# The Derived Category 2

I thought we would be able to move on to dg-categories today, but I was wrong. There was too much we didn’t do last time, plus an example might be nice. Recall that last time we took an arbitrary abelian category ${\mathcal{A}}$ and created the derived category ${D(\mathcal{A})}$ by making the category whose objects are complexes and morphisms are equivalence classes up to homotopy equivalence and we “invert” all quasi-isomorphisms (they become isomorphisms).

This post should be taken as just a smattering of remarks on this construction plus focusing in on the actual category we want to consider. Since the first step of the construction was to take the category of complexes we can actually alter this part of the construction in several natural ways and complete the next two steps without any problem. We called the category of complexes ${Kom(\mathcal{A})}$. Instead of using this whole category we might be interested in just the complexes that are bounded below (meaning ${A^i=0}$ for all ${i<<0}$) or bounded above or just plain bounded. These will be denoted ${Kom^+(\mathcal{A})}$, ${Kom^-(\mathcal{A})}$, or ${Kom^b(\mathcal{A})}$ respectively. Finishing the construction of the derived category in the same way with these gives ${D^+(\mathcal{A})}$, ${D^-(\mathcal{A})}$ and ${D^b(\mathcal{A})}$.

For the purposes of this post we will focus on the bounded derived category. The first remark is that if ${\mathcal{A}}$ has enough injectives, then in ${D^b(\mathcal{A})}$ for any ${A^\bullet}$ there is a complex ${I^\bullet}$ of injective objects such that they are isomorphic ${A^\bullet\simeq I^\bullet}$. I.e. any bounded complex is quasi-isomorphic to a complex of injectives and so if all we care about are objects up to isomorphism we may as well just work with complexes of injective objects.

In fact, something much stronger is true. Take ${\mathcal{I}\subset \mathcal{A}}$ to be the full subcategory of injective objects, then we have an equivalence of categories ${K^+(\mathcal{I})\simeq D^+(\mathcal{A})}$. Now let's shift our focus to ${D^b(Coh(X))}$ where ${X}$ is a scheme. Since this is the category that will appear the most we will simply denote it ${D^b(X)}$ and call it "the derived category of ${X}$" instead of the more accurate and cumbersome "the bounded derived category of coherent sheaves on $X$".

If you're paying attention you should be objecting to all of this post at this point. My remarks are useless here because ${Coh(X)}$ might be a nice abelian category, but it certainly doesn't have enough injectives. The fix for this is that in a sense that we'll make precise it is often good enough to take your complex of injectives in a category that has enough injectives as long as when you take cohomology it lands you back in the right place.

The exact statement is that if ${\mathcal{A}\subset \mathcal{B}}$ is a thick subcategory and any ${A}$ can be embedded in some ${I\in \mathcal{A}}$ which is injective as an object of ${\mathcal{B}}$ then we get an equivalence of categories ${D^b(A)\simeq D^b_\mathcal{A}(\mathcal{B})}$ where the notation ${D^b_\mathcal{A}(\mathcal{B})}$ means the full subcategory of ${D^b(\mathcal{B})}$ of complexes with cohomology in ${\mathcal{A}}$.

Now to not get in trouble at some point let's suppose our scheme ${X}$ is always a smooth projective variety over a field ${k}$ (this is the only case we'll consider in mirror symmetry so it's fine for these purposes). If we just take for a moment the category ${QCoh(X)}$ of quasi-coherent sheaves on ${X}$, then it is a standard theorem in algebraic geometry that any quasi-coherent sheaf ${\mathcal{F}}$ has an injective resolution ${0\rightarrow \mathcal{F}\rightarrow \mathcal{I}^\bullet}$ by quasi-coherent sheaves that are injective as ${\mathcal{O}_X}$-modules. If you think of the definition of the sheaf cohomology ${H^i(X, \mathcal{F})}$ in relation to the cohomology of the injective resolution complex it immediately shows that we can view ${D^b(QCoh(X))}$ as the full subcategory ${D^b_{QCoh}(Sh_{\mathcal{O}_X}(X))}$ of the bounded derived category of sheaves of ${\mathcal{O}_X}$-modules on ${X}$.

Even though this can't be done directly by some general theorem as in the quasi-coherent case we actually do get a theorem that parallels that case (again, warning: this cannot be done the same way since coherent things will not be injective). There is a natural functor ${D^b(X)\rightarrow D^b(QCoh(X))}$ and it induces an equivalence of ${D^b(X)}$ with the full subcategory ${D^b_{Coh}(QCoh(X))}$. This means that to work with the derived category of ${X}$ we can always replace (up to isomorphism) a complex of coherent sheaves by a complex of injective quasi-coherent sheaves whose cohomology is coherent.

There are so many interesting things we could talk about at this point (a derived form of Serre duality, how derived functors relate to this, when can you recover a variety from its derived category, Fourier-Mukai transforms, and so on…), but my goal is to keep trudging towards a statement of mirror symmetry and stopping to talk about all these things would lead us too far astray. Although, talking about a few of them might help some of you who might be in unfamiliar territory here feel a little more comfortable.

The last thing we'll look at today is an example. This might be completely unenlightening because I won't take the time to properly define everything, but we'll do it anyway. What is ${D^b(\mathbb{P}^n)}$ or even just ${D^b(\mathbb{P}^1)}$? We will say that the derived category ${D^b(X)}$ is generated by a collection of objects ${S}$ if the smallest full subcategory that contains ${S}$ and is closed under taking shifts and cones (we haven't defined that, but recall that the derived category satisfies this condition called being triangulated and those conditions just make sure the generated category is still triangulated) is ${D^b(X)}$ itself.

For some intuition, what this means is that any object of ${D^b(X)}$ can be reached by doing the two main operations of ${D^b(X)}$ a finite number of times using only the objects of ${S}$. In the case of projective space we get that ${D^b(\mathbb{P}^n)}$ is generated by just ${n+1}$ objects ${S=\{\mathcal{O}(n), \mathcal{O}(n-1), \ldots, \mathcal{O}(1), \mathcal{O}\}}$.

1. Can you say a little more (or give a reference) about the equivalence between $\mathcal{D}^b(\mathcal A)$ and $\mathcal{D}^b(\mathcal I)$? It would sound less surprising if by bounded you meant bounded cohomology. In a category where you have objects of infinite injective dimension, the naive approach to establishing an equivalence (inclusion in one direction, taking resolutions in the other) does not work, and you cannot produce triangulated functors which commute with taking cohomology, so I’m not sure how such a proof might even go.