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If $X$ is a noetherian scheme , then for any quasi-coherent sheaf $\mathscr{F}$ on $X$ that satisfies Serre's (S2) condition and an inclusion of an open subscheme $j:U\to X$ with complement of codimension at least 2 then we have $$j_{\star}j^{\star}\mathscr{F}\cong\mathscr{F}.$$ This is like the Hartogs phenomenon for functions of several complex variables. See for example the explanation here: Why does the (S2) property of a ring correspond to the Hartogs phenomenon?

My question is, is there a derived version of this? Namely if one works in the derived category and replace the $j_\star$ the derived push forward $Rj_\star$, when do we have $$Rj_\star j^\star\mathscr{F}\cong\mathscr{F}?$$

I want to know at least the case where $\mathscr{F}=\mathcal{O}_X$.

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This seems to be a dumb question as if $Rj_\star j^\star\cong Id$ then $j_\star$ is exact as we can always extend quasi-coherent sheaves on $U$ to a quasi-coherent one on $X$. I guess a better question should be why is $Rj_\star j^\star\mathscr{F}\cong \mathscr{F}$ not true... –  temp May 18 '12 at 7:00
    
Your non-derived statement is too broad. If $\mathcal F$ is the push-forward of the structure sheaf of a curve, it fails, just as if the ambient space were a curve. (and it fails if it's the sum of the structure sheaf of a curve with the structure sheaf of $X$, etc.) –  Ben Wieland May 18 '12 at 15:30
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Note: If $\mathscr F \neq \mathcal{O}_X$, the (S2) condition should be applied to $\mathscr F$, not to $X$. –  Charles Staats May 18 '12 at 20:15
    
@Charles Staats, oops, thanks for reminding me that, I had $\mathcal{O}_X$ in mind and I got the general setup wrong. –  temp May 19 '12 at 2:01
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1 Answer

up vote 6 down vote accepted

Unfortunately almost never (although it obviously holds if $j^* \mathcal{F} = 0$). In particular, it virtually never holds for $\mathcal{F} = \mathcal{O}_X$. I'm going to assume that $\mathcal{F}$ is coherent.

Let me give the quick answer first, then I'll explain it in more detail.

Quick answer: $R^i j_* j^* \mathcal{F}$ coincides with $H^{i+1}_{X \setminus U}(\mathcal{F})$ the $i+1$th cohomology with support at $Z = X \setminus U$ of $\mathcal{F}$. By basic facts about local cohomology, you can basically never expect this to vanish for all $i > 0$.

Detailed answer: Let's suppose that $X = \text{Spec} R$ for simplicity and that $U = X \setminus V(I)$ for some ideal $I$. Further assume that $j^* M = M|_U \neq 0$. Let $m$ be a minimal prime of $I$ such that $M_m \neq 0$. We now can replace $R$ by $R_m$ and $M$ by $M_m$.

Working locally, for any $R$-module $M$ we can identify $$ R^i j_* j^* M = R^i j_* (M|_U) = H^i(U, M|_U) = H^{i+1}_m(M) $$ for $i > 0$ where $H^*_m$ is local cohomology. This comes from the long exact sequence $$ H^i_m(M) \to H^i(X, M) = 0 \to H^i(U, M|_U) \to H^{i+1}_m(M) \to H^{i+1}(X, M) = 0. $$

Now, $H^{i+1}_m(M) \neq 0$ for any $i$ such that $\dim \text{Supp} M = i+1$ by Theorem 3.5.7 in Cohen-Macaulay Rings by Bruns and Herzog.

In particular, if $\dim \text{Supp} M \geq 2$, then we have non-vanishing $R^i j_* j^* M$. Now, if $\dim \text{Supp} M = 1$ you can still do something similar (the dimension can't be zero by our previous assumptions).

In this case, it follows that $H^1_m(M) \neq 0$ and in fact is infinitely generated. Thus $H^0(U, M|_U)$ is also infinitely generated since we have the exact sequence $$ M \to H^0(U, M|_U) \to H^1_m(M). $$

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Thanks, I just figured this out by myself. It already fails for $\mathbb{A}^2-0$. Too bad. –  temp May 18 '12 at 7:01
    
I guess a better question I can ask is, how likely is $Rj_\star j^\star \mathscr{F}\to \mathscr{F}$ a monomorphism in the derived category? –  temp May 18 '12 at 12:56
    
What do you mean by a monomorphism? –  Sasha May 18 '12 at 19:34
    
In the sense of category theory (where the category is the derived category), see en.wikipedia.org/wiki/Monomorphism –  temp May 19 '12 at 1:39
    
A monomorphism (in the categorical sense) in any triangulated category always splits as a direct summand embedding. This happens very rarely, so you can't count on this. –  Sasha May 19 '12 at 3:39
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