Let $X$ be a path-connected manifold (or a CW complex).

Let $\pi_1(X)$ be the fundamental group of $X$.

Let $\pi: \tilde X\longrightarrow X$ be a covering map.

For each $m\geq 0$, let $C_m(\tilde X)$ be the real chain group generated by all the $m$-cells of $\tilde X$.

Then $C_m(\tilde X)$ is a module over the real group algebra $\mathbb{R}(\pi_1(X))$.

Here $\pi_1(X)$ acts on $C_m(\tilde X)$ be the deck transformation.

Let $\rho: \pi_1(X)\longrightarrow O(n)$ be an orthogonal representation.

Then $\rho$ induces a ring homomorphism

$\mathbb{R}(\rho): \mathbb{R}(\pi_1(X))\longrightarrow \mathbb{R}(O(n))$.

Define the twisted chain complex by $C_m(\tilde X,\rho)=C_m(\tilde X)\otimes _{\pi_1(X)}\mathbb{R}^n$.


Question. Suppose $\mathbb{R}(\rho)$ is surjective. Does there always exist a representation $\rho$ such that the homology of the twisted chain complex is trivial? Whether or not could we add some hypothesis such that the answer is true? Are there any references?

Thanks for guidance!

  • 7
    $\begingroup$ Certainly not if $X$ is simply connected. $\endgroup$
    – abx
    Nov 19, 2020 at 14:38

1 Answer 1


There are finitely presented groups that do not have any non-trivial linear representations, so for these groups as fundamental group you are just asking whether the ordinary real homology of $X$ is trivial, which it usually won't be. As was pointed out in a comment the trivial group has this property, but there are plenty of infinite examples too.


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