Talk:Reidemeister torsion (Ex)

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 $r \neq 0$$<wikitex>; Since the chain complex in question is finite and levelwise projective we know that contractibility is implied by being acyclic. And we see directly that if $r \neq 0$ then the complex is exact, hence acyclic. To compute the Reidemeister torsion we need to construct a chain homotopy $s: C_\bullet \to C_{\bullet+1}$ such that $$ ds + sd = \id $$ Here we can take $s_0 : C_0 = \Q \to \Q\oplus \Q = C_1$ to be the map r \neq 0$ then the complex is exact, hence acyclic. To compute the Reidemeister torsion we need to construct a chain homotopy $s: C_\bullet \to C_{\bullet+1}$$s: C_\bullet \to C_{\bullet+1}$ such that

$\displaystyle ds + sd = \id$ $$\displaystyle ds + sd = \id$$

Here we can take $s_0 : C_0 = \Q \to \Q\oplus \Q = C_1$$s_0 : C_0 = \Q \to \Q\oplus \Q = C_1$ to be the map $1 \mapsto (0,1)$$1 \mapsto (0,1)$ as then

$\displaystyle s_{-1}(d_0(x)) + d_1(s_0(x)) = d_1(0,x) = x$ $$\displaystyle s_{-1}(d_0(x)) + d_1(s_0(x)) = d_1(0,x) = x$$

and then we need to choose $s_1: C_1 = \Q\oplus \Q \to \Q = C_2$$s_1: C_1 = \Q\oplus \Q \to \Q = C_2$ to be the map $(x,y) \mapsto \frac{x}{r}$$(x,y) \mapsto \frac{x}{r}$.

To compute the torsion of this complex we have to compute

$\displaystyle \lbrack d+s : C_{\mathrm{ev}} \to \C_{\mathrm{odd}} \rbrack \in K_1(\Q)$ $$\displaystyle \lbrack d+s : C_{\mathrm{ev}} \to \C_{\mathrm{odd}} \rbrack \in K_1(\Q)$$

which is given by

$\displaystyle (d+s)(x,y) = d_2(x) + s_0(y) = (rx,y)$ $$\displaystyle (d+s)(x,y) = d_2(x) + s_0(y) = (rx,y)$$

and hence the matrix is given by

$\displaystyle \begin{pmatrix} r & 0 \\ 0 & 1 \end{pmatrix}$ $$\displaystyle \begin{pmatrix} r & 0 \\ 0 & 1 \end{pmatrix}$$

which we can view as an element in $K_1(\Q)$$K_1(\Q)$.

\mapsto (0,1)$ as then $$ s_{-1}(d_0(x)) + d_1(s_0(x)) = d_1(0,x) = x $$ and then we need to choose $s_1: C_1 = \Q\oplus \Q \to \Q = C_2$ to be the map $(x,y) \mapsto \frac{x}{r}$. To compute the torsion of this complex we have to compute $$ \lbrack d+s : C_{\mathrm{ev}} \to \C_{\mathrm{odd}} \rbrack \in K_1(\Q) $$ which is given by $$ (d+s)(x,y) = d_2(x) + s_0(y) = (rx,y) $$ and hence the matrix is given by $$\begin{pmatrix} r & 0 \ 0 & 1 \end{pmatrix}$$ which we can view as an element in $K_1(\Q)$. r \neq 0 then the complex is exact, hence acyclic. To compute the Reidemeister torsion we need to construct a chain homotopy $s: C_\bullet \to C_{\bullet+1}$$s: C_\bullet \to C_{\bullet+1}$ such that

$\displaystyle ds + sd = \id$ $$\displaystyle ds + sd = \id$$

Here we can take $s_0 : C_0 = \Q \to \Q\oplus \Q = C_1$$s_0 : C_0 = \Q \to \Q\oplus \Q = C_1$ to be the map $1 \mapsto (0,1)$$1 \mapsto (0,1)$ as then

$\displaystyle s_{-1}(d_0(x)) + d_1(s_0(x)) = d_1(0,x) = x$ $$\displaystyle s_{-1}(d_0(x)) + d_1(s_0(x)) = d_1(0,x) = x$$

and then we need to choose $s_1: C_1 = \Q\oplus \Q \to \Q = C_2$$s_1: C_1 = \Q\oplus \Q \to \Q = C_2$ to be the map $(x,y) \mapsto \frac{x}{r}$$(x,y) \mapsto \frac{x}{r}$.

To compute the torsion of this complex we have to compute

$\displaystyle \lbrack d+s : C_{\mathrm{ev}} \to \C_{\mathrm{odd}} \rbrack \in K_1(\Q)$ $$\displaystyle \lbrack d+s : C_{\mathrm{ev}} \to \C_{\mathrm{odd}} \rbrack \in K_1(\Q)$$

which is given by

$\displaystyle (d+s)(x,y) = d_2(x) + s_0(y) = (rx,y)$ $$\displaystyle (d+s)(x,y) = d_2(x) + s_0(y) = (rx,y)$$

and hence the matrix is given by

$\displaystyle \begin{pmatrix} r & 0 \\ 0 & 1 \end{pmatrix}$ $$\displaystyle \begin{pmatrix} r & 0 \\ 0 & 1 \end{pmatrix}$$

which we can view as an element in $K_1(\Q)$$K_1(\Q)$.