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\tableofcontents

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\chapter{Torsion Invariants [Roman Sauer]}
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Torsion invariants fall into a class of so-called ``secondary invariants'' of topological spaces in the sense that they are only defined if a certain class of ``primary invariants'' (e.g. Betti numbers) vanish.
Often they reveal more subtle geometric information.
The following will contain a discussion of Whitehead and Reidemeister torsion.
Informally, corresponding primary invariants are Lefschetz numbers (Whitehead torsion) and the Euler characteristic (Reidemeister torsion).

\section{Review of Euler characteristic and Lefschetz numbers.}

\subsection{CW Complexes}

\begin{dfn*}
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  A (finite) \CmMark{CW-complex} is a hausdorff space with a decomposition $E$ into (finitely many) cells (space homeomorphic to some $\R^n$) such that for every $e \in E$ there is a continuous map $\phi_e \colon D^n \to X$ with $\phi_e \colon \mathring D^n \xrightarrow{\cong} e$ and $\Im(\phi_e|_{S^{n-1}}) \subset \bigcup_{f \in E, \dim f \leq n-1} f$.
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\end{dfn*}

\begin{expl*}
  \begin{enumerate}
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  \item Simplicial complexes, e.g. triangles, pyramids, etc.
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  \item But CW-complexes are more general, the following graph is CW for example:
    \begin{center}
      \begin{tikzpicture}
        \draw (0,0) to[bend left] (2,0);
        \draw (0,0) to[bend right] (2,0);
        \draw (2,0) to (3,0);
      \end{tikzpicture}
    \end{center}
    One can even attach a disc along its boundary to a single 1-cell.
  \end{enumerate}
\end{expl*}

\subsection{Euler characteristic}

\begin{dfn*}
  The Euler class $\chi(X)$ of a finite CW-complex $X$ is defined as $\chi(X) = \sum_{i \geq 0}(-1)^i \#(i\text{-cells of } X) \in \Z$.
\end{dfn*}

\begin{thm*}[Euler-Poincaré]
  \begin{align*}
    \chi(X) = \sum_{i \geq 0} (-1)^i b_i(X),
  \end{align*}
  where $b_i(X) = \rk_{\Z} H_i(X;\Z)$.
\end{thm*}

In particular, $\chi$ is a homotopy invariant.

\begin{proof}[``Proof'']
  $H_i(X;\Z) = H_i(C_{*}^{CW}(X))$, where $C_{*}^{CW}(X)$ is the cellular chain complex
  \begin{align*}
    \cdot \to C_{i+1}^{CW}(X)\xrightarrow{\partial} \underbrace{C_{i}^{CW}(X)}_{\cong \Z^{\# i\text{-cells}}} \xrightarrow{\partial} C_{i-1}^{CW}(X) \to \cdots
  \end{align*}
  Thus $\chi(C_{*}) := \sum_{i \geq 0} (-1)^i\rk_{\Z}(C_{i})$ and $\chi(C^{CW}(X)) = \chi(X)$.
  This boils down to
  \begin{align*}
    \chi(C_{*}) = \sum_{i \geq 0} \rk_{\Z}H_i(C_{*}) ( = \chi(H_{*}(C_{*}))].
  \end{align*}
  This is just additivity of the rank!
  Consider
  \begin{align*}
    C_1 \xrightarrow{\partial} C_0
  \end{align*}
  and note that we have the exact sequences $0 \to \Im \partial \to C_0 \to \underbrace{H_0}_{= C_0/\Im \partial} \to 0$ and $0 \to \underbrace{H_1}_{= \Ker \partial} \to C_1 \xrightarrow{\partial} \Im \partial \to 0$.

  Thus $\chi(C_{*)} = \rk_{\Z} C_0 - \rk_{\Z} C_1 = \rk_{\Z} \Im \partial + \rk_{\Z} H_0 - \rk_{\Z}H_1 - \rk_{\Z} \Im \partial = \rk H_0 - \rk H_1$, which completes the ``proof''.
\end{proof}

\subsection{Review of cellular homology}

Let $X$ be a CW-complex with cellular decomposition $E$.
Then we can consider the \CmMark{n-skeleton}
\begin{align*}
  X^n := \sum_{e \in E, \dim e \leq n} e,
\end{align*}
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which yields a filtration $X^0 \subset X^1 \subset \cdots \subset X$ such there is a push-out diagram
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\begin{equation*}
  \begin{tikzcd}
    \coprod S^{n-1} \ar{r} \ar[hook]{d} & X^{n-1} \ar[hook]{d} \\
    \coprod D^n \ar{r} & X^n
  \end{tikzcd}
\end{equation*}
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One could take this as an alternative definition of a CW-complex by a filtration with the push-out property.
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The cells can be recovered as connected components of $X^n\setminus X^{n-1}$.

We have
\begin{align*}
  C_i^{CW}(X) = H_i(X^i, X^{i+1}) \xleftarrow{\cong} H_i(\coprod D^i, \coprod S^{i-1}) \cong \bigoplus H_i(D^i, S^{i-1}) \cong \bigoplus \Z^{\# i\text{-cells}},
\end{align*}
where the first isomorphism $\leftarrow$ is given by excision.
The boundary maps $C_i^{CW}(X) \xrightarrow{\partial} C_{i-1}^{CW}(X) $ come from
\begin{align*}
  H_i(X^i,X^{i-1}) \to H_{i-1}(X^{i-1}) \to H_{i-1}(X^{i-1},X^{i-2}).
\end{align*}
Under this isomorphism, the matrix entry belonging to $(e,f)$ where $e$ is an $n$-cell, $f$ an $(n-1)$-cell is the \CmMark{degree} of the map.
\begin{align*}
  S^{i-1} \xrightarrow{\phi_e|_{S^{n-1}}} X^{i-1} \xrightarrow{\operatorname{proj}} X^{i-1}/(X^{i-1}\setminus f) \xleftarrow{\phi_f, \cong} D^{i-1}/S^{i-2} \cong S^{i-1}.
\end{align*}

\begin{expl*}
  Consider the torus as an identification square.
  We convince ourselves that the cellular chain complex is given as $\Z \to \Z \oplus \Z \to \Z$, where $1 \mapsto (0, 0)$, since it is described by a map $S^1 \to S^1$ traversing the 2-cell according to orientation has degree $0$.
\end{expl*}


\subsection{Lefschetz number}

Recall that a map $f \colon X \to P$ between CW-complexes is \CmMark{cellular}, if $f(X^i) \subset Y^i$ for all $i$.

\begin{thm*}[Cellular approximation]
  Any map between CW-complexes is homotopic to a cellular map.
\end{thm*}

\begin{dfn*}
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  The \CmMark{Lefschetz number} of a self-map $f \colon X \to X$ of a finite CW-complex is defined as
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  \begin{align*}
    \Lambda(f) = \sum_{i \geq 0} (-1)^i\tr C_i^{CW}(f) \in \Z.
  \end{align*}
\end{dfn*}

\begin{rem*}
  $\Lambda(\id_X) = \chi(X)$.
\end{rem*}

The following theorem yields a description of Lefschetz numbers by homology.
\begin{thm*}
  $\Lambda(f) = \sum_{i \geq 0}(-1)^i \tr H_i(f)$.
\end{thm*}
Thus, this number only depends on the homotopy class of $f$.

\begin{proof}
  Similar to the proof of Euler-Poincaré using the additivity of the trace, i.e. in the situation
  \begin{equation}
    \begin{tikzcd}[row sep=small]
      0 \ar{r} & A \ar{r} \ar{d}{a} & B \ar{r} \ar{d}{b} & C \ar{r} \ar{d}{c} & 0\\ 
      0 \ar{r} & A \ar{r} & B \ar{r} & C \ar{r} & 0
    \end{tikzcd}
  \end{equation}
  we have $\tr(b) = \tr(a) + \tr(c)$.
\end{proof}

\begin{thm*}
  If $f$ has no fixed point, then $\Lambda(f) = 0$.
\end{thm*}
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\begin{rem*}
  The converse is not true (think of counterexamples, e.g. $S^1 \wedge S^1$), although there is one in the case of simply-connected closed manifolds.
\end{rem*}
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\begin{proof}
  Let $X$ be metrizable and let $d$ be a metric.
  If $X$ is compact, there exists an $\varepsilon > 0$ with $d(f(x), x) > 3\varepsilon$.
  One can ``refine'' the CW-structure to a new one such that every cell has diameter $< \varepsilon$.
  By cellular approximation we can see that there exists a cellular map $g \colon X \to X$ with $g \simeq f$ and $d(g(x),f(x)) < \varepsilon$.
  Thus $g(\overline e) \cap \overline e = \emptyset$ for every cell $e$.
  Hence, the diagonal matrix entries of each $C_i^{CW}(g)$ are zero and thus $\Lambda(g) = \Lambda(f) = 0$.
\end{proof}
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\section{Whitehead torsion}

\subsection{Introduction/Motivation}

Given a homotopy equivalence $f \colon X \xrightarrow{\simeq} Y$ of finite CW-complexes, Whitehead torsion is an assignment $\tau(f) \in \Wh(\pi_1(Y))$ living in the so-called Whitehead group.

\begin{thm*}[Properties of Whitehead torsion]
  \begin{enumerate}[label=(\arabic*)]
  \item homotopy invariance
  \item\footnote{This is a deep theorem of Chapman.} If $f \colon X \to Y$ is a homeomorphism, then $\tau(f) = 0$.
  \item additivity:
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    A cellular push-out is a diagram
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    \begin{equation*}
      \begin{tikzcd}
        X_0 \ar{r}{f} \ar[hook]{d}{i} & X_2 \ar{d}\\
        X_1 \ar{r} & X
      \end{tikzcd}
    \end{equation*}
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    with $X_i$ be CW-complexes, where $f$ is cellular and $i$ is an inclusion of a sub-complex.
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    If the diagram
        \begin{equation*}
      \begin{tikzcd}
        X_0 \ar{rr} \ar[hook]{dd} \ar{rd}{f_0}[swap]{\simeq} &[0.8cm] &[-0.2cm] X_2 \ar[bend left=10]{rd}{f_2}[swap]{\simeq} &[0.8cm] \\
        & Y_0 \ar[near start]{rr}{\phi} \ar[near end,hook]{dd}{j} & & Y_2 \ar{dd}{i}\\
        X_1 \ar[crossing over]{rr} \ar[bend right=10]{rd}{f_1}[swap]{\simeq} & & X \ar[dashed]{rd}{f} \ar[leftarrow,crossing over]{uu} & \\
        & Y_1 \ar{rr}{\psi} & & Y
      \end{tikzcd}
    \end{equation*}
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    is a map of cellular push-outs such that $f_i$ are homotopy equivalences.
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    Then $f$ is a homotopy equivalence and
    \begin{align*}
      \tau(f) & = "\tau(f_1) + \tau(f_2) - \tau(f_0)"\\
      & = \psi_{*}(\tau(f_1))  + j_{*}(\tau(f_2)) - (\psi \circ i)_{*}(\tau(f_0)) \in \Wh(\pi_1(Y)).
    \end{align*}
   A similar additivity holds for the Lefschetz number.\footnote{Idea. $0 \to C_1(X_0) \to C_{*}(X_1) \oplus C_{*}(X_2) \to C_{*}(X) \to 0$ exact.}
 \item composition formula.
   If we have
   $\begin{tikzcd}
     X \ar{r}{f}[swap]{\simeq} & Y \ar{r}{g}[swap]{\simeq} & Z
   \end{tikzcd}$,
   then
   \begin{align*}
     \tau(g \circ f) & = "\tau(f) + \tau(g)"\\
     & = g_{*}(\tau(f) + \tau(g)) \in \Wh(\pi_1Z).
   \end{align*}
  \end{enumerate}
\end{thm*}

\begin{thm*}[s-cobordism theorem (Mazur, Barden, Stallings, Smale)]
  Let $M$ be a closed smooth manifold of dimension $\geq 5$.
  Let $(W, i, j)$ be an s-cobordism
  \begin{figure}[h!]
    \centering
    \begin{tikzpicture}[scale=0.8]
      
      \coordinate (L1) at (2,0);
      \coordinate (L2) at (2,3);

      \coordinate (LL1) at ($(L1)+(-4,0)$);
      \coordinate (LL2) at ($(L2)+(-4,0)$);
      
      \coordinate (R1) at (8,0.5);
      \coordinate (R2) at (8,2);

      \coordinate (RR1) at ($(R1)+(3,0)$);
      \coordinate (RR2) at ($(R2)+(3,0)$);

      % left inclusion
      
      \ellipsebetweenvert{LL1}{LL2}

      \node[below] at (LL1) {$M$};
     
      \draw[right hook->] ($(LL1) + (1.2,1.5)$) to node[above] {$i$} node[below]{$\simeq$} ($(L1) + (-1.2,1.5)$);

      % cobordism

      \ellipsebetweenvert{L1}{L2}
      \node[below] at (L1) {$M_0$};

      \draw[out=0,in=180] (L1) to (5,-1) to (R1);
      \draw[out=0,in=180] (L2) to (R2);

      \topgenus[0.35]{5,0}
      \topgenus[0.22]{6,1.2}

      \node at (6,3) {$W$};
      
      \ellipsebetweenvert[left]{R1}{R2}
      \node[below] at (R1) {$M_1$};

      % right inclusion

      \draw[left hook->] ($(RR1) + (-1,0.75)$) to node[above] {$j$} node[below]{$\simeq$} ($(R1) + (1,0.75)$);

      \ellipsebetweenvert{RR1}{RR2}
      
      \node[below] at (RR1) {$N$};
    \end{tikzpicture}
    % sketch: \includegraphics[width=0.8\textwidth]{img/1.png}
  \end{figure}
  
  i.e. $\partial W = M_0 \coprod M_1$ and $i \colon M \hookrightarrow W$, $j \colon N \hookrightarrow W$ are homotopy equivalences.
  Then $\tau(M \xrightarrow{i} W) = 0$ if and only if $(W,i_0,i_1)$ is trivial, i.e.
  \begin{figure}[h!]
    \centering
    \begin{tikzpicture}[every node/.style={scale=0.8}]
      \coordinate (LU1) at (2.4,1.5);
      \coordinate (LU2) at (2.4,2.5);
      
      \coordinate (RU1) at (5,1.3);
      \coordinate (RU2) at (5,2.3);

      \coordinate (LD1) at (2.4,0.5);
      \coordinate (LD2) at (2.4,-0.5);
      
      \coordinate (RD1) at (5,0.5);
      \coordinate (RD2) at (5,-0.5);

      % leftmost part
      
      \node at (0,3) {$\exists$};

      \node[scale=1.5] at (-1,1) {$M$};

      \draw[right hook->] (-0.4,1.2) to node[above] {$i$} (2,2);
      \draw[right hook->] (-0.4,0.8) to node[below] {$\operatorname{incl}$} (2,0);

      % upper cobordism

      \ellipsebetweenvert{LU1}{LU2}

      \draw[out=0,in=180] (LU1) to (3,1.6) to (RU1);
      \draw[out=0,in=180] (LU2) to (4,2.2) to (RU2);

      \ellipsebetweenvert[left]{RU1}{RU2}

      % arrow from the cobordism to the cylinder

      \draw[->] ($(LU1)!0.5!(RU1) + (0,-0.1)$) to node[left]{$\cong$} node[right]{diffeo} ($(LU1)!0.5!(RU1) + (0,-0.7)$);

      % cylinder

      \ellipsebetweenvert{LD1}{LD2}

      \draw (LD1) to (RD1);
      \draw (LD2) to (RD2);
      
      \ellipsebetweenvert[left]{RD1}{RD2}
    \end{tikzpicture}
    % sketch: \includegraphics[width=0.5\textwidth]{img/2.png}
  \end{figure}
\end{thm*}

This theorem implies the Poincaré conjecture in dimensions at least 6, which says that if $M$ is a closed (smooth) manifold that is homotopy equivalent to $S^n$, then $M$ is homeomorphic to $S^n$.

