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This is a follow up on my previous question for Lectures on Condensed Mathematics. I am reading 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).

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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    $\begingroup$ For 2, see Cor. 6.1(iii). In particular, $\mathbf R$ is very far from solid. As for $\mathbf Z[[T]]$, it is isomorphic to a product of copies of $\mathbf Z$, so that is clearly solid. For $\mathbf Z_p$ I think you can use the short exact sequence $0 \to (T-p)\mathbf Z[[T]] \to \mathbf Z[[T]] \to \mathbf Z_p \to 0$. The last equation of 6.4 follows immediately from Prop. 6.3. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Dec 16, 2020 at 18:36

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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 (Edit: no, I misread and thought the question was about filtered colimits. See comments) . 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.

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  • $\begingroup$ P.S.: if this answer doesn't make sense to you please do feel free to ask for clarification. $\endgroup$
    – Dustin Clausen
    Commented Dec 16, 2020 at 21:56
  • $\begingroup$ Thanks again! For 1. This follows by Solid is a reflective localization? For 3. I thought we want the case for "inverse system \projective of limits " rather for colimits. And in this case, I don't see why $\otimes$ commutes with it: i.e. in $\Bbb Z$, $\Bbb Q \otimes (\varprojlim \Bbb Z/p^n) = 0 \not= \Bbb Z_p \otimes \Bbb Q=\Bbb Q_p$. $\endgroup$
    – Bryan Shih
    Commented Dec 16, 2020 at 23:02
  • $\begingroup$ Sorry, in 3 I misread and thought you asked about filtered colimits. Filtered limits do not commute with solid tensor product in general. If they did, so would arbitrary products, by writing a product as a filtered limit of finite products which are direct sums. But you can see that if $M= \oplus \mathbb{Z}$ then $M \otimes \prod \mathbb{Z} = \oplus \prod \mathbb{Z}\neq \prod \oplus \mathbb{Z}$ when you have infinite product and direct sum. $\endgroup$
    – Dustin Clausen
    Commented Dec 17, 2020 at 6:42
  • $\begingroup$ Now i'm confused how we computed $\Bbb Z[[U]] \otimes^{L \blacksquare} \Bbb Z[[T]] $ : The text said 6.3 is used, but to me: $\Bbb Z[[U]] \not= \prod_{\Bbb Z} \Bbb Z$. but is an inverse limit. $\endgroup$
    – Bryan Shih
    Commented Dec 17, 2020 at 7:48
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    $\begingroup$ As a condensed abelian group, $\mathbb{Z}[[U]] = \prod_{\mathbb{Z}} \mathbb{Z}$, they are the same inverse limit of finite products. $\endgroup$
    – Achim Krause
    Commented Dec 17, 2020 at 8:51

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