Let $X$ be a manifold or a CW-complex.


Let


$\pi: \tilde X\longrightarrow X$


be a covering map.


Let $\pi\_1(X)$ be the fundamental group of $X$ and let $\rho: \pi\_1(X)\longrightarrow O(n)$ be an orthogonal representation.


Define the $\rho$-twisted chain complex of $\tilde X$ by


$C\_\*(\tilde X,\rho)=C\_\*(\tilde X)\otimes\_{\pi\_1(X)} \mathbb{R}^n$


where $\pi\_1(X)$ acts on $C\_\*(\tilde X)$ from the right by deck transformations and acts on $\mathbb{R}^n$ from the left by orthogonal transformations.


In the book: Lecture Notes in Algebraic Topology by
James F. Davis and
Paul Kirk, Chapter 5, the homology with local coefficients is defined as the homology of the $\rho$-twisted chain complex


$H\_\*(\tilde X,\rho)=H\_\*(C\_\*(\tilde X,\rho))$.


Question.


Can we add some additional hypothesis on $X$, the covering space $\tilde X$, and the covering map $\pi:\tilde X\longrightarrow X$ such that for such $X$ and $\tilde X$, we can always find an $n\geq 2$ and a $\rho$ satisfying that $H\_\*(\tilde X,\rho)$ is trivial?


Thanks for guidance.



triviality of homology with local coefficients
Maybe you are looking for something more interesting, but you can take $X=S^1$, universal cover $\tilde X$, and $\rho: {\mathbb Z}\to O(n)$ such that the image group has no fixed unit vectors in $R^n$. Then $H\_\*(\tilde X,\rho)=0$ (which is a nice exercise to work out if you are new to this material). A more challenging problem would be:


Construct a finite CW-complex $X$ such that for each $n\ge 2$ there exists a representation $\rho: \pi\_1(X)\to SO(n)$ with vanishing homology.


If you are interested in 3-dimensional topology, here are two classes of examples you should be aware of:


a. Suppose that $X$ is a closed connected orientable 3-manifold with finite nontrivial fundamental group $\pi$ and $\tilde X\to X$ is its universal covering. Then for each $\rho: \pi\to O(4)$ such that $\rho(\pi)$ has no fixed unit vectors, $H\_\*(\tilde X,\rho)=0$. (Examples of such $\rho$ are given by the fact that $\pi$ embeds in $SO(4)$ so that the image group acts freely on $S^3$.)


b. Suppose that $X$ is a closed connected orientable **arithmetic hyperbolic** 3-manifold and $\tilde X\to X$ is its universal covering. Then there exists a representation $\rho: \pi\_1(X)\to O(3)$ such that $H\_\*(\tilde X,\rho)=0$.


Edit. At the same time, there are spaces for which your property does not hold, for instance a space whose fundamental group admits only trivial orthogonal representations.


