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 Jan-Bernhard Kordaß committed Oct 25, 2016 1 2 \tableofcontents  Jan-Bernhard Kordaß committed Nov 22, 2016 3 \chapter{Torsion Invariants [Roman Sauer]}  Jan-Bernhard Kordaß committed Oct 25, 2016 4 5 6 7 8 9 10 11 12 13 14  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*}  Jan-Bernhard Kordaß committed Nov 23, 2016 15  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 $\Im(\phi_e|_{S^{n-1}}) \subset \bigcup_{f \in E, \dim f \leq n-1} f$.  Jan-Bernhard Kordaß committed Oct 25, 2016 16 17 18 19 \end{dfn*} \begin{expl*} \begin{enumerate}  Jan-Bernhard Kordaß committed Nov 23, 2016 20  \item Simplicial complexes, e.g. triangles, pyramids, etc.  Jan-Bernhard Kordaß committed Oct 25, 2016 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74  \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*}  Jan-Bernhard Kordaß committed Nov 23, 2016 75 which yields a filtration $X^0 \subset X^1 \subset \cdots \subset X$ such there is a push-out diagram  Jan-Bernhard Kordaß committed Oct 25, 2016 76 77 78 79 80 81 \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*}  Jan-Bernhard Kordaß committed Nov 23, 2016 82 One could take this as an alternative definition of a CW-complex by a filtration with the push-out property.  Jan-Bernhard Kordaß committed Oct 25, 2016 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 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*}  Jan-Bernhard Kordaß committed Nov 23, 2016 114  The \CmMark{Lefschetz number} of a self-map $f \colon X \to X$ of a finite CW-complex is defined as  Jan-Bernhard Kordaß committed Oct 25, 2016 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143  \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{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} we have $\tr(b) = \tr(a) + \tr(c)$. \end{proof} \begin{thm*} If $f$ has no fixed point, then $\Lambda(f) = 0$. \end{thm*}  Jan-Bernhard Kordaß committed Oct 25, 2016 144   Jan-Bernhard Kordaß committed Oct 25, 2016 145 146 147 \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*}  Jan-Bernhard Kordaß committed Oct 25, 2016 148   Jan-Bernhard Kordaß committed Oct 25, 2016 149 150 151 152 153 154 155 156 \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}  Jan-Bernhard Kordaß committed Oct 25, 2016 157   Jan-Bernhard Kordaß committed Oct 25, 2016 158   Jan-Bernhard Kordaß committed Nov 08, 2016 159 160 161 162 163 164 165 166 167 168 169 \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:  Jan-Bernhard Kordaß committed Nov 23, 2016 170  A cellular push-out is a diagram  Jan-Bernhard Kordaß committed Nov 08, 2016 171 172 173 174 175 176  \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*}  Jan-Bernhard Kordaß committed Nov 23, 2016 177  with $X_i$ be CW-complexes, where $f$ is cellular and $i$ is an inclusion of a sub-complex.  Jan-Bernhard Kordaß committed Nov 08, 2016 178 179 180 181 182 183 184 185 186  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*}  Jan-Bernhard Kordaß committed Nov 23, 2016 187  is a map of cellular push-outs such that $f_i$ are homotopy equivalences.  