\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{\ssx}[1][p]{\triangle^{#1}} \newcommand{\iosh}{i_{0\sharp}} \newcommand{\ish}{i_{1\sharp}} \newcommand{\ios}{i_{0*}} \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*{VIII.4 Euler characteristic and Lefschetz fixed point theorem} \begin{defn} Let $K$ be a finite simplicial complex. The Euler characteristic of $K$ is defined by\\ \begin{center} $\chi(K) := \sum (-1)^{p} rk (C_{p}(K)) = \sum (-1)^{p} \sharp \{\textrm{p-simplices of $K$ }\}.$\\ $\textrm{And let} \bet_{p}=rk (H_{p}(K)) = \textrm{the $p$-th Betti number} = rk (H_{p}(|K|)).$ \end{center} \end{defn} 참고로 앞에서 Betti number를 정의할 때 $H_{p}(K)$는 simplicial homology이고 $H_{p}(|K|)$는 singular homology이다. 이 정의가 자연스러운 이유는 simplicial homology와 singular homology사이에 natural equivalence가 있기 때문이다.\\ {\bf 1. $\chi(K) = \sum(-1)^{p} \bet_{p}$}\\ \begin{pf} \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & C_{p+1} \ar[r] & C_{p} \ar[r] & C_{p-1} \ar[r] & \cdots \hspace{2.0em} & (\ast)} \] 우선 위의 chain complex가 존재한다는 것은 아래와 같이 두개의 short exact sequence가 존재한다는 것과 동치이다.\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & Z_{p} \ar[r]^{i} & C_{p} \ar[r]^{\bd} & B_{p-1} \ar[r] & 0 \\ 0 \ar[r] & B_{p} \ar[r]_{i} & Z_{p} \ar[r] & H_{p} \ar[r] & 0 } \] 따라서,\\ $rk(C_{p}) = rk(Z_{p}) + rk(B_{p-1}) , \hspace{0.5em} rk(Z_{p}) = rk(B_{p}) + rk(H_{p})$\\ $\Rightarrow \chi = \sum (-1)^{p}rk(C_{p}) = \sum (-1)^{p} ( rk(H_{p}) + rk(B_{p}) + rk(B_{p-1})) \\ \hspace*{1.5em}= \sum (-1)^{p} rk(H_{p})$\\ \end{pf} {\bf Structure theorem of $R$-module}\\ Assume that $R$ is a P.I.D, $F$ is a free $R$-module of rank $s$ and $N$ is a submodule of $F$. Then there exists a basis $\{f_{1}, f_{2} , \cdots, f_{s} \}$ of $F$ and $d_{1}, d_{2}, \cdots d_{s} \in R$ such that \\ (1) nonzero elements of $\{d_{1}f_{1}, \cdots d_{s}f_{s}\}$ form a basis of $N$.\\ (2) $d_{1} | d_{2} | \cdots | d_{s}$\\ \begin{pf} Reference\\ Serge Lang, Algebra , pp. 521-522.\\ Jacobson, Basic Algebra 1, pp. 179-180.\\ \end{pf} \begin{cor} Let $G$ be an abelian group. Then $G$=free part $\bigoplus$ torsion part. And define $rk (G) := rk (\textrm{free part of $G$})$\\ \end{cor} \begin{cor} $rk (F) = rk (N) + rk (F/N)$.\\ In general,$rk (M) = rk (N) + rk (M/N)$ if\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & N \ar[r] & M \ar[r] & M/N \ar[r] & 0} \] is exact.