# Twisted torsions of reducible representations of knot groups

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.)