Homology

The category \(\Ch(\mathcal{A})\)

In [Category Theory] §Abelian Categories, we defined the category \(\Ch(\mathcal{A})\) of chain complexes, and in this section we examine more concretely that this is an abelian category. First, morphisms in \(\Ch(\mathcal{A})\) are given as follows.

Definition 1 For two chain complexes \(C_\bullet, D_\bullet\), a chain map \(f_\bullet:C_\bullet\rightarrow D_\bullet\) between them is a collection of maps \(f_n:C_n\rightarrow D_n\) making the following diagram

commute.

That is, that \(f_\bullet\) is a chain map means that for all \(n\),

\[f_{n-1}\circ d_n^C=d_n^D\circ f_n\]

holds. When there is no risk of confusion, we write both \(d^C\) and \(d^D\) simply as \(d\). Then, defining the category \(\Ch(\mathcal{A})\) by

• the collection of chain complexes of \(\mathcal{A}\), \(\obj(\Ch(\mathcal{A}))\),
• the collection of chain maps from chain complex \(C_\bullet\) to \(D_\bullet\), \(\Mor_{\Ch(\mathcal{A})}(C_\bullet,D_\bullet)\)

it is obvious that \(\Ch(\mathcal{A})\) becomes a category. Here, \(\id_{C_\bullet}\) is the chain map for which for all \(n\), \(\id_n:C_n\rightarrow C_n\) is defined as \(\id_{C_n}\), namely \(\id_\bullet:C_\bullet\rightarrow C_\bullet\), and the composition of chain maps is given by

\[g_\bullet\circ f_\bullet=(g_n\circ f_n)_{n\in\mathbb{Z}}\]

Moreover, the category \(\Ch(\mathcal{A})\) is an \(\Ab\)-category. This is because, given two chain maps \(f_\bullet,g_\bullet:C_\bullet\rightarrow D_\bullet\), defining their sum by

\[f_\bullet+g_\bullet=(f_n+g_n)_{n\in\mathbb{Z}}\]

makes \(\Mor_{\Ch(\mathcal{A})}(C_\bullet,D_\bullet)\) an abelian group, and given two chain complexes \(C_\bullet,D_\bullet\), it suffices to define their product \(C_\bullet\oplus D_\bullet\) as¹ the following chain complex

\[C_\bullet\oplus D_\bullet:\quad \cdots \longrightarrow C_{n+1}\oplus D_{n+1}\overset{d_{n+1}^C\oplus d_{n+1}^D}{\longrightarrow}C_n\oplus D_n\overset{d_{n}^C\oplus d_{n}^D}{\longrightarrow}C_{n-1}\oplus D_{n-1}\longrightarrow\cdots\]

The zero object is the following chain complex

\[\cdots\rightarrow 0\rightarrow 0\rightarrow 0\rightarrow\cdots\]

Now, to show that \(\Ch(\mathcal{A})\) is an abelian category, it suffices to define the kernel and cokernel of an arbitrary chain map \(f_\bullet:C_\bullet \rightarrow D_\bullet\). First, for each \(n\), since \(f_n:C_n \rightarrow D_n\) is a morphism in the abelian category \(\mathcal{A}\), it has \(i_n:\ker f_n\rightarrow C_n\) and \(p_n:D_n\rightarrow \coker f_n\). Consider the following diagram

Then, since

\[f_{n-1}\circ d_n\circ i_n=d_n\circ f_n\circ i_n=0\]

a morphism \(\ker f_n\rightarrow \ker f_{n-1}\) is naturally induced by the universal property of \(\ker f_{n-1}\). Similarly, \(\coker f_n\rightarrow \coker f_{n-1}\) is also defined, and these data constitute the two chain complexes

\[\ker(f_\bullet):\qquad \cdots\longrightarrow \ker f_{n+1}\longrightarrow \ker f_n\longrightarrow \ker f_{n-1}\longrightarrow\cdots\]

and

\[\coker(f_\bullet):\qquad \cdots\longrightarrow \coker f_{n+1}\longrightarrow \coker f_n\longrightarrow \coker f_{n-1}\longrightarrow\cdots\]

