Post 7: Perverse Sheaves Part 3

03 Jan 2024

In this post, I’ll talk about derived categories, the kind of category needed to define perverse sheaves.

For experienced readers: As in the previous discussions about categories, I will be ignoring set-theoretical size issues.

I want to start by addressing a personal misconception that I imparted onto my previous posts. Namely, I said that passing to the homotopy category $K(\mc A)$ was essentially optional for the development I was going for. However, it turns out that there is a more technical problem I was ignoring. While $K(\mc A)$ is still optional, describing the derived category without it is much more complicated.

Intuition

In Part 2, I discussed abelian categories and their homotopy categories. In particular, I defined a special type of morphism: the quasi-isomorphisms. These are the morphisms between chain complexes that induce isomorphisms in cohomology.

The derived category is essentially obtained by “making the quasi-isomorphisms into isomorphisms”. Let me begin with a familiar example.

Numbers? In my abstract math?

Suppose you have the integers, $\Z$, and you want to create the rational numbers, $\Q$. How do you do this formally? Even if you went through a whole undergraduate education, you might not know - it is often glossed over. We take for granted the fact that we can have fractions. Here’s one way to approach this formally:

To define addition and multiplication of fractions, we define the corresponding operations on pairs, and then check that these operations are well-defined up to equivalence. I won’t go through this, though.

Calculus of fractions

It turns out that if we take a “particularly nice” collection $S$ of morphisms in a category $\mc C$, we can play almost the same game in order to make a new category in which the morphisms in $S$ are isomorphisms. There are two common ways to denote this category, either $S^{-1}\mc C$ or $\mc C[ S^{-1} ]$. I’ll use the first one. Here’s how it is defined:

  1. The objects of $S^{-1}\mc C$ are the objects of $\mc C$.

  2. A roof diagram from $X$ to $Y$ is a pair $(f:A\to Y, s:A\to X)$ with $s\in S$.

  3. Two roof diagrams from $X$ to $Y$, $(f:A\to Y, s:A\to X)$ and $(g:A'\to Y,t:A'\to X)$, are equivalent if there are morphisms $a:A''\to A$ and $b:A''\to A’$ such that $s\circ a = t\circ b$ is in $S$ and $f\circ a=g\circ b$.

  4. A morphism between objects $X$ and $Y$ in $S^{-1}\mc C$ is an equivalence class of roof diagrams from $X$ to $Y$.

So, the idea you should have in mind is that a morphism in the new category is a fraction, where the denominator is an element of your special class of morphisms $S$. For instance, all the morphisms in $\mc C$ induce morphisms in $S^{-1}\mc C$ by taking the “denominator” to be an identity morphism (which we must assume is in $S$).

Similarly, the inverse of a morphism $s$ in $S$ is the “fraction” with “numerator” an identity morphism and “denominator” $s$.

Furthermore, the idea of composing morphisms is akin to adding fractions. How do you add fractions? First, you need to find a common denominator, and then you add the denominators. In much the same way, to compose roof diagrams, you need to find a roof diagram that “fits on top of them”, and then compose along the sides. I won’t discuss exactly what that means in this section, but you can take a peek in the formalities section if you want.

How do you know such a common roof diagram exists? Well, that was hidden in the phrase “particularly nice” in the beginning of this discussion. The ability to find such common roofs is a requirement if we are to form this new category in this manner! There are some other requirements as well, and I’ll talk about them in the formalities section.

You should be aware that you can define $S^{-1}\mc C$ for any collection of morphisms $S$. However, there is no simple description of the morphisms in the new category. You need arbitrary “zig-zags” instead of just roof diagrams. This is a reason for passing to $K(\mc A)$ in order to define the derived category.

The upshot

For understanding derived categories, or even for understanding perverse sheaves, you don’t need to think about the technicalities about composing morphisms - all that matters is that you can do it. In all the time I’ve worked with perverse sheaves, which admittedly isn’t that long, I’ve never had to break out a roof diagram.

Deriving your category

Now that we have the ability to localize, or “formally invert”, morphisms, we can finally define the derived category of an abelian category $\mc A$.

Yep, “that’s it”! Now, there’s all sorts of technicalities one should tend to, but if you want to enjoy your life, I suggest not worrying about it.

In case you forgot, or didn’t see, let me address the part about $\mc C$ being either $Ch(\mc A)$ or $K(\mc A)$. $K(\mc A)$ was constructed out of $Ch(\mc A)$, so you may wonder why one would bother going through $K(\mc A)$ if you only care about $D(\mc A)$. This is because the roof diagram method doesn’t work for $Ch(\mc A)$. You can still define localization without roof diagrams, but it is messier.

Let me also discuss, without proof, why we get the same category if we localize $Ch(\mc A)$ or $K(\mc A)$. While I didn’t define it this way, one can show that $K(\mc A)$ is actually a localization of $Ch(\mc A)$ at the collection of “(chain) homotopy equivalences”.1 Every homotopy equivalence is a quasi-isomorphism, so localizing at homotopy equivalences and then quasi-isomorphisms is the same as just localizing at quasi-isomorphisms.

Bounded variants

Finally, let me mention some full subcategories2 of the derived category that one also encounters for technical reasons that we’ll see soon. They are called the bounded below, bounded above, and bounded derived categories. In that same order, these are denoted by $D^+(\mc A)$, $D^-(\mc A)$, and $D^b(\mc A)$. Here are the definitions:

Induced maps in cohomology

Just like with $Ch(\mc A)$ and $K(\mc A)$, morphisms in $D(\mc A)$ have induced maps in cohomology. In particular, if a morphism $X\to Y$ is represented by a roof diagram $(f:A\to Y,s:A\to X)$, then for each integer $i$ we have a map $H^i(f)\circ H^i(s)^{-1}:H^i(X)\to H^i(Y)$. Notice that we are using the fact that $s$ is a quasi-isomorphism in order to take $H^i(s)^{-1}$.

