Local Coefficients, Twisted Homology, and Duality — Examples

Let $g$ be the positive loop and lift the vertex to $\tilde{v}\in \mathbb{R}$. Choose the lifted edge $\tilde{e}$ from $\tilde{v}$ to $g\tilde{v}$. Its boundary is $g\tilde{v}-\tilde{v}=\tilde{v}\cdot (g^{-1}-1)$ under [F1]'s right action. In the left fiber module, $g^{-1}m=T_{\overline{g^{-1}}}m=T_{g}m=Tm$. Tensoring therefore sends $m$ to $(T-1)m$.

There is one chain module $M$ in degrees one and zero and none elsewhere, so the kernel and cokernel of step 1.1 are exactly the displayed homology groups. Reversing the chosen loop yields $T^{-1}-1=-T^{-1}(T-1)$ and hence an isomorphic complex, so the answer is independent of the orientation convention. For $M=0$ all groups vanish; for $T=1$ the differential is zero and ordinary circle homology with coefficients in $M$ is recovered. No AC is used.

Compute the cellular differential, including $n=1$. [F1, F3, F5, F6] For $n=1$, the map $t\mapsto [\cos (\pi t):\sin (\pi t)]$ identifies $\mathbb{R}\mathrm{/}\mathbb{Z}$ with ${\mathbb{R}\mathbb{P}}^1$. Under [F3], the positive once-around loop is a generator $g$ of its infinite cyclic fundamental group. Choose the vertex lift at $0\in \mathbb{R}$ and the lifted open edge from $0$ to $1$. Its boundary is

The resulting complex has one copy of $\mathbb{Z}$ in each degree. For $0<k<n$, an even $k$ has zero outgoing differential and incoming image $2\mathbb{Z}$, giving $\mathbb{Z}\mathrm{/}2$; an odd $k$ has injective outgoing differential, giving zero. At degree zero, $d_1=2$ gives $\mathbb{Z}\mathrm{/}2$. At the top there is no incoming differential, so the kernel is $\mathbb{Z}$ for even $n$ and zero for odd $n$. This proves the table.

Compare with the orientation system in both ranges. [F2, F4, step 1.1, step 2.1] When $n=1$, the displayed identification with $\mathbb{R}\mathrm{/}\mathbb{Z}$ gives the projective line its usual circle orientation. Its orientation character is therefore trivial, whereas the positive generator acts by $-1$ on ${\mathbb{Z}}_{\mathrm{s}\mathrm{g}\mathrm{n}}$. Hence the sign system is not the orientation system, consistently with the zero top sign homology in step 2.1.

Give the rectangle the local orientation represented by the ordered coordinate directions $({\mathrm{\partial }}_x,{\mathrm{\partial }}_s)$. The gluing sends these to $({\mathrm{\partial }}_x,-{\mathrm{\partial }}_s)$, so one traversal of the core returns the negative local orientation. By [F1], its monodromy is $-1$.

Unwrap the gluing twice: $[0,2]\times [-1,1]\mathrm{/}(0,s)\sim (2,s)$ maps two-to-one onto $B$. This space is an annulus, and its core maps twice around the core of $B$. Hence the pulled-back monodromy is $(-1{)}^2=1$. Since the annulus retracts onto its core circle, the pulled-back rank-one local system is constant.

The two horizontal rectangle edges are joined into one boundary circle. Traversing this circle runs from $(0,1)$ to $(1,1)\sim (0,-1)$ and then from $(0,-1)$ to $(1,-1)\sim (0,1)$, so its image in the core has degree two. Its orientation monodromy is therefore $(-1{)}^2=1$. The collar extension in [F2] gives the same transport at boundary points, hence the restricted system is constant. The boundary circle is orientable even though $B$ is not. Empty-boundary and zero-ring cases are not part of this fixed example, and no AC is used.

The lifted boundary of the two-cell has $a_i$-coefficient $a_1^2\cdots a_{i-1}^2(1+a_i)$, obtained by differentiating the attaching word one letter at a time, or equivalently by grouping its two successive lifted incidences along $a_i$. Under the orientation action in [F1], each preceding square acts as $1$ and $1+a_i$ acts as $1-1=0$. Thus the twisted $d_2:\mathbb{Z}\to {\mathbb{Z}}^g$ is zero.

