Why would one expect a derived equivalence of categories to hold? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:56:04Z http://mathoverflow.net/feeds/question/17886 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17886/why-would-one-expect-a-derived-equivalence-of-categories-to-hold Why would one expect a derived equivalence of categories to hold? Chuck Hague 2010-03-11T18:03:58Z 2010-03-14T17:14:26Z <p>This question is perhaps somewhat soft, but I'm hoping that someone could provide a useful heuristic. My interest in this question mainly concerns various derived equivalences arising in geometric representation theory.</p> <p><strong>Background</strong></p> <p>For example, Bezrukavnikov, Mirkovic, and Rumynin have proved the following: Let $G$ be a reductive algebraic group over an algebraically closed field of positive characteristic. Then there is an equivalence between the bounded derived category of modules for the sheaf $\cal D$ of crystalline (divided-power) differential operators on the flag variety, and the bounded derived category of modules with certain central character for the enveloping algebra $\cal U$ of Lie($G$). What is interesting is that it is <em>not</em> true that this equivalence holds on the non-derived level: The category of $\cal D$-modules is not equivalent to the category of $\cal U$-modules with the appropriate central character. This is true in characteristic 0 (this is the Beilinson-Bernstein correspondence), but something is broken in positive characteristic: there are certain "bad" sheaves that are $\cal D$-modules which make the correspondence not hold.</p> <p>There are other results in geometric representation theory of this form. For example, Arkhipov, Bezrukavnikov, and Ginzburg have proved that there is an equivalence (in characteristic 0) between the bounded derived category of a certain block of representations for the quantum group associated to $G$, and the bounded derived category of $G \times \mathbb C^*$-equivariant sheaves on the cotangent bundle of the flag variety of $G$. Again, this equivalence does not hold on the non-derived level.</p> <p>In general, there are a number of results in geometric representation theory that hold on the derived level, but not the non-derived level.</p> <p><strong>Question</strong></p> <p>Here's my question: Why would one be led to expect that a derived equivalence holds, when the non-derived equivalence does not? It seems as though the passage to the derived level in some sense fixes something that was broken on the non-derived level; how does it do that?</p> http://mathoverflow.net/questions/17886/why-would-one-expect-a-derived-equivalence-of-categories-to-hold/17894#17894 Answer by JosÃ© Figueroa-O'Farrill for Why would one expect a derived equivalence of categories to hold? JosÃ© Figueroa-O'Farrill 2010-03-11T19:01:31Z 2010-03-11T19:01:31Z <p><strong>Disclaimer</strong>: I am not an expert.</p> <p>Curiously I just returned from a wonderful geometry seminar by Richard Thomas, who gave precisely a heuristic which addresses, perhaps if not your exact question, why you sometimes have to go to the derived category in order to get the equivalence.</p> <p>This is in the context of homological mirror symmetry à la Kontsevich. Roughly speaking (did I mention I was not an expert?) homological mirror symmetry is a conjectural equivalence between derived categories: the derived category of coherent sheaves in a Calabi-Yau manifold and the derived Fukaya category of the mirror. The latter makes sense for symplectic manifolds, whereas the former for complex manifolds. Hence in a sense homological mirror symmetry is relating symplectic geometry to complex geometry.</p> <p>Now symplectic geometry is very "floppy": symplectomorphisms are a dime a dozen and hence the Fukaya category has many autoequivalences, unlike the category of coherent sheaves due to the rigidity of complex geometry. Therefore one would not expect an equivalence of categories. What passing to the derived category does is to provide the extra autoequivalences which mirror symmetry requires.</p> <p>I'm sure others here can make this much more precise and I, for one, would enjoy reading their version. This is why I'm making this answer into community wiki.</p> http://mathoverflow.net/questions/17886/why-would-one-expect-a-derived-equivalence-of-categories-to-hold/17907#17907 Answer by Ben Webster for Why would one expect a derived equivalence of categories to hold? Ben Webster 2010-03-11T20:34:39Z 2010-03-12T01:50:20Z <p>One crude answer is that passing to derived functors fixes one obstruction to being an equivalence. Any equivalence of abelian categories certainly is exact (i.e. it preserves short exact sequences), though lots of exact functors are not equivalences (for example, think about representations of a group and forgetting the G-action). </p> <p>What derived functor does is fix this problem in a canonical way; you have to replace short exact sequences with exact triangles, but you get a functor which is your original "up to zeroth order," exact, and uniquely distinguished by these properties.