Jan-Bernhard Kordaß committed Nov 08, 2016 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443  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 $\Z/5 \cong \left$ 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*}  Jan-Bernhard Kordaß committed Dec 06, 2016 444   Roman Sauer committed Dec 05, 2016 445 \subsection{Whitehead torsion for chain complexes}  Jan-Bernhard Kordaß committed Nov 22, 2016 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464  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*}  Jan-Bernhard Kordaß committed Nov 23, 2016 465  For a continuous map $f \colon X \to Y$ we have the topological mapping cylinder.  Jan-Bernhard Kordaß committed Nov 22, 2016 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550  \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*}  Jan-Bernhard Kordaß committed Nov 23, 2016 551  is the full complex and thus contractible'' means short exact''.  Jan-Bernhard Kordaß committed Nov 22, 2016 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582  $C_1 \xrightarrow{\cong} C_0 \oplus C_2$ 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.  Jan-Bernhard Kordaß committed Nov 23, 2016 583  Thus $[A] = -[B]$ in $K_1(R)$ and $B$ is independent of the choice of $\gamma_{*}$, hence $A$ is independent of the choice of $\gamma_{*}$.  Jan-Bernhard Kordaß committed Nov 22, 2016 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 642 643 644 645 646 647 648 649 650 651 652 653 654 655 656 657 658 659 660 \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}  Jan-Bernhard Kordaß committed Oct 25, 2016 661   Jan-Bernhard Kordaß committed Dec 06, 2016 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702 703 704 705 706 707 708 709 710 711 712 713 714 715 716 717 718 719 720 721 722 723 724 725 726 727 728 729 730 731 732 733 734 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.  Jan-Bernhard Kordaß committed Dec 06, 2016 735 736 737 738 739 740 741 742 743 744 745 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}  Jan-Bernhard Kordaß committed Dec 06, 2016 746   Jan-Bernhard Kordaß committed Dec 06, 2016 747 748 749 750 751 752 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 771 772 773 774 775 776 777 778 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 $\tau(f)$.  Jan-Bernhard Kordaß committed Dec 06, 2016 779 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 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}  Jan-Bernhard Kordaß committed Dec 06, 2016 795 796 797 798 799 800 801 802 803 804 805 806 807 808 809 810 811 812 813 814 815 816 817 818 819 820 821 822 823 824 825 826 827 828 \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}  Jan-Bernhard Kordaß committed Dec 06, 2016 829  \begin{enumerate}[label=ad (\arabic*).,leftmargin=1.5cm]  Jan-Bernhard Kordaß committed Dec 06, 2016 830 831 832 833 834 835 836 837 838 839 840 841  \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.  benjamin.wassermann committed Dec 07, 2016 842  \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 $b_j$.  Jan-Bernhard Kordaß committed Dec 06, 2016 843 844 845 846 847 848 849 850 851 852 853 854  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*}  benjamin.wassermann committed Dec 07, 2016 855 856 857 858 859  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:  Jan-Bernhard Kordaß committed Dec 06, 2016 860 861  \begin{align*} \tau(\tilde X \hookrightarrow \tilde Y) = \tau(C_{*}^{\text{CW}}(\tilde Y, \tilde X))  benjamin.wassermann committed Dec 07, 2016 862  = \tau\left(\bigoplus_{j=1}^n \Z\pi \xrightarrow{A} \bigoplus_{j=1}^n \Z\pi\right) = a  