respectively. To show that \(\ker(f_\bullet)\) defined in this way is the kernel of \(f_\bullet\) in \(\Ch(\mathcal{A})\), one must show the following universal property

and for this again, it suffices to first obtain, for each \(n\), through the following diagram

the morphism \(j_n:A_n\rightarrow \ker f_n\), and then show that \(j_\bullet\) is a chain map. That is, it suffices to prove that the upper square of the following diagram

commutes. However, since a kernel is always a monomorphism, for this it suffices to show that

\[i_{n-1}\circ(j_{n-1}\circ d_n^A)=i_{n-1}\circ(d_n\circ j_n)\]

holds, which is obvious by definition, so \(\ker(f_\bullet)\) is indeed the kernel of \(f_\bullet\). Similarly, one can verify that \(\coker(f_\bullet)\) is the cokernel of \(f_\bullet\) in \(\Ch(\mathcal{A})\).

Finally, one must show that an arbitrary monomorphism \(f_\bullet\) is equal to \(\ker(\coker f_\bullet)\), and an arbitrary epimorphism \(f_\bullet\) is equal to \(\coker(\ker f_\bullet)\). This is done by first proving that \(f_\bullet\) being a monomorphism (resp. epimorphism) is equivalent to having, for all \(n\), that \(f_n\) is a monomorphism (resp. epimorphism), and then using the condition of an abelian category degree by degree and combining them. Since this process is almost identical to the above, we omit the detailed argument.

Homology

Fix a chain complex \(C_\bullet\) and define a new chain complex \(C_{\bullet-1}\) by

\[C_{\bullet-1}:\qquad\cdots\longrightarrow\underbrace{C_n}_\text{\small degree $n+1$}\overset{d_n}{\longrightarrow}\underbrace{C_{n-1}}_\text{\small degree $n$}\overset{d_{n-1}}{\longrightarrow}\underbrace{C_{n-2}}_\text{\small degree $n-1$}\longrightarrow\cdots\tag{1}\]

That is, \(C_{\bullet-1}\) is the chain complex obtained by shifting \(C_\bullet\) one position. Now define the chain map \(d_\bullet:C_\bullet\rightarrow C_{\bullet-1}\) as in the following diagram

Then \(\ker (d_\bullet)\) is well-defined, and considering the following diagram

from \(d_{n+1}\circ i_{n+1}=0\) and \(i_n\) being a monomorphism, we see that the differentials of \(\ker(d_\bullet)\) are all zero maps.

Similarly, \(\coker(d_\bullet)\) is also well-defined, and the differentials between them are also zero. Unlike the component of \(\ker(d_\bullet)\) of degree \(n\) examined above being \(\ker d_n\rightarrow C_n\), since the component of \(\coker(d_\bullet)\) of degree \(n\) is \(C_{n-1}\rightarrow \coker(d_n)\), there may be some confusion, so we shift degrees by one again and consider the chain complex \(\coker(d_{\bullet+1})\).

Now, regarding

\[\im(d_{\bullet+1})=\ker(C_\bullet\rightarrow \coker(d_{\bullet+1}))\]

and using the universal property of \(\ker(d_\bullet)\rightarrow C_\bullet\), we obtain a monomorphism

\[\im(d_{\bullet+1})\rightarrow\ker(d_\bullet)\]

and through this, we obtain a new chain complex

\[H_\bullet(C)=\ker(d_\bullet)/\im(d_{\bullet+1})\]

Explicitly, \(H_n(C)\) is induced by the following diagram

However, \(\ker(d_n)\rightarrow H_n\) is an epimorphism, and since \(\ker(d_n)\rightarrow\ker(d_{n-1})\) is \(0\), \(H_n(C)\rightarrow H_{n-1}(C)\) is also a zero map. This shows that (just as with \(\ker(d_\bullet)\) or \(\coker(d_\bullet)\)), \(H_\bullet(C)\) is not very meaningful as a chain complex, and instead, the individual components of \(H_\bullet(C)\), namely \(H_n(C)\in\obj(\mathcal{A})\), are the interesting objects.

