Post 5: Perverse Sheaves Part 2

26 Dec 2023

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In this post, I will discuss more on categories, abelian categories, chain complexes, and the homotopy category of an abelian category.

This post is long as-is, so it won’t have a “formal” section. Instead, that will take up the next post.

For experienced readers, I’m going to ignore set-theoretic issues (sorry ☹︎). Assume all categories are small or that we’re working in a Grothendieck universe, or whatever else floats your boat.

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Categories

Categories are collections of objects and morphisms¹ between them. If $f$ is a morphism from object $A$ to object $B$, then we write $f:A\to B$. Whenever you have morphisms $f:A\to B$ and $g:B\to C$, you can compose them to get a morphism $g\circ f:A\to C$. Sometimes mathematicians get lazy and write $gf$ instead of $g\circ f$. Composition is required to satisfy two conditions:

1. $h\circ(g\circ f)=(h\circ g)\circ f$. In other words, composition is associative.

2. For every object $A$, there is an identity morphism $1_A$ from $A$ to $A$, such that $1_A\circ f = f$ and $g\circ 1_a=g$.

A morphism $f:A\to B$ is an isomorphism if it has an inverse, i.e. a morphism $g:B\to A$ such that $f\circ g = 1_B$ and $g\circ f = 1_A$.

Some people might write $\text{ob}(\mc C)$ and $\text{mor}(\mc C)$ for the collections of objects and morphisms in $\mc C$, respectively. Then $A\in\text{ob}(\mc C)$ has a clear meaning: “$A$ is an object of $\mc C$”. However, I will take another convention, and write $A\in\mc C$ to mean “$A$ is an object of $\mc C$”. In many cases, the category one is working in is clear from context, so you might not even reference the category.

For objects $A$ and $B$ in category $\mc C$, we denote the set of morphisms from $A$ to $B$ by $\Hom_{\mc C}(A,B)$, or just $\Hom(A,B)$ if the category is clear from context.

Sometimes we say “category of [objects]” instead of saying, or at least “category of [objects] with these morphisms”, instead of “category with these objects and these morphisms”. Observe what I mean in the following examples:

• Set, the category of sets with morphisms being functions.

• Grp, the category of groups with morphisms being group homomorphisms.

• Top, the category of topological spaces with morphisms being continuous functions.

• $R$-mod, the category of (left) modules for a fixed ring $R$, with morphisms being $R$-linear functions.

It is important to note that not all categories can be thought of as consisting of “structures and structure-preserving functions between them”,² even though all of the examples above are of this form.

You might wonder if there is a category of categories. There is³, and the morphisms are called functors. We won’t need these in this post, but I will provide the definition since it is fundamental and will be used in future posts.

A functor $F:\mc C\to \mc D$ does two things. It sends any object $A\in\mc C$ to an object $F(A)\in\mc C$, and any morphism $f:A\to B$ in $\mc C$ to a morphism $F(f):F(A)\to F(B)$ in $\mc D$. Furthermore, $F$ must preserve identity morphisms and composition:

1. $F(1_A)=1_{F(A)}$

2. $F(g\circ f)=F(g)\circ F(f).$

There is also a notion of “morphism” between functors⁴, but we won’t need that yet.

Abelian categories

The next step towards derived categories are abelian categories. Essentially, abelian categories are categories where one can “do homological algebra”. For instance, you can define exact sequences and (left/right) exact functors, you can define chain complexes and their cohomology, and so on. It is fine if you haven’t learned homological algebra before – this is an introduction to the categorical side of it!

The most familiar⁵ example of an abelian category is $R$-mod. The categories Grp and Top are not abelian.

So, what is actually required for a category $\mc C$ to be abelian? Very roughly, you need the following:

1. It has a zero object, denoted $0$, with the property that for any $A\in\mc C$, there is only one morphism from $0$ to $A$ and only one morphism from $A$ to $0$. Note that there may be more than one zero object.⁶ The morphisms $0\to A$ and $A\to 0$ are also denoted by $0$.

2. You can “add” and “multiply” any finite number of objects to get a new object, and the results of addition and multiplication must be the same. This operation is typically denoted with $\oplus$.

3. Any morphism $f:A\to B$ has a “kernel” $k:K\to A$ and a “cokernel” $c:B\to C$, which are morphisms that are “universal with respect to the equations” $f\circ k=0$ and $c\circ f=0$. Often times, one just refers to the objects $K$ and $C$ as being the kernel and cokernel.

4. “Injective” morphisms are the kernel of some morphism, and “surjective” morphisms are the cokernel of some morphism.⁷

Now, I’ve intentionally made axioms 2 through 4 pretty vague. The reason for this is essentially that we won’t be using them explicitly, at least not to the extent that we really need to discuss the precise notions. For instance, you have define what a kernel and cokernel of a map of sheaves is, and they dont’t really bear any resemblance to the abstract categorical definitions of kernel and cokernel.

As a consequence of the first axiom, any two objects $A,B$ have a unique zero morphism between them, also denoted $0$, defined by the composition $A\to0\to B$. As an exercise, show that $0\circ f = 0$ and $g\circ 0=0$, with $0$ standing for any zero morphism. Then, use this to show that if $1_A$ is the zero morphism $A\to 0\to A$, then $A$ is a zero object.

As a consequence of the first three axioms, you can “enrich”⁸ $\mc C$ so that you can add two morphisms $f,g:A\to B$ to get a new morphism $f+g:A\to B$. In fact, this makes $\Hom(A,B)$ into an abelian group. The $0$ morphism is the group identity, and the morphism $-f$ denotes the group inverse of $-f$. There is one other requirement for this to be an “enrichment”, but we don’t need it.