As an exercise, show that this induced map is independent of the choice of roof diagram for the morphism.

Zero cohomology implies zero object

I want to mention a neat property of the derived category that might give you some intuition for why we have defined it.

In $K(\mc A)$ and $D(\mc A)$, we have a $0$ object, which we can explicitly give as a chain complex consisting of $0$ objects in $\mc A$.

In $K(\mc A)$, we can have a chain complex that has all of its cohomology objects being zero, but it is not a zero object. A particular example is the chain complex $0\to\Z\xr{\times 2}\Z\to \Z/2\Z\to 0$ of abelian groups, where the nonzero morphisms are multiplication by 2 and remainder of division by 2. The infinitely many zeroes on the ends of this complex have been suppressed.

On the other hand, if an object in $D(\mc A)$ has all its cohomology objects being zero, then it is a zero object. The proof is simple. There is always a zero map from the chain complex of zero objects to any other chain complex $A^\bullet$. If $A^\bullet$ has all cohomology objects zero, then this map induces isomorphisms in all the cohomology objects, which by definition, is an isomorphism in $D(\mc A)$. An object isomorphic to a zero object is a zero object, so we are done.

That’s it for the intuition section, but see below for a sneak peek at what’s coming up.

Formalities

Requirements for calculus of fractions

In order to define $S^{-1}\mc C$ with roof diagrams, one needs the following assumptions on $S$.

  1. (Identity) For all $X\in \mc C$, we have $1_X\in S$.

  2. (Composition) For all $f,g\in S$ such that $f\circ g$ is defined, we have $f\circ g\in S$.

  3. (Common Roof) For each pair of morphisms $f:A\to Y$ and $t:A'\to Y$, with $t\in S$, there exist morphisms $k:A''\to A'$ and $v:A''\to A$ with $v\in S$ and $f\circ v = t\circ k$.

  4. Let $f,g:X\to Y$. Suppose there is $s:Y\to Y'$ in $S$ such that $s\circ f = s\circ g$. Then there is $t:X'\to X$ in $S$ such that $f\circ s = g\circ s$.

To be completely honest, I’m not sure what the fourth assumption is used for. I would guess that it is used in showing that composition of roof diagrams is well-defined or associative up to equivalence.

Let me spell out how composition of roof diagrams is defined. Suppose we have roof diagrams from $X$ to $Y$ and from $Y$ to $Z$; say $s:A\to X$, $f:A\to Y$, $t:A'\to Y$, $g:A'\to Z$ with $s,t\in S$. We want to make a roof diagram from $X$ to $Z$, so we need $u:A''\to X$ and $h:A''\to Z$ with $u\in S$. Invoking the third assumption, we get morphisms $v:A''\to A$ in $S$ and $k:A''\to A'$. By assumption 2, $s\circ v$ is in $S$, so $s\circ v$ and $g\circ k$ give us a roof diagram from $X$ to $Z$ as desired.

As an exercise, show that the collection of quasi-isomorphisms in $Ch(\mc A)$ does not satisfy all of these assumptions. In this footnote3 I’ll outline an example given to me by “six” on Discord.

For a definition of localization without roof diagrams, see Section 1.1 here (opens in new tab).

Universal property of localization

The category $S^{-1}\mc C$ actually enjoys itself a universal property; c.f. my last post for a discussion on these. Namely, it is an initial pair $(S^{-1}\mc C, Q:\mc C\to S^{-1}\mc C)$ where the functor $Q$ sends any element of $S$ to an isomorphism.

Let me spell out the universal property. If $F:\mc C\to \mc D$ is a functor that sends any element of $S$ to an isomorphism, then there is a unique functor $\ol F:S^{-1}\mc C\to \mc D$ such that $F=\ol F\circ Q$.

What’s next?

We need one more algebraic tool before we can get to perverse sheaves. Namely, in order to define perverse sheaves, we need the concept of derived functors. The more classical point of view on derived functors is that they are a sequence of functors that “measure” the failure of a left or right exact functor to be exact. On the other hand, the more modern point of view is that they are extensions of left or right exact functors between abelian categories to functors between their derived categories.

Other than that intuition for what derived functors are, I can’t impart any intuition as to why we need them; it is just part of the definition of perverse sheaves.

Until next time!

Footnotes:

  1. A map $f:X\to Y$ is a homotopy equivalence if there is a map $g:Y\to X$ such that $f\circ g\sim 1_Y$ and $g\circ f\sim 1_X$. ↩

  2. This just means you restrict your objects and keep all the morphisms between them. ↩

  3. We work with $\mc A = $ AbGrp. Consider the complexes $X:$ $0\to \Z\to 0$ and $Y:$ $0\to\Z\to \Z \to 0$, with the second $\Z$ of the $Y$ at the same degree as the $\Z$ in the $X$. Let $f:X\to Y$ be the map with $f^0=1_{\Z}$, and let $g$ be the zero map. Then the unique map $s$ from $Y$ to the complex of zeroes is a quasi-isomorphism and $s\circ f = s\circ g$. However, there is no quasi-isomorphism $t$ with $f\circ t = g\circ t$; otherwise, the zero map would be an automorphism of $\Z$. ↩