Jan-Bernhard Kordaß committed Dec 06, 2016 863 864 865 866 867  \end{align*} for $[A] \in \Wh(\pi)$. \end{enumerate} \end{proof}  Jan-Bernhard Kordaß committed Dec 13, 2016 868 869 870 871 872 873 874 875 876 877 878 879 880 881 882 883 884 885 886 887 888 889 890 891 892 893 894 895 896 897 898 899 900 901 902 903 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 937 938 939 940 941 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 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*}  benjamin.wassermann committed Jan 19, 2017 1017 1018 In some sense this is a topological version of the s-cobordism Theorem.  Jan-Bernhard Kordaß committed Dec 13, 2016 1019   benjamin.wassermann committed Jan 18, 2017 1020 1021 1022 1023 1024 1025 1026 1027 1028 1029 1030 1031 1032 1033 1034 1035 1036 1037 1038 1039 1040 1041 1042 1043 1044 1045 1046 1047 1048 1049 1050 1051 1052 1053 1054 1055 1056 1057 1058 1059 \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 t . (z_1,z_2) := (\xi_pz_1,\xi_p^qz_2) 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?} \\  benjamin.wassermann committed Jan 19, 2017 1060 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}.  benjamin.wassermann committed Jan 18, 2017 1061 1062 1063  \subsection{A $CW$-structure for $L(p,q)$}  benjamin.wassermann committed Jan 19, 2017 1064 1065 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 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}, 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 t.(x,y,z) := f ( t. f^{-1}(x,y,z)). 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  benjamin.wassermann committed Jan 18, 2017 1066 1067 1068  \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 $X^0 := \{ t^k.e_0: k=0,\dots,p-1 \}$.  benjamin.wassermann committed Jan 19, 2017 1069 \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 $e_1 := \{f(0,e^{is}): s \in [0,\frac{2\pi}{p}] \}$.  benjamin.wassermann committed Jan 18, 2017 1070 1071 1072 1073 1074 \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}  benjamin.wassermann committed Jan 19, 2017 1075 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:  benjamin.wassermann committed Jan 18, 2017 1076 1077 1078 1079 1080 1081  \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}  benjamin.wassermann committed Jan 19, 2017 1082 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):  benjamin.wassermann committed Jan 18, 2017 1083 1084 1085 1086 1087 1088 1089 1090 1091 1092 1093 1094 1095 1096 1097 1098 1099 1100 1101 1102 1103 1104 1105 1106 1107 1108 1109 1110 1111 1112 1113 1114 1115 1116 1117 1118 1119 1120 1121 1122 1123 1124 1125 1126 1127 1128 1129 1130 1131 1132 1133 1134 1135 1136 1137 1138 1139 1140 1141 1142 1143 1144 1145 1146 1147 1148 1149 1150 1151 1152 1153 1154 1155 1156 1157 1158 1159 1160 1161 1162 1163 1164 1165 1166  \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 f_1(e_1) = \sum_{j=0}^{q'ar-1} (t')^{r'j} e_1'. \label{chain} 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 f_2 = f_1 + \alpha(\gamma-1) \label{eq1} 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 f_3 = (\sum_{i=0}^{a-1} \gamma^i)f_2 + \beta (\sum_{i=0}^{p-1} \gamma^i) \label{eq2} 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}  benjamin.wassermann committed Jan 19, 2017 1167 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.  