I would appreciate either an explanation or a reference for what is going on here.


Let $f : X \rightarrow Y$ be a morphism of algebraic varieties. The derived projection formula implies that for a sheaf $\mathcal{F}$ on $Y$, we have

$$Rf_\ast Lf^\ast \mathcal{F} \cong Rf_\ast \mathcal{O}_X \otimes^L \mathcal{F}.$$

Suppose for example that $R^i f_\ast \mathcal{O}_X = 0$ when $i>0$, and is $\mathcal{O}_Y$ when $i=0$. It's natural to guess that the cohomology of the right hand side is just $\mathcal{F}$ in degree 0 and 0 otherwise.


Is this true? More generally, is there a spectral sequence calculating the cohomology of the composite of a right derived- with a left derived functor?

Is there an exact sequence of terms of low degree, as there is for the composite of two right derived functors?


1 Answer 1


The answer is yes. Now, concerning the general question - it is not important here that one functor is right derived and the other is left derived. You just have triangulated functors between triangulated categories. Also you have t-structures on all categories which give you filtrations on all objects. And the spectral sequence controls their interaction.

To be more precise, let $F:T_1 \to T_2$ and $G:T_2 \to T_3$ be two triangulated functors. You want to apply $G\circ F$ to a pure (with only one nontrivial cohomology) object $X \in T_1$ and compute the cohomology of $G(F(X))$. First, let us denote by $Y_i$ the $i$-th cohomology of $F(X)$. In other words, there is a filtration on $F(X)$ with factors $Y_i$. Now we apply the functor $G$. We obtain an object $G(F(X))$ with a filtration with factors $G(Y_i)$. Now we want to compute the cohomology of this object. These are computed by the spectral sequence. So, you see, there is nothing special about the functors, the question is about filtered objects in triangulated categories.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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