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Since the chain complex in question is finite and levelwise projective we know that contractibility is implied by being acyclic. And we see directly that if then the complex is exact, hence acyclic.
To compute the Reidemeister torsion we need to construct a chain homotopy such that
Here we can take to be the map as then
and then we need to choose to be the map .
To compute the torsion of this complex we have to consider the class
which is given by
and hence the matrix of in the standard basis is given by
whose class in is the same as the class of since
where
stands for equality in
and
is the elementary matrix with
at the
entry.
\mapsto (1,0)$ as then
$$ s_{-1}(d_0(x)) + d_1(s_0(x)) = d_1(x,0) = x $$
and then we need to choose $s_1: C_1 = \Q\oplus \Q \to \Q = C_2$ to be the map $(x,y) \mapsto \frac{y}{r}$.
To compute the torsion of this complex we have to consider the class
$$ \lbrack d+s : C_{\mathrm{ev}} \to C_{\mathrm{odd}} \rbrack \in K_1(\Q) $$
which is given by
$$ (d+s)(x,y) = d_2(x) + s_0(y) = (y,rx) $$
and hence the matrix of $d+s$ in the standard basis is given by
$$A:=\begin{pmatrix} 0 & 1 \ r & 0 \end{pmatrix},$$
whose class in $K_1(\Q)$ is the same as the class of $(-r)$ since
$$ A\simeq E_{12}^{-1}E_{21}^{1}AE_{21}^{-r}=\begin{pmatrix} -r & 0 \ 0 & 1 \end{pmatrix} \simeq (-r) $$ where $\simeq$ stands for equality in $K_1(\Q)$ and $E_{ij}^q$ is the elementary matrix with $q\in\Q$ at the $(i,j)$ entry.
r \neq 0 then the complex is exact, hence acyclic.
To compute the Reidemeister torsion we need to construct a chain homotopy
such that
Here we can take to be the map as then
and then we need to choose to be the map .
To compute the torsion of this complex we have to consider the class
which is given by
and hence the matrix of in the standard basis is given by
whose class in is the same as the class of since
where
stands for equality in
and
is the elementary matrix with
at the
entry.