Commit 7debe7fd authored by benjamin.wassermann's avatar benjamin.wassermann
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......@@ -1014,7 +1014,8 @@ We say that $(Y,X)$ and $(Z,X)$ are \CmMark{equivalent}, if $Y \doglegarrowright
is an isomorphism of abelian groups.
\end{thm*}
In some sense this is a topological version of the s-cobordism theorem.
In some sense this is a topological version of the s-cobordism Theorem.
\section{Lens spaces}
......@@ -1056,29 +1057,29 @@ yields $H_1(L(p,q),\mathbb Z) \cong \pi_1(L(p,q)) = \mathbb Z_p$, while {\itshap
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?} \\
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}.
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}.
\subsection{A $CW$-structure for $L(p,q)$}
Let $p,q$ be two coprime integers, $pr: S^3 \to L(p,q)$ the induced covering projection. Throughout this subsection, we will denote with $\pi$ the group of deck transformations $deck(pr) \equiv \pi_1(L(p,q)) \equiv \mathbb Z_p \equiv < t| t^p=1 >$. Therefore, finding an appropriate cell-structure on $L(p,q)$ amounts to finding a corresponding cell-structure on $S^3$ that is invariant under the action of $\pi$. For this purpose, it will be useful to make the following identification: Let $\mathbb R^3_+$ be the one-point compactification of $\mathbb R^3$ with added point $\infty$, and let $f: S^3 \to \mathbb R^3_+$ be a {\itshape stereographic projection} homeomorphism, defined by \begin{equation} f(z_1,z_2) = \begin{cases} (\frac{x_1}{1-x_4},\frac{x_2}{1-x_4},\frac{x_3}{1-x_4}) & x_4 \neq 1 \\ \infty & x_4 = 1 \end{cases},\end{equation} where we use the convention $z_1 = x_1 + ix_2$, $z_2 = x_3 + ix_4$ with $x_j \in \mathbb R$ for all $j=1,2,3,4$.
There is a canonical action of $\pi$ on $\mathbb R^3_+$, given by the push-forward of the action of $\pi$ on $S^3$ by $f$, i.e, we set \begin{equation} t.(x,y,z) := f ( t. f^{-1}(x,y,z)). \end{equation} It is the unique action on $\mathbb R^3_+$ making the map $f$ $\pi$-equivariant, thus it is the deck group action of the covering projection $pr \circ f^{-1}: \mathbb R^3_+ \to L(p,q)$. We will now define a $\pi$-equivariant $CW$-structure on $R^3_+$, starting with
Let $p,q$ be two coprime integers, $pr: S^3 \to L(p,q)$ the induced covering projection. Throughout this subsection, we will denote with $\pi$ the group of deck transformations $deck(pr) \equiv \pi_1(L(p,q)) \equiv \mathbb Z_p \equiv < t| t^p=1 >$. Therefore, finding an appropriate cell-structure on $L(p,q)$ amounts to finding a corresponding cell-structure on $S^3$ that is invariant under the action of $\pi$. For this purpose, it will be useful to make the following identification: Let $(\mathbb R^3)^*$ be the one-point compactification of $\mathbb R^3$ with added point $\infty$, and let $f: S^3 \to (\mathbb R^3)^*$ be a {\itshape stereographic projection} homeomorphism, defined by \begin{equation} f(z_1,z_2) = \begin{cases} (\frac{x_1}{1-x_4},\frac{x_2}{1-x_4},\frac{x_3}{1-x_4}) & x_4 \neq 1 \\ \infty & x_4 = 1 \end{cases},\end{equation} where we use the convention $z_1 = x_1 + ix_2$, $z_2 = x_3 + ix_4$ with $x_j \in \mathbb R$ for all $j=1,2,3,4$.
There is a canonical action of $\pi$ on $(\mathbb R^3)^*$, given by the push-forward of the action of $\pi$ on $S^3$ by $f$, i.e, we set \begin{equation} t.(x,y,z) := f ( t. f^{-1}(x,y,z)). \end{equation} It is the unique action on $(\mathbb R^3)^*$ making the map $f$ $\pi$-equivariant, thus it is the deck group action of the covering projection $pr \circ f^{-1}: (\mathbb R^3)^* \to L(p,q)$. We will now define a $\pi$-equivariant $CW$-structure on $(\mathbb R^3)^*$, starting with
\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 $X^0 := \{ t^k.e_0: k=0,\dots,p-1 \}$.
\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 $e_1 := \{f(0,e^{is}): s \in [0,\frac{2\pi}{p}] \}$.
\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 $e_1 := \{f(0,e^{is}): s \in [0,\frac{2\pi}{p}] \}$.
\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}
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:
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:
\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}
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):
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):
\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.
......@@ -1163,17 +1164,17 @@ Using this Theorem, the following statements can be easily shown:
\subsection{Reidemeister-Torsion}
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 and $f: S \to R$ a ring homomorphism.
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.
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 $R$
is commutative, taking the determinant of a matrix with entries in $R$ induces a well-defined homomorphism \begin{equation*} \det: \tilde{K_1}(R) \to R^{\times} / \pm 1 \end{equation*} which is even an isomorphism whenever $R$ is a field. We will only be interested in the case $S = \mathbb Z[G]$ for some group $G$. Then, we write $\tilde{G} \subset R^\times$ for the subgroup generated by $\det(f(G))$ and $-1$. By slight abuse of notation, we will also denote the induced composition \begin{equation*} \tilde{K_1}(R) \to R^{\times} / \pm 1 \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 $R^{\times} / \tilde{G}$.
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 $R^{\times} / \tilde{G}$.
\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
{\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.
{\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.
\end{dfn*}
We will illustrate the general setting for applying Reidemeister-Torsion, which will also clarify the choice of the target group.
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 cells turns
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
$(C_*(\tilde{X}) \otimes_\rho R)$ 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*}
......@@ -1199,14 +1200,12 @@ Then $\rho'(t) = \xi^a \neq 1$. From our previously established work, we see tha
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
in $\mathbb C^{\times} /< \pm \{1, \xi, \xi^2, \dots,\xi^{p-1} \} >$. In particular, if $f$ is a simple homotopy equivalence, then $|\xi^a - 1||\xi^{ar} - 1| = |\xi - 1||\xi^{r'} - 1|$.
in $\mathbb C^{\times} /< \pm \{1, \xi, \xi^2, \dots,\xi^{p-1} \} >$. In particular, if $f$ is a {\itshape simple} homotopy equivalence, $|\xi^a - 1||\xi^{ar} - 1| = |\xi - 1||\xi^{r'} - 1|$.
\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*}
\chapter{Harmonic Maps [Andy Sanders]}
Also consider the notes \url{www.mathi.uni-heidelberg.de/~asanders/harmonicmaps.htm}.
......
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