The proof of this implication is basically along these lines:
Pick disjoint embedded $n$-disks in $M$ and remove them.
The result is a manifold $W$ with two boundary components.
Consider the s-cobordism theorem for this manifold as depicted in the figure below, where $f$ is a diffeomorphism of $(n-1)$-spheres.
\begin{figure}[h!]
  \centering
  \begin{tikzpicture}
    \coordinate (U1) at (0,3);
    \coordinate (U2) at (2.2,3);
    
    \coordinate (D1) at (-0.2,0);
    \coordinate (D2) at (1.8,0);

    % bordism on the left
    
    \ellipsebetweenhor{U1}{U2}

    \draw[out=270,in=90] (U1) to (0.3,1.8) to (D1);
    \draw[out=270,in=90] (U2) to (2.3,1.7) to (D2);
    
    \ellipsebetweenhor[upper]{D1}{D2}

    \node[left] at (0,1.5) {$W$};

    % arrows and lower disk

    \draw[bend left=15,->] (2.8,3.3) to node[above] {$f$} (4.2,3.3);

    \draw[->] (3,1.5) to node[above] {$\cong$} (4,1.5);

    \draw[left hook->] (2.5,-1) to (1.3,-0.5);
    \draw[right hook->] (4.2,-1) to (5.5,-0.5);

    \draw[pattern=north west lines,pattern color=gray] (3.35,-1) ellipse (0.6 and 0.3);
    \node at (3.8,-0.4) {$D^n$};

    % cylinder on the right

    \ellipsebetweenhor{5,3}{7,3}

    \draw (5,3) to (5,0);
    \draw (7,3) to (7,0);
    
    \ellipsebetweenhor[upper]{5,0}{7,0}
    
    \node[right] at (7,1.5) {$S^{n-1} \times [0,1]$};
      
  \end{tikzpicture}
  % sketch: \includegraphics[width=0.7\textwidth]{img/3.png}
\end{figure}

By filling top and bottom, the Poincaré conjecture is implied, if we can extend a diffeomorphism $f \colon S^{n-1} \xrightarrow{\cong} S^{n-1}$ to a homeomorphism $F \colon D^n \xrightarrow{\cong} D^n$.
This can be done by the so-called Alexander trick ($F(t x) = tf(x)$ for $t \in [0,1], x\in S^{n-1}$).


\subsection{Whitehead group and lower K-theory}

Let $R$ be a unital ring.
Then
\begin{align*}
  K_0(R) := \left< G \mid R \right>_{\text{ab}}
\end{align*}
with generators $G = \{$ isomorphism classes $[P]$ of fin. gen. projective $R$-modules $\}$ and relations $R = \{\ [P_1] = [P_0] + [P_2]$ whenever $0 \to P_0 \to P_1 \to P_2 \to 0$ is exact $\}$.
(Recall that a direct summand of in a free $R$-module is called a \CmMark{projective module}.)
This can be understood as a kind of universal dimension for projective $R$-modules.
\begin{align*}
  K_1(R) := \left< G \mid R \right>_{\text{ab}}
\end{align*}
with generators $G = \{$ conjugacy classes $[f]$ of automorphisms $f \colon P \to P$ of fin. gen. projective $R$-modules $\}$ and relations $R$ given as follows.

\begin{enumerate}[label=(\roman*)]
\item Every commuting diagram 
  \begin{equation*}
  \begin{tikzcd}
    0 \ar{r} & P_0 \ar{r} \ar{d}{f_0}[swap]{\cong} & P_1 \ar{r} \ar{d}{f_1}[swap]{\cong} & P_2 \ar{r} \ar{d}{f_2}[swap]{\cong} & 0\\
    0 \ar{r} & P_0 \ar{r} & P_1 \ar{r} & P_2 \ar{r} & 0.
  \end{tikzcd}
\end{equation*}
gives rise to a relation $[f_1] = [f_0] + [f_2]$.
\item $f,g \colon P \xrightarrow{\cong} P$ yield a relation $[f \circ g] = [f] + [g]$.
\end{enumerate}

This can be understood as an attempt to define a universal determinant of an automorphism.

There is a more common definition of $K_1(R)$ in terms of the general linear groups with coefficients in $R$.
Recall that $\GL(R) = \colim_{n \to \infty} \GL_n(R)$ with respect to the inclusion $\GL_n(R) \hookrightarrow \GL_{n+1}(R)$ to the upper left block.
Then
\begin{align*}
  K_1(R) = \GL(R)_{\text{ab}} = \GL(R)/[\GL(R),\GL(R)].
\end{align*}

The so-called \CmMark{Whitehead lemma} states that $[\GL(R),\GL(R)] = E(R)$, where $E(R)$ is the subgroup of $\GL(R)$ generated by all elementary upper triangular matrices with ones on the diagonal.
As a consequence, if $R$ is a field then the determinant defines an isomorphism $\det \colon K_1(R) \xrightarrow{\cong} \R\setminus\{0\}$.

To see the equivalence of these two definitions, we can use the map
\begin{align*}
  \GL(R) & \to K_1(R)\\
  A & \mapsto [R^n \to R^n, \ x \mapsto Ax]
\end{align*}
and the fact that is descents to $\GL(R)_{\text{ab}} \to K_1(R)$.

The inverse homomorphism is given by $R^n \cong \{ P \oplus Q \xrightarrow{f \otimes \id} P \oplus Q \mid f \text{ iso } \}$.

\begin{dfn*}
  Let $\Gamma$ be a group.
  Then the \CmMark{Whitehead group} is defined as
  \begin{align*}
    \Wh(\Gamma) := \coker(\Gamma \times \{\pm 1\} \to K_1(\Z[\Gamma]), \ (\gamma, \pm 1) \mapsto \pm[\gamma])
  \end{align*}
\end{dfn*}

\begin{expl*}
  The Whitehead group $\Wh(\{1\})$ is trivial, since the determinant yields an isomorphism $K_1(\Z) \xrightarrow{\det} \{\pm 1\}$.

  It is a conjecture that torsion-free groups $\Gamma$ have vanishing Whitehead group.

  Assertion: $\Wh(\Z/5) \cong \Z$.
  Here only prove that $\Wh(\Z/5)$ is infinite.
  We have a map $\phi_{*} \colon K_1(\Z[\Z/5]) \to K_1(\C)$ induced by $\Z[\Z/5] \xrightarrow{\phi} \C, \ t \mapsto \xi$, where $\Z/5 \cong \left<t\right>$ and $\xi = \exp(2\pi i/5) \in \C$.
  Thus
  \begin{equation*}
    \begin{tikzcd}
      K_1(\Z[\Z/5]) \ar{r}{\phi_{*}} \ar{d} & K_1(\C) \ar{r}{\det} & \C^{\times} \ar{r}{|\blank|} & \R_{> 0}\\
      \Wh(\Z/5) \ar[bend right=10]{rrru}{\tau} & & &
    \end{tikzcd}
  \end{equation*}
  One can see that $1 - t - t^{-1}$ is a unit in $\Z[\Z/5]$, since $(1 - t - t^{-1})( - t^2 - t^3) = 1$ and thus $\tau([1 - t - t^{-1}]) \neq 1$
\end{expl*}

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\subsection{Whitehead torsion for chain complexes}
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In the following let us repeat some preliminaries on chain complexes.
Let $R$ be a (not necessarily commutative) ring.

Let $f_{*} \colon C_{*} \to D_{*}$ be an $R$-chain map.
The \CmMark{mapping cylinder} $\cyl(f_{*})$ is an $R$-chain complex with p-th differential
\begin{align*}
  C_p \oplus C_{p-1} \oplus D_p
  \xrightarrow{%
    \begin{pmatrix}
      c_p & -\id & 0\\
      0 & -c_{p-1} & 0\\
      0 & f_{p-1} & f
    \end{pmatrix}}
  C_{p-1} \oplus C_{p-2} \oplus D_{p-1},
  \quad
\end{align*}

\begin{rem*}
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  For a continuous map $f \colon X \to Y$ we have the topological mapping cylinder.
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  \begin{equation*}
    \begin{tikzcd}
      X \ar{r}{f} \ar[hook]{d}{i_0} & Y \ar[hook]{d}\\
      X \times [0,1] \ar{r} & \cyl(f)
    \end{tikzcd}
  \end{equation*}
  If $X,Y$ are CW-complexes and $f$ is cellular, then
  \begin{align*}
    \cyl(C_{*}(f) \colon C_{*}(X) \to C_{*}(Y)) = C_{*}(\cyl(f)). 
  \end{align*}
\end{rem*}

The \CmMark{mapping cone} $\cone(f_{*} \colon C_{*} \to D_{*})$ is a quotient of $\cyl(f_{*})$ by the obvious copy of $C_{*}$, so its differential is
\begin{align*}
  C_{p-1} \oplus D_+
  \xrightarrow{%
    \begin{pmatrix}
      -c_{p-1} & 0 \\
      f_{p-1} & d_p
    \end{pmatrix}}
  C_{p-2} \oplus D_{p-1}
\end{align*}

\begin{rem*}
  Again there is a topological analogue, the topological mapping cone $\cone(f) = \cyl(f)/X \times \{1\}$.
  These are related via
  \begin{align*}
    \cone_i(C_{*}(f)) = C_i(\cone(f)) \text{ for } i > 0.
  \end{align*}
\end{rem*}

The \CmMark{suspension} $\Sigma C_{*}$ of an $R$-chain complex $C_{*}$ is a chain complex with $p$-th differential
\begin{align*}
  C_{p-1} \xrightarrow{-c_{p-1}} C_{p-2},
\end{align*}
which is isomorphic to a quotient of $\cone(\id_{C_{*}})$ by $C_{*}$.

We have two exact sequences
\begin{align*}
  & 0 \to C_{*} \to \cyl(f_{*}) \to \cone(f_{*}) \to 0
  & 0 \to D_{*} \to \cone(f_{*}) \to \Sigma C_{*} \to 0
\end{align*}

\begin{dfn*}
  An $R$-chain complex $C_{*}$ is \CmMark{finite}, if $|C_p| = 0$ for $p >> 0$ and each $C_p$ is finitely generated.
  It is called \CmMark{projective}, if each $C_p$ is projective; \CmMark{free}, if each $C_p$ is free, and \CmMark{based free}, if each $C_p$ is based free with a preferred basis.
\end{dfn*}

\begin{rem*}
  Let $f_{*} \colon C_{*} \to D_{*}$ be a chain map between projective chain complexes.
  Then the following statements are equivalent
  \begin{enumerate}
  \item $f_{*}$ is a homology isomorphism (i.e. $H_i(f_{*})$ is an isomorphism for all $i$)
  \item $f_{*}$ is a chain homotopy equivalence
  \item $\cone(f_{*})$ is contractible (i.e. $\cone(f_{*}) \simeq 0$).
  \end{enumerate}
  This can be seen from the following sequence (together with the fundamental lemma in homological algebra to show the equivalence of the first two statements).
  \begin{equation*}
    \begin{tikzcd}
      0 \ar{r} & C_{*} \ar{r} & \cyl(f) \ar[<->]{d}{\simeq} \ar{r} & \cone(f_{*}) \ar{r} & 0\\
      & & D_{*} & &
    \end{tikzcd}
  \end{equation*}
  A short exact sequence of chain complexes induces a long exact sequence in homology associated to $C_{*}$, which implies $\cone(f_{*}) = 0 \rightsquigarrow H_{*}(\cone(f_{*})) = 0\rightsquigarrow f_{*}$ is homology isomorphism.
\end{rem*}

\begin{lemma*}
  Let $C_+$ be a based free, finite $R$-chain complex that is contractible.
  Let $\gamma_p \colon C_p \to C_{p+1}$ for $p \in \Z$ be a chain contraction, i.e.
  \begin{align*}
    c_{p+1} \circ \gamma_p + \gamma_{p-1} \circ c_p = \id - 0.
  \end{align*}
  Then the $R$-homomorphism $(c_{*} + \gamma_{*}) \colon C_{\text{odd}} \to C_{\text{ev}}$ (where $C_{\text{odd}} = \bigoplus_p C_{2p+1}$ and $C_{\text{ev}} = \bigoplus_p C_{2p}$) is an isomorphism.
  Let $A$ be its representing matrix.
  Its class $[A] \in K_1(R)$ is independent of the choice of $\gamma_{*}$.
\end{lemma*}

\begin{expl*}
  Let $C_p = 0$ unless $i \in \{0,1,2\}$.
  Then
  \begin{equation*}
    \begin{tikzcd}
      0 \ar{r} & C_2 \ar{r}[swap]{c_2} & C_1 \ar{r}[swap]{c_1} \ar[bend right]{l}[swap]{\gamma_1} & C_0 \ar{r} \ar[bend right]{l}[swap]{\gamma_0} & 0
    \end{tikzcd}
  \end{equation*}
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  is the full complex and thus ``contractible'' means ``short exact''.
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  $C_1 \xrightarrow{\cong} C_0 \oplus C_2$ via $x \mapsto c_1 (x) + \gamma_1(x)$ with inverse $C_0 \oplus C_2 \xrightarrow{\cong} C_1, (x,y) \mapsto \gamma_0(x) + c_2(y)$.
  
