Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is a question on geometric tilting theory. On smooth projective variety it is possible to define in general tilting object as perfect complex that satisfy some properties, but are there examples of tilting objects that are actually complexes? All examples I know are vector bundles obtained as sums of exceptional sequences. What would be examples of different nature?

Edit: if we denote variety $X$ then tilting object $T$ in $D(X)$ is a perfect complex (or from the point of view of abstract triangulated categories "compact object") such that $T$ generates $D(X)$ and $Ext^i(T,T)=0$ for $i \neq 0$.

share|improve this question
    
Could you add the definition of a tilting complex for concreteness? –  Benjamin Antieau Mar 5 '13 at 19:14

2 Answers 2

up vote 3 down vote accepted

The answer to your more specific question is yes, there are tilting objects that involve complexes not quasi-isomorphic to any vector bundle. Consider the blowup $X$ of $\mathbb{P}^2$ at a single point $p$. Then, Orlov showed that there is a semiorthogonal decomposition $D^b(X)=\langle e,O_X,O_X(1),O_X(2)\rangle$, where $O_X(i)$ is the pullback of $O_{\mathbb{P}^2}(i)$, and where $e$ can be taken to be $i_*O_E(-1)$, where $i:E\rightarrow X$ is the inclusion of the exceptional divisor. The objects $e,O_X(i)$ are in fact all exceptional, so their direct sum is a tilting object for $D^b(X)$. However, $e$ is not quasi-isomorphic to a vector bundle, because it is supported along the exceptional divisor.

share|improve this answer
    
dear benjamin, +1, I now await your third answer where you exhibit a tilting object with cohomology concentrated not just in degree zero! –  Jacob Bell Mar 9 '13 at 21:38
    
thanks for the comment –  Jacob Bell Mar 11 '13 at 13:31
    
Exceptional sequences of line bundles having non-zero cohomology certainly exist (for instance, see Hille-Perling 2006). By tensoring with a suitable line bundle, we assume that the structure sheaf $O_X$ appears as an element of the sequence. However, the direct sum of these line bundles is not a tilting complex, as usually defined. The existence of the cohomology is an obstruction to finding an equivalence $D^b(X)\cong D^b(A)$ where $A$ is an ordinary associative algebra. An arbitrary exceptional sequence determines an equivalence $D^b(X)\cong D^b(A)$, but where $A$ is a dg algebra. –  Benjamin Antieau Mar 11 '13 at 17:29

One trivial example is the derived category of a central simple algebra $D(A)$ where the class of $A$ in the Brauer group is non-zero. A non-zero simple left ideal $I$ of $A$ is a tilting complex, but the endomorphism algebra of $I$ is not the base field $k$ but rather $D$, the division algebra Morita equivalent to $A$. So, it is not an exceptional object. Similar examples arise in the derived categories of Severi-Brauer varieties.

share|improve this answer
    
could you expand your comment on Brauer-Severi varieties please? –  Jacob Bell Mar 5 '13 at 23:18
    
A Severi-Brauer variety is a twisted form of projective space over a field. The first example is a smooth projective genus $0$ curve without any points. Such a curve $C$ is the SB variety associated to a quaternion algebra $D$. The derived category of $C$ has a semiorthogonal decomposition $<e_1,e_2>$ where $Hom(e_1,e_1[n])=0$ if $n\neq 0$ and $k$ if $n=0$, while $Hom(e_2,e_2[n])=0$ if $n\neq 0$ and $D$ if $k=0$. Thus, $e_2$ is not an exceptional object in the classical sense, but the object $e_1\oplus e_2$ is a tilting complex. This was worked out in detail by Marcello Bernardara in 2009. –  Benjamin Antieau Mar 6 '13 at 3:36
    
I'm sorry, but I'm nor familiar with SB varieties, but thank for bringing this papers to my attention, it looks like beautiful and interesting result, definitively something that I want to learn, but I need some time first on SB varieties. –  Sasha Pavlov Mar 9 '13 at 12:30

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.