\documentclass[12pt ]{article} \setlength{\textwidth}{14 true cm} \setlength{\textheight}{20 true cm} \usepackage{hangul} \usepackage{amscd,amsmath} \usepackage{amsfonts} \usepackage{amssymb,theorem} \usepackage{longtable} \usepackage{floatflt} \usepackage{texdraw, epic, pstcol ,pstricks ,pst-3d, pst-poly,pst-grad,pst-node, pst-text} \usepackage{floatflt} \usepackage[arrow,matrix]{xy} \usepackage{graphicx} \newcommand{\Aff}{\mbox{\it Aff}} \newcommand{\aff}{\mbox{\it aff}} \newcommand{\alp}{\alpha} \newcommand{\bet}{\beta} \newcommand{\del}{\delta} \newcommand{\gam}{\gamma} \newcommand{\vep}{\varepsilon} \newcommand{\eps}{\epsilon} \newcommand{\lam}{\lambda} \newcommand{\kap}{\kappa} \newcommand{\sig}{\sigma} \newcommand{\ome}{\omega} \newcommand{\Gam}{\Gamma} \newcommand{\Ome}{\Omega} \newcommand{\Sig}{\Sigma} \newcommand{\Del}{\Delta} \newcommand{\Lam}{\Lambda} \newcommand{\disi}{\displaystyle{\sum_{i=1}^n}} \newcommand{\disj}{\displaystyle{\sum_{j=1}^m}} \newcommand{\rb}{\mathbb{R}} \newcommand{\zb}{\mathbb{Z}} \newcommand{\bd}{\partial} \newcommand{\key}[1]{\emph{\textbf{#1}}} \newcommand{\rh}[1]{\widetilde{H}(#1)} \newcommand{\hh}{\widetilde{H}} \newcommand{\hhp}{\widetilde{H}_p} \newcommand{\sn}{S^{n-1}} \newcommand{\snn}{S^{n}} \newcommand{\dn}{D^{n}} \newcommand{\ttl}{T^{2}} \newcommand{\is}{i_{1*}} \newcommand{\sm}{S^n-s^r} \newtheorem{thm}{정리} \newtheorem{cor}[thm]{따름정리} \newtheorem{lem}[thm]{보조정리} \newtheorem{prop}[thm]{명제} \newtheorem{cl}{Claim} {\theorembodyfont{\rm} \newtheorem{ex}{예} \newtheorem{que}{질문} \newtheorem{notation}{Notation}[section] \newtheorem{defn}{정의} \newtheorem{rem}{주} \newtheorem{note}{Note} } \renewcommand{\thenote}{} \renewcommand{\therem}{} \newenvironment{pf}{{\bf 증명}}{\hfill\framebox[2mm]{}} \begin{document} \parindent=0cm \section*{I.3 Boundary identification of two manifolds $M_{1}, M_{2}$ with $f:\bd M_{1} \overset{\cong}{\rightarrow} \bd M_{2}$} {\bf Fact} Let $M$ be a topological (or $\mathcal{C}^{\infty}$) manifold with $\bd M \neq \emptyset$. Then there exists an open neighborhood $U$ of $\bd M$ such that $U \underset{\underset{h}{\leftarrow}}{\cong} \bd M \times [0,1)$ with $h(x,0) = x , \hspace{0.3em} x \in \bd M$.\\ $U$ is called a collar.\\ {\bf The idea of proof}\\ Use local collar and splice together using Partition of Unity.\\ (See Vick(homology theory) for topological case and Milnor($h$-cobordism) for $\mathcal{C}^{\infty}$ case.)\\ Let $(X,A) = (M_{1}, \bd M_{1})$ (a collared pair), $Y=M_{2}$ and let $f: \bd M_{1} \rightarrow \bd M_{2}$ be a homeomorphism. Then we obtain a manifold $W=M_{1} \bigcup_{f} M_{2}$ identifying the boundaries of $M_{1}$ and $M_{2}$ via $f$.