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

Let $X$ be a smooth projective variety over a field, than tilting object $T$ on $X$ is a perfect complex that is a compact generator of the derived category $\operatorname{D}(QCoh(X))$ and satisfies condition $Ext^{i}(T, T)=0$ for $i \neq 0$. Tilting algebra $A=End(T)$ is a finite dimensional algebra of finite global dimension. Main result of such geometric tilting theory is an equivalence of triangulated categories $D(X)$ and $D(A)$.

Are there examples of two not isomorphic smooth projective varieties with tilting objects $(X, T)$ and $(X', T')$ such that $End_X(T) \cong End_{X'}(T')$?

In particular this would imply that $\operatorname{D^b}(X) \simeq \operatorname{D^b}(X')$, so by results of Bondal and Orlov $\omega_X$ can't be (anti-)ample, because in this case $X \cong X'$. Moreover, I suppose that if $\omega_X$ is ample than $X$ could not admit a tilting object, but I don't know how to prove this.

share|cite|improve this question
Related to your last paragraph: there are examples of non-isomorphic varieties which are derived equivalent. The derived equivalences are then called Fourier-Mukai transforms. For example, in dimension 3, all crepant resolutions of a variety with terminal singularities are derived equivalent. There is lots of information in the chapter "Derived categories of coherent sheaves on algebraic varieties" by Yukinobu Toda in Triangulated Categories, edited by Holm, Joergensen, and Rouquier, including some discussion of tilting objects, but an answer to your question didn't obviously follow for me. – Hugh Thomas Jan 10 '13 at 23:48

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.