# The Derived Category 1

I promised some sort of an explanation of Kontsevich mirror symmetry. Going in there are two things to know. First, it is a conjecture so it hasn’t been established yet. Second, it isn’t even really a well-formulated conjecture. There are cases where there is a firm conjecture, but there are other cases where part of the conjecture is to figure out what the conjecture should be.

The other thing I should point out is that this is sometimes called “homological mirror symmetry”. The general form is that a certain derived category should be equivalent to a certain other category. This means that for at least the first two posts of this series I will only be describing two different categories and it won’t be at all obvious why this should be related to the mirror symmetry I alluded to last time.

Today we’ll discuss what the derived category of an abelian category is. In your head you can just keep thinking that our category is the abelian category of coherent sheaves on some sufficiently nice scheme or variety, since this is what will come up in mirror symmetry. I see absolutely no reason why I shouldn’t do the construction in the more general case of an aribtrary abelian category since it doesn’t introduce any extra difficulties (and for algebraists you can just think of ${R}$-modules or something).

The notation in the literature varies greatly. I’m going to follow Huybrechts. Let ${\mathcal{A}}$ be an abelian category. One basic related category is to form ${Kom(\mathcal{A})}$ which is the category of complexes. The objects are chain complexes ${A^\bullet}$ and morphisms ${f: A^\bullet \rightarrow B^\bullet}$ are just maps ${A^i\rightarrow B^i}$ for all ${i}$ that commute with the morphisms in the complex. This is again an abelian category and there is an embedding ${\mathcal{A}\rightarrow Kom(\mathcal{A})}$ as a full subcategory by ${A\mapsto (\cdots \rightarrow 0 \rightarrow A \rightarrow 0 \rightarrow 0 \rightarrow \cdots)}$ where ${A}$ sits in degree ${0}$.

Given any object ${(A^\bullet, d^\bullet)}$ in ${Kom(\mathcal{A})}$ (i.e. any complex of objects from ${A}$) we can take cohomology ${H^i(A^\bullet)=ker(d^i)/im(d^{i-1})}$. The cohomology lies in ${\mathcal{A}}$. Given a morphism of complexes ${f:A^\bullet \rightarrow B^\bullet}$ there is an induced map on each cohomology ${H^i(f): H^i(A^\bullet)\rightarrow H^i(B^\bullet)}$. If ${f}$ induces an isomorphism for all ${i}$ we say that ${f}$ is a quasi-isomorphism.

Loosely speaking the derived category ${D(\mathcal{A})}$ is a category (not necessarily abelian) in which quasi-isomorphisms become isomorphisms. One way to construct it is by proving the existence of a functor ${Q: Kom(\mathcal{A})\rightarrow D(\mathcal{A})}$ that takes quasi-isomorphisms to isomorphisms and ${Q}$ satisfies a certain universal property. This isn’t so easy to work with so we’ll describe ${D(\mathcal{A})}$ in a different way.

Recall that two morphisms ${f,g: A^\bullet \rightarrow B^\bullet}$ are homotopy equivalent if there is a collection of maps ${h^i: A^i\rightarrow B^{i-1}}$ such that ${f^i-g^i=h^{i+1}\circ d_A^i+d_B^{i-1}\circ h^i}$. We can form ${K(\mathcal{A})}$ often called the homotopy category of ${\mathcal{A}}$ where the objects are still just complexes, but a morphism of complexes is an equivalence class where the relation is being homotopy equivalent.

The functor ${Q}$ factors through ${K(\mathcal{A})}$. To describe the derived category ${D(\mathcal{A})}$ I’ll just tell you the objects and the morphisms. The objects are still just complexes, i.e. the objects of ${K(\mathcal{A})}$. The morphisms are trickier to describe. A morphism ${A^\bullet \rightarrow B^\bullet}$ is a choice of (equivalence class of) quasi-isomorphism ${C^\bullet \rightarrow A^\bullet}$ plus a map (in ${K(\mathcal{A})}$) ${C^\bullet \rightarrow B^\bullet}$.

This may look weird at first, but recall that all we are really trying to do is make all our quasi-isomorphisms actual isomorphisms. Thus when giving a morphism ${A^\bullet \rightarrow B^\bullet}$ you may need to pick something isomorphic ${C^\bullet}$ before defining the map, and then since ${Q}$ factors through ${K(\mathcal{A})}$ everything must be done only up to homotopy equivalences.

The derived category is no longer abelian, but it is what is known as a triangulated category. We will not talk about the general definition of one of these, since it is a sort of ad hoc, difficult, poorly behaved definition. Some people might tell you that the reason for the bad behavior is actually a result of forgetting too much stuff. You are actually decategorifying (an ${\infty}$-delooping, whatever that means) a stable ${(\infty, 1)}$-category which is a very nicely behaved structure and you may as well just work with the nicer thing.

Now whether or not you actually believe this (I’m on the fence myself) there is a much nicer, easier to understand way to say this for the derived category of coherent sheaves on a projective variety. We’ll talk about this next time, because it is actually a necessary part of the statement of Kontsevich Mirror Symmetry.

1. I’m just going to repeat a standard piece of propaganda for $(\infty, 1)$-categories: they take care of the fact that triangles are not functorial. Namely, the whole problem is that triangles come from homotopy limits (or colimits) and these are not functorial in the *homotopy* category (which is too crude to see these). Though this lack of functoriality can be taken care of using dg-categories, I believe. I wonder if this can be used to simplify a whole lot of proofs (in BBD, for instance, a lot of effort has to be spent on making triangles functorial by use of $t$-structures (under one mild vanishing hypothesis, you get functoriality of triangles even on the homotopy category — I should think of what that means $\infty$-categorically). Another consequence of this newfound functoriality is that the higher-categorical version of the derived category actually has a nice universal property!