\\ {\bf Examples}\\ {\bf 1.(Sphere)}\\ Let $f:\sn \rightarrow \sn$ be a homeomorphism.\\ Then $W = \dn \bigcup_{f} \dn \cong \snn$(homeomorphism). But, not diffeomorphism, in general, for $\mathcal{C}^{\infty}$-category.\\ \begin{pf} Let $\bar{f} : \dn \overset{\cong}{\rightarrow} \dn$ be an extension of $f$.\\(Such extension always exists, e.g., a "radial extension".)\\ \[ \xymatrix @=1em @*[c] { % \dn \coprod \dn \ar[d]^{p} \ar[r]^{\bar{f}\coprod id.} & \dn \coprod \dn \ar[d]^{p} & & x,f(x) \ar[r] \ar[d] & f(x),f(x)\ar[d]\\ \dn \bigcup_{f} \dn \ar@{.>}[r] & \dn \bigcup_{id.} \dn = \snn & & x \sim f(x) & f(x) \sim f(x) } \] 왼쪽의 그림에서 점선의 map은 오른쪽 그림과 같이 $\dn \bigcup_{f} \dn$의 $x \sim f(x) , x \in \bd D^{n}(= S^{n-1})$가 $\dn \bigcup_{id.} \dn$의 $f(x) \sim f(x)$로 보내지므로 잘 정의된다. 따라서 자명한 continuous, one-to-one 그리고 onto조건이 더해져서 homeomorphism이 된다. \end{pf} Exactly same argument shows the following.\\ {\bf Theorem A}\\ Given $M_{i}$ with $\bd M_{i} \neq \emptyset$, if $f,g : \bd M_{1} \overset{\cong}{\rightarrow} \bd M_{2}$ such that $g^{-1}\circ f : \bd M_{1} \rightarrow \bd M_{1}$ can be extended to a homeomorphism $M_{1} \rightarrow M_{1}$, then $M_{1} \bigcup_{f} M_{2} \cong M_{1} \bigcup_{g} M_{2}$.\\ \begin{pf} \[ \xymatrix @=1em @*[c] { % M_{1} \coprod M_{2} \ar[d]^{p} \ar[r]^{\overline{g^{-1}\circ f} \coprod id.} & M_{1} \coprod M_{2} \ar[d] ^{p} & \hspace*{1.0em} & x,f(x) \ar[d] \ar[r] & g^{-1}\circ f(x), f(x) \ar[d] \\ M_{1} \bigcup_{f} M_{2} \ar@{.>}[r]^{\phi} & M_{1} \bigcup_{g} M_{2} & \hspace*{1.0em} & x \sim f(x) & f(x) \sim f(x) } \] 왼쪽의 그림에서 점선 $\phi$는 오른쪽 그림에서 보듯이 위와 같은 이유로 잘 정의된다. 또한 $f^{-1} \circ g = (g^{-1}\circ f)^{-1}$이므로 extension이 여전히 유효하고 같은 방법으로 $\overline{f^{-1} \circ g} = (\overline{g^{-1} \circ f})^{-1}$인 $\psi = \phi^{-1}$를 얻을 수 있다.\\ \end{pf} {\bf 2.(Lens space)}\\ Let $M_{i} = D^{2} \times S^{1}$ with $\bd M_{i} = S^{1} \times S^{1} = \ttl$. Consider $f : \ttl \overset{\cong}{\rightarrow} \ttl$ and what $M_{1} \bigcup_{f} M_{2}$ is.\\ {\bf Theorem B}\\ Let $M_{i}$ with $\bd M_{i} \neq \emptyset$ be manifolds and $f,g : \bd M_{1} \overset{\cong}{\rightarrow} \bd M_{2}$ be homeomorphisms isotopic to each other. Then $M_{1} \bigcup_{f} M_{2} \cong M_{1} \bigcup_{g} M_{2}$.\\ \begin{pf} By Theorem A, it suffices to show the following statement : \\ {\bf claim} $f:\bd M \rightarrow \bd M$ isotopic to $id.$. Then $f$ can be extended to a homeomorphism $\bar{f} : M \rightarrow M$.