Talk:Reidemeister torsion (Ex)

Revision as of 12:46, 31 August 2013 (view source) Markus Land (Talk | contribs) (Created page with "<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 \ne...")		Latest revision as of 18:22, 3 September 2013 (view source) Carlos Moraga Ferrandiz (Talk | contribs) m (A twist in the definition of the contraction is corrected. An easy calculation in K_1 is added.)
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	To compute the Reidemeister torsion we need to construct a chain homotopy $s: C_\bullet \to C_{\bullet+1}$ such that		To compute the Reidemeister torsion we need to construct a chain homotopy $s: C_\bullet \to C_{\bullet+1}$ such that
	$$ ds + sd = \id $$		$$ ds + sd = \id $$
−	Here we can take $s_0 : C_0 = \Q \to \Q\oplus \Q = C_1$ to be the map $1 \mapsto (0,1)$ as then	+	Here we can take $s_0 : C_0 = \Q \to \Q\oplus \Q = C_1$ to be the map $1 \mapsto (1,0)$ as then
−	$$ s_{-1}(d_0(x)) + d_1(s_0(x)) = d_1(0,x) = x $$	+	$$ s_{-1}(d_0(x)) + d_1(s_0(x)) = d_1(x,0) = x $$
−	and then we need to choose $s_1: C_1 = \Q\oplus \Q \to \Q = C_2$ to be the map $(x,y) \mapsto \frac{x}{r}$.	+	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 compute	+	To compute the torsion of this complex we have to consider the class
−	$$ \lbrack d+s : C_{\mathrm{ev}} \to \C_{\mathrm{odd}} \rbrack \in K_1(\Q) $$	+	$$ \lbrack d+s : C_{\mathrm{ev}} \to C_{\mathrm{odd}} \rbrack \in K_1(\Q) $$
	which is given by		which is given by
−	$$ (d+s)(x,y) = d_2(x) + s_0(y) = (rx,y) $$	+	$$ (d+s)(x,y) = d_2(x) + s_0(y) = (y,rx) $$
−	and hence the matrix is given by	+	and hence the matrix of $d+s$ in the standard basis is given by
−	$$\begin{pmatrix} r & 0 \\ 0 & 1 \end{pmatrix}$$	+	$$A:=\begin{pmatrix} 0 & 1 \\ r & 0 \end{pmatrix},$$
−	which we can view as an element in $K_1(\Q)$.	+	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.
	</wikitex>		</wikitex>

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Latest revision as of 18:22, 3 September 2013

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 $<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 $s: C_\bullet \to C_{\bullet+1}$ such that

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

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

which is given by

and hence the matrix of $d+s$ in the standard basis is given by

whose class in $K_1(\Q)$ is the same as the class of $(-r)$ since

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.

\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 $s: C_\bullet \to C_{\bullet+1}$ such that

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

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

which is given by

and hence the matrix of $d+s$ in the standard basis is given by

whose class in $K_1(\Q)$ is the same as the class of $(-r)$ since

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.