# Homology with twisted coefficients of the circle

$$\DeclareMathOperator{\Aut}{Aut} \newcommand{\p}{\pi_1X} \newcommand{\Z}{\mathbb{Z}} \newcommand{\XX}{\tilde{X} }$$ Let $$X$$ be a path connected and locally path connected topological space and $$\rho:\p\to \Aut(A)$$ a representation of the fundamental group into an abelian group $$A$$. We define $$H_k(X;A_\rho),$$ the homology of $$X$$ with coefficients in $$A$$ twisted by $$\rho$$, by taking the homology of the chain complex

$$C_{\bullet}(\tilde{X}) \;\underset{\Z[\p]}\otimes\; A$$

where:

• $$\Z[\p]$$ is the group ring of the fundamental group

• $$\tilde{X}$$ is the universal cover of $$X$$

• $$C_{\bullet}(\XX)$$ is the singular chain complex of $$\tilde{X}$$, endowed with the natural right $$\Z[\p]$$-module structure given by the monodromy action. That is: any $$g\in \p$$ acts naturally on $$\XX$$ as a deck transformation $$g:\XX\to\XX$$. Thus given a $$k$$-simplex $$\sigma: \Delta^k\to \XX$$ we define $$\sigma \cdot g := g\circ\sigma : \Delta^k\to \XX \to \XX$$

• The $$\mathbb{Z}[\p]$$-module structure on $$A$$ is given by the representation $$\rho$$.

I am trying to compute $$H_1(X;A_\rho)$$ when $$X=S^1$$ and $$A=\Z_3$$, with twist given by the nontrivial representation $$\rho:\mathbb{Z}\to \Aut(\Z_3)$$, i.e. the map given by $$\rho(1) = \begin{cases} 0 \mapsto 0 \\ 1 \mapsto 2 \\ 2 \mapsto 1 \end{cases}$$

I have the geometrical picture of what's going on in my mind, but I'm having troubles with the algebra.

What I did so far:

The CW structure of $$S^1$$ can be pulled back to $$\XX \cong \mathbb{R}$$, and we obtain the cellular chain complex (as $$\Z[\p]$$-module) $$0\to \Z[\p] \to \Z[\p] \to 0,$$ and since $$\Z[\p] \cong \Z[t, t^{-1}]$$ the above is $$0\to \Z[t, t^{-1}] \to \Z[t, t^{-1}] \to 0.$$ The boundary map is given by multiplication by $$t-1$$.

I need help:

I can't figure out what happens to the boundary map when we apply to the above chain complex the functor $$-\;\underset{\Z[\p]}\otimes\;\Z_3$$

Can you help me with that, and thus give me the final result for this twisted homology groups?

Moreover, can this be generalized easily to an arbitrary abelian group $$A$$ and any representation $$\rho:\p\to \Aut(A)$$?

• Note that $S^1$ has a CW structure given by one 0-cell and one 1-cell, and this pulls back to a CW structure on $\mathbb R = \tilde S^1$. Note also that $\mathbb Z[\pi_1 S^1] = \mathbb Z [\mathbb Z] = \mathbb Z[t,t^{-1}]$. Now, what $\mathbb Z[t, t^{-1}]$ module is $S_0(\mathbb R)$, and $S_1(\mathbb R)$ if we use cellular chains? Now, compute the boundary map. Hint: almost all of the work can be done without knowing the representation $\mathbb Z \to \Aut(A)$. Jun 20, 2013 at 7:56
• Also, the only automorphism $\mathbb Z /2 \to \mathbb Z/2$ is the identity, so every representation $G\to \Aut (\mathbb Z/2)$. Jun 21, 2013 at 6:02
• ....is trivial. Jun 21, 2013 at 21:10
• Hi Justin, thanks a lot for your help! I will look into it next from Monday on, because now I'm too busy preparing another exam Jun 22, 2013 at 12:12
• Is the action of the fundamental group of $X$ on its universal conver $\tilde{X}$, that you describe, on the right, or on the left?
– Bill
Sep 2, 2014 at 21:28

$$\DeclareMathOperator{\Aut}{Aut} \newcommand{\p}{\pi_1X} \newcommand{\Z}{\mathbb{Z}} \newcommand{\XX}{\tilde{X} }$$Ok, I finally figured it out! So, the cellular chain complex of $$\XX$$ is given by $$0\longrightarrow \Z[t, t^{-1}] \overset{\delta}\longrightarrow \Z[t, t^{-1}] \longrightarrow 0,$$ where the boundary map $$\delta$$ is just multiplication by $$t-1$$, for geometrical reasons.
The new boundary map $$D:\Z_3\to\Z_3$$ which we obtain afte tensoring with $$\Z_3$$ over $$\Z[t,t^{-1}]$$ can be computed by looking at the sequence of maps $$\Z_3 \overset{\cong}\longrightarrow \Z[t,t^{-1}]\underset{\Z[t,t^{-1}]}\otimes\Z_3 \; \overset{\delta\otimes id}\longrightarrow \Z[t,t^{-1}]\underset{\Z[t,t^{-1}]}\otimes\Z_3 \overset{\cong}\longrightarrow \Z_3,$$ where the first map is $$a\mapsto 1\otimes a$$ and last map is $$1\otimes a \mapsto a$$. Of course the point is that we need reduce to the form $$1\otimes a$$ before applying the last map. Therefore we get $$a \mapsto 1\otimes a \mapsto (t-1)\otimes a = 1 \otimes (ta - a) \mapsto ta - a.$$ Hence by a direct computation $$D$$ turns out to be the identity of $$\Z_3$$: $$D(0) = 0, \quad D(1)=t\cdot1 - 1 = 2-1 = 1, \quad D(2)=t\cdot 2 - 2 = 1-2 \equiv 2 \mod 3.$$
Therefore we conclude that the homology groups of $$S^1$$ with coefficients in $$\Z_3$$ twisted by the nontrivial $$\rho$$ are trivial: $$H_0(S^1\Z_3)_\rho \cong H_1(S^1\Z_3)_\rho \cong 0$$