# triviality of homology with local coefficients

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.

• 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. Nov 25, 2020 at 14:40
• @CalvinMcPhail-Snyder: It is sometimes useful to consider intermediate covering spaces in this setting, but I agree, the universal one is most common. Nov 25, 2020 at 14:45

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.
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 $$H_*(\tilde X,\rho)=0$$.