benjamin.wassermann committed Jan 18, 2017 1168 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 $R$  benjamin.wassermann committed Jan 19, 2017 1169 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 $R^{\times} / \tilde{G}$.  benjamin.wassermann committed Jan 18, 2017 1170 1171 1172  \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  benjamin.wassermann committed Jan 19, 2017 1173 {\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.  benjamin.wassermann committed Jan 18, 2017 1174 1175 1176 \end{dfn*} We will illustrate the general setting for applying Reidemeister-Torsion, which will also clarify the choice of the target group.  benjamin.wassermann committed Jan 19, 2017 1177 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  benjamin.wassermann committed Jan 18, 2017 1178 1179 1180 1181 1182 1183 1184 1185 1186 1187 1188 1189 1190 1191 1192 1193 1194 1195 1196 1197 1198 1199 1200 1201 1202 $(C_*(\tilde{X}) \otimes_\rho R)$ 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  benjamin.wassermann committed Jan 19, 2017 1203 in $\mathbb C^{\times} /< \pm \{1, \xi, \xi^2, \dots,\xi^{p-1} \} >$. In particular, if $f$ is a {\itshape simple} homotopy equivalence, $|\xi^a - 1||\xi^{ar} - 1| = |\xi - 1||\xi^{r'} - 1|$.  benjamin.wassermann committed Jan 18, 2017 1204 1205 1206 1207 1208 \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*}  Jan-Bernhard Kordaß committed Feb 07, 2017 1209 1210 1211 1212 1213 1214 1215 1216 1217 1218 1219 1220 1221 1222 1223 1224 1225 1226 1227 1228 1229 1230 1231 1232 1233 1234 1235 1236 1237 1238 1239 1240 1241 1242 1243 1244 1245 1246 1247 1248 1249 1250 1251 1252 1253 1254 1255 1256 1257 1258 1259 1260 1261 1262 1263 1264 1265 1266 1267 1268 1269 1270 1271 1272 1273 1274 1275 1276 1277 1278 1279 1280 1281 1282 1283 1284 1285 1286 1287 1288 1289 1290 1291 1292 1293 1294 1295 1296 1297 1298 1299 1300 1301 1302 1303 1304 1305 1306 1307 1308 1309 1310 1311 1312 1313 1314 1315 1316 1317 1318 1319 1320 1321 1322 1323 1324 1325 1326 1327 1328 1329 1330 1331 1332 1333 1334 1335 1336 1337 1338 1339 1340 1341 1342 1343 1344 1345 1346 1347 1348 1349 1350 1351 1352 1353 1354 1355 1356 1357 1358 1359 1360 1361 1362 1363 1364 1365 1366 1367 1368 1369 1370 1371 1372 1373 1374 1375 1376 1377 1378 1379 1380 1381 1382 1383  % 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}  Jan-Bernhard Kordaß committed Oct 25, 2016 1384 1385 \chapter{Harmonic Maps [Andy Sanders]}  Jan-Bernhard Kordaß committed Nov 08, 2016 1386 Also consider the notes \url{www.mathi.uni-heidelberg.de/~asanders/harmonicmaps.htm}.  Jan-Bernhard Kordaß committed Nov 08, 2016 1387   Jan-Bernhard Kordaß committed Oct 25, 2016 1388 \section{Basics of harmonic maps}  Jan-Bernhard Kordaß committed Nov 08, 2016 1389 1390 In the following let every manifold be oriented (for integration safety reasons).  Jan-Bernhard Kordaß committed Dec 06, 2016 1391   Jan-Bernhard Kordaß committed Oct 25, 2016 1392 1393 1394 1395 1396 1397 1398 1399 1400 1401 1402 1403 1404 1405 1406 1407 1408 1409 1410 \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*}  Jan-Bernhard Kordaß committed Nov 08, 2016 1411  \dop q(s,t) = q(\nabla s,t) + q(s, \nabla t).  