You might be bogged down by all of the information so far. Take a break if you need to, but don’t get discouraged!

Chain complexes

Before I start this, a disclaimer: There are a lot of terms that start with the word “chain”. Some people will often remove the word “chain”, even from the term “chain complex”! I was trying not to do this, but I stopped trying at a certain point in writing. Sorry!

For any abelian category $\mc A$, one can form an abelian category $Ch(\mc A)$ of chain⁹ complexes. A chain complex is a “diagram” like this:

\[\cdots\xr{d^{-2}} A^{-1}\xr{d^{-1}} A^0\xr{d^0} A^1\xr{d^1}\cdots\]

The $A^i$ are objects in $\mc A$, and the differentials $d^i$ are required to satisfy $d^i\circ d^{i-1}=0$. One might denote a chain complex by $(A^\bullet, d^\bullet)$, or just $A^\bullet$, or even just $A$. I’ll stick to $A^\bullet$ for the sweet spot between clarity and brevity. Furthermore, one might suppress the indices on the differentials, but I won’t do that. If one is dealing with several chain complexes, then you might put a subscript on the differentials, i.e. $d^i_A$ for the differentials of $A^\bullet$.

A morphism of chain complexes is called a chain map. A chain map $f:A^\bullet \to B^\bullet$ is a collection of morphisms $f^i:A^i\to B^i$ that “commute” with the differentials: $d^i_B\circ f^i=f^{i+1}\circ d^i_A$. As an exercise, what is the identity morphism of a chain complex? What is the composition of two chain maps? $Ch(\mc A)$ is also abelian; what is the $0$ object and what are the $0$ morphisms?

Cohomology

An important part of a chain complex $A^\bullet$ is its cohomology. One could even argue that this is what mathematicians in this area care the most about.

The property $d^i\circ d^{i-1}=0$ can be thought of as saying $\im d^{i-1}\subset \ker d^i$, where im means image and ker means kernel. In other words, $d^{i-1}$ sends something to something that gets sent to $0$ by $d^i$. This doesn’t strictly make sense in an arbitrary abelian category, but with a little work you can make it precise. If you are working in $R$-mod, then the image and kernel of morphisms are defined as usual, as is the quotient of a module by a submodule.

You can then define the quotient $(\ker d^i)/(\im d^{i-1})$, called the $i$th cohomology of $A^\bullet$, denoted by $H^i(A^\bullet)$. Note that this is an object of $\mc A$.

It turns out that chain maps $f:A^\bullet \to B^\bullet$ “induce” maps¹⁰ in cohomology, $H^i(f):H^i(A^\bullet)\to H^i(B^\bullet)$. This leads to a key component in defining derived categories: quasi-isomorphisms. A chain map $f$ is a quasi-isomorphism if $H^i(f)$ is an isomorphism for all $i$.

The homotopy category

In constructing the derived category of an abelian category $\mc A$, there is an optional step of constructing the homotopy category, denoted $K(\mc A)$. In fact, I probably won’t even be exploiting the one reason¹¹ I can think of for constructing this category. Nevertheless, it exists, and it is useful. The less experienced reader may feel free to skip this subsection.

There are two ways to construct $K(\mc A)$, but I will only fully discuss one of them. See this footnote¹² for a brief indication of the other one.

Two chain maps $f,g:A^\bullet\to B^\bullet$ are said to be chain homotopic if there is a collection of morphisms $s^i:A^i\to B^{i-1}$ such that $f^i-g^i=d_B^{i-1}\circ s^i + s^{i+1}\circ d^i_A$. The collection of morphisms, sometimes denoted $s:f\to g$, is called a chain homotopy. If $f$ and $g$ are chain homotopic, we write $f\sim g$. Chain homotopy is an equivalence relation.

While the definition may seem opaque, it is a useful notion because of the following fact: homotopic maps induce the same maps in cohomology. However, it is not true that if two maps induce the same maps in cohomology, then they are homotopic. Try and find a counterexample! See this footnote¹³ for a starting hint and this footnote¹⁴ for an outline.

In any case, if we were to “identify” chain homotopic maps, i.e. treat $f$ and $g$ as “equal” if $f\sim g$, then we wouldn’t lose information about induced maps in cohomology. This leads to the definition of the homotopy category.

We define the homotopy category $K(\mc A)$ as the category of chain complexes, where the morphisms are chain homotopy equivalence classes of chain maps.¹⁵ If $f$ is chain map, one denotes the corresponding equivalence class by $[f]$, or sometimes just $f$; I will use the first notation. The identity morphisms in $K(\mc A)$ are the classes $[1_A]$. Composition is defined¹⁶ as follows: $[g]\circ [f]=[g\circ f]$.

It is important to note that $K(\mc A)$ can fail to be abelian. I would say it is “usually” not abelian, with about 80% certainty.

A morphism in $K(\mc A)$ is called a quasi-isomorphism if one, and therefore¹⁷ all, of its representatives is a quasi-isomorphism.

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Preview of derived categories

I want to close out this post with a few “preview” remarks on derived categories.

As mentioned previously, the quasi-isomorphisms are a key ingredient in defining the derived category. In particular, we are going to “formally invert” the quasi-isomorphisms. The idea is that we enlarge the collection of morphisms by adding inverses to quasi-isomorphisms. While this sounds simple, you have to take into account that you will also include all the new possible compositions of morphisms. In particular, defining the composition of morphisms in the derived category is a bit of a mess.

I also want to address a mistake I made in my last post. I said that the objects of $D(\mc A)$ were equivalence classes of chain complexes. This is not true! The objects of $D(\mc A)$ are in fact just chain complexes.

Well, that’s all for now. See you next time!

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Footnotes: