Let $X$ be a smooth projective variety (over the field of complex numbers). Let $T$ be strong generator of $D^b(X)$ : this means that every object in $D^b(X)$ can be obtained in a given finite number of steps from $T$ using the following operations : shift, direct sums, taking direct summands and cones.


Let $F,G : D^b(X) \rightarrow D^b(X)$ be two exact auto-equivalences and let $\rho : F \rightarrow G$ be a natural transformation. I assume that $\rho(T) : F(T) \rightarrow G(T)$ is zero. Are there some conditions on $T$, $F$,$G$ and $\rho$ (the weakest, the best) which would allow me to prove that $\rho$ vanishes as a natural transformation?


Thanks a lot!



Vanishing natural transformation and strong generator
This isn't really an answer to the question, but an example to show how badly things can go wrong.


Let $M$ be any indecomposable object of $D^b(X)$. The functor $\operatorname{Hom}(M,-)$, from $D^b(X)$ to vector spaces, has a subfunctor consisting of all maps from $M$ that are not split monomorphisms. Let $S\_M$ be the quotient functor, so if $M$ has multiplicity $d$ in a direct sum decomposition of $N$, then $S\_M(N)$ is a $d$-dimensional vector space.


Now let $S$ be the [Serre functor](http://ncatlab.org/nlab/show/Serre+functor) on $D^b(X)$, so that $\operatorname{Hom}\left(N\_1,S(N\_2)\right)$ is naturally dual to $\operatorname{Hom}(N\_2,N\_1)$.


For each object $N$, taking a map $\alpha:N\to N$ to the trace of $S\_M(\alpha)$ gives an element of the dual of $\operatorname{Hom}(N,N)$, and the corresponding maps $N\to S(N)$ give a natural transformation $\rho\_M$ from the identity functor to $S$ which vanishes on all objects $N$ that do not have $M$ as a direct summand.


Whatever strong generator $T$ you choose for $D^b(X)$, you can choose $M$ so that it is not a direct summand of $T$, showing that there is no strong generator that can detect natural transformations from the identity functor to the Serre functor.


