Knowlpedia

Homology module

The nth homology H_n(C) = ker(d_n)/im(d_{n+1}) of a chain complex of modules.

Let RR be a and let

⋯→dn+1Cn→dnCn−1→dn−1⋯\cdots \xrightarrow{d_{n+1}} C_n \xrightarrow{d_n} C_{n-1} \xrightarrow{d_{n-1}} \cdots

be a of , i.e. dn∘dn+1=0d_n\circ d_{n+1}=0 for all nn.

The nnth cycles and nnth boundaries of C∙C_\bullet are the submodules

Zn(C∙):=ker⁡(dn)⊆Cn,Bn(C∙):=im⁡(dn+1)⊆Cn.Z_n(C_\bullet) := \ker(d_n)\subseteq C_n, \qquad B_n(C_\bullet) := \operatorname{im}(d_{n+1})\subseteq C_n.

Since dn∘dn+1=0d_n\circ d_{n+1}=0, one has Bn(C∙)⊆Zn(C∙)B_n(C_\bullet)\subseteq Z_n(C_\bullet). The nnth homology module is

Hn(C∙):=Zn(C∙)/Bn(C∙).H_n(C_\bullet) := Z_n(C_\bullet) / B_n(C_\bullet).
Equivalent characterizations

Equivalently, Hn(C∙)=0H_n(C_\bullet)=0 for all nn iff C∙C_\bullet is an .

Properties
  • A f:C∙→D∙f:C_\bullet\to D_\bullet induces RR-linear maps Hn(f):Hn(C∙)→Hn(D∙)H_n(f):H_n(C_\bullet)\to H_n(D_\bullet) for all nn.
  • chain maps induce the same maps on homology.
Examples

Example 1: Two-term complex over Z\mathbb Z

Consider the chain complex C∙C_\bullet with C1=ZC_1=\mathbb Z, C0=ZC_0=\mathbb Z, and d1:Z→Zd_1:\mathbb Z\to\mathbb Z given by multiplication by nn, with all other Ci=0C_i=0. Then

H1(C∙)=ker⁡(×n)=0,H0(C∙)=coker⁡(×n)≅Z/nZ.H_1(C_\bullet)=\ker(\times n)=0,\qquad H_0(C_\bullet)=\operatorname{coker}(\times n)\cong \mathbb Z/n\mathbb Z.

Example 2: Detecting exactness

Let 0→A→uB→vC→00\to A\xrightarrow{u} B\xrightarrow{v} C\to 0 be a of RR-modules, viewed as a chain complex

0→A→uB→vC→00\to A \xrightarrow{u} B \xrightarrow{v} C \to 0

concentrated in degrees 2,1,02,1,0. Then the sequence is exact iff H2=H1=H0=0H_2=H_1=H_0=0, i.e. iff the complex is .

Example 3: Tor\mathrm{Tor} as homology (concrete computation)

Let M=Z/nZM=\mathbb Z/n\mathbb Z. A of MM over Z\mathbb Z is

0→Z→×nZ→Z/nZ→0.0\to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n\mathbb Z \to 0.

Tensoring with Z/mZ\mathbb Z/m\mathbb Z gives the chain complex

0→Z/mZ→×nZ/mZ→0.0\to \mathbb Z/m\mathbb Z \xrightarrow{\times n} \mathbb Z/m\mathbb Z \to 0.

Its homology is

H1≅ker⁡(×n:Z/m→Z/m)≅Z/gcd⁡(m,n)Z,H_1 \cong \ker(\times n:\mathbb Z/m\to \mathbb Z/m)\cong \mathbb Z/\gcd(m,n)\mathbb Z,

so

Tor1Z(Z/n,Z/m)≅Z/gcd⁡(m,n)Z.\mathrm{Tor}_1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)\cong \mathbb Z/\gcd(m,n)\mathbb Z.

(Here we are using the identification H1(P∙⊗−)≅Tor1H_1(P_\bullet\otimes -)\cong \mathrm{Tor}_1; see .)