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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.)

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This should be true, here is a (hoperfully correct) sketch of proof: the Reidemeister torsion is a continuous function on the set of acyclic representations (as it can be computed from determinants of matrices whose coefficients vary continuously with the representation). On the other hand one can simultaneously conjugate any subset of upper triangular matrices arbitrarily close to diagonal matrices. So, if both the upper-triangular and diagonal representations are acyclic, the torsion is the same for them.

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  • $\begingroup$ Thanks! I think I have an algebraic proof of this fact, but it's nice to know it's true for other reasons. $\endgroup$ – Calvin McPhail-Snyder Oct 28 '19 at 18:51

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