  Let $\tilde\gamma$ be another chain contraction
  \begin{equation*}
    \begin{tikzcd}
      C_0 \oplus C_2 \ar{r}{\cong}[swap]{(\tilde \gamma_1, c_2)} & 
      C_1 \ar{r}{\cong}[swap]{(c_1,\gamma_1)} & C_0 \oplus C_2    
    \end{tikzcd}
  \end{equation*}
  Let $x + y \in C_0 \oplus C_2$.
  Then
  \begin{align*}
    \tilde \gamma_0(x) + c_2(y) \mapsto \ & c_1\tilde \gamma_0(x) + \gamma_1 \tilde\gamma_0(x) + \gamma_1 c_2(y)\\
    & = (x - \underbrace{\tilde\gamma_2c_0(x))}_{=0} + \gamma_1\tilde\gamma_0(x) + (y - \underbrace{c_3\gamma_2(y)}_{=0})\\
    & = x + y + \gamma_1 \tilde\gamma_0(x),
  \end{align*}
  which is represented by a matrix $\begin{pmatrix} 1 & * \\ 0 & 1 \end{pmatrix}$.
\end{expl*}

\begin{proof}
  Let $\gamma_{*}, \delta_{*}$ be two chain contractions of $C_{*}$.
  Then we consider
  \begin{align*}
    (c_{*} + \delta_{*})_{\text{odd}} \colon C_{\text{odd}} \xrightarrow{A} C_{\text{ev}}
    \text{ and }
    (c_{*} + \delta_{*})_{\text{ev}} \colon C_{\text{ev}} \xrightarrow{B} C_{\text{odd}}
  \end{align*}
  represented by matrices $A$ and $B$.
  Define $\mu_n = (\gamma_{n+1} - \delta_{n+1}) \circ \delta_n$ and $\nu_n = (\delta_{n+1} - \gamma_{n+1}) \circ \gamma_n$.
  Then $(\id + \mu_{*})_{\text{odd}}$, $(\id + \nu_{*})_{\text{ev}}$ and $(c_{*} + \gamma_{*})_{\text{odd}} \circ (\id + \mu_{*})_{\text{odd}} \circ (c_{*} + \delta_{*})_{\text{ev}}$ are represented by upper triangular matrices with ones on the diagonal.

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  Thus $[A] = -[B]$ in $K_1(R)$ and $B$ is independent of the choice of $\gamma_{*}$, hence $A$ is independent of the choice of $\gamma_{*}$.
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\end{proof}

\begin{dfn*}
  \begin{enumerate}[label=(\roman*)]
  \item For a contractible, based free, finite $R$-chain complex $C_{*}$ define
    \begin{align*}
      \tau(C_{*}) := [(c_{*} + \gamma_{*})_{\text{odd}}] \in K_1(R)
    \end{align*}
    (for some or every) choice of $\gamma_{*} \colon C_{*} \simeq 0$).
  \item Let $f_{*} \colon C_{*} \to D_{*}$ be a chain homotopy equivalence of based free, finite $R$-chain complexes.
    The \CmMark{Whitehead torsion} of $f_{*}$ is
    \begin{align*}
      \tau(f_{*}) := \tau(\cone(f_{*})) \in K_1(R).
    \end{align*}
  \end{enumerate}
\end{dfn*}

We say that a short exact sequence of based free modules
\begin{align*}
  0 \to A \xrightarrow{j} B \xrightarrow{p} C \to 0
\end{align*}
is \CmMark{based exact}, if $\text{basis}_B = B_1 \coprod B_2$, such that $B_1 = j(\text{basis}_{A)}$ and $p(B_2) = \text{basis}_C$.

\begin{lemma*}
  Consider the following diagram with based exact rows.
\begin{equation*}
  \begin{tikzcd}
    0 \ar{r} & C_{*}' \ar{r} \ar{d}{f_{*}}[swap]{\simeq} & D_{*}' \ar{r} \ar{d}{g_{*}}[swap]{\simeq} & E_{*}' \ar{r} \ar{d}{h_{*}}[swap]{\simeq} & 0\\
    0 \ar{r} & C_{*} \ar{r} & D_{*} \ar{r} & E_{*} \ar{r} & 0
  \end{tikzcd}
\end{equation*}
Then $\tau(g_{*}) = \tau(f_{*}) + \tau(h_{*})$.
\end{lemma*}

\begin{proof}
  We have the following diagrams with exact rows and columns.
  \begin{equation*}
    \begin{tikzcd}
      & 0 \ar{d} & 0 \ar{d} & 0 \ar{d} & \\
      0 \ar{r} & C_{*}' \ar{r} \ar{d}{\simeq} & D_{*}' \ar{r} \ar{d}{\simeq} & E_{*}' \ar{r} \ar{d}{\simeq} & 0\\
      0 \ar{r} & \cyl(f_{*}) \ar{r} \ar{d} & \cyl(g_{*}) \ar{r} \ar{d}  & \cyl(h_{*}) \ar{r} \ar{d} & 0\\
      0 \ar{r} & \cone(f_{*}) \ar{r} \ar{d} & \cone(g_{*}) \ar{r} \ar{d} & \cone(h_{*}) \ar{r} \ar{d} & 0\\
      & 0 & 0 & 0
    \end{tikzcd}
  \end{equation*}
  We may assume that the short exact sequence given by the columns 
  \begin{align}\label{eq:ses-1}
    0 \to C_{*} \xrightarrow{j_{*}} D_{*} \xrightarrow{p_{*}} E_{*} \to 0
  \end{align}
  is a based exact sequence of contractible based exact sequence of contractible based free, finite chain complexes and then have to prove that $\tau(D_{*}) = \tau(C_{*}) + \tau(E_{*})$.

  The sequence \eqref{eq:ses-1} splits as chain complexes.\footnote{This heavily depends on contractibility and is not true for arbitrary chain complexes.}
  Let $e_*$ be a contraction of $E_{*}$ and let $\sigma_i \colon E_i \to D_i$ be a split of $p_i$ for all $i$.
  Set $s_i \colon E_i \to D_i, s_i := d_{i+1} \circ \sigma_{i+1} \circ \varepsilon_i + \sigma_i \circ \varepsilon_{i-1} \circ e_i$.
  
  Claim: $s_{*}$ is a chain map and $p_{*} \circ s_{*} = \id_{E_{*}}$.

  Hence we obtain an isomorphism of chain complexes
  \begin{align*}
    (j_{*},s_{*}) \colon C_{*} \oplus E_{*} \xrightarrow{\cong} D_{*},
  \end{align*}
  which has a corresponding matrix of the form $\begin{pmatrix}\Id & * \\ 0 & \Id \end{pmatrix}$.

  We can finish the proof with the following remark.
  If $u_{*} \colon C_{*} \to D_{*}$ is a chain isomorphism of then
  \begin{align*}
    \tau(C_{*}) - \tau(D_{*}) = \sum_p (-1)^p [u_p] \in K_1(R).
  \end{align*}
  This can be shown by transporting a chain contraction $\gamma_{*}$ for $C_{*}$ to one for $D_{*}$ via $u_{*}$, which yields a diagram:
  \begin{equation*}
    \begin{tikzcd}
      C_{\text{odd}} \ar{r} \ar{d}{u_{\text{odd}}}[swap]{\simeq} & C_{\text{ev}} \ar{d}{u_{\text{ev}}}[swap]{\simeq}\\
      D_{\text{odd}} \ar{r} & D_{\text{ev}}.
    \end{tikzcd}
  \end{equation*}
\end{proof}

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Recall that last time we proved the following Lemma.

\begin{lemma*}
  \begin{enumerate}[label=(\arabic*)]
  \item Additivity
    \begin{equation*}
      \begin{tikzcd}
        0 \ar{r} & C_{*}' \ar{r} \ar{d}{f_{*}}[swap]{\simeq} & D_{*}' \ar{r} \ar{d}{g_{*}}[swap]{\simeq} & E_{*}' \ar{r} \ar{d}{h_{*}}[swap]{\simeq} & 0\\
        0 \ar{r} & C_{*} \ar{r} & D_{*} \ar{r} & E_{*} \ar{r} & 0
      \end{tikzcd}
    \end{equation*}
    Then $\tau(g_{*}) = \tau(f_{*}) + \tau(h_{*})$.
  \item Homotopy invariance.
    If $f_{*} \simeq g_{*} \colon C_{*} \xrightarrow{\simeq} D_{*}$, then $\tau(f_{*}) = \tau(g_{*})$.
  \item Composition formula.
    $\tau(g_{*} \circ f_{*}) = \tau(g_{*}) + \tau(f_{*})$.
  \end{enumerate}
\end{lemma*}

\begin{proof}
  \begin{enumerate}[label=ad (\arabic*),leftmargin=1.5cm]
    \setcounter{enumi}{2}
  \item If $h_{*} \colon f_{*} \simeq g_{*}$, then we have an isomorphism
    \begin{align*}
      \cone(f_{*})  \colon C_{*-1} \oplus D_{*} \xrightarrow{\begin{pmatrix}\id & 0 \\ h_{*-1} & \id\end{pmatrix}} C_{*-1} \oplus D_{*} = \cone(g_{*}).
    \end{align*}
    Thus
    \begin{align*}
      & \tau(f_{*}) - \tau(g_{*}) = \tau(\cone(f_{*})) - \tau(\cone(g_{*}))\\
      & \qquad = \sum_{p \geq 0} (-1)^p \begin{pmatrix}\id & 0 \\ h_{*-1} & \id\end{pmatrix} = 0 \in K_1(R).
    \end{align*}
  \end{enumerate}
\end{proof}


\subsection{Whitehead torsion of maps between CW-complexes}

Let $X,Y$ be connected finite CW-complexes and let $f \colon X \xrightarrow{\simeq} Y$ be a homotopy equivalence.
Further pick base points $x \in X, y = f(x) \in Y$ and set $\pi := \pi_1(Y,y)$.
We identify $\pi_1(X,x)$ with $\pi$ via $\pi_1(f)$ and considering the universal covers we obtain the diagram
\begin{equation*}
  \begin{tikzcd}
    \tilde X \ar{r}{\tilde f}[swap]{\simeq} \ar{d}{\pr_X} & \tilde Y \ar{d}{\pr_Y}\\
    X \ar{r}{f}[swap]{\simeq} & Y
  \end{tikzcd}
\end{equation*}
There is a unique lift $\tilde f$ with $\tilde f(\tilde x) = \tilde y$.
Further, $\tilde f$ is $\pi$-equivariant.
$\tilde X$ caries a CW structure $\tilde X^n = \pr_X^{-1}(X^n)$.

Thus $C^{\text{CW}}_{*}(\tilde f) \colon C_{*}^{\text{CW}}(\tilde X) \xrightarrow{\simeq} C_{*}^{\text{CW}}(\tilde Y)$, where $C_{n}^{CW}(\tilde X) = H_n(\tilde X^n, \tilde X^{n-1})$, are chain homotopy equivalences of $\Z\pi$-chain complexes.
We obtain a $\Z\pi$-basis of $C_n^{\text{CW}}(\tilde X)$ by choosing ($\pi$-)push-outs:
\begin{equation*}
  \begin{tikzcd}
    \coprod_{I_n} \pi \times S^{n-1} \ar{r} \ar[hook]{d} & \tilde X^{n-1} \ar[hook]{d} \\
    \coprod_{I_n} \pi \times D^n \ar{r} & \tilde X^n
  \end{tikzcd}.
\end{equation*}
This yields a cellular basis
\begin{align*}
  C_n^{\text{CW}}(\tilde X) = H_n(\tilde X^n, \tilde X^{n-1}) \xleftarrow{\cong} \bigoplus_{I_n} H_n(\pi \times (D^n, S^{n-1})) = \bigoplus_{I_n}\Z\pi
\end{align*}
The matrix of base change when we take another push-out looks like
\begin{align*}
  \bigoplus_{I_n} \Z\pi
  \xrightarrow{%
    \begin{pmatrix}\pm g_1 & & \\ & \ddots &\\ & & \pm g_n\end{pmatrix}P}
  \bigoplus_{I_n} \Z\pi,
\end{align*}
where $P$ is a permutation matrix.
Hence we obtain a well-defined element $\tau(C_{*}^{\text{CW}}(\tilde f)) \in \Wh(\pi)$ independent of the choices of cellular bases of $\tilde X, \tilde Y$.

It may still depend on the choice of basepoints.
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Let $y,y' \in Y$ be connected by two paths $\alpha$ and $\beta$ as in the figure below.

\begin{figure}[h!]
  \centering
  \begin{tikzpicture}
    \fill (0,0) circle (0.05);
    \fill (5,1) circle (0.05);
    \draw (0,0) node[left]{$y$} to[out=0,in=215] node[below] {$\beta$} (5,1) node[right]{$y'$};
    \draw (0,0) to[out=100,in=174] node[above]{$\alpha$} (5,1);
  \end{tikzpicture}
\end{figure}
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Then $\pi_1(Y,y) \xrightarrow{\alpha_{*}, \beta_{*}} \pi_1(Y,y')$ differ by an inner automorphism of $\pi_1(Y,y')$, which induces the identity on $\Wh(\pi_1(Y,y'))$.
So there is a canonical isomorphism
\begin{align*}
  \phi_{y,y'} \colon \Wh(\pi_1(Y,y)) \xrightarrow{\cong} \Wh(\pi_1(Y,y'))
\end{align*}
induced by any choice of path from $y$ to $y'$.
This yields a basepoint free definition of the Whitehead group.

\begin{dfn*}
  $\Wh(\pi_1(Y)) := \colim_{y \in Y} \Wh(\pi_1(Y,y)) = \coprod_{y \in Y}\Wh(\pi_1(Y,y))/\sim$
  for $z \sim \phi_{y,y'}$.
\end{dfn*}
One verifies that $\tau(C_{*}^{\text{CW}}(\tilde f)) \in \Wh(\pi(Y))$ is independent of all choices of basepoints.

\begin{dfn*}
  Let $X \xrightarrow{f} Y$ be a homotopy equivalence of finite CW-complexes.
  Define
  \begin{align*}
    \Wh(\pi(Y)) := \bigoplus_{C \in \pi_0(Y)} \Wh(\pi(C))
  \end{align*}
  and
  \begin{align*}
    \tau(f) := \left(\tau(C_{*}^{\text{CW}}(\tilde f \colon \tilde{f^{-1}(C)} \to \tilde C))\right)_{C \in \pi_0(Y)},
  \end{align*}
  which is called the \CmMark{Whitehead torsion} of $f$.
\end{dfn*}

\begin{rem*}
  The elementary properties of $\tau(f)$ stated in the beginning of chapter 2 are now direct consequences of the lemma in section 2.3.
\end{rem*}

In the following we will discuss the topological meaning of $\tau(f)$.
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For this let $S^{n-2} \subset S_+^{n-1} \subset S^{n-1} \subset D^n$, where $S_+^{n-1}$ denotes the upper hemisphere.

\begin{figure}[h!]
  \centering
  \begin{tikzpicture}
    \draw [thick,domain=0:180] plot ({cos(\x)}, {sin(\x)});
    \node[above] at (0,1) {$S^1_+$};
    \draw[pattern=north west lines,pattern color=gray] (0,0) circle (1);
    \fill (-1,0) circle (0.05) node[left]{$S^0$};
    \fill (1,0) circle (0.05);

    \draw[pattern=north west lines,pattern color=gray] (3,-1) to[out=60,in=202] (6,0) to[out=100,in=30] (3.5,1) to[out=210,in=85] (3,-1);
    \draw[thick] (3,-1) to[out=60,in=202] node[below] {$X$} (6,0);
  \end{tikzpicture}
\end{figure}

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\begin{equation*}
  \begin{tikzcd}
    S_+^{n-1} \ar{r}{q} \ar[hook]{d}{\simeq} & X \ar[hook]{d}{\simeq} \\
    D^n \ar{r}{\bar q} & Y
  \end{tikzcd}
\end{equation*}
Such a homotopy equivalence $X \to Y$ as in the diagram is called \CmMark{elementary expansion}.

