# Homology with twisted coefficients of the circle


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 $\rho:\p\to \Aut(A)$?

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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 $\mathbb Z \to \Aut(A)$. – Justin Young Jun 20 '13 at 7:56
Also, the only automorphism $\mathbb Z /2 \to \mathbb Z/2$ is the identity, so every representation $G\to \Aut (\mathbb Z/2)$. – Justin Young Jun 21 '13 at 6:02
....is trivial. – Justin Young Jun 21 '13 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 – Abramo Jun 22 '13 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 '14 at 21:28

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: $$H_0(S^1\Z_3)_\rho \cong H_1(S^1\Z_3)_\rho \cong 0$$