Let $L$ be a link in $S^3$ and $\rho : \pi_1(S^3 \setminus L) \to \operatorname{SL}_2(\mathbb C)$ be a representation of its knot group. If the twisted homology $H^\rho(S^3 \setminus L)$ is acyclic, we can obtain the twisted Reidemeister torsion $\tau(S^3 \setminus L, \rho)$, which is closely related to the twisted Alexander polynomial. (One can also do this for other matrix groups than $\operatorname{SL}_2(\mathbb C)$.)
Thinking of $\rho$ as a two-dimensional representation of $\pi_L := \pi_1(S^3 \setminus L)$, I am interested in the case where this representation is reducible, but possibly not indecomposable. After choosing the right basis this is the same as asking that the matrices of $\rho$ are always of the form $$ \begin{pmatrix} \kappa & \epsilon \\ 0 & \kappa^{-1} \end{pmatrix} $$ for some $\kappa \ne 0$. If the $\epsilon$ can all be chosen to be zero, then the representation is decomposable and $\tau(S^3 \setminus L, \rho)$ is (a square of) the ordinary Reidemeister torsion.
There are examples where $\rho$ is reducible but not indecomposable. However, in those cases it seems that the indecomposability doesn't matter. More formally, one can obtain another representation $\bar \rho$ of $L$ by taking each meridian $x$ of $L$ and setting the upper-right entry of $\bar \rho(x)$ to be zero. (For a basis-independent definition, we are just replacing the meridians with diagonal matrices with the same eigenvalues.) Then $$ \tau(S^3 \setminus L, \rho) = \tau(S^3 \setminus L, \bar \rho) $$ Is there a proof of this fact? Is it true?
This seems natural, because the torsion is a sort of determinant and the matrices above are upper-triangular. However, I am not aware of a proof or counterexample. (It's not totally obvious, because the twisted Burau representation used to define the torsion isn't necessarily upper-triangular even when $\rho$ is.)