\\ \end{cor} \begin{pf} \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% & 0 \ar[d] & 0 \ar[d] & 0 \ar[d] & \\ 0 \ar[r] & K \ar[d] \ar[r]^{=} & K \ar[d] \ar[r] & 0 \ar[d] \ar[r] & 0 \\ 0 \ar[r] & p^{-1}(N) \ar[d] \ar[r] & F \ar[d]^{p} \ar[r] & \clubsuit \ar[d]^{f} \ar[r] & 0\\ 0 \ar[r] & N \ar[d] \ar[r] & M \ar[d] \ar[r] & M/N \ar[d] \ar[r] & 0\\ & 0 & 0 & 0 & } \] 위의 commutative diagram을 살펴보면 nine lemma에 의해서 $f$가 isomorphism이 되고 따라서, $rk (M) = rk (F) - rk (K) = rk (p^{-1}(N)) + rk (\clubsuit) - rk (K) = rk (N) + rk (M/N)$이 성립한다.\\ \end{pf} \begin{cor} If ($\ast$) is acyclic (i.e., exact), then $\sum(-1)^{p} rk (C_{p}) = 0$. \end{cor} {\bf Examples}\\ $\chi(S^{n}) = 1 +(-1)^{n} = \begin{cases} 0 & \textrm{n is odd} \\ 2 & \textrm{n is even} \end{cases}$\\ $\hspace*{1.5em}(\because H_{0}(S^{n}) = H_{n}(S^{n}) = \zb)$\\ $\chi(\sum_{g}) = 1 - 2g + 1 = 2 - 2g\\ \hspace*{1.5em}(\because H_{0}(\sum_g) = \zb, H_{1}(\sum_g) = \zb^{2g}, H_{2}(\sum_{g}) = \zb)$\\ \begin{pspicture}(-0.2,-0.2)(4.5,3.0)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] %1줄% \rput(0,0){ \rput(0,0.5){$\chi$} \psellipse[fillstyle=solid,fillcolor=lightgray](0.7,0.5)(0.5,0.35) \pscircle[fillstyle=solid,fillcolor=white](0.4,0.5){0.1} \pscircle[fillstyle=solid,fillcolor=white](0.7,0.5){0.1} \pscircle[fillstyle=solid,fillcolor=white](1.0,0.5){0.1} \rput(1.5,0.5){=} \rput(1.8,0.5){$\chi$} \pscircle(2.3,0.5){0.2}\pscircle(2.7,0.5){0.2}\pscircle(3.1,0.5){0.2} \rput(3.5,0.5){=} \rput(4.0,0.5){-2}} %2줄% \rput(0,1.0){ \rput(0,0.5){$\chi$} \psellipse[fillstyle=solid,fillcolor=lightgray](0.7,0.5)(0.5,0.35) \pscircle[fillstyle=solid,fillcolor=white](0.4,0.5){0.15} \pscircle[fillstyle=solid,fillcolor=white](1.0,0.5){0.15} \rput(1.5,0.5){=} \rput(1.8,0.5){$\chi$} \pscircle(2.3,0.5){0.2}\pscircle(2.7,0.5){0.2} \rput(3.5,0.5){=} \rput(4.0,0.5){-1}} %3줄% \rput(0,2.0){ \rput(0,0.5){$\chi$} \psellipse[fillstyle=solid,fillcolor=lightgray](0.7,0.5)(0.5,0.35) \pscircle[fillstyle=solid,fillcolor=white](0.7,0.5){0.2} \rput(1.5,0.5){=} \rput(1.8,0.5){$\chi$} \pscircle(2.3,0.5){0.2} \rput(3.5,0.5){=} \rput(4.0,0.5){0}} \end{pspicture} {\bf 2.