A map $r \colon Y \to X$ such that $r \circ j = \id_X$ is called \CmMark{elementary collapse} ($r$ will be a homotopy inverse of $j$).

\begin{rem*}
  If $X$ is a CW-complex and $q(S^{n-2}) \subset X^{n-2}$ with $q(S^{n-1}_+) \subset X^{n-1}$, then $Y$ inherits a natural CW-structure.
  Then $Y$ is the result of attaching an $(n-1)$-cell and then an $n$-cell to $X$.
\end{rem*}

\begin{dfn*}
  Let $f \colon X \to Y$ be a map of finite CW-complexes.
  We call $f$ a \CmMark{simple homotopy equivalence}, if it is homotopic to a zig-zag of elementary expansions and collapses, i.e. there exists maps $f(i)$ such that
  \begin{equation*}
    \begin{tikzcd}
      X = X(0) \ar{r}{f(0)} \ar[bend right]{rrr}{f} & X(1) \ar{r} & \cdots \ar{r}{f(n-1)} & X(n) = Y
    \end{tikzcd},
  \end{equation*}
  commutes up to homotopy, where each $f(i)$ is an elementary expansion or collapse.
\end{dfn*}

\begin{thm*}
  \begin{enumerate}[label=(\arabic*)]
  \item A homotopy equivalence $f \colon X \to Y$ is simple if and only if $\tau(f) = 0$.
  \item For every element $a \in \Wh(\pi(X))$ there is a homotopy equivalence $X \xrightarrow{f} Y$ such that $f_{*}^{-1}(\tau(f)) \in \Wh(\pi(X))$.
  \end{enumerate}
\end{thm*}

\begin{proof}
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  \begin{enumerate}[label=ad (\arabic*).,leftmargin=1.5cm]
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  \item We will only prove the direction ``simple $\implies \tau(f) = 0$''.
    We may assume that $f$ is an elementary expansion
    \begin{equation*}
      \begin{tikzcd}
        0 \ar{r} & C_{*}(\tilde X) \ar{r}{f_{*}} & C_{*}(\tilde Y) \ar{r} & C_{*}(\tilde Y, \tilde X) \ar{r} & 0\\
        0 \ar{r} & C_{*}(\tilde X) \ar{r}{\id} \ar{u}{\id} & C_{*}(\tilde X) \ar{r} \ar{u}{f_{*}} & 0 \ar{r} \ar{u}{0} & 0
      \end{tikzcd}
    \end{equation*}
    Thus, by additivity, we have $\tau(f) = \tau(C_{*}(\tilde Y, \tilde X))$.
    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.
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  \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 $b_j$.
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    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}
        f_1 \\ \vdots \\ f_n
      \end{pmatrix}
      = A \cdot 
      \begin{pmatrix}
        b_1 \\ \vdots \\ b_n
      \end{pmatrix}
      \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*}
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    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:
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    \begin{align*}
      \tau(\tilde X \hookrightarrow \tilde Y) = \tau(C_{*}^{\text{CW}}(\tilde Y, \tilde X))
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      = \tau\left(\bigoplus_{j=1}^n \Z\pi \xrightarrow{A} \bigoplus_{j=1}^n \Z\pi\right) = a
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    \end{align*}
    for $[A] \in \Wh(\pi)$.
  \end{enumerate}  
\end{proof}

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The reverse statement ``$\tau(f) = 0 \implies f $ simple'' follows from a geometric description of the Whitehead group.
To this end, we need some lemmas about elementary collapsed and expansions.

\textbf{Notation.}
Write
\begin{itemize}
\item $X \nearrow Y$, if there is a sequence of elementary expansions from $X$ to $Y$.
\item $X \searrow Y$, if there is a sequence of elementary collapses from $X$ to $Y$.
\item $X \doglegarrowright Y$, if there is a sequence of elementary expansions or collapses from $X$ to $Y$.
\end{itemize}

\begin{lemma*}[The relativity principle.]
  Given cellular pushouts
\begin{equation*}
  \begin{tikzcd}
    X \ar[hook]{d} \ar{r}{f} & Y\ar[hook]{d}  & X \ar{r}{f}\ar[hook]{d} & Y \ar[hook]{d}\\
    Z \ar{r} & W & Z' \ar{r} & W'
\end{tikzcd}
\end{equation*}
where we assume that $Z \doglegarrowright Z'$ rel X.
Then $W \doglegarrowright W'$ rel $Y$ (by the ``same'' sequence of elementary expansions or collapses).
\end{lemma*}

\begin{lemma*}[The cylinder lemma]
  Let $f \colon X \to Y$ be cellular and let $A \subset X$ be a subcomplex.
  Then the inclusion $\cyl(f|_A) \hookrightarrow \cyl(f)$ is a composition of elementary expansions.
  
  (Special case: $A = \emptyset, Y \hookrightarrow \cyl(f)$.)
\end{lemma*}

\begin{proof}
  First consider the case $X = D^n \cup_q A$ for $q \colon S^{n-1} \to A$, i.e. $X$ is obtained by gluing an $n$-ball to $A$.
  This yields a pushout
  \begin{equation*}
    \begin{tikzcd}
      S^{n-1} \times [0,1] \cup_{S^{n-1} \times \{0\}} D^n \times \{0\} \ar{r} \ar[hook]{d}[swap]{\simeq} & \cyl(f|_A) \ar[hook]{d}[swap]{\simeq}\\
      D^n \times [0,1] \ar{r} & \cyl(f).
    \end{tikzcd}
  \end{equation*}
  For the left hand side of the diagram, we have $(D^n \times [0,1], S^{n-1} \times [0,1] \cup_{S^{n-1} \times \{0\}} D^n \times \{0\}) \cong (D^{n+1}, S^n_+)$, where $S^n_+$ is a hemisphere.
  Thus the pushout describes one elementar expansion.
\end{proof}

\begin{lemma*}[Relative isomorphism lemma]
  If we have
  \begin{equation*}
    \begin{tikzcd}
      & Y_1 \ar{dd}{\cong}[swap]{h} \\
      X \ar[hook]{ru} \ar[hook]{rd} \\
      & Y_2
    \end{tikzcd},
  \end{equation*}
  where $h$ is a CW isomorphism, then $Y_1 \doglegarrowright Y_2$ rel $X$.
\end{lemma*}

\begin{proof}
  By the cylinder lemma we have $X \times [0,1] \cup Y_2 \times \{1\} = \cyl(h|_X) \nearrow \cyl(h)$.
  By the same proof, $X \times [0,1] \cup Y_1 \times \{0\} \nearrow \cyl(h)$.

  Applying the relativity principle to $\pr \colon X \times [0,1] \to X$
  \begin{equation*}
    \begin{tikzcd}
      X \times [0,1] \ar{r}{\pr} \ar[hook]{d} & X \ar[hook]{d} & X \times [0,1] \ar{r}{\pr} \ar[hook]{d} & X \ar[hook]{d} \\
      \cyl(h|_X) \ar{r} \ar[bend right]{rr}[swap]{\doglegarrowright} & Y \ar[bend right,dashed]{rr}[swap]{\doglegarrowright}  & X \times [0,1] \cup Y_1 \times \{0\} \ar{r} & Y
    \end{tikzcd}
  \end{equation*}
\end{proof}

\begin{lemma*}[Homotopy lemma]
  Let cellular maps $f, g \colon K \to L$ be homotopic.
  Then $\cyl(f) \doglegarrowright \cyl(g)$ rel $K \cup L$ (top, bottom).
\end{lemma*}

\begin{proof}
  Let $H$ be a cellular homotopy with $H_0 = f, H_1 = g$.
  We have to show that
  \begin{align*}
    \cyl(H_0) \cup K \times [0,1] \nearrow \cyl(H) \nwarrow \cyl(H_1) \cup K \times [0,1].
  \end{align*}
  This is implied by the general fact for $X = K \times [0,1]$ and $X_0 = K \times \{0\}$ and $f = H$.
  \begin{enumerate}
  \item[($*$)] If $f \colon X \to Y$ is cellular and $X \supset X_0 \nearrow X$, then $X \cup \cyl(f|_{X_0}) \nearrow \cyl(f)$.
  \end{enumerate}
  Now apply the relativity principle with respect to $\pr \colon K \times [0,1] \to K$
  \begin{equation*}
    \begin{tikzcd}
      K \times [0,1] \ar{r}{\pr} \ar[hook]{d} & K \ar{d} & K \times [0,1] \ar{r}{\pr} \ar[hook]{d} & K \ar[hook]{d} \\
      \cyl(H_0) \cup K \times [0,1] \ar{r} \ar[bend right]{rr}[swap]{\doglegarrowright} & \cyl(H_0) & \cyl(H_1) \cup K \times [0,1] \ar{r} & \cyl(H_1)
    \end{tikzcd}
  \end{equation*}
  Thus, $\cyl(f) \doglegarrowright \cyl(g)$.
\end{proof}

Let $(Y,X)$ and $(Z,X)$ be paris of finite CW complexes such that $X \overset{\simeq}{\hookrightarrow} Y$ and $X \overset{\simeq}{\hookrightarrow} Z$ are homotopy equivalences.
We say that $(Y,X)$ and $(Z,X)$ are \CmMark{equivalent}, if $Y \doglegarrowright X$ rel $X$.

\begin{dfn*}
  The \CmMark{geometric Whitehead group} $\Wh^{\text{geo}}(X)$ of $X$ is the set of equivalence classes of such pairs $(Y,X)$ of finite CW complexes with $X \overset{\simeq}{\hookrightarrow} Y$.

  $\Wh^{\text{geo}}(X)$ carries the structure of an abelian group.
  \begin{enumerate}
  \item Abelian addition: $[Y,X] + [Z,X] = [Y \cup_X Z, X]$
    This is well-defined since for $(Y',X) \sim (Y,X)$
    \begin{equation*}
      \begin{tikzcd}
        X \ar[hook]{r}{\simeq} \ar[hook]{d}{\simeq} & Z \ar[hook]{d}{\simeq}  & X \ar{r}{\simeq} \ar[hook]{d}{\simeq} & Z \ar[hook]{d} \\ 
        Y \ar[hook]{r} \ar[bend right]{rr}[swap]{\doglegarrowright} & Y \cup_X Z & Y' \ar{r} & Y' \cup_X Z
      \end{tikzcd}
    \end{equation*}
    Thus, $(Y \cup_X Z, X) \sim (Y' \cup_X Z, X)$.
  \item Zero element: $[X,X]$.
  \item Inverse: Take a pair $(Y,X)$ and let $D \colon Y \to X$ be a cellular strong deformation retract.
    
    \begin{figure}[h!]
      \centering
      \caption{TODO: Add figure for $2\cyl(D)$.}
    \end{figure}

    Claim: $[2\cyl(D),X] = -[Y,X]$.
    \begin{align*}
      [2 \cyl(D),X] + [Y,X] = [2 \cyl(D) \cup_X Y, X] = [\cyl(i \circ D) \cup \tilde{\cyl}(D), X]
    \end{align*}
    for $i \circ D \colon Y \to Y \simeq \id$, the homotopy lemma yields $\cyl(i \circ D) \doglegarrowright Y \times [0,1]$ rel $Y \times \{0\} \cup Y$
    \begin{align*}
      = [Y \times [0,1], \tilde{\cyl}(D),X]
    \end{align*}
    $Y \times [0,1] \searrow X \times [0,1] \cup Y \times \{0\}$ which follows from $(*)$ for $\id_Y$.
    \begin{align*}
      = [Y \times [0,1] \cup \tilde{\cyl}(D), X].
    \end{align*}
    
    $\tilde \cyl(D) \searrow X \times [-1,0]$ by the cylinder lemma for $D$.
    \begin{align*}
      = [X \times [-1,1], X]
      = [X,X]
      = 0
    \end{align*}
  \end{enumerate}
\end{dfn*}

\begin{thm*}
  \begin{align*}
    \Wh^{\text{geo}}(X) \xrightarrow{\cong} \Wh(\pi(X)),
    \quad
    [Y,X] \mapsto i_{*}^{-1}(\tau(i \colon X \hookrightarrow Y))
  \end{align*}
  is an isomorphism of abelian groups.
\end{thm*}

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In some sense this is a topological version of the s-cobordism Theorem.

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\section{Lens spaces}

\begin{dfn*}
Let $p,q \in \mathbb N$ be two {\itshape coprime} integers. Identify $S^3$ with the 
subset $\{(z_1,z_2) \in \mathbb C^2: |z_1|^2 + |z_2|^2 = 1 \} \subset \mathbb C^2$. There is a
$\mathbb Z_p$-action on $S^3$, given by \begin{equation}t . (z_1,z_2) := (\xi_pz_1,\xi_p^qz_2)\end{equation} with $\xi_p := e^{2\pi i/p}$ and $t$ the generator of $\mathbb Z_p$.
The ($3$-dimensional) {\itshape Lens space} $L(p,q)$ is defined as the quotient of $S^3$ under this action.
\end{dfn*}

\begin{expl*}
\begin{enumerate}
\item If $p = 1$, then the induced action on $S^3$ is obviously trivial. Thus, for all $q \in \mathbb N$, we get $L(1,q) \cong S^3$.
\item If $p = 2$, $q = 1$, the induced action of $\mathbb Z_2$ on $S^3$ is generated by the antipodal map $z \mapsto -z$. Thus, $L(2,1) \cong \mathbb {RP}^3$. 
\end{enumerate}
\end{expl*}

The following is easy to verify and will motivate the main content of this subsection:

\begin{lemma*}
Let $p,q \in \mathbb N$ be chosen as above. Then $L(p,q)$ is a closed (i.e compact, connected and without boundary) orientable $3$-dimensional manifold.
Further, let $p',q' \in \mathbb N$ be another pair of comprime integers. Then $L(p,q)$ and $L(p',q')$ have the 
same homotopy groups/(Co-)homology groups/cohomology ring if and only if $p = p'$. 
\end{lemma*}

\begin{proof}
Since $p,q$ are coprime, it is easily verified that that induced $\mathbb Z_p$-action on $S^3$ is free, 
thus properly discontinuously, since $\mathbb Z_p$ is finite. Moreover, the action is also orientation-preserving, showing
that $L(p,q)$ is indeed an orientable $3$-dimensional manifold, which is closed since $S^3$ is already closed.
Moreover, as $S^3$ is simply-connected, it is the univeral cover of $L(p,q)$. Elementary covering theory then yields
\begin{equation*} \pi_n(L(p,q)) = \begin{cases} \mathbb Z_p & n = 1 \\ \pi_n(S^3) & n \neq 1 \end{cases} \end{equation*}
showing that the homotopy groups of $L(p,q)$ only depend on $p$. \\
To compute the (integer) homology groups, note first that we immediately get both $H_0(L(p,q),\mathbb Z) = H_3(L(p,q),\mathbb Z) = \mathbb Z$ and
$H_n(L(p,q),\mathbb Z) = 0$ for $n \geq 4$, as $L(p,q)$ is closed, connected and an orientable $3$-manifold. Moreover, Abelianization of the fundamental group
yields $H_1(L(p,q),\mathbb Z) \cong \pi_1(L(p,q)) = \mathbb Z_p$, while {\itshape Poincaré-Duality} and the {\itshape Universal Coefficient Theorem} give us $H_2(L(p,q),\mathbb Z) \cong H^1(L(p,q),\mathbb Z) \cong Hom_{\mathbb Z}(\mathbb Z_p, \mathbb Z) = 0$. We summarize:
\begin{equation*} H_n(L(p,q),\mathbb Z) = \begin{cases} \mathbb Z & n = 0,3 \\ \mathbb Z_p & n = 1 \\ 0 & else \end{cases}.\end{equation*} This shows that also the homology groups of $L(p,q)$ only depend on $p$. By applying once again Poincaré-Duality, one can easily show that the same holds true for the cohomology groups of $L(p,q)$ with the (cup-product) ring structure always being trivial. This finishes the proof of the Lemma. 
\end{proof}