\\ {\bf Proof of claim}\\ \psset{unit=0.7cm} \begin{center} \begin{pspicture}(-1.0,-2.0)(8.0,4.0)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \rput(0,0){ \pspolygon[fillstyle=solid,fillcolor=lightgray](0,0)(1,0)(1,4)(0,4) \pscircle[fillstyle=solid,fillcolor=white](0.5,2.0){0.2} \pspolygon[fillstyle=solid,fillcolor=gray](1,0)(2,0)(2,4)(1,4) \psline[linecolor=red](0.7,2.0)(2,2) \psdots(2,2) \rput(2.4,2){$x$} \psline{->}(2.3,0.5)(2.0,1.0) \rput(2.5,0.3){$\small{\bd M}$} \rput(1.3,-0.8){$\small{\bd M \times I}$} \psline(1,-0.1)(1.5,-0.3)(2,-0.1) } \rput(5.0,0){ \pspolygon[fillstyle=solid,fillcolor=lightgray](0,0)(1,0)(1,4)(0,4) \pscircle[fillstyle=solid,fillcolor=white](0.5,2.0){0.2} \pspolygon[fillstyle=solid,fillcolor=gray](1,0)(2,0)(2,4)(1,4) \psline[linecolor=red](0.7,2.0)(1.0,2.0)(2,3.5) \psdots(2,3.5) \rput(2.5,3.5){$f(x)$} \psline{->}(2.3,0.5)(2.0,1.0) \rput(2.5,0.3){$\small{\bd M}$} \rput(1.3,-0.8){$\small{\bd M \times I}$}\psline(1,-0.1)(1.5,-0.3)(2,-0.1) } \psline[linestyle=dashed]{->}(2.4,-0.5)(5.0,-0.5) \rput(3.5,-1.0){$\exists \bar{f}$} \end{pspicture} \end{center} such that $\bar{f}(x,t) = (f_{t}(x),t)$on a collar neighborhood $U \cong M \times [0,1)$ and define $\bar{f} = id.$ on the complement of $U$. \end{pf} {\bf Question : } What are homeomorphisms of $T^{2}$ up to isotopy?\\ {\bf Answer : } Classical result $\Rightarrow Gl(2,\zb)$.\\ In fact, $Homeo(T^{2}) \overset{\phi}{\twoheadrightarrow} Aut(H_{1}(T^{2}))(=Gl(2,\zb))$\\ \hspace*{7.5em} $f \mapsto f_{\ast}$\\ with $ker(\phi) = Homeo_{0}(T^{2}) =$ homeomorphism homotopic(isotopic) to $id.$.\\ $\therefore \mathcal{H}(T^{2}) / \mathcal{H}_{0}(T^{2}) \cong Gl(2,\zb)$.\\ Show $\phi$ is onto ; Given $g \in Gl(2,\zb)$, view $g : \rb^{2} \to \rb^{2}$ preserving the integral lattice. $\Rightarrow \bar{g} : T^{2} \to T^{2}$ is induced.\\ 일반적으로 closed surface 에서 "homotopy $\Rightarrow$ isotopy"가 성립하며 $ker(\phi)= Homeo_{0}(T^{2})$는 다음 exercise 2(3)을 이용하여 $\tilde{f}$와 $id.$사이에 straight line homotopy를 줄 수 있고, 이것을 $T^{2}$상에서의 homotopy 로 내릴 수 있다.\\ {\bf HW 7} Prove the followings.\\ \begin{prop} \[ \xymatrix @=1em @*[c] { % \widetilde{X} & \overset{\tilde{f}}{\rightarrow} & \widetilde{X'} && \forall \gam \in \Gamma = \textrm{deck group of $\widetilde{X}$},\\ \downarrow^{p} && \downarrow^{p} & \Leftrightarrow & \exists ! \gam' \in \Gamma' \textrm{such that}\\ X& \overset{\exists f}{\dashrightarrow} & X' && \widetilde{f} \circ \gam = \gam' \circ \widetilde{f}} \] where $p$ is a regular covering. \end{prop} \begin{cor} (1) \[ \xymatrix @=1em @*[c] { % \widetilde{X} \ar[d]^{p} \ar[r]^{\widetilde{f}}_{\cong} & \widetilde{X'} \ar[d]^{p'} & \Leftrightarrow & C_{\widetilde{f}} : \Gam \overset{\cong}{\rightarrow} \Gamma'\\ X \ar@{.