}More generally, given a diagram of homomorphisms of finitely generated modules over P.I.D, \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & A \ar[d]^{\phi' = \phi|} \ar[r]^{i} & B \ar[d]^{\phi} \ar[r]^{p} & C \ar[d]^{\bar{\phi}} \ar[r] & 0 & \textrm{: exact}\\ 0 \ar[r] & A \ar[r]_{i} & B \ar[r]_{p} & C \ar[r] & 0 & } \] where $B$ is free, we have\\ \begin{center} $tr (\phi) = tr (\phi') + tr (\bar{\phi})$ \end{center} where $tr (\bar{\phi})$ is taken on free part $FC$ of $C$.\\ 특히, 여기서 $\phi$가 identity인 경우 1의 내용과 동일하다.\\ \begin{pf} Let $\{b_{1}, \cdots , b_{k}, b_{k+1}, \cdots , b_{n}\}$ be a basis for $B$ such that $\{d_{1}b_{1}, \cdots , d_{k}b_{k}\}$ is a basis for $A$. And $\phi(b_{j}) := \sum r_{ij}b_{i}$. Then\\ Matrix of $\phi = (r_{ij}) = \begin{pmatrix} \ast_{1} & \ast \\ 0 & \ast_{2} \end{pmatrix}$.\\ where $\ast_{1}$ is the matrix of $\phi|$ and $\ast_{2}$ is the matrix of $\bar{\phi}$ on $FC = <\bar{b_{k+1}}, \cdots , \bar{b_{n}}>$.\\ {\bf Note}\\ $\phi|(d_{j}b_{j}) = \phi(d_{j}b_{j}) = \sum_{i} r_{ij}d_{j}b_{i}$인데 한편, $\phi|(d_{j}b_{j}) = \sum a_{ij}d_{i}b_{i}$ 이라 두면, $a_{ij}d_{i} = r_{ij}d_{j}$가 성립한다. 따라서, $tr (\phi|) = \sum a_{ii} = \sum r_{ii}$이다.\\ Note내용에 따라 위의 증명이 완성된다. \\ \end{pf} {\bf 3.Hopf trace formula}\\ Let $C$ be a finitely generated chain complex (of $R$-module), i.e., each $C_{p}$ is finitely generated. And $\phi : C \to C$ be a chain map. Then\\ \begin{center} $\sum (-1)^{p}tr (\phi_{p},C_{p}) = \sum (-1)^{p}tr (\phi_{p*},H_{p}(C)/T_{p}(C))$ \end{center} 여기서 $T_{p}(C)$는 $H_{p}(C)$의 Torsion part이다.\\ \begin{pf} \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & C_{p+1} \ar[d]^{\phi_{p+1}} \ar[r] & C_{p} \ar[d]^{\phi_{p}} \ar[r] & C_{p-1} \ar[d]^{\phi_{p-1}} \ar[r] & \cdots \\ \cdots \ar[r] & C_{p+1} \ar[r] & C_{p} \ar[r] & C_{p-1} \ar[r] & \cdots} \] 위의 commutative diagram에서 1의 경우와 마찬가지로 아래 두개의 commutative diagram을 얻는다. \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & Z_{p} \ar[d]^{\phi_{p}^{'}} \ar[r] & C_{p} \ar[d]^{\phi_{p}} \ar[r] & B_{p-1} \ar[d]^{\phi_{p-1}^{''}} \ar[r] & 0\\ 0 \ar[r] & Z_{p} \ar[r] & C_{p} \ar[r] & B_{p-1} \ar[r] & 0 } \] \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & B_{p} \ar[d]^{\phi_{p}^{''}} \ar[r] & Z_{p} \ar[d]^{\phi_{p}^{'}} \ar[r] & H_{p} \ar[d]^{\phi_{p*}} \ar[r] & 0\\ 0 \ar[r] & B_{p} \ar[r] & Z_{p} \ar[r] & H_{p} \ar[r] & 0 } \] 그리고 각각의 diagram에서 $tr (\phi_{p}) = tr (\phi_{p}^{'}) + tr (\phi_{p-1}^{''})$ 와 $tr (\phi_{p}^{'}) = tr (\phi_{p}^{''}) + tr (\phi_{p*})$임을 알 수 있고, 따라서\\ $tr (\phi_{p}) = tr (\phi_{p*}) + tr (\phi_{p}^{''}) + tr (\phi_{p-1}^{''})\\ \Rightarrow \sum (-1)^{p}tr (\phi_{p}) = \sum (-1)^{p}tr (\phi_{p*})$가 성립한다.