Consequently, if $q,q'$ are two integers, coprime to some other integer $p$, we have shown that the two spaces $L(p,q)$ and $L(p,q')$ are completely indistinguishable with regards to elementary homotopy invariants. This motivates the {\itshape main question} of this section, which we will solve in full generality:
\\
{\bfseries Are $L(p,q)$ and $L(p,q')$ homotopy equivalent? If so, are they even homeomorphic?} \\
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An integral part in solving both questions is the construction and close inspection of a particular {\itshape CW-structure} on $L(p,q)$, or, equivalently, a $\mathbb Z_p$-equivariant CW-structure on $S^3$, which will highly depend on the choice of $q$. Along with some simple techniques from classic homotopy theory, this will suffice to answer the first question. In order to answer the second question, we will need to introduce another object, the so-called {\itshape Reidemeister-Torsion}, which can be defined on any lens space and is shown to be a more sensitive invariant of such, namely, a {\itshape simple homotopy invariant}. 
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\subsection{A $CW$-structure for $L(p,q)$} 

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Let $p,q$ be two coprime integers, $pr: S^3 \to L(p,q)$ the induced covering projection. Throughout this subsection, we will denote with $\pi$ the group of deck transformations $deck(pr) \equiv \pi_1(L(p,q)) \equiv \mathbb Z_p \equiv < t| t^p=1 >$. Therefore, finding an appropriate cell-structure on $L(p,q)$ amounts to finding a corresponding cell-structure on $S^3$ that is invariant under the action of $\pi$. For this purpose, it will be useful to make the following identification: Let $(\mathbb R^3)^*$ be the one-point compactification of $\mathbb R^3$ with added point $\infty$, and let $f: S^3 \to (\mathbb R^3)^*$ be a {\itshape stereographic projection} homeomorphism, defined by \begin{equation} f(z_1,z_2) = \begin{cases} (\frac{x_1}{1-x_4},\frac{x_2}{1-x_4},\frac{x_3}{1-x_4}) & x_4 \neq 1 \\ \infty & x_4 = 1 \end{cases},\end{equation} where we use the convention $z_1 = x_1 + ix_2$, $z_2 = x_3 + ix_4$ with $x_j \in \mathbb R$ for all $j=1,2,3,4$. 
There is a canonical action of $\pi$ on $(\mathbb R^3)^*$, given by the push-forward of the action of $\pi$ on $S^3$ by $f$, i.e, we set \begin{equation} t.(x,y,z) := f ( t. f^{-1}(x,y,z)). \end{equation} It is the unique action on $(\mathbb R^3)^*$ making the map $f$ $\pi$-equivariant, thus it is the deck group action of the covering projection $pr \circ f^{-1}: (\mathbb R^3)^* \to L(p,q)$. We will now define a $\pi$-equivariant $CW$-structure on $(\mathbb R^3)^*$, starting with 
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\begin{enumerate}
\item {\bfseries the $0$-skeleton $X^0$}: We set this as the $\pi$-orbit of the point $e_0 := (0,0,0)$, i.e $X^0 := \{ t^k.e_0: k=0,\dots,p-1 \}$. 
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\item {\bfseries The $1$-skeleton $X^1$}: Note first that $X^0$ lies completely in the ($(\mathbb R^3)^*$-) closure of the $z$-axis, whose pre-image under $f$ is simply $\{0\} \times S^1 \subset S^3$. With that in mind, we define $X^1$ as the $\pi$-orbit of the arc $e_1 := \{f(0,e^{is}): s \in [0,\frac{2\pi}{p}] \}$. 
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\item {\bfseries The $2$-skeleton $X^2$}: We define this as the $\pi$-orbit of the 2-cell $e_2$, which we set to be the closure of the half-plane $\{ (x,y,z): y = 0, x \geq 0 \}$. 
\item {\bfseries The $3$-skeleton $X^3$}: Finally, the $3$-skeleton is defined as the $\pi$-orbit of the 3-cell $e_3$, which we set to be the "slice of cake" between $e_2$ and $t.e_2$, i.e the closure
of the space $\{(x,y,z): x + iy = r e^{is}: r \geq 0 \cap s \in [0,\frac{2\pi}{p}] \}$. 
\end{enumerate}

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This defines a $\pi$-CW structure on $(\mathbb R^3)^*$ with one $\pi$-equivariant cell in each positive dimension up to $3$. Endowing each cell with the canonical orientation induced by a preferred fundamental class of $(\mathbb R^3)^*$, and identifying each $n$-cell $e_n$ ($n=0,\dots,3)$ with its corresponding generator in the induced {cellular chain complex}, we can see the following properties:
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\begin{enumerate} \item $X^1$ is the closure of the $z$-axis. Moreover, let $r \in \mathbb Z_p$ be the inverse of $q \mod p$. Then \begin{align*} \partial e_1 &= f(0,\xi_p) - f(0,1) \\&= f(0,\xi_p^{qr}) - f(0,1) \\ &= f( t^r.(0,1)) - f(0,1) \\ &= t^r.e_0 - e_0 = (t^r - 1).e_0 \end{align*}
\item The boundary of $e_2$ is all of $X_1$, each $1$-cell is traversed exactly once by the boundary map, and the induced orientations match up. More precisely, we have $\partial e_2 = \sum_{k=0}^{p-1} t^{rk} e_1 = \sum_{k=0}^{p-1} t^k e_1$. 
\item We have $\partial e_3 = t.e_2 - e_2$. 
\end{enumerate}

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All in all, this implies that the corresponding cellular {\itshape $\mathbb Z[\pi]$-module} chain complex of $(\mathbb R^3)^*$ looks as follows (the appearing boundary maps are multiplication by the denoted element):
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\begin{align}
  0 \to \mathbb Z[\pi] \xrightarrow{t - 1} \mathbb Z[\pi] \xrightarrow {\sum_{k=0}^{p-1} t^k} \mathbb Z[\pi] \xrightarrow{t^r - 1} \mathbb Z[\pi] \to 0.
	\label{equivariant}
\end{align}

Finally, we get our first Theorem of this section:

\begin{thm*}[Homotopy classification of Lens Spaces]
Let $q,q'$ be integers coprime to some integer $p$ and let $t \in \pi_1(L(p,q)) =: \pi$, $t' \in \pi_1(L(p,q')) := \pi'$ be preferred generators of the respective groups as above. Let $r,r'$ be the $\mod p$ inverse elements of $q$ and $q'$ respectively. Suppose $f: L(p,q) \to L(p,q')$ is a map with $\pi_1(f)(t) = (t')^a$. Assume $a$ and $p$ are coprime. Then:
\begin{enumerate}
\item
\begin{enumerate} \item $q \cdot deg(f) \equiv q' a^2 \mod p$,
\item $f$ is a homotopy equivalence $\Leftrightarrow deg(f) = \pm 1$.
\end{enumerate}
\item Furthermore, if there exists $a \in \mathbb N$ with $q'a^2 \cong \pm q \mod p$, then there exists a map $g: L(p,q) \to L(p,q')$ with $deg(g) = \pm 1$. 
\end{enumerate}
\end{thm*}

\begin{proof}

\begin{enumerate}
\item (a): By the {\itshape Cellular Approximation Theorem}, we may assume that both $f$ and its lift $\tilde{f}: S^3 \to S^3$ are cellular maps and that $\tilde{f}(e_0) = e_0'$. For simplicity, we write $L := L(p,q)$ and $L' := L(p,q')$. \\ 
Let $\hat{e_k} := pr(e_k)$ and $\hat{e_k}' := pr'(e_k')$ be the {\itshape unique} $k$-cells of $L$, resp. $L'$ ($k=0,\dots,3$).
 Following the argument from above, we see that $\hat{e_1}$ represents the loop $t^r \in \pi$, and likewise $\hat{e_1}'$ represents $(t')^{r'} \in \pi'$. Since $f$ is cellular and $\pi_1(f)$ takes $t$ to $(t')^a$, it follows that $\pi_1(f)(\hat{e_1}) = (t')^{ar} = (t')^{q'r'ar} = (\hat{e_1}')^{q'ar}$, so the induced {\itshape $\mathbb Z$-chain map} $f_*: C_1(L,\mathbb Z) \to C_1(L',\mathbb Z)$ takes $\hat{e_1}$ to $q'ar\hat{e_1}'$.
Therefore, it follows that the chain map $f_1: C_1(S^3, \mathbb Z[\pi]) \to C_1(S^3,\mathbb Z[\pi'])$ induced by the lift $\tilde{f}$ takes $e_1$ to a sum of $q'ar$ translates of $e_1'$. More precisely
 \begin{equation} f_1(e_1) = \sum_{j=0}^{q'ar-1} (t')^{r'j} e_1'. \label{chain}\end{equation}
To avoid confusion and make matters easier, let $\Gamma := \mathbb Z [\mathbb Z_p] = \mathbb Z[\gamma]/(\gamma^p-1)$ and identify $\mathbb Z[\pi]$ with $\Gamma$ by the unique isomorphism determined by $t \mapsto \gamma^a$. 
Similarly, identify $\mathbb Z[\pi']$ with $\Gamma$ via $t' \mapsto \gamma$. With these identifications, the two equivariant chain complexes of $S^3$ induced by the action of $\pi$,$\pi'$ and the chain map $f_*$ induced by $\tilde{f}$ between them are given by the diagram 
    \begin{equation*}
      \begin{tikzcd}
        0 \ar{r} & \Gamma \ar{r}{\delta_3}\ar{d}{f_3} & \Gamma \ar{r}{\delta_2}\ar{d}{f_2} & \Gamma \ar{r}{\delta_1}\ar{d}{f_1} & \Gamma \ar{r}\ar{d}{f_0} & 0 \\
         0 \ar{r} & \Gamma \ar{r}{\delta_3'} & \Gamma \ar{r}{\delta_2'} & \Gamma \ar{r}{\delta_1'} & \Gamma \ar{r} & 0. 
      \end{tikzcd}
    \end{equation*}
As a consequence of \ref{equivariant}, the boundary maps in the above diagram are each multiplication by an element as follows.
 \begin{align*} \delta_1 &= \gamma^{ar} - 1 \\
\delta_2 &= \sum_{j=0}^{p-1} \gamma^{aj} = \sum_{j=0}^{p-1} \gamma^{j} \\
\delta_3 &= \gamma^{a}-1 \\
\delta_1'&= \gamma^{r'} - 1 \\
\delta_2'&= \sum_{j=0}^{p-1} \gamma^j - 1 \\
\delta_3'&= \gamma - 1. \\
\end{align*}
Since each $f_i$ is again simply mulitiplication by an appropriate element of $\Gamma$, we may identify each $f_i$ with that element. 
Since $\tilde{f}(e_0) = e_0'$ and $\pi_1{f}(t) = (t')^a$, it follows that $f_0 = Id_{\Gamma}$. From Equation \ref{chain}, we see that $f_1 = \sum_{j=0}^{q'ar-1} (\gamma')^{r'j}$.
Since $\delta_2'  f_2 = f_1  \delta_2 \in \Gamma$, we get that $(f_2-f_1)\sum_{j=0}^{p-1} \gamma^j = \alpha (\gamma^p - 1)$ for some $\alpha \in \Gamma$ when regarding each element as a polynomial in $\mathbb Z[\gamma]$.
Thus \begin{equation} f_2 = f_1 + \alpha(\gamma-1) \label{eq1} \end{equation} as elements in $\Gamma$. 
Similarly, $\delta_3' f_3 = f_2 \delta_3$, i.e $(\gamma - 1)f_3 = f_2(\gamma^{a} -1)$, so that $(\gamma - 1)f_3 = f_2(\sum_{i=0}^{a-1} \gamma^i)(\gamma -1)$. Hence there exists $\beta \in \Gamma$, such that 
\begin{equation} f_3 = (\sum_{i=0}^{a-1} \gamma^i)f_2 + \beta (\sum_{i=0}^{p-1} \gamma^i) \label{eq2} \end{equation}
Let $\epsilon: \Gamma \to \mathbb Z$ be the {\itshape augmentation homomorphism} defined by sending $\eta := \sum_{i=0}^n a_i \gamma^i$ to $\epsilon(\eta) := \sum_{i=0}^n a_i$. By inducting over the degree of $\eta$ (when regarded as a polynomial in $\mathbb Z[\gamma]$), one can easily show that $\epsilon(\eta) = 0$ implies that $\eta \sum_{i=0}^{p-1} \gamma^i = 0 \in \Gamma$. Therefore $\epsilon$ induces an isomorphism \begin{align*} H_3(C_*(S^3, \mathbb Z[\pi])) &\cong \ker \delta_3 = span < \sum_{i=0}^{p-1} \gamma^i > \to \mathbb Z \\ \eta \sum_{i=0}^{p-1} \gamma^i &\mapsto \epsilon(\eta) \end{align*}
and this is {\itshape precisely} the isomorphism identifying $\ker \delta_3$ with $H_3(C_*(S^3, \mathbb Z[\pi]))$. Identifying $\ker \delta_3'$ with $H_3(C_*(S^3,\mathbb Z[\pi']))$ in the very same way, we conclude that since $f_3(\sum_{i=0}^{p-1} \gamma^i) = \epsilon(f_3)( \sum_{i=0}^{p-1} \gamma^i) = deg(\tilde{f})(\sum_{i=0}^{p-1} \gamma^i)$, we have $\epsilon(f_3) = deg(\tilde{f})$. Using Equations \ref{eq1} and \ref{eq2}, we deduce \begin{align*} deg(\tilde{f}) &= \epsilon(f_3) \\ &= \epsilon((\sum_{i=0}^{a-1} \gamma^i)f_2 + \beta (\sum_{i=0}^{p-1} \gamma^i)) \\ &= a\epsilon(f_2) + \epsilon(\beta)p \\ &= a\epsilon(f_1 + \alpha(\gamma-1)) + \epsilon(\beta)p \\ &= a(q'ar)+\epsilon(\beta)p \equiv a^2q'r \mod p \end{align*}
Since both covering maps $\pr: S^3 \to L$ and $\pr': S^3 \to L'$ have degree $p$, we use multiplicativity of the degree to finally conclude that $\deg(f) = \deg(\tilde{f}) \equiv a^2q'r \mod p$, whence $q\deg(f) \equiv q'a^2 \mod p$, as claimed. \\
(b): If $f$ is a homotopy equivalence, then the induced $\mathbb Z$-linear map $H_3(f): H_3(L(p,q),\mathbb Z) \to H_3(L(p,q'),\mathbb Z)$ is an isomorphism, hence $deg(f) = \pm 1$. 
Conversely, suppose that $deg(f) = \pm 1$. Let $\tilde{f}: S^3 \to S^3$ be the lift of $f$ on the universal cover, so that the following diagram commutes: 
\begin{equation*}
  \begin{tikzcd}
     S^3 \ar{r}{\tilde{f}} \ar{d}{pr} & S^3 \ar{d}{pr'} \\
     L(p,q) \ar{r}{f} & L(p,q')
  \end{tikzcd}
\end{equation*}
By multiplicativity of the degree, we get $deg(pr')deg(\tilde{f}) = deg(f)deg(pr)$, i.e $p*deg(\tilde{f}) = \pm p$, from which follows that $deg(\tilde{f}) = \pm 1$. 
This shows that $H_3(\tilde{f})$ is an isomorphism. Iterated applications of the {\itshape Hurewicz-Theorem} then yield that $\pi_n(\tilde{f}): \pi_n(S^3) \to \pi_n(S^3)$ is an 
isomorphism as well for all $n \in \mathbb N$ (using the fact that $H_n(S^3, \mathbb Z) = 0$ for all $n \geq 4$, as well as $\pi_n(S^3) = 0$ for $n=1,2$). Therefore $\pi_n(f): \pi_n(L(p,q)) \to \pi_n(L(p,q'))$ is an isomorphism for all $n \geq 2$ (using the fact that, in this case, the covering maps induce isomorphisms of respective homotopy groups). Since $\pi_1(f)$ is an isomorphism by assumption, {\itshape Whitehead's Theorem} yields that $f$ is indeed a homotopy equivalence, proving the converse.
\item As explained before, we may identify the (oriented) $1$-skeleton $L(p,q)^{(1)}$ with the generating loop $t^r \in \pi_1(L(p,q))$, and likewise $L(p,q')^{(1)}$ with $(t')^{r'}$. Define $g_1: L(p,q)^{(1)} \to L(p,q')^{(1)}$ by taking $t$ to $t'^a$ (i.e, wrapping the loop $t$ $a$-times around $t'$). Let $\alpha: S^1 \to L(p,q)^{(1)}$ be the attaching map of the (unique) 2-cell of $L(p,q)$. Since $\alpha$ is already null-homotopic in $L(p,q)^{(2)}$, so is the composition map $g_1 \circ \alpha$. Therefore, there exists an extension $g_2: L(p,q)^{(2)} \to L(p,q')$ of $g_1$, which can be extended even further to a map $g': L(p,q) \to L(p,q')$, since $\pi_2(L(p,q')) = \pi_2(S^3) = 0$. 
Evidently, the relation $q'a^2 \equiv \pm q \mod p$ implies that $a$ and $p$ are coprime. Hence, by the proven assertion $1 (a)$, we have $deg(g') \equiv \pm 1 \mod p$. \\
 We can now modify $g'$ to a map $g$ with $deg(g) = \pm 1$ by iterating the following method as often as necessary: Choose some point $x \in L(p,q)$ and an embedded closed $3$
-ball $B$ containing $x$
%Choose an inscribed $3$-ball $B_2 \ni x$ with $\overline{B_2} \subset B_1$
. Contracting the boundary $\partial B$ to a point, we obtain a quotient map $\iota: L(p,q) \to L(p,q) \vee S^3$. By precomposing the covering map $pr': S^3 \to L(p,q')$ with an appropriate rotation and, possibly, a reflection, we can construct a map of the the form $g' \vee h: L(p,q) \vee S^3 \to L(p,q')$ with the property that $deg(h) = \pm p$. It follows that the map $g'' := (g' \vee h) \circ \iota: L(p,q) \to L(p,q')$ has degree $deg(g'') = deg(g') \pm p$. Since $B$ can be chosen far away from $L(p,q)^{(1)}$, we have $\pi_1(g') = \pi_1(g'')$. Thus, after sufficiently many iterations, we obtain the desired map $g: L(p,q) \to L(p,q')$ with $deg(g) = \pm 1$ and $\pi_1(g)(t) = t'^a$. This proves $2)$.
\end{enumerate}
\end{proof}

