Commit ce5c6561 authored by benjamin.wassermann's avatar benjamin.wassermann

Update contents.tex

parent 9f0dbdbf
\tableofcontents
\chapter{Torsion Invariants [Roman Sauer]}
......@@ -840,7 +839,7 @@ A map $r \colon Y \to X$ such that $r \circ j = \id_X$ is called \CmMark{element
But this is a very simple CW-complex with differential $\id \colon \Z\pi \to \Z\pi$ (for $\pi = \pi_1(Y)$) from degree $n$ to degree $n-1$ and thus $\tau(C_{*}(\tilde Y,\tilde X)) = 0$.
The converse is more complicated.
\item Let $A \in \GL_n(\Z\pi)$ ($n \geq 3$) and let $X' = X \vee \bigvee_{j = 1}^nS^{n-1}$, where we denote the $j$-th inclusion of $S^{n-1}$ into the wedge by $b_j$.
\item Let $A \in \GL_n(\Z\pi)$ ($n \geq 3$) be an element representing $a \in \Wh(\pi(X))$ and let $X' = X \vee \bigvee_{j = 1}^nS^{n-1}$, where we denote the $j$-th inclusion of $S^{n-1}$ into the wedge by $b_j$.
We attach $n$ $n$-cells to $X'$ via attaching maps $f_j \colon S^{n-1} \to X'$ (relevant: $[f_j] \in \i_{n-1}(X')$ which yields $Y$.
\begin{align*}
\begin{pmatrix}
......@@ -853,10 +852,14 @@ A map $r \colon Y \to X$ such that $r \circ j = \id_X$ is called \CmMark{element
\ni \bigoplus_{j = 1}^n \pi_{n-1}(X')
\text{ for } b_j \in \pi_{n-1}(X') \curvearrowleft \pi_1(X') = \pi_1(X) = \pi.
\end{align*}
One sees by closer inspection that
One sees by closer inspection that the relative chain complex $C_{*}^{\text{CW}}(\tilde Y, \tilde X)$ has only two non-trivial chain groups in two consecutive dimensions, with
boundary map realized by the $\Z\pi$-isomorpism $A$. Thus, $C_{*}^{\text{CW}}(\tilde Y, \tilde X)$ is acyclic, and since $\pi_1(\tilde Y,\tilde X) = 0$,
we get from the {\itshape relative Hurewicz theorem} that $\pi_k(\tilde Y,\tilde X) = 0$ for all $k \in \mathbb N$. Since $n \geq 3$, we have $\pi_1(Y,X) = 0$, so we can conclude
that $\pi_k(X) \cong \pi_k(Y)$ for all $k$, where the isomorphism is realised by the map $\pi_k(X \hookrightarrow Y)$. By {\itshape Whitehead's theorem}, $X \hookrightarrow Y$ is
a homotopy equivalence, so we can indeed compute its Whitehead torsion. We get:
\begin{align*}
\tau(\tilde X \hookrightarrow \tilde Y) = \tau(C_{*}^{\text{CW}}(\tilde Y, \tilde X))
= \tau\left(\bigoplus_{j=1}^n \Z\pi \xrightarrow{A} \bigoplus_{j=1}^n \Z\pi\right)
= \tau\left(\bigoplus_{j=1}^n \Z\pi \xrightarrow{A} \bigoplus_{j=1}^n \Z\pi\right) = a
\end{align*}
for $[A] \in \Wh(\pi)$.
\end{enumerate}
......
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