\\ \end{pf} {\bf Note} If $\phi$= id. $\Rightarrow tr (\phi_{p})= rk (C_{p})$ and 1 is a special case.\\ \begin{defn} Let $K$ be a finite simplicial complex and $f :|K| \to |K|$. \\ $\lam(f) := \sum (-1)^{p}tr (f_{*},H_{p}(|K|)/T_{p}(|K|)) = \textrm{Lefschetz number of $f$}$\\ \end{defn} {\bf 4.Lefschetz fixed point theorem}\\ Let $K$ be a finite simplicial complex and $f : |K| \to |K|$.\\ If $\lam(f) \neq 0$, then $f$ has a fixed point.\\ 우선 $S^{n}$의 경우를 살펴보자. $H_{0}(S^{n}) \to H_{0}(S^{n})$은 identity이고, $H_{n}(S^{n}) \to H_{n}(S^{n})$에서 $tr (f_{*})=deg f$이므로 $\lam(f)= 1+ (-1)^{n} deg f$가 되어 $\lam(f) \neq 0$이라는 것은 $deg f \neq (-1)^{n+1}$이라는 것이다. 그리고 이 경우는 이미 앞에서 다루었다.\\ \begin{pf} Assume $f$ has no fixed point and we will prove $\lam(f)=0$. Since $|K|$ is compact, there exists $\eps > 0$ such that $d(x,f(x)) > \eps, \forall x \in |K|$. May assume $mesh(K) < \eps$ by passing to a subdivision.\\ Let $g : K^{'} \to K$ be a simplicial approximation of $f$. Recall $\forall \sig \in K^{'}, g(\sig)$ and $f(\sig)$ lie in the same simplex $\tau \in K$. Then, $g(\sig) \cap \sig = \emptyset$. If not, $g(\sig) \cap \sig \neq \emptyset, g(\sig) \cap \tau \neq \emptyset , f(\sig) \subset \tau$ and $ diam(\tau) <\eps. \Rightarrow d(x,f(x)) < \eps$ for $x \in \sig \cap g(\sig)$. It is a contradiction.\\ Now consider $\phi_{p} : C_{p}(K^{'}) \overset{g_{\sharp}}{\to} C_{p}(K) \overset{sd}{\to} C_{p}(K^{'})$. Then $tr (\phi_{p})=0$ , since the coefficient of $\sig$ in $\phi_{p}(\sig)$ is 0.\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \phi_{*} : & H_{p}(K^{'}) \ar[d]^{\eta_{*}^{\cong}} \ar[r]^{g_{*}} & H_{p}(K) \ar[d]^{\eta_{*}^{\cong}} \ar[r]^{sd_{*}} & H_{p}(K^{'}) \ar[d]^{\eta_{*}^{\cong}} & \textrm{commutes}\\ & H_{p}(|K|) \ar[r]_{g_{*}=f_{*}} & H_{p}(|K|) \ar[r]_{sd_{*}=id} & H_{p}(|K^{'}|) & } \] 위의 diagram에서 오른쪽 부분의 commutativity는 다음의 diagram으로부터 당연하다. \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% C_{p}(K) \ar[r]^{sd} & C_{p}(K^{'})&\\ \triangle_{p}(K) \ar[d] ^{\theta} \ar[u]^{\mu} \ar[r]^{sd} & \triangle_{p}(K^{'}) \ar[d]^{\theta} \ar[u]^{\mu}& \textrm{commutes}\\ S_{p}(|K|) \ar[r]^{sd} & S_{p}(|K^{'}|) & } \] $\Rightarrow tr (\phi_{*}) = tr (f_{*})$ and $\lam(f) = \sum (-1)^{p} tr(f_{*}) = \sum (-1)^{p}tr (\phi_{*}) = \sum (-1)^{p}tr (\phi_{p}) = 0$ \end{pf} \begin{cor} (A generalization of Brouwer fixed point theorem)\\ Let $|K|$ be acyclic (i.e. $\widetilde{H}(|K|) = 0$) and $f:|K| \to |K|$.