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In the setting described in Bernstein, Beilinson, and Deligne, associated to a scheme $X$, a closed subscheme $i: Z \to X$ and its open complement $j: U \to X$ we have six functors between the corresponding derived categories of étale sheaves $i^*, i_*, i^!, j_!, j^*, j_*$ (they are all derived but I will ommit the L's and R's).

My question is: if $X$ is a variety over a field $k$ with structural morphism $f: X \to k$ and $F$ is an object of $D^b(k)$, then is the canonical morphism $f^*F \to i_*i^*f^*F \oplus j_*j^*f^*F$ ever a monomorphism in the triangulated category $D^b(X)$?

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What is $a^*$? . –  Piotr Achinger Mar 10 '12 at 5:29
    
My guess is $a=f$ –  Rami Mar 10 '12 at 5:35
    
Isn't it true that $j^*$, $i^*$, $a^*$ and $j_*$ are exact? Also $a^* F$ is flat, so $i^* a^* F$ is just the usual (non-derived) pull-back? Also, $F$ is a direct sum of its cohomology. That is, you are basically asking if $\mathcal{O}_X\to \mathcal{O}_Z\oplus\mathcal{O}_U$ is a monomorphism. Is that right? –  Piotr Achinger Mar 10 '12 at 5:46
    
Since sheaves on $X$ are a full subcategory of $D^b(X)$, if this is a monomorphism, then it is still a monomorphism in the category of sheaves on $X$. This I think may fail if $X$ has an embedded point, e.g. $X = \mathrm{Spec } k[x,y]/(xy, y^2)$ and $Z = \{0\}$ with reduced closed subscheme structure. –  Piotr Achinger Mar 10 '12 at 5:52
    
Also, it fails if you can find a vector bundle $E$ such that $H^1(X, E)\neq 0$ but $H^1(U, E|_U) = 0$ and $H^1(Z, E|_Z)=0$. This happens probably quite often, but I don't know if such examples exist in general. –  Piotr Achinger Mar 10 '12 at 6:06
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2 Answers

up vote 1 down vote accepted

Let me gather my comments into an answer. Take $X$ to be a smooth projective curve, $Z$ a closed point, $U$ its complement, $F=k$. We are asking if the map $\mathcal{O}_X\to \mathcal{O}_Z\oplus \mathcal{O}_U$ is a monomorphism in the derived category.

Take $E=\omega_X^{-1}$. Let $E[-1]\to \mathcal{O}_X$ be the map corresponding to the nontrivial element of $Ext^1(\omega_X^{-1}, \mathcal{O}_X) = H^1(X, \omega_X)=k$. Then since $H^1(U, E|_U)$ and $H^1(Z, E|_Z)$ are both zero (since $Z$ and $U$ are affine, the composition $E[-1]\to \mathcal{O}_X\to \mathcal{O}_Z\oplus\mathcal{O}_U$ is zero.

It may also happen that the map $\mathcal{O}\to \mathcal{O}_Z\oplus\mathcal{O}_U$ is not a monomorphism even in the category of sheaves, e.g. $X = \mathrm{Spec} k[x,y]/(xy, y^2)$ be the line with an embedded point, $Z$ be the reduced point 0 and $U$ the complement. Then the nilpotent global section $x$ of $\mathcal{O}_X$ maps to zero in both $\mathcal{O}_Z$ and $\mathcal{O}_U$.

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What about $X=\mathbb P^1_k$, $Z$ a point, $U=\mathbb A^1_k$ the open complement, and $F=Q_\ell$ (constant sheaf) ?

Then the map $$ Hom_{D(X)}(f^*F[-2],f^*F) \longrightarrow Hom_{D(X)}(f^*F[-2],i_*i^*f^*F) \oplus Hom_{D(X)}(f^*F[-2],j_*j^*f^*F)$$ gets identified with the restriction map $$ H^2(\mathbb P^1_k, Q_\ell) \longrightarrow H^2(pt, Q_\ell) \oplus H^2(\mathbb A^1_k, Q_\ell)$$ which is clearly non injective.

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This is precisely the counterexample I got from Achinger's comment. I feel like he should get the reputation points, but I'm not really sure how to do that. To be fair, you were the one who typed it up nice. Can you share them somehow? –  name Mar 10 '12 at 16:10
    
Ah OK, I have just discovered the "show 2 more comments" button and I now understand that you got it before my answer. Well, in any case I don't care about reputation, so you should give it to Piotr. For that, he should write up an official "answer" I guess, so you can validate it as THE answer. –  Jef Mar 10 '12 at 18:42
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