>}[r]^{\exists f}_{\cong} & X'&&} \] (2) 위에서 $X=X'$인 경우 좌변 $\Leftrightarrow \widetilde{f} \in N_{G}(\Gamma), $ when $G=Aut(\widetilde{X})$.(The type of the automorphisms depends on the category.)\\ (3) Universal covering case의 경우에는\\ $\widetilde{f} \in C_{G}(\Gamma)$ ( = Centralizer) $\Leftrightarrow f_{\ast} = id. : \pi_{1}(X) \rightarrow \pi_{1}(X)$ when $\widetilde{f}$ fixes a base point.\\ \end{cor} Q. Suppose ${M_i}^3$ be a 3-manifold with $\bd M_i = T^2, \hspace{1em} i= 1,2.$ \\ When is $M_1 \underset{f}{\cup} M_2 \cong M_1 \underset{g}{\cup} M_2 ,$ where $f, g: T^2 \rightarrow T^2 $ are homeomorphism?\\ A. Theorem A $\Rightarrow$ If $g^{-1}\cdot f : T^2 \rightarrow T^2$ can be extended to a homeomorphism: $M_1 \rightarrow M_1$, then $M_1 \underset{f}{\cup} M_2 \cong M_1 \underset{g}{\cup} M_2 .$\\ \begin{prop} Suppose $M_1 = D^2 \times S^1 ,$ a solid torus. Then $h : \bd M_1 = T^2 \rightarrow T^2 $ can be extended to $\bar{h} : M_1 \rightarrow M_1$ iff $h_* : H_1 (T^2 ) = \mathbb{Z}\cdot m \oplus \mathbb{Z}\cdot l \rightarrow H_1(T^2)$ is of the form $\left(\begin {array}{cc} \pm 1 & c\\ 0 & \pm 1 \end{array}\right)$ , equivalently, $h$ sends meridian to meridian.\end{prop} \begin{pf} $(\Rightarrow )$ \[ \xymatrix @M=1ex @C=3em @R=1em @*[c]{% H_1(T^2 ) = \mathbb{Z}\oplus\mathbb{Z} = \ar[r]^<<<<<<{h_* : \cong}\ar[d]^{i_*} & \mathbb{Z}\ar[d]\oplus\mathbb{Z} & m \ar@{|->}[d]\ar@{|->}[r] & am+bl\ar@{|->}[d] \\ H_1(D^2 \times S^1 ) = \mathbb{Z} = \ar[r]^<<<<<<{\bar{h_*}}& \mathbb{Z} & 0 \ar@{|->}[r] & bl } \] $\Rightarrow b=0$ and $a = \pm 1 (\because h_* : m \mapsto am :\cong )$\\ $\Rightarrow h_*= \left(\begin{array}{cc} \pm 1 & c\\ 0 & d \end{array}\right) $ and $h_*$ is invertible.\\ $\Rightarrow h_*= \left(\begin{array}{cc} \pm 1 & c\\ 0 & \pm 1 \end{array}\right)$(복호동순 아님)\\ ($\Leftarrow)$ May assume $h$ is a "linear map" through isotopy in the radial direction. And it is enough to check for $h = \left(\begin{array}{cc}1&n\\0&1\end{array}\right) , \left(\begin{array}{cc}1&0\\0&-1\end{array}\right) , $ and $\left(\begin{array}{cc}-1&0\\0&1\end{array}\right)$ , and each of these is clearly a restriction of homeomorphism of a solid torus. \end{pf} \begin{cor} $M_1 = D^2 \times S^1 , \bd M_2 = T^2$ and let $h:\bd M_1 \overset{\cong}{\rightarrow} \bd M_2 .$\\ Then the topological type of $M_1 \underset{h}{\cup} M_2$ depends only on $|h_*(m)|$, i.e., if $h_* (m) = \pm h'_* (m),$ then $M_1\underset{h}{\cup}M_2 \cong M_1\underset{h'}{\cup}M_2. $\end{cor} \begin{pf} Suppose $h, h' : \bd M_1 \overset{\cong}{\rightarrow} \bd M_2$ s.t. $h_* (m) = \pm h'_*(m).