Using this Theorem, the following statements can be easily shown:

\begin{expl*}
\begin{enumerate}
\item $L(5,1)$ and $L(5,2)$ {\itshape are not} homotopy equivalent
\item $L(7,1)$ and $L(7,2)$ {\itshape are} homotopy equivalent.
\item A Lens Space $L(p,q)$ admits an orientation-reversing homotopy equivalence if and only if there exists an integer $a$, such that $(a^2+1)q \equiv 0 \mod p$. 
\end{enumerate}
\end{expl*}

\subsection{Reidemeister-Torsion}

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Let $S$ be a ring and $(C_*,\partial_*)$ a finite, based free $S$-chain complex (with basis $\{b_i\}$). Further, let $R$ be a commutative ring with unit and $f: S \to R$ a ring homomorphism. 
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Then one can produce the finite $R$-chain complex $(C_* \otimes_{f} R, \partial_* \otimes id_R)$, which we assume to be based with the canonical choice $\{b_i \otimes 1 \}$. Note that since $R$
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is commutative, taking the determinant of a matrix with entries in $R$ induces a well-defined homomorphism \begin{equation*} \det: K_1(R) \to R^{\times} \end{equation*} which is even an isomorphism whenever $R$ is a field. From now on, we restrict to the case $S = \mathbb Z[G]$ for some group $G$ and assume $f(G) \subset R^{\times}$. We can therefore identify $f(G)$ with its corresponding subset of invertible one by one matrices in $GL(R)$. This allows us to define $\tilde{G} \subset R^\times$, the subgroup generated by $\det([f(G)])$ (which is simply $f(G)$ again) and $-1$. Well aware of the ambiguity, we will also denote the induced composition \begin{equation*} K_1(R) \to R^{\times} \to R^{\times} / \tilde{G} \end{equation*} simply by $\det$. In the above setting, unless specifically stated otherwise, we will henceforth always assume $\det$ to have image in $R^{\times} / \tilde{G}$.
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\begin{dfn*}
Let $G$, $S = \mathbb Z[G]$, $(C_*,\partial_*)$, and $R$ be as above. Let $f: S \to R$ be a ring homomorphism, such that $(C_* \otimes_{f} R, \partial_* \otimes id_R)$ is acyclic. The 
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{\itshape Reidemeister-Torsion of $C_*$ with respect to $f$} is defined as \begin{equation*} \Delta_f(C_*) := \det(\tau(C_* \otimes_{f} R)) \in R^{\times}/ \tilde{G}, \end{equation*} where $\tau(C_* \otimes_f R) \in \tilde{K_1}(R)$ denotes the usual torsion, defined in Section 2.
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\end{dfn*} 

We will illustrate the general setting for applying Reidemeister-Torsion, which will also clarify the choice of the target group. 
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Let $X$ be a finite CW-complex and $\rho: \mathbb Z[\pi_1(X)] \to R$ a ring homomorphism into a commutative ring $R$. A choice of lifts and orientations for each cell turns
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$(C_*(\tilde{X}) \otimes_\rho R)$ into a based free, finite $R$-chain complex. If it is additionally acyclic, $\Delta_\rho(C_*(\tilde{X}))$ is well-defined and doesn't depend on the ordering of the basis. Moreover $\Delta_\rho(C_*(\tilde{X}))$ is even independent of the choice of lifts and orientations, since different choices of lifts and orientations correspond to a basis permutation by elements in $\rho(G)$ and possibly changing signs, thus giving rise to the same element in $R^{\times} / \widetilde{\pi_1(X)}$. This allows us to simply write \begin{equation*}
\Delta_\rho(X) = \Delta_\rho(C_*(\tilde{X})). \end{equation*}

The following Lemma is an easy consequence of the additivity of torsion, as shown in Section 2: 

\begin{lemma*}
Let $X$ and $Y$ be two finite, connected CW-complexes and $f: X \to Y$ a homotopy equivalence. Suppose that $\rho: \mathbb Z[\pi_1(Y)] \to R$ is a ring homomorphism. 
Let $\rho': \mathbb Z[\pi_1(X)] \to R$ be the ring homomorphism induced by $\rho$ and $f$, and let $\rho_*: Wh(\pi_1(Y)) \to R^{\times} / \widetilde{\pi_1(Y)}$ be the induced
group homomorphism. If both $C_*(\tilde{X}) \otimes_{\rho'} R$ and $C_*(\tilde{Y}) \otimes_{\rho} R$ are acyclic, we have 
\begin{equation*} \Delta_{\rho'}(X) - \Delta_{\rho}(Y) = \rho_*(\tau(f)), \end{equation*}
where $\tau(f) \in Wh(\pi_1(Y))$ denotes the usual {\itshape Whitehead-torsion} of $f$. In particular, if $f$ is a simple homotopy equivalence (for example, if $f$ is a homeomorphism), then $\Delta_{\rho'}(X) = \Delta_{\rho}(Y)$. 
\end{lemma*}

\begin{expl*}
\begin{enumerate}
\item
Let $f: L(p,q) \to L(p,q')$ be a homotopy equivalence, $r,r'$ the respective inverse elements of $q,q'$ mod $p$, and $t,t'$ the preferred generators of the fundamental groups. We know that $\pi_1(f)(t) = (t')^a$ for some $a$ with $q' \cdot a^2 \equiv \pm q \mod p$ . Let $\xi$ be a $p$-th root of unity, $\xi \neq 1$, and let $\rho: \mathbb Z[\pi_1(L(p,q'))] \to \mathbb C$ be the unique ring homomorphism determined by $t' \mapsto \xi$. 
Then $\rho'(t) = \xi^a \neq 1$. From our previously established work, we see that $C_*(\widetilde{L(p,q)}) \otimes_{\rho'} \mathbb C$ is the acyclic complex of the form \begin{equation*}  \begin{tikzcd}
     0 \ar{r} & \mathbb C \ar{r}{\xi^a-1} & \mathbb C \ar{r}{0} & \mathbb C \ar{r}{\xi^{ar}-1} & \mathbb C \ar{r} & 0,
  \end{tikzcd} \end{equation*}
	while $C_*(\widetilde{L(p,q')}) \otimes_{\rho} \mathbb C$ is also acyclic of the form 
	\begin{equation*}  \begin{tikzcd}
     0 \ar{r} & \mathbb C \ar{r}{\xi-1} & \mathbb C \ar{r}{0} & \mathbb C \ar{r}{\xi^{r'}-1} & \mathbb C \ar{r} & 0.
  \end{tikzcd} \end{equation*}
Using an obvious chain contraction in each complex, we can easily compute that $\Delta_{\rho'}(L(p,q)) = (\xi^a - 1)(\xi^{ar} -1)$, while $\Delta_{\rho}(L(p,q')) = (\xi - 1)(\xi^{r'} - 1)$, as elements
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in $\mathbb C^{\times} /< \pm \{1, \xi, \xi^2, \dots,\xi^{p-1} \} >$. In particular, if $f$ is a {\itshape simple} homotopy equivalence, $|\xi^a - 1||\xi^{ar} - 1| = |\xi - 1||\xi^{r'} - 1|$. 
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\item We have seen already that $L(7,1)$ and $L(7,2)$ are homotopy equivalent. If they were also homeomorphic, than the previous example would imply that $|\xi^a - 1|^2 = |\xi - 1||\xi^4 - 1|$.
Since $a \equiv \pm 2 \mod 7$, $a=3,4$ are the only possible choices, for both of which one can compute by elementary methods that the Equation does not hold true. Thus, $L(7,1)$ and $L(7,2)$ are not simple homotopy equivalent and, in particular, not homeomorphic. 
\end{enumerate}
\end{expl*}

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% 2017-02-07

Recall that we considered $D_{*} \simeq 0$ a finite dimensional Hilbert space and observed
\begin{align*}
  \ln |\det \rho(D_{*})| = -\frac{1}{2} \sum_{p \geq 0} (-1)^p \ln \det \Delta_p.
\end{align*}
We want to have this in the $\infty$-dimensional setting, specifically
\begin{align*}
  D_{*} = l^2 \Gamma \otimes_{\Z\Gamma} C_{*}(\tilde M),
  \quad
  \Gamma = \pi_1M
\end{align*}
provided $b_p^{(2)}(M) = \dim_{\Gamma}(\ker(\Delta_p^{(n)})) = 0$ for all $p \geq 0$.
This actually happens in quite a few interesting cases, as highlighted by the following conjecture.

\begin{conj*}[Hopf-Singer]
  Every closed odd-dimensional aspherical manifold is $l^2$-acyclic.
  (It is verified, e.g., for locally symmetric spaces.)
\end{conj*}

We need to define a determinant in the $\infty$-dimensional setting.


\subsection{Digression on spectral calculus}

Let $A \colon l^2\Gamma^n \to l^2\Gamma^n$ be a bounded positive equvariant operator.
Then we can exhibit the following map
\begin{align*}
  \{ \text{polyn. functions on } [0, \|A\|] \} & \to \{ \text{ bdd., $\Gamma$-equiv. operators } l^2\Gamma \to l^2\Gamma\} =: L(\Gamma)\\
  P & \mapsto P(A),
\end{align*}
where $L(\Gamma)$ is called \CmMark{von Neumann algebra} of $P$.