\\ Then, $f$ has a fixed point.\\ \end{cor} $\because H_{0}(|K|)= \zb \Rightarrow \lam(f)=1 \neq 0$\\ {\bf Example} 1. $f: S^{n} \to S^{n} \Rightarrow \lam(f)=1+(-1)^{n} deg f$.\\ 2. $f \simeq id. : |K| \to |K| \Rightarrow \lam(f) = \lam(id.) = \chi(|K|)$\\ If $\chi(|K|) \neq 0 \Rightarrow f \simeq id. $has a fixed point.\\ $S^{2n}$에 nonvanishing vector field가 있다고 가정하자. 그러면, 이 vector field를 따라서 flow를 줄 수 있고, flow를 따라서 identity와 homotopic한 map을 생각할 수 있다. 그런데 위의 예는 fixed point가 없는 이 map의 존재성이 $S^{2n}$의 Euler characteristic이 0이어야 함을 얘기하는 것이므로 모순이다. 따라서 $S^{2n}$에는 nonvanishing vector field가 존재할 수 없다.\\ 마찬가지로 compact smooth manifold $M$상에 nonvanishing vector field가 존재하면, $\chi(M) = 0$이 되어야 한다. 실제로는 역도 성립하고 Euler characteristic은 nonvanishing vector field가 존재하기 위한 abstraction으로도 파악될 수 있다.\\ \newpage {\bf 숙제 4}\\ \begin{center} \begin{pspicture}(-0.5,-0.5)(5.5,5.5)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] % Torus1 % \rput(1,1){\psellipse(0,0)(1,.7) } % genus1 % \rput(1,1){\psarc(0,0.9){1}{240}{300}\psarc(0,-0.95){1}{70}{110} } % map f% \psline{->}(2.1,1.2)(2.7,1.2) \rput(2.5,0.8){$f=\bar{F}$} % Torus2 % \rput(4,1){\psellipse(0,0)(1,.7) } % genus2 % \rput(4,1){\psarc(0,0.9){1}{240}{300}\psarc(0,-0.95){1}{70}{110} } % plane 1 % \rput(1,4){\psline{->}(0,-1.2)(0,1.2) \psline{->}(-1.2,0)(1.2,0) \pscircle(0,0){0.03}\pscircle(0.3,0){0.03}\pscircle(0.6,0){0.03}\pscircle(0.9,0){0.03} \pscircle(-0.3,0){0.03}\pscircle(-0.6,0){0.03}\pscircle(-0.9,0){0.03} \pscircle(0,0.3){0.03}\pscircle(0.3,0.3){0.03}\pscircle(0.6,0.3){0.03}\pscircle(0.9,0.3){0.03} \pscircle(-0.3,0.3){0.03}\pscircle(-0.6,0.3){0.03}\pscircle(-0.9,0.3){0.03} \pscircle(0,0.6){0.03}\pscircle(0.3,0.6){0.03}\pscircle(0.6,0.6){0.03}\pscircle(0.9,0.6){0.03} \pscircle(-0.3,0.6){0.03}\pscircle(-0.6,0.6){0.03}\pscircle(-0.9,0.6){0.03} \pscircle(0,0.9){0.03}\pscircle(0.3,0.9){0.03}\pscircle(0.6,0.9){0.03}\pscircle(0.9,0.9){0.03} \pscircle(-0.3,0.9){0.03}\pscircle(-0.6,0.9){0.03}\pscircle(-0.9,0.9){0.03} \pscircle(0,-0.3){0.03}\pscircle(0.3,-0.3){0.03}\pscircle(0.6,-0.3){0.03}\pscircle(0.9,-0.3){0.03} \pscircle(-0.3,-0.3){0.03}\pscircle(-0.6,-0.3){0.03}\pscircle(-0.9,-0.3){0.03} \pscircle(0,-0.6){0.03}\pscircle(0.3,-0.6){0.03}\pscircle(0.6,-0.6){0.03}\pscircle(0.9,-0.6){0.03} \pscircle(-0.3,-0.6){0.03}\pscircle(-0.6,-0.6){0.03}\pscircle(-0.9,-0.6){0.03} \pscircle(0,-0.9){0.03}\pscircle(0.3,-0.9){0.03}\pscircle(0.6,-0.9){0.03}\pscircle(0.9,-0.9){0.03} \pscircle(-0.3,-0.9){0.03}\pscircle(-0.6,-0.9){0.03}\pscircle(-0.9,-0.9){0.03} \rput(1.1,1.1){$\rb^{2}$}} %plane2% \rput(4,4){\psline{->}(0,-1.2)(0,1.2) \psline{->}(-1.2,0)(1.2,0) \pscircle(0,0){0.03}\pscircle(0.3,0){0.03}\pscircle(0.6,0){0.03}\pscircle(0.9,0){0.03} \pscircle(-0.3,0){0.03}\pscircle(-0.6,0){0.03}\pscircle(-0.9,0){0.03} \pscircle(0,0.3){0.03}\pscircle(0.3,0.3){0.03}\pscircle(0.6,0.3){0.03}\pscircle(0.9,0.3){0.03} \pscircle(-0.3,0.3){0.03}\pscircle(-0.6,0.3){0.03}\pscircle(-0.9,0.3){0.03} \pscircle(0,0.6){0.03}\pscircle(0.3,0.6){0.03}\pscircle(0.6,0.6){0.03}\pscircle(0.9,0.6){0.03} \pscircle(-0.3,0.6){0.03}\pscircle(-0.6,0.6){0.03}\pscircle(-0.9,0.6){0.03} \pscircle(0,0.9){0.03}\pscircle(0.3,0.9){0.03}\pscircle(0.6,0.9){0.03}\pscircle(0.9,0.9){0.03} \pscircle(-0.3,0.9){0.03}\pscircle(-0.6,0.9){0.03}\pscircle(-0.9,0.9){0.03} \pscircle(0,-0.3){0.03}\pscircle(0.3,-0.3){0.03}\pscircle(0.6,-0.3){0.03}\pscircle(0.9,-0.3){0.03} \pscircle(-0.3,-0.3){0.03}\pscircle(-0.6,-0.3){0.03}\pscircle(-0.9,-0.3){0.03} \pscircle(0,-0.6){0.03}\pscircle(0.3,-0.6){0.03}\pscircle(0.6,-0.6){0.03}\pscircle(0.9,-0.6){0.03} \pscircle(-0.3,-0.6){0.03}\pscircle(-0.6,-0.6){0.03}\pscircle(-0.9,-0.6){0.03} \pscircle(0,-0.9){0.03}\pscircle(0.3,-0.9){0.03}\pscircle(0.6,-0.9){0.03}\pscircle(0.9,-0.9){0.03} \pscircle(-0.3,-0.9){0.03}\pscircle(-0.6,-0.9){0.03}\pscircle(-0.9,-0.9){0.03} \rput(1.1,1.1){$\rb^{2}$}} %map F% \psline{->}(2.0,3.6)(3.0,3.6) \rput(2.5,3.4){$F$} %down% \psline{->}(1.0,2.6)(1.0,1.9) \psline{->}(4.0,2.6)(4.0,1.9) \end{pspicture} \end{center} 우선 $\rb^{2}$에 격자$\{(m,n)| m,n \in \zb\}$가 주어져 있다고 하자. 그리고 격자점들을 보존하는 linear map $F (\in GL(2,\zb))$를 생각하자. 이제 torus의 universal covering을 이 $\rb^{2}$라고 할 때, $\rb^{2}$사이의 map $F$는 $x+\{(m,n)\}$을 $F(x)+\{(m,n)\}$으로 보내는 map이 되므로 torus 사이의 map $f=\bar{F}$를 induce한다. 이때, $\lam(f)$는 어떻게 되는가?\\ \end{document}