Let $X$ be a scheme and $U$ be an open subscheme. The proof of the Thomason-Trobaugh Theorem implies that under some mild assumptions, for any perfect complex $F$ on $U$, we have that $F\oplus F[1]$ can be extended to a perfect complex on $X$. I'm just wondering whether there exists examples where $F$ is a perfect complex on $U$ but $F$ itself cannot be extended to $X$? I've found an example when $X$ is the cone $xy-z^{2}=0$ and $U$ is the complement of the origin. Is there an example for smooth $X$?
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$\begingroup$ It would be interesting if you added your own example, for completeness $\endgroup$– QfwfqCommented Jun 23, 2018 at 23:16
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$\begingroup$ My example is the following. The open subscheme $U$ is the smooth locus of the cone, let $F$ be the line bundle on $U$ corresponds to the divisor of a straight line through the origin. If there were a perfect complex $\mathcal{P}$ on $X$ such that its restriction to $U$ is isomorphic to $F$, then it implies that $F$ is the restriction of a line bundle on $X$, namely $det(\mathcal{P})$. But this is not the case, $F$ cannot be extended to a line bundle on $X$ $\endgroup$– user105178Commented Jun 23, 2018 at 23:46
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All perfect complexes on an open $U$ of a smooth variety $X$ extend to perfect complexes on $X$. In fact, I believe a much more general statement is true: For an open substack $U$ of a separated Noetherian algebraic stack $X$, all bounded complexes of coherent sheaves on $U$ extend to bounded complexes of coherent sheaves on $X$. Since for smooth varieties, bounded complexes of coherent sheaves are all quasi-isomorphic to a perfect complex, this implies my first claim.
For a single coherent sheaf, this is Corollary 15.5 in Champs Algébriques, Laumon-Moret-Bailly (see also 27.22 in the Stacks Project). Here is a sketch of a proof for a general complex (for simplicity, take $f:U\rightarrow X$ an open immersion of varieties). The key claims are:
- Every quasi-coherent sheaf $\mathcal{F}$ on $X$ is an filtered colimit of coherent subsheaves $\mathcal{F}_i,$ $i$ in some index set.
and
- If $\mathcal{F}$ and $\mathcal{F}_i$ are as in 1), then for any coherent sheaf $\mathcal{G}$, a map $\mathcal{G}\rightarrow\mathcal{F}$ factors through one of the $\mathcal{F}_i.$
Given these two claims, the proof proceeds as follows. Take a bounded complex $C$ of coherent sheaves on $U$. Denote the (non-derived) pushforward of $C$ along $f$ by $f_*C$. We will find a subcomplex $C'$ of $f_*C$ of coherent sheaves, whose pullback along $f$ still gives the original complex $C$. This subcomplex will be constructed one term at a time, starting from the left. So assume that we have around found $C'$ with $C'_i\subseteq f_*C$ coherent for $i<0$ and $f^*C'\cong C.$ Now we need to refine our subcomplex to a subcomplex $C''$ with $C''_0$ coherent.
To do this, express $C'_0$ as a filtered colimit of coherent subsheaves $C'_{0,i}$ using Claim 1. By Claim 2, the map $C_0\rightarrow f^*C'_0$ factors through some $f^*C'_{0,i}$. Also by Claim 2, the map $C'_{-1}\rightarrow C'_0$ factors through some $C'_{0,j}.$ Because our colimit is filtered, we can take some $C'_{0,k}$ which contains both $C'_{0,i}$ and $C'_{0,j}.$ Then take $C''_i$ to be $C'_i$ for $i\neq 0$ and to be $C'_{0,k}$ for $i=0.$ Iterating this procedure gives us the desired bounded complex of coherent sheaves on $X$.
Claim 1 turns out be more annoying to write out a formal proof for than I thought... so I'll just refer you to 27.22.3 in the Stacks Project. Here is a proof of Claim 2 though:
There exist a finite collection of sections $s_i\in\mathcal{G}(U_i),$ $U_i$ an affine open, that together generate $\mathcal{G}$ as a $\mathcal{O}_X$-module. The image of each of these sections is then contained in some $\mathcal{F}_i$, and so by filteredness there will be some $\mathcal{F}_i$ that contains the images of all the $s_i$. The map $\mathcal{G}\rightarrow\mathcal{F}$ will factor through this $\mathcal{F}_i$.
