most recent 30 from http://mathoverflow.net 2013-05-20T18:00:00Z http://mathoverflow.net/feeds/question/109909 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109909/derived-push-forward-and-pull-back Wajcha 2012-10-17T13:41:37Z 2012-10-17T14:13:06Z

<p>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 $$j^*j_*F=F\otimes j^*j_*O_D?$$</p>

http://mathoverflow.net/questions/109909/derived-push-forward-and-pull-back/109911#109911 Sasha 2012-10-17T14:13:06Z 2012-10-17T14:13:06Z

<p>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 $j^*j_*F$.</p>