\Bbb Z[S] \simeq \underline{Hom}(C(S, \Bbb Z ), \Bbb Z) $?

  1. In pg 33 line 4 how is $Hom(C(S,\Bbb Z), \Bbb Z)$ identified with measure on $S$?

  1. In proof of Proposition 5.7, there was the following equivalence,

$$ RHom(\prod_J \Bbb R, \Bbb Z) \simeq RHom_{\Bbb R} (\prod_J \Bbb R, R\underline{Hom}(\Bbb R, \Bbb Z) )=0$$

why is this true and what exactly does $RHom_{\Bbb R}$ mean? Is there some adjunction happening here from $\Bbb Z$ modules of $Cond(Set)$ to $\Bbb R$-modules of $Cond(Set)$?

I would appreciate if there are some related references for the general set up in 2 and 3.

For 1, I would like to compute the $T$ points for $T$ extremely. disconnected. What i don't see is an easy expression for lhs.
$$\Bbb Z[S] (T)= \bigoplus_{C(T,S)}\Bbb Z$$ Conversely for rhs we have

$$\underline{Hom}(C(S,\Bbb Z), \Bbb Z)(T)=Hom(\Bbb Z[T] \otimes C(S,\Bbb Z), \Bbb Z) $$ This doesn't seem an easy expression to handle too.


1 Answer 1

  1. Correct, as both sides are the $S$-indexed direct sum of copies of $\mathbb{Z}$. For the LHS this holds by the universal property of $\mathbb{Z}[S]$, and for the RHS note that $C(S,\mathbb{Z}) = \prod_S \mathbb{Z} = \oplus_S \mathbb{Z}$ which lets you calculate.

  2. By definition, one could say. It's reasonable to define a $\mathbb{Z}$-valued measure on a profinite set to be an element of $Hom(C(S,\mathbb{Z}),\mathbb{Z})$. You could also define a measure to be a finitely additive assignment of an integer to each clopen subset of S. You can consider the indicator functions of clopen subsets to see the equivalence.

  3. $RHom_{\mathbb{R}}(M,N)$ means a complex calculating Ext's from M to N in the category of $\mathbb{R}$-modules in condensed abelian groups. The equality holds because both are the same as $RHom_{\mathbb{R}}(\mathbb{R} \otimes_\mathbb{Z}^L \prod_J \mathbb{R},\mathbb{Z})$ by some adjunctions.

There's also another way of looking at 3. Abstractly, the claim is that if $A$ is a condensed algebra and $N$ a condensed abelian group with $\underline{RHom}(A,N)=0$, then $RHom(M,N)=0$ whenever $M$ has an $A$-module structure. By adjunction, the hypothesis is equivalent to saying that $RHom(A\otimes_{\mathbb{Z}}^L M,N)=0$ for all condensed abelian groups $M$. If $M$ has an $A$-module structure, then $M$ is a retract of $A\otimes^L M$ by the action map $A\otimes^LM\rightarrow M$ in one direction and the unit $M=\mathbb{Z}\otimes^L M\rightarrow A\otimes^LM$. Thus we deduce $RHom(M,N)=0$ as claimed.


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