$ Then $(h^{-1}h')_* (m ) = \pm m .$ Proposition and Theorem A $\Rightarrow M_1\underset{h}{\cup}M_2 \cong M_1\underset{h'}{\cup}M_2. $\end{pf}\\ \textbf{Lens Space}\\ $L(p,q) = D^2 \times S^1 \underset{h}{\cup} D^2\times S^1 , \hspace{0.5em} h(m) = qm+pl $, where $p$ and $q$ are relatively prime.\footnote{Because $h_* $ is homeomorphism, determinant of $h_*$ is $\pm 1$. Then $qs -rp = \pm 1$, i.e., $p$ and $q$ are relatively prime.}\\ Note. $L(1,0 ) = S^3\\ \hspace*{2.7em} L(0,1) = S^2\times S^1 \\ \hspace*{2.7em} L(2,1) = \mathbb{R}P^2$\\ \textbf{숙제 8.} (1) $L(p,q) \cong L(p,-q) \cong L(-p,q) \cong L(-p,-q) \cong L(p,q+kp)\hspace{0.5em} \forall k\in \mathbb{Z}\\ \hspace*{3.7em}(2) \pi _1 (L(p,q)) = \mathbb{Z}/p$ (Use Van-Kampen Theorem.)\\ \textbf{Homology of $L(p,q)$} \[ \xymatrix @M=1ex @C=2em @R=1em @*[c]{% \cdots \ar[r]& H_q (S^1\times S^1 )\ar[r]^>>>>>{(i_*, h_* )} & H_q (D^2\times S^1 )\oplus H_q(D^2\times S^1 ) \ar[r]& H_q(L)\ar[r] & \cdots } \] $\Rightarrow H_q(L) = 0$ if $q\geq 4$. \[ \xymatrix @M=1ex @C = 0.6em @R=1em @*[c]{% 0 \ar[r]& H_3(L)\ar[r]& H_2(S^1\times S^1)\ar[r]&\ar[r] 0 &H_2(L)\ar[r]& H_1(S^1\times S^1)\ar[r] & \mathbb{Z}\oplus\mathbb{Z}\ar[r]&H_1(L)\ar[r]&\tilde{H_0}(S^1\times S^1) } \], where $h_* = \left(\begin{array}{cc} q&r\\ p&s \end{array}\right)\Rightarrow $\\ $(i_*, h_* ) = \left(\begin{array}{cc} 0&1\\p&s\end{array}\right) :H_1 (S^1\times S^1 ) = \mathbb{Z}\oplus \mathbb{Z} \rightarrow H_1 (D^2\times S^1)\oplus H_1(D^2\times S^1) = \mathbb{Z}\oplus \mathbb{Z}$\\ (1) $p \neq 0 \Rightarrow det \neq 0$ and $(i_* ,h_*)$ is injective.$\Rightarrow$\[ \xymatrix @M=1ex @C = 1em @R=1em @*[c]{% \ar[r] & H_2 (L)\ar[rr]\ar[dr] & & H_1 (S^1\times S^1)\ar[r] & \\ & & 0\ar[ur] & & } \] $\Rightarrow H_2(L) = 0 $ and $H_1 (L) = \mathbb{Z}\oplus \mathbb{Z} /im (i_* , h_*) = \mathbb{Z}/p$\\ (2) $p=0 \Rightarrow h(m) = qm \overset{h :isomorphism}{\Rightarrow} q = \pm 1$ (meridian$\mapsto$ meridian)\\ $H_2(L) = ker \left(\begin{array}{cc}0&1\\0&s \end{array}\right) = \mathbb{Z}$ and $H_1 (L) = cok \left(\begin{array}{cc} 0&1 \\0 &s\end{array}\right) =\mathbb{Z}$\\ In this case, $L = S^2\times S^1$.\\ Remark. (1) $p \neq p' \Rightarrow L(p,q) \ncong L(p',q' )\\ \hspace*{4.1em}(2)$ Fact. $L(p,q)\cong L(p,q' ) \Leftrightarrow q' = \pm q^{\pm 1} (mod\hspace{0.3em} p)\\ \hspace*{8.3em} L(p,q) \simeq L(p,q') \Leftrightarrow q q' = \pm m^2 (mod \hspace{0.3em}p)$ for some m\\ (Ref. M. Cohen. A course in simple homotopy theory)\\ \end{document}