Lemma. First homology of a short chain complex over a PID [LNMK]

Let

C2→∂2C1→∂1C0C_2 \xrightarrow{\partial_2} C_1 \xrightarrow{\partial_1} C_0

be a chain complexDefinitionChain complex2026-10-04 · Lîm Tsú-thuàn of free finite rank RR-modules over a principal ideal domain RR. Then

H1(C∙)≅Rb⊕R/(a1)⊕⋯⊕R/(an)H_1(C_\bullet) \cong R^b \oplus R/(a_1) \oplus \cdots \oplus R/(a_n)

where a1,…,an∈Ra_1, \dots, a_n \in R are the invariant factors of ∂2\partial_2 and where b=rankker⁡∂1−nb = \text{rank} \ker \partial_1 - n.

Example Real projective plane [local-0]

Take R=ZR = \mathbb{Z} and the real projective plane RP2\mathbb{R}P^2 as a Δ\Delta-complexDefinitionDelta complex2026-10-04 · Lîm Tsú-thuàn: two vertices v,wv, w, three edges a,b,ca, b, c, and two triangles U,LU, L.

figure tex6340

Order the vertices of each triangle [v0,v1,v2][v_0, v_1, v_2] so that v0v1=cv_0 v_1 = c in both, v1v2=a,v0v2=bv_1 v_2 = a, v_0 v_2 = b in UU, and v1v2=b,v0v2=av_1 v_2 = b, v_0 v_2 = a in LL. Then

∂2U=a−b+c∂2L=−a+b+c\partial_2 U = a - b + c \qquad \partial_2 L = -a + b + c

In the bases U,LU, L and a,b,ca, b, c, the Smith normal form of ∂2\partial_2 is

(1−1−1111)→R2+R1, R3−R1(1−10002)→C2+C1, R2↔R3(100200)\begin{pmatrix} 1 & -1 \\ -1 & 1 \\ 1 & 1 \end{pmatrix} \xrightarrow{R_2 + R_1,\ R_3 - R_1} \begin{pmatrix} 1 & -1 \\ 0 & 0 \\ 0 & 2 \end{pmatrix} \xrightarrow{C_2 + C_1,\ R_2 \leftrightarrow R_3} \begin{pmatrix} 1 & 0 \\ 0 & 2 \\ 0 & 0 \end{pmatrix}

so the invariant factors are a1=1,a2=2a_1 = 1, a_2 = 2 and n=2n = 2. For bb, we have ∂1a=∂1b=w−v\partial_1 a = \partial_1 b = w - v and ∂1c=0\partial_1 c = 0, so ∂1\partial_1 has rank 11

rankker⁡∂1=rank C1−rank ∂1=3−1=2\text{rank} \ker \partial_1 = \text{rank}\ C_1 - \text{rank}\ \partial_1 = 3 - 1 = 2

Hence, b=2−2=0b = 2 - 2 = 0. Now applied the theorem

H1(RP2)≅Z2−2⊕Z/(1)⊕Z/(2)≅Z/2H_1(\mathbb{R}P^2) \cong \mathbb{Z}^{2-2} \oplus \mathbb{Z}/(1) \oplus \mathbb{Z}/(2) \cong \mathbb{Z}/2