Definition 2 For an arbitrary chain complex

\[C_\bullet:\quad\cdots\overset{d_{n+2}}{\longrightarrow} C_{n+1}\overset{d_{n+1}}{\longrightarrow} C_n\overset{d_n}{\longrightarrow} C_{n-1}\overset{d_{n-1}}{\longrightarrow}\cdots\]

, the object in \(C_\bullet\) of \(n\)-cycles is given by \(Z_n(C)=\ker (d_n)\). Also, we call \(\im(d_{n+1})\) the object in \(C_\bullet\) of \(n\)-boundaries, and write it as \(B_n(C)\). At this time, for \(C_\bullet\), the \(n\)-th homology is defined by \(H_n(C)=Z_n(C)/B_n(C)\).

Similarly, for a cochain complex, \(n\)-cocycles, \(n\)-coboundaries, and \(n\)-th cohomology are defined.

\(H_\bullet\) itself is a functor from \(\Ch(\mathcal{A})\) to \(\Ch(\mathcal{A})\), but each component \(H_n\) defines a functor from \(\Ch(\mathcal{A})\) to \(\mathcal{A}\).

Proposition 3 For any \(n\), \(H_n\) is a functor from \(\Ch(\mathcal{A})\) to \(\mathcal{A}\).

Proof

For an arbitrary chain map \(f_\bullet:C_\bullet\rightarrow D_\bullet\), applying §Diagram chasing, ⁋Lemma 4 shows that \(f_n\) induces \(Z_n(C)\rightarrow Z_n(D)\) and \(B_n(C)\rightarrow B_n(D)\), respectively. Therefore \(f_n\) induces

\[H_n(f):H_n(C)\rightarrow H_n(D)\]

. Checking that the data obtained this way form a functor is all straightforward.

Double complexes and translation

Finally, we define double complexes. Intuitively, this can be thought of as a chain complex consisting of the abelian category \(\Ch(\mathcal{A})\), that is, an object of \(\Ch(\Ch(\mathcal{A}))\). The chain complex \(C_{\bullet-1}\) of (1) examined above and the chain map below it are examples of this.

However, when doing calculations later, there are times when having the signs of each row alternate yields cleaner results.² Adopting this sign convention, we define as follows.

Definition 4 A double complex consisting of objects of an abelian category \(\mathcal{A}\) is an object equipped with two maps

\[d^h:C_{p,q}\rightarrow C_{p-1,q},\qquad d^v:C_{p,q}\rightarrow C_{p,q-1}\]

satisfying

\[(d^h)^2=(d^v)^2=d^vd^h+d^hd^v=0\]

The condition \(d^vd^h+d^hd^v=0\) means \(d^vd^h=-d^hd^v\), so double complexes are not objects of \(\Ch(\Ch(\mathcal{A}))\); however, defining from \(d^v\) the expression

\[f_{p,q}=(-1)^pd_{p,q}^v:C_{p,q}\rightarrow C_{p,q-1}\]

allows us to treat them as objects of \(\Ch(\Ch(\mathcal{A}))\).

Definition 5 For a double complex \(C_{p,q}\), the total complex \(\Tot(C)_\bullet\) is defined in each degree \(n\) by

\[(\Tot(C))_n = \bigoplus_{p+q=n} C_{p,q}\]

and its differential is given by

\[d = d^h + d^v : (\Tot(C))_n \rightarrow (\Tot(C))_{n-1}\]

