Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

All functors are derived and all categories are bounded derived categories of coherent sheaves. Suppose that we have got an inclusion of a smooth divisor $j:D\rightarrow X$ in a smooth projective variety. Is it true that $$j^*j_*F=F\otimes j^*j_*O_D?$$

share|improve this question
I should add a few remarks: 1) This is not the well known $j_*j^*F=F\otimes j_* O_D$ 2) If it is not well known it is probably wrong, but there is always a triangle: $F\otimes O_D(-D)[1]\rightarrow j^*j_*F \rightarrow F$ by Bondal, Orlov "Semiort. decomp." Lemma 3.3 3) We can calculate RHS by restricting and tensoring the resolution $0\rightarrow O(-D)\rightarrow O\rightarrow j_*O_D\rightarrow 0$ As, after restricting to $D$ the map is zero, we see that RHS is a sum of $F$ and $F\otimes O_D(-D)[1]$. In particular it fits to the previous triangle (that maybe does not split). –  Wajcha Oct 17 '12 at 13:50
4) If $F$ is an injective coherent sheaf then LHS and RHS fit into the triangle: $(LvR)HS\rightarrow F\rightarrow F\otimes O_D(-D)[2]$ with the last morphism being zero, as these are injective sheaves in diffrent shifts. 5)There is a morphism from LHS to RHS induced by a morphism from $j_* F$ to $j_*(F\otimes j^*j_*O_D)=j_*F\otimes j_*O_D$, that comes from $O_X\rightarrow j_*O_D$. –  Wajcha Oct 17 '12 at 13:58
6) It is true on the (not nec. full) subcategory $j^*(D(X))$ –  Wajcha Oct 17 '12 at 14:00

1 Answer 1

up vote 7 down vote accepted

The answer is no. The simplest example that I know is $X = P^3$, $D = P^1\times P^1$, $F = O(0,1)$. In this case $j^*j_*O_D = O_D \oplus O_D(-2)[1]$ and hence $F\otimes j^*j_*O_D = F \oplus F(-2)[1]$, while $j^*j_*F$ fits into a triangle $$ F(-2)[1] \to j^*j_*F \to F $$ which is not split. To see this note that $j_*F$ has a resolution of the form $$ 0 \to O_X(-1)^2 \to O_X^2 \to j_*F \to 0, $$ which gives a distinguished triangle $$ O_D(-1,-1)^2 \to O_D^2 \to j^*j_*F. $$ It follows easily from this that $Hom(F,j^*j_*F) = 0$, which shows that $F$ is not a direct summand of $j^*j_*F$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.