In Huybrechts' book FourierMukai Transforms in Algebraic Geometry there is an exercise (Exercise 1.7 and 1.8) to prove the statement that the derived category $D^b(X)$ of a CalabiYau variety $X$ has no nontrivial semiorthogonal decompositions. Yet also, it is later stated that it is known that every toric variety admits a full exceptional collection. Since every full exceptional collection gives rise to a semiorthogonal decomposition (Example 1.60) and there exist toric CalabiYau varieties (though not smooth projective ones), these statements seem to contradict each other. Is someone able to explain the proof of this exercise and why it requires smoothness or projectivity?
$\begingroup$
$\endgroup$
4

$\begingroup$ the exercise, it seems, says "CalabiYau manifold", not "CalabiYau variety" (this does not substantially address your question but shows that there is no typo) $\endgroup$– user74900Commented Mar 9, 2019 at 19:18

1$\begingroup$ Directly above the exercise he writes: "CalabiYau manifolds, i.e. varieties with trivial canonical bundle", which shows that he is not assuming smooth or projective (this is the first mention of CalabiYau in this book) $\endgroup$– JeffCommented Mar 9, 2019 at 19:48

$\begingroup$ The standard proof of this exercise very much relies on smoothness + properness, in that the key point is Serre duality. $\endgroup$– dhyCommented Mar 9, 2019 at 19:52

$\begingroup$ @dhy is it possible get away with gorensteinnes+properness? $\endgroup$– user74900Commented Mar 9, 2019 at 20:01
Add a comment

1 Answer
$\begingroup$
$\endgroup$
For both statements properness is essential. The proof of the exercise (a theorem of Bridgeland) is based on Serre duality, and the construction of an exceptional collection (by Kawamata) is inductive with the base of induction given by the case of a weighted projective space. And in projective case no toric variety is CalabiYau, so there is no contradiction.