For convenience of discussion we have written the differential somewhat concisely, but unpacking this reveals that actual computation is somewhat more complicated than the above formula. For example, suppose a double complex \(C_{p,q}\) is given, and for convenience assume that only where \(p,q\geq 0\) is this complex not \(0\). Then in the total complex \(\Tot(C)_\bullet\) of this double complex, to compute the differential \(d:\Tot(C)_2\rightarrow \Tot(C)_1\) one must proceed as follows. First, an element of \(\Tot(C)_2\) has the form \((x_{0,2}, x_{1,1}, x_{2,0})\),³ and the differential of this element becomes the following element containing all the information of

\[d^h(x_{0,2})+d^v(x_{0,2})=d^v(x_{0,2}), \quad d^h(x_{1,1})+d^v(x_{1,1}),\quad d^h(x_{2,0})+d^v(x_{2,0})=d^h(x_{2,0})\]

namely

\[(d^v(x_{0,2})+d^h(x_{1,1}), d^v(x_{1,1})+d^h(x_{2,0}))\in\Tot(C)_1=C_{0,1}\oplus C_{1,0}\]

That is, \(\Tot(C)\) packages the double complex \(C_{p,q}\) into a single complex, and in doing so simultaneously considers all components with total degree \(p+q=n\).

To show that this is a complex, one must show \(d^2=0\), which follows from the formula

\[(d^h + d^v)^2 = (d^h)^2 + d^vd^h + d^hd^v + (d^v)^2\]

Here \((d^h)^2=0\) and \((d^v)^2=0\), and by the double complex condition \(d^vd^h + d^hd^v=0\), the remaining two terms also cancel.

Meanwhile, since we used the above sign convention in defining double complexes, the \(p\)-th translation of \(C\), \(C[p]\), is also the chain complex given by

\[C[p]_n=C_{n+p}\]

and its differential must be defined as \((-1)^pd\). For instance, \(C[-1]\) means the same chain complex as \(C_{\bullet-1}\) defined above, but with the signs of each differential reversed.

In any case, regardless of these newly created signs, it is obvious that

\[H_n(C[p])=H_{n+p}(C)\]

and moreover, if \(f:C\rightarrow D\) is a chain map, then defining \(f[p]\) by the formula

\[f[p]_n=f_{n+p}\]

we can form \(f[p]:C[p]\rightarrow D[p]\). That is, translation is a functor from \(\Ch(\mathcal{A})\) to itself.

Truncation

Finally, let an arbitrary chain complex \(C_\bullet\) be given. For any integer \(n\), the chain complex defined by the formula

\[(\tau_{\geq n}C)_i=\begin{cases}0&\text{if $i < n$}\\ Z_n&\text{if $i=n$}\\ C_i&\text{if $i > n$}\end{cases}\]

and the same differential as \(C_\bullet\), denoted \((\tau_{\geq n}C)_\bullet\), is called the intelligent truncation. Then

\[H_i(\tau_{\geq n}C)=\begin{cases}0&\text{if $i < n$}\\ H_i(C)&\text{if $i\geq n$}\end{cases}\]

Similarly, \((\tau_{< n}C)_\bullet\) can also be defined. Meanwhile, the brutal truncation is defined by the formula

\[(\sigma_{\geq n}C)_i=\begin{cases}0&\text{if $i < n$}\\ C_i&\text{otherwise}\end{cases}\]

The definition itself may seem more natural for \((\sigma_{\geq n}C)_\bullet\), but looking at the \(n\)th homology one can verify that the intelligent truncation gives a better operation.

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References

[Wei] C.A. Weibel. An Introduction to Homological Algebra. Cambridge Studies in Advanced Mathematics. Cambridge University Press, 1995.
[Hu] S.T. Hu, Introduction to homological algebra. University Microfilms, 1979.

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1. Here \(\oplus\) is understood as the coproduct in \(\mathcal{A}\). Since one can prove that the product and coproduct of finite objects coincide in an abelian category, in fact \(A\oplus B\cong A\times B\). ↩

2. For example, for the Dolbeault operator \(d=\partial+\bar{\partial}\), the condition \(d^2=0\) is equivalent to \(0=\partial^2+(\partial\bar{\partial}+\bar{\partial}\partial)+\bar{\partial}^2\), and considering the bidegree we know that \(\partial\bar{\partial}+\bar{\partial}\partial=0\). ↩

3. Since we assumed for convenience that only where \(p,q\geq 0\) is \(C_{p,q}\neq 0\), terms such as \(x_{-1,3}\) can be ignored. ↩