MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question comes from the first reduction step of Theorem 4.2 of Kawamata's paper on K-equivalent implies D-equivalent on toroidal varieties. But my question has little to do with this theorem.

A variety $X$ is called toroidal variety if for every closed point $x\in X$, there is a closed point $p$ in an affine toric variety $W$, such that there exists an isomorphism of complete local rings:$$\widehat{\mathcal{O}_{W,p}} \cong \widehat{\mathcal{O}_{X,x}}.$$

Suppose one wants to prove the derived equivalent of $D^b(Coh(X))$ and $D^b(Coh(Y))$, where $X,Y$ are toroidal varieties. Why it is enough to reduce to the case where $X,Y$ are toric varieties?

Kawamata gave an explanation which is unclear for me. He claimed "since the set of all the point sheaves span the derived categories, our assertion can be proved analytic locally." I know that $k(x)$ form a spanning class of $D^b(Coh(X))$, and I guess he want to use the result that

If $F: D_1 \to D_2$ is a functor between triangulated categories, and $F$ has left and right adjoints, if $\Omega$ is a spanning class of $D_1$, then $F$ is an equivalent iff $Hom(A,B[l]) \cong Hom(F(A),F(B)[l]), \forall A, B \in \Omega.$

However, I cannot see why this result can be applied locally in our case.

share|cite|improve this question
That isn't classically the definition of toroidal varieties. One also requires that the ideal of the boundary restricts locally in these analytic identifications with the toric boundary ideal of the corresponding toric variety. Your definition is that of locally toric varieties, a much weaker and less restrictive class of varieties. – HNuer Jul 14 '13 at 22:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.