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I am watching Scholze's and Clausen's masterclass on Condensed Mathematics and I don't understand or can find any references on something they said.

You have a resolution

$$ \dots \to \mathbb{Z}[\mathbb{R}^2] \to \mathbb{Z}[\mathbb{R}] \to \mathbb{R} \to 0$$

They refer to the spectral sequence arising from this complex to later talk about sheaf cohomology, but I am not sure what they mean. Maybe they use the filtration induced by truncation?

A similar thing is referenced in their notes (see Corollary 4.8), where there is a resolution and in that case it is mentioned that this is obtained by applying RHom, but I am not sure how you get a spectral sequence then either.

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    $\begingroup$ Exactly, he means the spectral sequence for $Ext^i(-,A)$ coming from the "stupid" filtration on the resolution (without the $\mathbb{R}$ term) induced by truncation. If there were only two terms in the resolution, this would be the same as the induced long exact sequence on $Ext^i(-,A)$. $\endgroup$
    – Dustin Clausen
    Commented Dec 17, 2020 at 21:02
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    $\begingroup$ You can also think of it like this. If it were a projective resolution, we could use it to calculate $Ext^i(\mathbb{R},A)$ in terms of $Hom(P_i,A)$. The spectral sequence is saying that even when the terms are not projective, you can still in some sense calculate $Ext^i(\mathbb{R},A)$ in terms of not just $Hom(P_i,A)$, but $RHom(P_i,A)$. I think this plus the special case where there are only 2 terms mentioned above give the "feel" for this sort of thing. $\endgroup$
    – Dustin Clausen
    Commented Dec 17, 2020 at 21:07
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    $\begingroup$ @DustinClausen Is this what "homological cavemen" like myself might call hyper(co)homology? i.e. take a projective resolution of each $P_i$ and then take the (co)homology of the total complex? $\endgroup$
    – Yemon Choi
    Commented Dec 17, 2020 at 21:48
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    $\begingroup$ Yes, Yemon. I don't know what it's called, but that's what it is. $\endgroup$
    – Dustin Clausen
    Commented Dec 18, 2020 at 9:31

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