Let $X$ be a quasi-compact and quasi-separated scheme, and $U\subseteq X$ be a quasi-compact open subscheme. Then we can consider $Rj_*\mathcal{O}_U$ the (derived) pushforward of the structure sheaf of $U$. This is a coconnective sheaf of rings (in fact it is locally of Tor amplitude in $(-∞,0]$, since locally it can be represented as a finite limit of sheaves of the form $\mathcal{O}_X[f^{-1}]$).

Q: Can $Rj_*\mathcal{O}_U$ be represented as a filtered colimit of uniformly bounded above perfect complexes?

Note: I am using the homological grading convention, so, e.g., for me every connective sheaf is bounded below.

The result is true if $X$ has an ample family of line bundles, because then I could just use the colimit of Koszul complexes, but I'd like it to be true in more generality.

  • 4
    $\begingroup$ I am not sure how useful it is, but you can try the following standard construction. Let $\tilde{X}$ be the blowup of the complement of $U$ in $X$. Then the map $j$ decomposes as $\pi \circ i$, where $i \colon U \to \tilde{X}$ and $\pi$ is the blowup. The point is that $i$ is affine, so $Ri_* = i_*$, and $\pi$ is projective, so you can use a relative Cech complex to compute $R\pi_*$. $\endgroup$
    – Sasha
    Jun 27, 2019 at 10:47
  • $\begingroup$ @DenisNardin $p_*$ does not preserve perfect complexes under further hypotheses (finite Tor-amplitude, or domain and codomain regular). If you can choose $i$ to be a regular immersion (I guess this should be possible but not sure without checking) then the projection of the blowing up would be lci so that would suffice. $\endgroup$ Jun 27, 2019 at 15:07
  • $\begingroup$ @crystalline Ugh.. I misunderstood theorem 2.5.4 in Thomason-Trobaugh $\endgroup$ Jun 27, 2019 at 15:26
  • $\begingroup$ I mistyped (you understood but just to clarify), I meant it does not preserve perfect complexes without further hypotheses $\endgroup$ Jun 27, 2019 at 16:39
  • $\begingroup$ @crystalline It seems to me that if I manage to prove that $R\pi_*\mathcal{O}$ is perfect, the proof is complete though (since $U$ is an effective Cartier divisor in $\tilde X$). Do you agree? $\endgroup$ Jun 28, 2019 at 6:09


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