# Representing $j_*\mathcal{O}_U$ as filtered colimit of perfect complexes

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.

• 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 $R\pi_*$. Jun 27, 2019 at 10:47
• @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. Jun 27, 2019 at 15:07
• @crystalline Ugh.. I misunderstood theorem 2.5.4 in Thomason-Trobaugh Jun 27, 2019 at 15:26
• I mistyped (you understood but just to clarify), I meant it does not preserve perfect complexes without further hypotheses Jun 27, 2019 at 16:39
• @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? Jun 28, 2019 at 6:09