Group Homology Transfer and Low-Degree Exact Sequences: Examples

Put $R=\mathbb{Z}{C}_m$ and $N=1+g+\cdots +g^{m-1}$. Use one copy of R in each nonnegative degree, augmentation $ϵ(\sum a_i{g}^i)=\sum a_i$, differential multiplication by $g-1$ in odd degrees and by N in positive even degrees. Since $(g-1)N=g^m-1=0$ and $ϵ(g-1)=0$, this is an augmented chain complex.

For $a={\sum }_{i=0}^{m-1}{a}_i{g}^i$, the coefficient of $g^i$ in $(g-1)a$ is $a_{i-1}-a_i$ (indices modulo m). Thus its kernel consists exactly of constant coefficient vectors, namely $\mathbb{Z}N$. Also $Na=({\sum }_i{a}_i)N$, so the image of multiplication by N is $\mathbb{Z}N$ and its kernel is $\ker ϵ$. If $\sum a_i=0$, then $a={\sum }_{i=1}^{m-1}{a}_i(g^i-1)=(g-1){\sum }_{i=1}^{m-1}{a}_i(1+\cdots +g^{i-1})$. Hence $\ker ϵ=(g-1)R$, proving exactness in every degree. This also covers m=1: the two maps are zero and identity and the displayed sums are empty.

Each term is free, so step 2.1 makes this complex $P\to \mathbb{Z}$ a projective resolution of the trivial left $R$-module. Let $P^{\mathrm{s}\mathrm{u}\mathrm{p}}\to \mathbb{Z}$ be the supplied resolution used in [F2]. By [F3], $P$ and $P^{\mathrm{s}\mathrm{u}\mathrm{p}}$ are chain-homotopy equivalent over $\mathbb{Z}$. The additive functor $\mathbb{Z}{\otimes }_R-$ carries the comparison maps and their homotopies to comparison maps and homotopies, so $H_{\ast }(\mathbb{Z}{\otimes }_{R}P)\cong H_{\ast }(\mathbb{Z}{\otimes }_R{P}^{\mathrm{s}\mathrm{u}\mathrm{p}})=H_{\ast }(C_m;\mathbb{Z})$. Tensoring $P$ makes multiplication by $g-1$ zero and multiplication by N multiplication by m. Thus degree zero is $\mathbb{Z}$, every odd degree has kernel $\mathbb{Z}$ modulo $m\mathbb{Z}$, and every positive even degree has kernel of $m:\mathbb{Z}\to \mathbb{Z}$, which is zero.

Multiplication by m on $\mathbb{Z}\mathrm{/}m\mathbb{Z}$ is zero, and on the zero groups it is zero. This explicitly verifies the conclusion of F1. Degree zero is excluded: $m\cdot 1\mathrm{\ne }0$ in $\mathbb{Z}$. For m=1 all positive groups vanish.

Every element of the free group on one generator has a unique integer exponent, so identify F with $\mathbb{Z}$. Reduction modulo m is onto and has kernel $m\mathbb{Z}$. All elements of F commute, hence $[F,R]=0$, $F_{\mathrm{a}\mathrm{b}}=\mathbb{Z}$, and $(C_m{)}_{\mathrm{a}\mathrm{b}}=C_m$. F1 therefore gives $0\to H_2(C_m;\mathbb{Z})\stackrel{j}{\to }m\mathbb{Z}\stackrel{\iota }{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}m\mathbb{Z}\to 0$, where $\iota$ is the actual subgroup inclusion.

The inclusion has zero kernel. Exactness gives $\mathrm{im}j=0$, while j is injective, so $H_2(C_m;\mathbb{Z})=0$. Under the isomorphism $\mathbb{Z}\to m\mathbb{Z}$, $a\mapsto ma$, the middle map becomes multiplication by m; its image is exactly the kernel of reduction modulo m. For m=1 this map is identity and the final quotient is zero.

For triples with first two coordinates (a,b), (u,v), (x,y), the extra central terms in the two associative products are $av+(a+u)y$ and $uy+a(v+y)$, which are equal. The identity is (0,0,0) and the inverse of (a,b,c) is $(-a,-b,-c+ab)$ by multiplication on both sides. Projection onto the first two coordinates is an onto homomorphism with central kernel $N=\{(0,0,c)\}$. Thus these formulas really give the stated group extension.

The action on A is trivial and N is central, so d is a crossed homomorphism and conjugation fixes it. Choose $\alpha (a,b)=(a,b,0)$ and $\eta (a,b)=0$. They are normalized and $\alpha (q)\ast d-d=0=\delta \eta (q)$. Multiplication gives $\alpha (a,b)\alpha (u,v)=(a+u,b+v,av)$, so the factor set, as an element of N identified with $\mathbb{Z}$, is $f((a,b),(u,v))=av$. F1 yields $F=-f$, not f. These explicit maps supply every choice required for this instance of F1 and F2.

The cochain F vanishes if either input is zero. Its cocycle identity is $-uy+(a+u)y-a(v+y)+av=0$ for the three inputs in step 1.1. With trivial action on the abelian quotient, every coboundary has the form $b(q)+b(r)-b(q+r)$ and is symmetric in q,r. But $F((1,0),(0,1))=-1$, whereas $F((0,1),(1,0))=0$. Thus F is not a coboundary and its class is nonzero.