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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!

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2 Answers 2

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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.

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  • $\begingroup$ Thanks for this answer as the equivalences I am really interested in are indeed the identity and the Serre functor! Still, I am not sure I understand the last bit. To make your example really an example, you must be able to prove that your natural transformation from the identity to the Serre functor is not zero. Your example is for isntace a non-example on an elliptic curve, as all natural transformations from the identity to the Serre functor vanish. Do you know some examples of smooth projective varieties for which the natural transformations you defined is not zero? $\endgroup$
    – Libli
    Aug 30, 2015 at 18:15
  • $\begingroup$ @Libli My natural transformation is non-zero on $M$. $\endgroup$ Aug 30, 2015 at 18:18
  • $\begingroup$ Ouuups sorry I made a stupid mistake with my example of the elliptic curve (I confused $id[2]$ and $id[1]$). So I understand now your example. It is great! Thank you so much! $\endgroup$
    – Libli
    Aug 30, 2015 at 19:21
  • $\begingroup$ This wonderful extremely enjoyable answer shows that there are many natural transformations $\rho : 1 \to S$. As far as I can see these examples are not $2$-morphisms in the $2$-category of triangulated categories, i.e., they are not compatible with distinguished triangles, i.e., not compatible with the shift functor. $\endgroup$ Aug 31, 2015 at 1:20
  • $\begingroup$ @CountDracula Yes. Though you can take the sum of the $\rho_M$ for an arbitrary set of isomorphism classes of objects $M$ (this makes sense since on each object of $D^b(X)$ all but finitely many of the $\rho_M$ vanish), and if you do this with a set of objects closed under the shift then you'll produce a natural transformation compatible with the shift functor. $\endgroup$ Aug 31, 2015 at 8:28
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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).

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  • $\begingroup$ Thank you very much for your answer. It's nice example! Though I would have liked some conditions (on the generator for instance) to guarantee that the transformation vanishes. $\endgroup$
    – Libli
    Aug 30, 2015 at 18:07

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