# Homology with local coefficients and flat bundles

I have a problem in understanding Prop.1.10 in https://pages.uoregon.edu/sadofsky/691/sseq-local-coefficients.pdf My interpretation is as follows:

Let $$X$$ be a smooth n-manifold and $$U\to X$$ be its universal cover. Notice that $$U$$ is a $$\pi_1 X$$ principal bundle. Let $$\rho: \pi_1 X\to Aut(S)$$ be a an action of $$\pi_1 X$$ on the group $$S$$. Then we can form the associated bundle $$\tilde{X}=U\times_\rho S$$. The proposition 1.10 says that $$H_*(X,S) \simeq H_*(\tilde{X},\mathbb{Z})$$ where the LHS is the homology with twisted coefficients in $$S$$.

Problem: If $$S=V$$ is a vector space then $$\tilde{X}=U\times_{\rho} V$$ is the flat bundle associated to the flat connection induced by $$\rho.$$ This is a vector bundle over $$X$$ hence it retract over $$X$$ therefore $$H_*(X,V) \simeq H_*(\tilde{X},\mathbb{Z})\simeq H_*(X, \mathbb{Z}).$$ Which is not true for any flat connection.

• The highlighted proposition 1.10 is false as stated. – Moishe Kohan Jan 9 '20 at 18:16
• @MoisheKohan, am I interpreting it wrongly? Or is the proposition in the document false as well? – Warlock of Firetop Mountain Jan 10 '20 at 9:27

What is true is that if $$A = \Bbb Z[\pi_1(X)/H]$$ as a $$\pi_1(X)$$-module, and $$p: X' \to X$$ is the covering space associated to the subgroup $$H \subset \pi_1(X)$$, then $$\mathcal A = \tilde X \times_{\pi_1 X} A$$ gives a local coefficient system over $$X$$ (if you like to think of a local coefficient system as being a bundle of abelian groups over $$X$$). Then we have $$H_*(X;\mathcal A) \cong H_*(X'; \Bbb Z).$$
• @WarlockofFiretopMountain Sorry, I don't, I picked up most of this by brute-force computation. It might be a nice exercise for you to compute the homology of the circle equipped with a local coefficient system with fiber $\Bbb C^n$, and $1 \in \pi_1 S^1$ acting by $A \in GL_n(\Bbb C)$. Also, if you use the definition of the local-coefficient chain complex from bundle of groups, I suggest you show explicitly that we have an isomorphism of chain complexes $C_*(X;\mathcal A) \cong C_*(X';\Bbb Z)$, it's not too hard and is revealing. – Jacque Lemure Jan 11 '20 at 18:22