The torsion of a link complement $S^3 \setminus L$ is defined in terms of the twisted chain complex $C\_\*(S^3 \setminus L; \rho)$, where $\rho : S^3 \setminus L \to \operatorname{GL}\_k(\mathbb{k})$ is a representation of the link complement into the group of matrices over a field $\mathbb{k}$. (For the usual Reidemeister torsion/Alexander polynomial, $\rho$ is the representation sending each meridian to a variable $t \in \operatorname{GL}\_1(\mathbb{Q}(t))$, but more generally we can consider nonabelian representations.)


For the torsion to be defined, the complex $C\_\*(S^3 \setminus L; \rho)$ needs to be acyclic.
There are two conditions usually given in the literature. One is that there is some meridian $x$ with $\det(\rho(x) - I\_k)$ nonzero. Another is related to the reduced Burau representation. If $L$ is represented as the closure of a braid $\beta$ on $n$ strands, then we have $n$ meridians $x\_1, \dots, x\_n$ corresponding to the strands of $\beta$. The second condition says that $C\_\*(S^3 \setminus L; \rho)$ is acyclic if $\det(\rho(x\_1 \cdots x\_n) - I\_k)$ is nonzero.


Either of these conditions is sufficient, but is either necessary? Are they equivalent?



Equivalence of conditions for torsions of links to be defined
I'm not sure that $\det(\rho(x)-I\_k) \neq 0$ is sufficient for $C\_\*(S^3 \setminus L;\rho)$ to be acyclic. Write $X\_L:=S^3 \setminus L$. Let $\omega \in S^1 \setminus \lbrace 1 \rbrace$ and take the one-dimensional representation $\rho \colon \pi\_1(X\_L) \to \mathbb{C}^\times$ that maps each meridian to $\omega$. This way $\det(\rho(x)-I\_1)=\omega-1 \neq 0$. However, I believe that $H\_1(X\_L;\rho)$ is zero if and only if $\Delta\_L(\omega) \neq 0$ (the dimension of $H\_1(X\_L;\rho)$ is the nullity $\eta\_L(\omega$)). Consequently, if $\omega$ is a root of $\Delta\_L$, then I think that $C\_\*(X\_L;\rho)$ is not acyclic.


So I think that the condition is necessary but not sufficient: the complex is acyclic if and only if all the $H\_i(X\_L;\rho)$ vanish. Your condition basically controls $H\_0(X\_L;\rho)$. If the complex is acyclic, then the $H\_0$ must vanish which should imply that $\det(\rho(x)-I\_k) \neq 0$.


Now to braids. From now on, $L=\widehat{\beta}$ is a braid closure and $\rho \colon F\_n \to GL\_k(\mathbb{k})$ is a representation of the free group that extends to a representation of $\pi\_1(X\_L).$ For the same reason as above, I'm not sure the condition is sufficient. I don't think it is necessary either. Assume $L$ is a knot, that $\beta$ has $n=3$ strands and that $\omega^3=1$. I am taking the same $1$-dimensional representation as above. Now $C\_\*(X\_L;\rho)$ is acyclic ($\det(\rho(x)-I\_1)=\omega-1 \neq 0$ and $\Delta\_L(\omega) \neq 0$ because for a knot $K$, $\Delta\_K(t)$ does not have roots that are prime powers of unity). However, by my choice of $\omega$, I have $\det(\rho(x\_1x\_2x\_3)-I\_1)=w^3-1=0$.


In particular, the conditions are not equivalent: for that example $\det(\rho(x)-I\_k) \neq 0$ but $\det(\rho(x\_1x\_2\cdots x\_n)-I\_k) = 0$.


Finally, a topological remark. Write $D\_n$ for the $n$ times punctured disc. The free group $\pi\_1(D\_n)$ has the $x\_i$ as its generators, so morally $\partial D\_n$ "is" $x\_1x\_2\cdots x\_n$. Now $X\_L$ can be obtained from the exterior of the closure of the braid in the solid torus, by adding on an extra solid torus. Equivalently, $X\_L=X\_{\widehat{\beta} \cup \partial D\_n} \cup (D^2 \times \partial D\_n)$. The torsion of $D^2 \times \partial D\_n$ is your $\det(\rho(x\_1x\_2\cdots x\_n)-I\_k)$. Now the resulting Mayer-Vietoris sequence with twisted coefficients relates the acyclicity of $C\_\*(X\_K;\rho)$ to $\det(\rho(x\_1x\_2\cdots x\_n)-I\_k)$.


