Computations in condensed mathematics II, page 43

Question

This is a follow up on my previous question for Lectures on Condensed Mathematics. I am ahead at page 43. But it is not directly clear to me from the results that:

1. How do we know $\Bbb Z[[T]], \Bbb R, \Bbb Z_p$ are solid modules?

2. How did we obtain that $\Bbb R^{L \blacksquare}=0?$

3. For the last equation 6.4 to holds, does $-\otimes^{L\blacksquare}-$ commute with filtered limits in each variable?

Some elaborations would help. I'd also like to know what are the formal aspects of the deductions (which I suppose is the great part this new category).

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Basically all I know which are solid: are

1. those of the form $(-)^{L\blacksquare}$,
2. Those which are local objects. Characterization 5.8.
3. $\prod_I \Bbb Z$, the cpt. projective generators.

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Answer 1

Let me basically repeat what R. says in the comments.

For 1, $\mathbb{Z}[[T]]$ and $\mathbb{Z}_p$ are solid becuase $\mathbb{Z}$ is solid and solid abelian groups are closed under all limits and colimits. But $\mathbb{R}$ is not solid, see 2.

For 2, the first point is that $\mathbb{R}$ is pseudocoherent as a condensed abelian group, i.e. the following equivalent conditions are satisfied:

1. $Ext^i(\mathbb{R},-)$ commutes with filtered colimits for all i;
2. $\mathbb{R}$ admits a projective resolution where the terms are compact projective condensed abelian groups.

(The equivalence of 1 and 2 is valid in any abelian category generated by compact projectives. It's a good exercise in homological algebra if you'd like to try.)

The pseudocoherence of $\mathbb{R}$ follows from the short exact sequence $\mathbb{Z}\rightarrow\mathbb{R}\rightarrow\mathbb{R}/\mathbb{Z}$. Indeed, $\mathbb{Z}$ is clearly pseudo-coherent (it is projective and compact) and the Breen-Deligne resolution implies that any compact abelian group is psuedo-coherent. Equivalent condition 1) shows that pseudocoherent modules have the 2 out of 3 property in short exact sequences, so this implies that $\mathbb{R}$ is pseudo-coherent.

The second point is that for a pseudocoherent condensed abelian group $M$ the derived solidification of $M$ identifies with $\underline{RHom}(\underline{RHom}(M,\mathbb{Z}),\mathbb{Z})$. Indeed, using condition 2 one reduces to checking that when $M=\mathbb{Z}[S]$ for $S$ extr. disconnected we have that $\underline{RHom}(\mathbb{Z}[S],\mathbb{Z})$ lives in degree $0$ and $\underline{RHom}(-,\mathbb{Z})$ on it is the derived solidification of $\mathbb{Z}[S]$. But these we verified in producing the solid theory (they follow from Specker's theorem and the definition of solidification).

Thus it suffices to see that $\underline{RHom}(\mathbb{R},\mathbb{Z})=0$. But this was verified in the proof of the existence of the solid theory. It follows from the Breen-Deligne resolution and the fact that $\mathbb{R}$ is contractible (so its cohomology with $\mathbb{Z}$-coefficients vanishes).

For 3, the answer is yes. For ordinary tensor product this is clear, and it follows for derived tensor product because filtered colimits are exact so they have vanishing higher derived functors.