Jan-Bernhard Kordaß committed Oct 25, 2016 1412 1413 1414 1415 1416 1417 1418 1419 1420 1421 1422 1423 1424 1425 1426 1427 1428 1429 1430 1431 1432 1433 1434 1435 1436 1437 1438 1439 1440 1441 1442 1443 1444 1445 1446 1447 1448 1449 1450 1451 1452 1453 1454 1455 1456 1457 1458 1459 1460 1461 1462 \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]  Jan-Bernhard Kordaß committed Nov 23, 2016 1463  An integration by parts argument shows that $\delta^{\nabla}$ exists and, when $\nabla$ is a metric connection, then  Jan-Bernhard Kordaß committed Oct 25, 2016 1464 1465 1466 1467 1468 1469 1470 1471 1472 1473 1474 1475 1476 1477 1478 1479 1480 1481 1482 1483 1484 1485 1486 1487 1488 1489 1490 1491 1492 1493 1494 1495 1496 1497 1498 1499 1500 1501 1502 1503 1504 1505 1506 1507 1508 1509 1510 1511 1512 1513 1514 1515 1516 1517 1518 1519 1520 1521 1522 1523 1524 1525 1526 1527 1528 1529 1530 1531 1532 1533 1534 1535 1536 1537 1538 1539  \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  Jan-Bernhard Kordaß committed Nov 08, 2016 1540 1541 1542  = - \int_M \left< \nu, \tau(f)\right> \dop V_g, \end{align*} where $(*)$ follows by a calculation in local co-ordinates.  Jan-Bernhard Kordaß committed Oct 25, 2016 1543 1544 1545 1546 1547 1548 1549 1550 1551 1552 1553 1554 1555 1556 1557 1558 1559 1560 1561 1562 1563 1564 \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 $\tau(f) = (\Delta_gf^{\gamma} + \Gamma_{\alpha\beta}^{\gamma}(f) \partial_if^{\alpha}\partial_jf^{\beta}g^{ij})$.  Jan-Bernhard Kordaß committed Nov 08, 2016 1565 1566 1567 1568 1569 1570 \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*}  Jan-Bernhard Kordaß committed Nov 23, 2016 1571 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 $N$.}  Jan-Bernhard Kordaß committed Nov 08, 2016 1572 1573 1574 1575 1576 1577 1578 1579 1580 1581 1582 1583 1584 1585 1586 1587 1588 1589 \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 committed Nov 22, 2016 1590  ($=0$ if $\{g_{ij}\}$ is constant) and $\{g_{ij}\} = g_{11} = f(\partial_t,\partial_t) = \dop t^2(\partial_t,\partial_t) = 1$.  Jan-Bernhard Kordaß committed Nov 08, 2016 1591 1592 1593 1594  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*}  Jan-Bernhard Kordaß committed Nov 23, 2016 1595  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 $E(\eta)|_a^b = \frac{1}{2}\int_a^b\|\dot\eta\|^2\dop t$.  Jan-Bernhard Kordaß committed Nov 08, 2016 1596 1597 1598 1599 1600 1601 1602 1603 1604 1605 1606 1607 1608 1609 1610 1611 1612 1613 1614 1615 1616 1617 1618 1619 1620 1621 1622 1623 1624 1625 1626 1627 1628 1629 1630 1631 1632 1633 1634 1635 1636 1637 1638 1639 1640 1641 1642 1643 1644 1645 1646 1647 1648 1649 1650 1651 1652 1653 1654 1655 1656 1657 1658 1659 1660 1661 1662 1663 1664 1665 1666 1667 1668 1669 1670 1671 1672 1673 1674 1675 1676 1677 1678 1679 1680 1681 1682 1683 1684 1685 \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}  Jan-Bernhard Kordaß committed Nov 23, 2016 1686 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 $E \to (M,g)$.  Jan-Bernhard Kordaß committed Nov 08, 2016 1687 1688 1689 1690 1691 1692 1693 1694 1695 1696 1697 1698 1699 1700 1701 1702 1703 1704 1705 1706 1707 1708 1709 1710 1711 1712 1713 1714 1715 1716 1717 1718 1719 1720 1721 1722 1723 1724 1725 1726 1727 1728 1729 1730 1731 1732 1733 1734 1735 1736 1737 1738 1739 1740 1741 1742 1743 1744 1745  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 \Omega^p(M,E), \quad \omega \mapsto \tr_g(\hat\nabla \omega). \end{align*} Recall that the \CmMark{Hodge Laplacian} was the operator \begin{align*} \dop^{\nabla} \colon \Omega^p(M,E) \to \Omega^{p+1}(M,E), \quad \alpha \otimes u \mapsto \dop \alpha \otimes u + (-1)^p \alpha \wedge \nabla u. \end{align*} With respect to the $L^2$-pairing $\beta \otimes v \mapsto \int_Mg(\alpha,\beta) a(u,v)\dop V_g$ it has a formal adjoint \begin{align*} \delta^{\nabla}\colon \Omega^{p+1}(M,E) \to \Omega^p(M,E). \end{align*} The Hodge Laplacian is the degree preserving operator given by $\dop^{\nabla} \circ \delta^{\nabla} + \delta^{\nabla} \circ \dop^{\nabla} =: \Delta_a$. The \CmMark[Bochner-Lichnerowicz formula]{(generalized) Bochner-Lichnerowicz formula} is given by \begin{align*} \nabla_a \omega = - \nabla^2\omega + S_{\omega}. \end{align*} for $S_{\omega} \in \Omega^p(M,E)$ with \begin{align*} S_{\omega}(X_1, \cdots, X_p) = \sum_{k = 1}^p\sum_{i = 1}^m(-1)^k(R^{\tilde \nabla}(e_i, X_k)\omega) (e_i,X_1, \ldots, \hat X_k, \ldots, X_n) \end{align*} for $X_i \in \T_pM$, $m = \dim M$ and $\{e_i\}$ an orthonormal frame around $p$.\footnote{Hat ($\hat X_k$), as always, means to omit the k-th term.} \begin{cor*} Let $f \colon (M,g) \to (N,h)$ be harmonic. Then \begin{align*} \Delta_ge(f) = \|B_f\|^2 - \sum_{ij}\underbrace{h(R^h(f_{*}e_i,f_{*}e_j) f_{*}e_j, f_{*}e_i))}_{= \lambda \sec(e_i,e_j)} + \sum_ih(f_{*}(\Ric^g(e_i)),f_ke_i) \end{align*} for an orthonormal frame $\{e_i\}$. \end{cor*} The key observation for an application of this is that, if $\Ric^g$ is a positive operator, then the latter sum is positive. \begin{thm*}[Eells-Sampson] Let $(M,g)$ be a closed with non-negative Ricci curvature and let $(N,h)$ have non-positive sectional curvature. \begin{enumerate}[label=(\roman*)] \item Then any harmonic map $f \colon (M,g) \to (N,h)$ is totally geodesic, i.e. $\nabla \dop f = B_f = 0$. \item If $\Ric^g$ is positive at any point, then $f$ is constant. \item If the sectional curvature of $(N,h)$ is strictly negative, then $f$ is constant or $f(M)$ is closed geodesic. \end{enumerate} \end{thm*} \begin{proof} The first statement easily follows from the corollary and $\int_M \left< \nabla u, \nabla v\right> \dop V_g = - \int_M \Delta u \cdot v \dop V_g$. The second is also not that hard and the last requires some work. \end{proof}  Jan-Bernhard Kordaß committed Nov 22, 2016 1746 1747 1748 1749 1750 1751 1752 1753 1754 1755 1756 1757 1758 1759 1760 1761  \section{The Eells-Sampson existence theorem} \textbf{Story:} Given two manifolds $M,N$, is there a best map in a given free homotopy class $\beta \in [M,N]$, where $[M,N]$ denotes the free homotopy classes of smooth maps. From now on, best'' means harmonic with respect to some riemannian metric. \begin{expl*} If $M = S^n$, then it is a theorem that every homotopy class $\gamma \in [S^1,N]$ (for $N$ closed) admits a harmonic representative $\gamma \colon S^1 \to (N,h)$, i.e. is a closed geodesic. \end{expl*} \begin{expl*} What about $\dim(M) \geq 2$. In this case it depends on the curvature of $(N,h)$. Consider the flat torus $\mathbb T^2$ and the round sphere $\mathbb S^2$.  Jan-Bernhard Kordaß committed Nov 23, 2016 1762  For a degree 1 map $\mathbb T^2 \to \mathbb S^2$ there is no harmonic map in the homotopy class (see the book by Lin on geometry of harmonic maps).  