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Aug 31, 2015 at 8:28 comment added Jeremy Rickard @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.
Aug 31, 2015 at 1:20 comment added Count Dracula 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.
Aug 30, 2015 at 19:21 vote accept Libli
Aug 30, 2015 at 19:21 comment added Libli 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!
Aug 30, 2015 at 18:18 comment added Jeremy Rickard @Libli My natural transformation is non-zero on $M$.
Aug 30, 2015 at 18:15 comment added Libli 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?
Aug 30, 2015 at 18:09 vote accept Libli
Aug 30, 2015 at 18:12
Aug 30, 2015 at 10:15 history answered Jeremy Rickard CC BY-SA 3.0