This homeomorphism extends to bounded Borel functions on $[0,\|A\|]$ and we obtain a \emph{spectral measure}.
By the Riesz representation theorem, there exists a unique Borel probability measure $\mu_A$ supported on $[0,\|A\|]$ such that
\begin{align*}
  \int_{\R} f \dop\mu_{A} = \tr_{\Gamma}(f(A)).
\end{align*}

\begin{expl*}
  Let $\Gamma = \{1\}$ and $A \colon \C^n \to \C^n$ be a positive matrix we can express as $A = S^{-1} \diag(\lambda_1, \ldots, \lambda_n) S$.
  For a bounded Borel function, we have $f(A) = S^{-1}\diag(f(\lambda_1), \ldots, f(\lambda_n))$.
  In this case, $\tr_{\Gamma}$ is the ordinary trace and the spectral probability measure obtained is
  \begin{align*}
    \mu_A = \frac{1}{n} \left(\sum_{i = 1}^n \delta_{\lambda_i}\right).
  \end{align*}
  Here we have
  \begin{align*}
    \ln \det A = \sum_{i=1}^n \ln(\lambda_i) = n\int_{o^+}^{\infty}\ln(\lambda) \dop \mu_A(\lambda),
  \end{align*}
  which motivates the following general definition.
\end{expl*}

\begin{dfn*}
  The \CmMark{Fuglede-Kadison determinant} $\det^{(2)}(A)$ is defined as
  \begin{align*}
    {\textstyle \det^{(2)}(A)} := \exp \left( \int_{0^+}^{\infty} \ln(\lambda) \dop \mu_A(\lambda) \right) \in \R,
  \end{align*}
  provided $\int_0^{\infty} \ln(\lambda) \dop \mu_A$ exists.
  In this case we say that $A$ is \CmMark{determinant class}.
\end{dfn*}

\begin{rem*}
  If $A$ is right multiplication by a matrix in $M_n(\Z[\Gamma])$ and $\Gamma$ is sofic, then $A$ is determinant class (Elek-Szabo).
\end{rem*}

\begin{dfn*}
  Let $M$ be a finite CW-complex.
  Assume that $\Gamma = \pi_1M$ is sofic and that $M$ is $l^2$-acyclic.
  Then the \CmMark{$l^2$-torsion} $\rho^{(2)}(M)$ is defined as
  \begin{align*}
    \rho^{(2)}(M) = -\frac{1}{2} \sum_{p \geq 0} (-1)^p p \cdot \ln {\textstyle \det^{(2)}}(\Delta_p^{(2)} \colon l^2 \Gamma \otimes_{\Z\Gamma} C_p(\tilde M) \to l^2 \Gamma \otimes_{\Z\Gamma} C_p(\tilde M)).
  \end{align*}
\end{dfn*}

\begin{rem*}
  Note that the following remarks are actually propositions.
  \begin{itemize}
  \item $l^2$-torsion is a homotopy invariant
  \item Let $M$ be a closed hyperbolic $3$-manifold.
    Then $\rho^{(2)}(M) = \frac{1}{6\pi} \operatorname{vol} (M, g_{\text{hyp}})$.
    This implies, in particular, the homotopy invariance of the hyperbolic volume, which also follows non-trivially by Mostow rigidity.
  \end{itemize}
\end{rem*}


\subsection{Approximation}

\begin{thm*}[Lück]
  Let $M$ be a finite CW-complex with $\Gamma = \pi_1M$ residually finite.
  Let $\Gamma = \Gamma_0 > \Gamma_1 > \Gamma_2 > \cdots$ be a \CmMark{residual chain}, i.e. $[\Gamma: \Gamma] = 0$, $\Gamma_i \vartriangleleft \Gamma$, $\bigcup \Gamma_i \{1\}$.
  Then
  \begin{align*}
    b_p^{(2)}(M) = \lim_{i \to \infty} \frac{b_p(M_i)}{[\Gamma : \Gamma_i]},
  \end{align*}
  where $\Gamma_i\backslash M = M_i$.
\end{thm*}

The following is a bold conjecture (that probably is only true up to a few more adjectives added, e.g. (arithmetic) hyperbolic manifold).

\begin{conj*}
  Let $M^{2n+1}$ be a closed odd-dimeninsional aspherical manifold with residually finite $\Gamma = \pi_1M$ and let $(\Gamma_i)$ be a residual chain in $\Gamma$.
  Then $\rho^{(2)}(M) = \lim_{i \to \infty} \frac{\ln |\operatorname{tors} H_n(M_i; \Z)|}{[\Gamma : \Gamma_i]}$.
\end{conj*}

In particular, this would tell us that, if $M$ is a closed hyperbolic 3-manifold, then
\begin{align*}
  \frac{1}{6\pi} \operatorname{vol}(M) = \lim_{i \to \infty} \frac{\ln |\operatorname{tors}(\Gamma_i)_{\text{ab}}|}{[\Gamma : \Gamma_i]}.
\end{align*}
(There is a result of quite similar nature if one consideres suitably twisted coefficients by Bergeron-.)


\begin{proof}[Sketch of Lück's approximation theorem]
  
  Consider the Laplace operators
  \begin{align*}
    & \Delta^{(2)}_p \colon l^2 \Gamma \otimes_{\Z\Gamma} C_p(\tilde M) \cong l^2\Gamma^n \to l^2\Gamma^n\\
    & \Delta_p(i) \colon C_p(M_i) \cong \C[\Gamma/\Gamma_i]^n \to \C[\Gamma/\Gamma_i]^n.
  \end{align*}
  Both come from a multiplication with $A \in M_n(\Z[\Gamma])$ (resp. $A_i \in M_n(\Z[\Gamma/\Gamma_i])$), where $A_i$ is the ``reduction'' of $A$ mod $\Gamma_i$.

  We want to show that
  \begin{align*}
    b_p^{(2)}(M) = \tr_{\Gamma}(\pr_{\Ker A}) = \lim_{i \to \infty} \tr_{\Gamma/\Gamma_i}(\pr_{\Ker A_i}) = \lim \frac{b_p(M_i)}{[\Gamma:\Gamma_i]}.
  \end{align*}
  This follows from spectral calculus, by using the equalities $\tr_{\Gamma}(\pr_{\Ker A}) = \mu_A(\{0\})$ and $\lim_{i \to \infty} \tr_{\Gamma/\Gamma_i}(\pr_{\Ker A_i}) = \lim_{i \to \infty} \mu_{A_i}(\{0\})$.

  To prove the equality on this level it is important to consider the spectral measure in a neighbourhood of $\{0\}$.

  Recall that measures $\nu_i$ on the real line \CmMark[weak convergence of measures]{converge weakly}, if $\int_{\R}f\dop \nu_i \to \int_{\R} f\dop \nu$ converges for all $f \in C_c(\R)$.
  (In good situations, it is enough to check this for monomials.)

  To check this for $(A_i)$, we regard
  \begin{align*}
    \int_{\R} x^m \dop \mu_A = \tr_{\Gamma}(A^m)
    \text{ and }
    \int_{\R} x^m \dop \mu_{A_i} = \tr_{\Gamma/\Gamma_i}(A_i^m).
  \end{align*}
  As the right-hand-sides are purely combinatorial, we get $\tr_{\Gamma/\Gamma_i}(A^m) \to \tr_{\Gamma}(A^m)$ and conclude $\mu_{A_i} \to \mu_A$ weakly.

  Caveat: For the Borel probability measure on $\R$ we have $\nu_i = i \chi_{(0,\frac{1}{i}]}\dop \lambda \to \nu = \delta_{\{0\}}$ weakly, but $0 = \nu(\{0\}) \not\to \delta_{\{0\}}(\{0\}) = 1$.

  Without loss of ideas (needed for the general case), we consider
  \begin{align*}
    A_i \colon \C[\Gamma/\Gamma_i] \to \C[\Gamma/\Gamma_i].
  \end{align*}
  coming from a matrix over $\Z[\Gamma/\Gamma_i]$.
  Further, fix $i$, let $n = [\Gamma : \Gamma_i]$ and denote by $0 = \lambda_1 = \cdots \lambda_m < \lambda_{m+1} \leq \dots \leq \lambda_n$ the eigenvalues of $A_i$.
  The characteristic polynomial of $A_i$ is given by $p(z) = z^mq(z)$ for $q \in \Z[t]$ and this yields
  \begin{align*}
    \lambda_{m+1} \cdots \lambda_n - q(0) \geq 1.
  \end{align*}

  Small eigenvaolues: $N(\varepsilon) = \#$ eigenvalues of $A_i$ in $(0,\varepsilon)$.
  We get $1 \leq \lambda_{m+1} \cdots \lambda_n \leq \varepsilon^{N(\varepsilon)}\|A_i\|^n \leq \varepsilon^{N(\varepsilon)} \cdot \text{const}^n$.
  Thus taking a logarithm, we obtain $\frac{N(\varepsilon)}{n} \leq \frac{\text{const}}{|\log \varepsilon|}$ (independent of $i$).
  Since $\mu_{A_i}((0,\varepsilon)) = \frac{N(\varepsilon)}{n}$, we have found an upper bound for it.

  Basic measure theory shows that weak convergence yields $\limsup \mu_{A_i}(W) \leq \mu_A(W)$ for a closed set $W$ and $\liminf \mu_{A_i}(U) \geq \mu_A(U)$ for $U$ open.

  Now consider
  \begin{align*}
    \liminf \mu_i(\{0\}) & = \liminf(\mu_i([0,\varepsilon)) - \mu((0,\varepsilon)))\\
    & \geq \liminf(\mu_i((-\varepsilon,\varepsilon)) - \frac{\text{const}}{|\log \varepsilon|}\\
    & \geq \mu((\varepsilon, \varepsilon)) - \frac{\text{const}}{|\log \varepsilon|}\\
    & \geq \mu(\{0\}) - \frac{\text{const}}{|\log \varepsilon|}.
  \end{align*}
  As this does not depend on $i$, it shows the claim.
\end{proof}


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\chapter{Harmonic Maps [Andy Sanders]}

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Also consider the notes \url{www.mathi.uni-heidelberg.de/~asanders/harmonicmaps.htm}.
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\section{Basics of harmonic maps}
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In the following let every manifold be oriented (for integration safety reasons).

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\subsection{Background differential geometry}

Let $E \to M$ be an $\R$-vector bundle over $M$ (second countable, hausdorff manifold) of rank $r$.
A \CmMark{connection} $\nabla$ on $E$ is an $\R$-linear map
\begin{align*}
  \nabla \colon \Omega^0(E) \to \Omega^0(\T^{*}M \otimes_{\R} E) =: \Omega^1(M,E),
  s \mapsto \nabla_{\blank} s
\end{align*}
where $\Omega^0(E)$ denotes smooth sections in $E$, such that
\begin{enumerate}
\item $\nabla_{X+Y}s = \nabla_Xs + \nabla_Ys$,
\item $\nabla_X(s+s') = \nabla_X s + \nabla_Xs'$
\item $\nabla_{fX} s = f\nabla_Xs$
\item $\nabla_X(fs) = f\nabla_Xs + X(f) s$.
\end{enumerate}

Let $q$ be an inner product on $E$.
We say that $\nabla$ is a \CmMark{metric connection} for $q$, if for all $s,t \in \Omega^0(E)$ we have
\begin{align*}
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  \dop q(s,t) = q(\nabla s,t) + q(s, \nabla t).
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\end{align*}

\begin{expl*}
  Let $(M,g)$ be a riemannian manifold with tangent bundle $E = \T M$ and Levi-Civita connection $\nabla$ of $g$.

  Let $X,Y \in \Omega^0(M)$ be vector fields, i.e. $X = X^i \frac{\partial}{\partial x^i}$ and $Y = Y^j \frac{\partial}{\partial x^j}$ in local co-ordinates.
  (Abbreviate $\partial_i$ for $\frac{\partial}{\partial x^i}$.)
  \begin{align*}
    \nabla_XY = \nabla_{X^i\partial_i} Y^i\partial_i
    = X^i(\nabla_{\partial_i}Y^i\partial_i)
    = X^i(\partial_iY^i\partial_i + Y^i\nabla_{\partial_i}\partial_i)
    = X^i(\partial_iY^i\partial_i + Y^i\Gamma_{ij}^k\partial_i)
  \end{align*}
  where $\Gamma_{ij}^k = g^{km}(\partial_ig_{im} + \partial_j g_{im} - \partial_mg_{ij})$ for $g_{ij} = g(\partial_i,\partial_j)$ and $g^{km}$ is the $km$-entry of $g^{-1}$.
\end{expl*}

Out of $E$ one can build another bundle $E^{*} = \Hom(E,\R)$ and given another vector bundle $F$, one can build $\Hom(E,F)$, 

\begin{dfn*}
  Let $(E,\nabla) \to M$ be a vector bundle with a connection over $M$.
  The space of \CmMark{$p$-forms} on $m$ with values in $E$ is the $C^{\infty}(M)$-module $\Omega^p(M,E) = \Omega^0(M,\bigwedge^p\T^{*}M \otimes E)$.
  Elements $\alpha$ in $\Omega^p(M,E)$ have representations as linear combination of $\alpha_{i_1,\cdots,i_p}\dop x^{i_1} \wedge \cdots \wedge \dop x^{i_p} \otimes (s_1, \cdots s_p)$.
\end{dfn*}

\begin{dfn*}
  The exterior covariant derivative is the map given by extension of
  \begin{align*}
    \dop^{\nabla} \colon \Omega^p(M,E) & \to \Omega^{p+1}(M,E),\\
    \alpha \otimes u & \mapsto \dop^{\nabla}(\alpha \otimes u) = \dop \alpha \otimes u + (-1)^p \alpha \wedge \nabla u
  \end{align*}
  for $\alpha \in \bigwedge^p\T^{*}M$, $u \in \Omega^0(E)$.
\end{dfn*}

We want to define an inner product on $\Omega^p(M,E)$.
For this, fix a metric $g$ on $M$ and let $(E,\nabla,q) \to M$ be a vector bundle with metric and connection over $M$.
\begin{align*}
  \left< \alpha \otimes u, p \otimes v\right>
  = \int_M g(\alpha,p) q(u,v) \dop v_g
\end{align*}
is a number.
(For this integral to be finite, assume $M$ is compact or work with compactly supported sections.)

\begin{dfn*}
  The \CmMark{exterior covariant codifferential}\footnote{non-standard notation} is the formal $L^2$-adjoint of $d$
  \begin{align*}
    \delta^{\nabla} \colon \Omega^p(M,E) \to \Omega^p(M,E)
  \end{align*}
  such that $\left< \dop^{\nabla}(\alpha \otimes u), \beta \otimes v\right> = \left<\alpha \otimes u, \delta^{\nabla}(\beta \otimes v)\right>$.
\end{dfn*}

\begin{rem*}[Fact]
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  An integration by parts argument shows that $\delta^{\nabla}$ exists and, when $\nabla$ is a metric connection, then
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  \begin{align*}
    \delta^{\nabla} \colon \Omega^1(M,E) \to \Omega^0(M,E),
    \
    \alpha \otimes u \mapsto -\tr_g(\nabla^{\T^{*} \otimes E} \alpha \otimes u),
  \end{align*}
  where for $\Omega^1(M,E) \to \Omega^0(M, \T^{*}M \otimes \T^{*}M \otimes E)$, we can take a trace with the metric by choosing an orthonormal basis.
\end{rem*}

\begin{dfn*}
  A \CmMark{harmonic $p$-form} with values in $E$ is an element $\omega_i \in \Omega^p(M,E)$ such that $\delta^{\nabla} = \delta^{\nabla} \omega = 0$.
  As a matter of fact this is equivalent to $\Delta \omega = 0$ for $\Delta := \delta^{\nabla} \circ \dop^{\nabla} + \dop^{\nabla} \circ \delta^{\nabla}$ (Consider $\left<\Delta \omega, \omega\right>$ and utilize the obvious stuff).
\end{dfn*}


\subsection{Definition of harmonic maps of 1st variation formula}

Let $(M,g)$ and $(N,h)$ be two riemannian manifolds and let $f \colon M \to N$ be a smooth map.
Then $\dop f \colon \T M \to \T N$ is an element $\dop f \in \Omega^0(\Hom(\T M, f^{*}\T N)) = \Omega^0(\T^{*}M \otimes f^{*}\T N)$.