</p> <p>So, what BMR do is take a functor which is not even exact (and thus obviously not an equivalence), and show that the lack of exactness is "the only problem" for this being an equivalence.</p> <p><strong>EDIT:</strong> Let me just add, from a more philosophical perspective, that derived equivalences are just a lot more common. There are just more of them out in the world. Given an algebra A, Morita equivalences to A are classified essentially by projective generating A-modules, whereas derived Morita equivalences of dg-algebras are in bijection with all objects in the derived category of $A-mod$ which generate (in the sense that nothing has trivial Ext with them): you look at the dg-Ext algebra of the object with itself. If you have an interesting algebra (say, a finite dimensional one of wild representation type), there are a lot more of the latter than the former in a very precise sense. Of course, the vast majority of these are completely uncomputable an tell you nothing, but there are enough of them in the mix to make things interesting.</p> http://mathoverflow.net/questions/17886/why-would-one-expect-a-derived-equivalence-of-categories-to-hold/17910#17910 Answer by Liran Shaul for Why would one expect a derived equivalence of categories to hold? Liran Shaul 2010-03-11T20:40:47Z 2010-03-11T20:40:47Z <p>The following might be useful for your question:</p> <p><a href="http://arxiv.org/pdf/math.RA/9810134" rel="nofollow">http://arxiv.org/pdf/math.RA/9810134</a></p> http://mathoverflow.net/questions/17886/why-would-one-expect-a-derived-equivalence-of-categories-to-hold/18010#18010 Answer by Geordie Williamson for Why would one expect a derived equivalence of categories to hold? Geordie Williamson 2010-03-12T20:51:48Z 2010-03-14T17:14:26Z <p>Another motivation is more basic. For example, in the modular representation theory of finite groups it is often the case that one has two blocks $A$ and $B$ of two different groups and one suspects (for example) that both blocks have the same number of simple modules (an example of this is given by Alperin's conjecture).</p> <p>Now, suppose that $A$ and $B$ are Morita equivalent. Then not only do $A$ and $B$ have the same number of simple modules, but specifying a Morita equivalence gives a bijection between the simple modules.</p> <p>Often, however, $A$ and $B$ are not Morita equivalent, but rather derived equivalent. The Grothendieck groups of $D^b(A)$ and $D^b(C)$ have a basis given by the classes of the simple modules of $A$ and $B$. A derived equivalence induces an isomorphism between the Grothendieck groups, and hence $A$ and $B$ have the same number of simple modules if they are derived equivalent. Note now, however, that the derived equivalence does not induce a bijection between simple modules, because simple modules need not correspond under the derived equivalence.</p> <p>As an example of this approach is Broué's abelian defect group conjecture (which predicts a derived equivalence between certain $A$ and $B$). It implies Alperin's conjecture (in the abelian defect case), but provides a structural reason for the equality (and also implies much more structural results about characters, "perfect isometries" ...).</p> <p>Hence, one may search for a derived equivalence to give a structure explanation for various concrete numerical equalities.</p> <p>(I think another example of this is given by Bezrukavnikov's use of perverse coherent sheaves on the nilpotent cone to explain some numerical equivalences observed by Vogan, but I don't know much about this.)</p> http://mathoverflow.net/questions/17886/why-would-one-expect-a-derived-equivalence-of-categories-to-hold/18014#18014 Answer by Greg Muller for Why would one expect a derived equivalence of categories to hold? Greg Muller 2010-03-12T21:37:40Z 2010-03-12T21:37:40Z <p>I asked a similar question to Daniel Huybrechts some time ago, in the form of "If I have a derived equivalence between two varieties, what is this telling me about the relation between the two varieties?" </p> <p>His answer was that this should be crudely regarded as saying that each of the varieties is the moduli space for a (sufficiently interesting) moduli problem on the other variety. This is definitely the 'Fourier-Mukai' perspective, that says to understand a derived equivalence, see where the skyscraper sheaves of points go under the equivalence.</p> <p>It doesn't answer your question in general, but it makes a useful heuristic in some geometric cases.</p>