Vanishing natural transformation and strong generator 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!
 A: The "obvious" method to attempt a proof is an induction argument where you induct on the number of operations from your list of four operations needed to produce a given object, and show that the morphism is zero on this object.
This argument is fine for the first three operations (I think - maybe with some mild assumptions) but you run into a problem with cones.
Here's a counterexample demonstrating the problem, though in this case $X$ is not actually smooth projective.
Let's work with $X = \operatorname{Spec} k[t]/t^2$ and let's use $D^b(Coh)$, not the category of perfect complexes. Then $k$ is an element of this category, and generated, because $\mathcal O_X$ is an extension of two copies of $k$. There is a natural transformation, multiplication by $t$, from the identity functor to the identity functor, that is zero on this object but is nonzero in general.
The problem is that a natural transform can be zero on two objects $A ,B$ with a map $A \to B$ , but nonzero on their mapping cone.
I don't see any reasonable condition you can place that will avoid this obstruction, because you have to guarantee that whenever certain maps vanish ( $\rho_A: F(A) \to G(A)$ and $\rho_B: F(B) \to G(B)$ ) some higher homotopies also vanish (the homotopy making the natural transformation diagram commute).
A: 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 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.