Next, the metrics $g,h$ induce a metric on $\T^{*}M \otimes f^{*}\T N$.

\begin{dfn*}
  The energy density of $f \colon M \to N$ is $e(f) := \frac{1}{2} \left< \dop f, \dop f\right>_{\T^{*}M \otimes f^{*}\T N} = \frac{1}{2} \|\dop f\|^2$.
\end{dfn*}

Choose co-ordinates $\{x^i\}$ in $M$ and $\{y^i\}$ in $N$.
With respect to these, we have
\begin{align*}
  \frac{1}{2} \|\dop f \|^2 = \frac{1}{2}y^{ij} \partial_if^{*}\partial_jf^{\beta}h_{\alpha\beta}(f).
\end{align*}

\begin{dfn*}
  The \CmMark{Dirlichlet energy} is given by
  \begin{align*}
    E \colon C^2_0(M,N) \to \R,
    \
    f \mapsto \int_M e(f) \dop V_g.
  \end{align*}
  A \CmMark{critical map} (or \CmMark{stationary map}) is a map $f \colon M \to N$ such that for all compactly supported $F \colon M \times (-\varepsilon, \varepsilon) \to N$ $C^2$-map (variation of $f$) with $F(x,0) = f(x)$ we have that
  \begin{align}\label{eq:first-variation}
    \delta E(\nu) := \left.\frac{\dop}{\dop t} E(F) \right|_{t = 0} = 0
  \end{align}
  for $\nu = \frac{\dop}{\dop t} F|_{t = 0} \in \Omega^0(f^{*}\T N)$.
  The \cref{eq:first-variation} is called \CmMark{first variation in the direction of $\nu$}.
\end{dfn*}

\begin{dfn*}
  The map $f \colon (M,g) \to (N,h)$ is called \CmMark{harmonic}, if it is a critical point for the Dirlichlet energy.
\end{dfn*}

\begin{dfn*}
  Let $\dop f \in \Omega^1(M, f^{*}\T N)$ then $\nabla \dop f \in \Omega^0(M, \T^{*}M \otimes \T^{*}M \otimes E)$.
  
  The \CmMark{second fundamental form} of $f$ is $\nabla \dop f := B_f$, which is a symmetric $2$-tensor on $M$.
\end{dfn*}

\begin{dfn*}
  The \CmMark{tension field} of $f$ is the trace of $B_f$: $\tau(f) := \tr_g(B_f) \in \Omega^0(M,f^{*}\T N)$.
\end{dfn*}

\begin{thm*}[1st variation of $E$]
  Let $F \colon M \times (\varepsilon, \varepsilon) \to N$ a variation of $f$ and let $\nu = \frac{\dop}{\dop t}F|_{t = 0}$.
  Then
  \begin{align*}
    \delta E(\nu) = \frac{\dop}{\dop t}E(F)|_{t = 0} = - \int_M \left<\tau(f), \nu\right> \dop v_g.
  \end{align*}
\end{thm*}

\begin{proof}
  The variation $F \colon M \times (-\varepsilon,\varepsilon) \to N$ yields a pullback connection on $F^{*}\T N$, which shows
  \begin{align*}
    \frac{\dop}{\dop t}E(F)|_{t = 0} & = \frac{1}{2} \int_M \frac{\dop}{\dop t}\left<\dop F, \dop F\right> \dop V_g|_{t = 0}
    = \int_M \left<\nabla_{\frac{\partial}{\partial t}}\dop F, \dop F\right> \dop V_g|_{t = 0}\\
    & = \int_M \left<\nabla^{f^{*}\T N}\nu, \dop f\right> \dop V_g
    \overset{(*)}{=} \int_M \left<\nu, \delta^{\nabla^{f^{*}\T N}} \dop f\right> \dop V_g\\
    & = -\int_M \left< \nu, \tr_g(\nabla \dop f) \right> \dop V_g
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    = - \int_M \left< \nu, \tau(f)\right> \dop V_g,
  \end{align*}
  where $(*)$ follows by a calculation in local co-ordinates.
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\end{proof}

\begin{cor*}[Fundamental theorem of the calculus of variations]
  A $C^2$-map $f \colon (M,g) \to (N,h)$ is harmonic if and only if $\tau(f) = 0$.
\end{cor*}

What does $\tau(f) = 0$ look like?

Fix local co-ordinates $\{x^i\}$ on $M$ and $\{y^j\}$ on $N$.
Then $\dop f = \partial_if^{alpha} \dop x^i \otimes \frac{\partial}{\partial y^{\alpha}}$ and thus
\begin{align*}
  \nabla \dop f
  & = \nabla \partial_i f^{\alpha} \dop x^i \otimes \frac{\partial}{\partial y^{\alpha}}
  = \partial_j\partial_i f^{\alpha} \dop x^j \otimes \dop x^i \otimes \frac{\partial}{\partial y^{\alpha}}
  + \partial_if^{\alpha} \nabla \dop x^i \otimes \frac{\partial}{\partial y^{\alpha}}\\
  & = A + \partial_i f^{\alpha}(\nabla \dop x^i \otimes \frac{\partial}{\partial y^{\alpha}} + \dop x^i \otimes \nabla \frac{\partial}{\partial y^{\alpha}})\\
  & = A + \partial_i f^{\alpha}( -\Gamma^i_{jk} \dop x^i \otimes \dop x^k \otimes \frac{\partial}{\partial y^{\alpha}} + \dop x^i \partial_j f^{\beta} \Gamma^{\gamma}_{\alpha\beta} \frac{\partial}{\partial y^{\gamma}})\\
  & = \partial_i \partial_jf^{\gamma} \Gamma_{ij}^k \partial_k f^{\gamma} + \Gamma_{\alpha\beta}^{\gamma}(f) \partial_jf^{\alpha}\partial_if^{\beta)} \dop x^i \otimes \dop x^j \otimes \frac{\partial}{\partial y^j}.
\end{align*}
Thus $\tau(f) = (\Delta_gf^{\gamma} + \Gamma_{\alpha\beta}^{\gamma}(f) \partial_if^{\alpha}\partial_jf^{\beta}g^{ij})$.


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\section{Example and the Bochner formula (a glimpse of rigidity)}

Recall that above we considered $C^2$-maps $f \colon (M,g) \to (N,h)$ with tension field
\begin{align*}
  \tau(f) := \tr_g(\nabla \dop f) = 0 \in \Omega^0(M,f^*\T N).
\end{align*}
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In local co-ordinates $\{x^i\}$ on $M$ and $\{y^{\alpha}\}$ on $N$ this means \footnote{Use roman indices for the $M$ and Greek ones for $N$.} 
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\begin{align*}
  \tau(f)^{\gamma} \frac{\partial}{\partial y^{\gamma}} = (\Delta_g f^{\gamma} + \tilde \Gamma_{\alpha\beta}^{\gamma}(f) \partial_if^{\alpha}\partial_jf^{\beta}g^{ij})\partial_{\gamma} = 0,
\end{align*}
where $\tilde \Gamma$ are the Christoffel symbols on $(N,h)$.

\begin{expl*}
  \begin{enumerate}[label=\Roman*.]
  \item Let $(M,g) = (\R, \dop t^2)$ and let $\eta \colon \R \to (N,h)$.
  From above we know that the Laplace-Beltrami operator here reads
  \begin{align*}
    \Delta_gf = g^{ij} (\partial_i\partial_jf - \Gamma_{ij}^k\partial_kf)
    = g^{ij}(\partial_i\partial_jf)
    = \partial_t^2f
  \end{align*}
  for
  \begin{align*}
    \Gamma_{ij}^k = \frac{g^{km}}{2}(\partial_ig_{im} + \partial_jg_{im} - \partial_m(g_{ij})
  \end{align*}
Jan-Bernhard Kordaß's avatar
Jan-Bernhard Kordaß committed
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  ($=0$ if $\{g_{ij}\}$ is constant) and $\{g_{ij}\} = g_{11} = f(\partial_t,\partial_t) = \dop t^2(\partial_t,\partial_t) = 1$.
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  Hence
  \begin{align*}
    \tau(\eta)^{\gamma}\partial_{\gamma} = (\ddot \eta^{\gamma} + \tilde \Gamma_{\alpha\beta}^{\gamma}(\eta) \dot\eta^{\alpha}\dot\eta^{\beta})\partial_{\gamma} = 0,
  \end{align*}
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  which is if and only if $\eta$ is a geodesic, i.e. the covariant derivative along $M$ of the curves speed vanishes: $\frac{\Dop}{\dop t} \dot \eta = 0$ and thus $E(\eta)|_a^b = \frac{1}{2}\int_a^b\|\dot\eta\|^2\dop t$.
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\item Now let $f \colon (M,g) \to \R$.
  Here $\tau(f) = \Delta_gf = 0$.
  \begin{prop*}
    If $M$ is closed, then the energy harmonic functions are constant.
  \end{prop*}
  \begin{proof}
    By Green's theorem (integration by parts)
    \begin{align*}
      \int_M \underbrace{g(\nabla f, \nabla f)}_{=\|\nabla f\|^2} \dop V_g = - \int \Delta_g f \cdot f \dop V_g = 0
    \end{align*}
    for $\dop V_g = \sqrt{\det (\{g_{ij}\})} \dop x^1 \wedge \cdots \wedge \dop x^n$.
    Thus $\|\nabla f\|^2 = 0$ and $f$ must be constant.
  \end{proof}
  In our example, this shows $\Delta_g f = \lambda f$.
\item Let $f \colon (M,g) \to (N,h)$ be an isometric immersion, i.e. $\dop f$ is injective and $g = f^{*}h = h(\dop f, \dop f)$.
  Then we have
  \begin{align*}
    e(f) & = \frac{1}{2} \|\dop f\|^2
    = \frac{1}{2}h_{\alpha\beta} \partial_if^{\alpha}\partial_jf^{\beta}g^{ij}
    = \frac{1}{2}\partial_if^{\alpha}\partial_jf^{\beta}h(\partial_{\alpha},\partial_{\beta}) g^{ij}\\
    & = \frac{1}{2}h(\partial_if^{\alpha}\partial_{\alpha},\partial_jf^{\beta}\partial_{\beta})g^{ij}
    = \frac{1}{2}h(\dop f(\partial_i),\dop f(\partial_j)) g^{ij}
    = \frac{1}{2}g_{ij}g^{ij}
    = \frac{m}{2}
  \end{align*}
  and hence $E(f) = \frac{m}{2}\Vol(f)$, where $\Vol(f) = \int_M\dop V_{f^{*}h} = \int_M\dop V_g$.
  This shows that $f$ is critical for $E$ if and only if $f$ is critical for $\Vol \colon \Imm(M,N) \to \R_+$.
  The latter is clearly if and only if $f$ is a \textbf{minimal submanifold}.

  Examples of minimal submanifolds in $\R^3$ include the 2-plane, or the helicoid.
\end{enumerate}
\end{expl*}

\subsection{Composition laws for harmonic maps}

Consider the composition
\begin{align*}
  (M,g) \xrightarrow{f} (N,h) \xrightarrow{u} (Z,b).
\end{align*}
In general, if $f,u$ are harmonic, this needs not be harmonic again, which can be considered ``a bug or a feature''.
\begin{align*}
  B_{u \circ f}(X,Y) = B_u(\dop f(X), \dop f(Y)) + \dop u (B_f(X,Y))
\end{align*}
for $X,Y \in \T_pM$ and thus $B_{u \circ f} = \nabla^{\T^{*}M \otimes (u \circ f)^{*}\T N}(\dop(u \circ f))$.
Hence $\tau(u \circ f) = \dop (\tau(f)) + \tr_g(f^{*}B_u)$.

If $f$ is harmonic, then $\tau(u \circ f) = \tr_g(f^{*}B_u)$.
\begin{prop*}
  If $f \colon M \to N$, is harmonic and $u \colon N \to Z$ is totally geodesic, i.e. $B_u = 0$.
  Then $u \circ f$ is harmonic.
\end{prop*}

What if $u \colon N \to \R$ is a function and $f$ is harmonic?
Then
\begin{align*}
  \tau(u \circ f) = \tr_g(f^{*}B_u)
  = \tr_g(f^{*}(\Hess(u))
  = \sum_{i = 1}^n f^{*}(\Hess(u)) (E_i,E_i).
\end{align*}
Recall that a function $u \colon (N,h) \to \R$ is convex, if $\Hess(u)$ is positive definite.
If $f$ is harmonic and $u$ is convex, then $\tau(u \circ f) = \nabla_g u \circ f \geq 0$ (these are called \CmMark{subharmonic functions}).

\begin{thm*}
  A map is harmonic if and only if it pulls back germs of convex functions to germs of subharmonic functions.
\end{thm*}

There are various useful applications of the ``synthetic view'' on harmonic functions (e.g. Gromov-Shane).

\begin{thm*}
  Suppose $(M,g)$ is closed, connected and $(N,h)$ is $1$-connected with non-positive curvature.
  Then every harmonic map $f \colon (M,g) \to (N,h)$ is constant.
\end{thm*}

\begin{proof}
  The distance function $N \to \R_{\geq 0}, x \mapsto \dop_N(p,x)^2$ for every $p \in N$ is actually smooth and strictly convex, e.g. $\dop_{\R^n}(0,x)^2 = x_1^2 + \cdots + x_n^2$.

  In case $f$ is harmonic, we have
  \begin{align*}
    \Delta_gu \circ f = \tau(u \circ f) = \tr_g(f^{*}B_u) \geq 0
  \end{align*}
  and
  \begin{align*}
    -\int \| \dop(u \circ f)\|^2 \dop V_g = \int_M \Delta_gu \circ f \dop V_g \geq 0.
  \end{align*}
  Thus $\|\dop (u \circ f)\| = 0$ and hence $u \circ f$ is constant.
\end{proof}


\subsection{Bochner formulas}

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Let $(E,\nabla,a)$ be a riemannian vector bundle, i.e. $a$ is a metric on $E$, $\nabla$ is a connection on $E$ preserving $a$ ($\nabla a = 0$) and there is a vector bundle projection map $E \to (M,g)$.
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Let $\omega \in \Omega^p(M,E)$ and let $\nabla$ be a connection on $\Omega^p(M,E)$.
\begin{align*}
  \hat\nabla \colon \Omega^p(M,E) \to \Omega^p(M,\T^{*}M \otimes \T^{*}M \otimes E),
  \quad
  \omega \mapsto ( (X,Y) \mapsto \nabla_X\nabla_Y\omega - \nabla_{\nabla_XY}\omega )
\end{align*}
The \CmMark{trace Laplacian} is the operator
\begin{align*}
  \nabla^2 \colon \Omega^p(M,E) \to \