Jan-Bernhard Kordaß committed Nov 22, 2016 1763 1764 1765 1766 1767 1768 1769 1770 1771 \end{expl*} \begin{thm*}[Eells-Sampson 1964] Let $(M,g), (N,h)$ be closed manifolds and $h$ with non-positive sectional curvature. Then given any $f \colon M \to N$ $C^2$-map there exists a harmonic map $u \colon (M,g) \to (N,h)$ such that $u$ is freely homotopic to $f$. \end{thm*} Try to take $\tau(u) = 0$ for some $u \sim f$.  Jan-Bernhard Kordaß committed Nov 23, 2016 1772 In this approach $E \colon C^2(M,N) \to \R, f \mapsto \frac{1}{2} \int_M \|\dop f\|^2 dV_g$ such that $E(f_n) \to \inf_{f \in C^2}E(f)$, we would have to weaken to topology considering Sobolev spaces $W^{1,2}(M,N)$  Jan-Bernhard Kordaß committed Nov 22, 2016 1773 1774  The other approach using gradient flow goes as follows.  Jan-Bernhard Kordaß committed Nov 23, 2016 1775 Try to solve initial value problem (IVP).  Jan-Bernhard Kordaß committed Nov 22, 2016 1776 1777 1778 1779 1780 1781 1782 1783 1784 1785 1786 1787 1788 1789 1790 1791 1792 1793 1794 1795 1796 1797 1798 1799 1800 1801 1802 1803 1804 1805 1806 1807 1808 1809 1810 1811 1812 1813 1814 1815 1816 1817 1818 1819 1820 1821 1822 Let $f \colon M \times (0,\infty) \to N$, such that $\frac{\partial f}{\partial t} = \tau(f_t)$ and $f(\blank, 0) = f$. Recall the first variation of every $f_t \colon M \to N, \frac{\dop}{\dop t}f_t|_{t = 0} = 0$. Then $\delta E(\nu) = \frac{\dop}{\dop t}E(f_t)|_{t=0} = -\int_M \left<\tau(f),\nu\right>dV_g = -Q(\tau(f),\nu)$, where $\left<\blank,\blank\right>$ is the inner product on $f^{*}\T N$ induced by $h$. If we manage to solve $\frac{\partial f}{\partial t} = \tau(f)$, then \begin{align*} \frac{\dop}{\dop t} E(f_t)|_{t = t_0} = \int_M \left<\tau(f),\tau(f_t) \right> \dop V_g \leq 0 \end{align*} and equal to zero if and only if $\tau(f_{t_0}) = 0$. $\frac{\partial f^{\gamma}}{\partial t} = \Delta_gf^{\gamma} + \Gamma_{\alpha\beta}^{\gamma}(f)\partial_if^{\alpha}\partial_if^{\beta}g^{il}$. \subsection{1st short time existence} \begin{thm*} Suppose $f \colon M \to N$ is a $C^2$-map. Then there exists a $T_{\text{max}} > 0$ such that (IVP) \begin{align*} \frac{\partial f_t}{\partial t} = \tau(f_t) \text{ and } f_0 \equiv f \end{align*} has a solution on $[0, T_{\text{max}}]$. If $T_{\text{max}} < \infty$, then \begin{align*} \limsup_{t \nearrow T, x \in M}(f_t) = + \infty. \end{align*} \end{thm*} Note that there is no assumption on the curvature. \subsection{Need another Bochner formula} Let $(N,h)$ has non-positive sectional curvature and let $M$ be an $m$-dimensional manifold. Then we can calculate \begin{align*} & \frac{\partial}{\partial t} e(f_t) - \Delta_ge(f_t)\\ & \quad = -\underbrace{\|B_{f_t}\|^2}_{=\nabla \dop f_t} - \sum_{i=1}^n h(\sum_{j = 1}^m \dop f_t(\Ric^g(e_{i},e_j)e_j),\dop f_t(e_i))\\ & \qquad + \underbrace{\sum_{i,j = 1}^m h(R^h(\dop f_t(e_i), \dop f_t(e_j))\dop f_t(e_j),\dop f_t(e_i))}_{ \leq m}, \end{align*} where $\Ric^g \colon \T M \otimes \T M \to \R$ is the Ricci tensor, $R^h$ is the full curvature tensor of $(N,h)$ and $\{e_1, \ldots, e_m\}$ is an orthonormal frame of $N$. The latter summand is $\operatorname{const} \sec^h(\operatorname{span}(\dop f_t(e_i),\dop f_t(e_j)))$.  Jan-Bernhard Kordaß committed Nov 22, 2016 1823 We continue\footnote{TODO: According to Andy, who erased this part very quickly, there should be some mistake somewhere here...}  Jan-Bernhard Kordaß committed Nov 22, 2016 1824 1825 1826 1827 1828 1829 1830 1831 1832 1833 1834 1835 1836 1837 1838 1839 1840 1841 1842