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


$\pi: \tilde X\longrightarrow X$

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

$H_*(\tilde X,\rho)=H_*(C_*(\tilde X,\rho))$.


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.

  • 2
    $\begingroup$ I assume you mean $H_*( X,\rho)=H_*(C_*(\tilde X,\rho))$? And usually I would think about this for $\tilde X$ the universal cover of $X$. But maybe you're coming at this from a different angle than me. $\endgroup$ Nov 25, 2020 at 14:40
  • 1
    $\begingroup$ @CalvinMcPhail-Snyder: It is sometimes useful to consider intermediate covering spaces in this setting, but I agree, the universal one is most common. $\endgroup$ Nov 25, 2020 at 14:45

1 Answer 1


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.


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