5
$\begingroup$

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.

$\endgroup$
2
  • 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

5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.