Each lifted one-cell has endpoint incidence $a_i-1$, which evaluates to $-2$. Hence $d_1:{\mathbb{Z}}^g\to \mathbb{Z}$ is $(x_1,\dots ,x_g)\mapsto -2{\sum }_i{x}_i$. In particular $H_2(N_g;\mathcal{O})=\ker d_2=\mathbb{Z}$; also $H_1(N_g;\mathcal{O})\cong {\mathbb{Z}}^{g-1}$ and $H_0(N_g;\mathcal{O})\cong \mathbb{Z}\mathrm{/}2$. These include $g=1$, where the middle group is zero.

There is a canonical pairing $\mathcal{O}\otimes \mathcal{O}\to \underset{‾}{\mathbb{Z}}$: after choosing either generator $o_x$, send $o_x\otimes o_x$ to $1$ and extend bilinearly. Replacing $o_x$ by $-o_x$ changes both factors, so the map is independent of the choice; the two monodromy signs cancel, so it commutes with transport. Apply [F3] first with the constant system and then with $\mathcal{L}=\mathcal{O}$. The first target is $\mathcal{O}$, and the second is $\mathcal{O}\otimes \mathcal{O}\cong \underset{‾}{\mathbb{Z}}$, giving the two displayed families. Step 2.1 verifies the top group directly. Degrees outside $0,1,2$ vanish, and no AC beyond that already assumed by [F3] is introduced.

Give $S^1$ one vertex and one oriented edge. If $g$ denotes the positive loop, a lift of the edge has boundary $g\tilde{v}-\tilde{v}$. Under the published right-chain convention this is $\tilde{v}\cdot (g^{-1}-1)$, while the corresponding left fiber action of $g^{-1}$ is the specified monodromy $T$. Thus [F1] gives the two-term complex $0\to \mathbb{Z}\stackrel{T-1}{\to }\mathbb{Z}\to 0$. For $T=-1$, its differential is $-2$, with zero kernel and cokernel $\mathbb{Z}\mathrm{/}2$. For $T=1$, its differential is zero, so both degree-one and degree-zero groups are $\mathbb{Z}$.

The two systems have isomorphic stalk $\mathbb{Z}$ at every point but different loop transport and different homology. Hence stalk data without monodromy cannot replace a local system. The sign of the differential could be $+2$ after reversing the chosen cell, with the same kernel and cokernel. No other degrees occur and no AC is used.

Lift one positive circuit of the base interval in the mapping-torus model $(F\times [0,1])\mathrm{/}(x,1)\sim (h(x),0)$. Its endpoint identification on the fiber is $h$, so [F1] and [F2] give monodromy $h_{\ast }$ on $H_q(F;R)$. Therefore the correct base groups retain this local system rather than replacing it by a constant copy of its stalk.

Let $F=S^1$ and let $h$ be a reflection. On $H_1(S^1;\mathbb{Z})=\mathbb{Z}$, $h_{\ast }=-1$. Give the base circle one vertex and one edge. Its lifted edge boundary evaluates through [F3] to $T-1:\mathbb{Z}\to \mathbb{Z}$. For the correct monodromy $T=-1$, this is multiplication by $-2$, so the $q=1$ row is $H_0(S^1;{\mathbb{Z}}_{-1})=\mathbb{Z}\mathrm{/}2$ and $H_1(S^1;{\mathbb{Z}}_{-1})=0$. If monodromy is discarded, $T=1$ makes the differential zero, giving $H_0=\mathbb{Z}$ and $H_1=\mathbb{Z}$ instead.

In the $q=0$ row the reflection acts trivially on $H_0(S^1;\mathbb{Z})$, so the untwisted and correct rows agree there; the discrepancy is specifically caused by monodromy, not by the fiber groups. This comparison computes only the proposed coefficient rows and does not invoke or assert convergence of a Serre spectral sequence. No AC is used.