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.

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

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Question. Does there always exist a representation $\rho$ such that the homology of the twisted chain complex is trivial?

Thanks for guidance!

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!