\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[all]{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{\vph}{\varphi} \newcommand{\Sig}{\Sigma} \newcommand{\Del}{\Delta} \newcommand{\Lam}{\Lambda} \newcommand{\rb}{\mathbb{R}} \newcommand{\zb}{\mathbb{Z}} \newcommand{\bd}{\partial} \newcommand{\cc}{\mathcal{C}} \newcommand{\dd}{\mathcal{D}} \newcommand{\hh}{\widetilde{H}} \newcommand{\tphi}{\widetilde{\phi}} \newcommand{\hhp}{\widetilde{H}_p} \newcommand{\hhq}{\widetilde{H}_q} \newcommand{\hp}{H_p} \newcommand{\hph}{H^{p}} \newcommand{\cpc}{C^{p}(\cc;G)} \newcommand{\snn}{S^{n+1}} \newcommand{\htx}{\textrm{Hom}} \newcommand{\etx}{\textrn{Ext}} \newcommand{\dlm}{\underset{\longrightarrow}{\lim}} \newcommand{\ch}{\check{H}} \newcommand{\co}{\overset{\circ}{V}} \newcommand{\cu}{\overset{\circ}{U}} \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]{}\\} \newenvironment{pf2}{{\bf Proof}}{\hfill\framebox[2mm]{}} \begin{document} \parindent=0cm \section*{VI.2 Lefschetz Duality} \textbf{Induced orientation on the boundary}\\ \textbf{1.} Let $M$ be an $R$-orientable manifold with boundary $\bd M \neq\phi\\ x\in\bd M, W$ a coordinate neighborhood of $x\cong \mathbb{R}^n_+=\{x\in\mathbb{R}^n\hspace{0.2em}|\hspace{0.2em}x_n\geq 0\}\\ \hspace*{3em}\supset V $ a coordinate neighborhood of $x\cong D^n_+=\{x\in D^n\hspace{0.2em}|\hspace{0.2em}x_n\geq 0\}.$\\ Want to extent the orientation on $\overset{\circ}{V}$ to $\bd V$ locally first in a compatible way to obtain a global extension :\\ orientation $\in \Gamma\co \overset{\cong}{\underset{j_{\co}}{\longleftarrow}} H_n(M, M-\co)\\ \overset{Want\hspace{0.2em} "\bd"}{\longrightarrow} $ orientation $\in \Gamma(\bd V)\overset{\cong}{\underset{j_{\bd\co}}{\longleftarrow}}H_{n-1}(\bd M, \bd M-\bd V)$:\\ Note that $j_{\co}$ is $\cong$ since \[ \xymatrix @C=2em @*[c]{% \textrm{orientation}\in \Gamma \co & H_n(M, M-\co)\ar[l]^{j_{\co}} &\\H_n(\mathbb{R}^n,\mathbb{R}^n-\co) \ar[u]^{j :\cong}& H_n(\rb^n_+,\rb^n_+ - \co)\ar[u]^{excision:\cong}\ar[l]^{excision :\cong} }\] and similarly $ \Gamma(\bd V) \overset{\cong}{\underset{j_{\bd V} }{\longleftarrow}} H_{n-1}(\bd M, \bd M-\bd V)$\\ Consider the long exact sequence of the triple $(M,M-\co , M-V)$ and note that $M-V$ is a strong deformation retract of $M$. \[\scriptsize{ \xymatrix @M=1ex @C=2em @R=1em @*[c]{% & & \Gamma\co \ar@{.>}[dddl]& &\\ 0\ar@{=}[r] & H_n(M, M-V)\ar[r] & H_n(M,M-\co)\ar[r]^{\bd:\cong}\ar[u]^{\cong} & H_{n-1}(M-\co, M-\co-\bd V) \ar[r]& 0\\ & & & H_{n-1}(U,U-\bd V)\ar[u]^{excision:\cong}\ar[d]^{\bd W \textrm{is a s.d.r. of} U}\\ & R\cong\Gamma(\bd V)&\ar[l]^{\cong} H_{n-1}(\bd M, \bd M-\bd V)& H_{n-1}(\bd W, \bd W-\bd V)\ar[l]^{excision:\cong} }} \] Here $U=(W-\co)\cap "\{x\in \mathbb{R}^n | 0\leq x_n < 1\}."$\\ Note the construction of $\bd_V : \Gamma\co \rightarrow \Gamma(\bd V)$ is compatible:\\ i.e., let $x\in U\subset V$, where $x\in \bd M$ and $U$ is a half ball neighborhood of $x$ contained in $V$ , then \[\scriptsize{ \xymatrix @M=1ex @C=0.7em @R=0.7em @*[c]{% &&& &\Gamma\co\ar[ddllll]^{\bd_V:\cong}\ar[dddd]^{r}& && &\ar[llll]\ar[dddd]\ar[ld]^{\bd:\cong}H_n(M,M-\co)\\ && && & && H_{n-1}(M-\co, M-V) \ar[dddd]& \\ \Gamma(\bd V)\ar[dddd]^{r:retraction}&&& & & &H_{n-1}(\bd M, \bd M-\bd V) \ar[llllll]^{\cong}\ar[dddd]\ar[ru]^{excision:\cong}& & \\ & & \textrm{want commutitivity } & && & &\\ && \textrm{of left face}&&\Gamma\cu \ar[ddllll]^{\bd_U:\cong}& && & H_{n-1}(M, M-\cu) \ar[llll]\ar[ld]^{\bd:\cong}\\ & && & &&&H_n(M-\cu, M-U) &\\ \Gamma(\bd U)&& & & &&H_{n-1}(\bd M, \bd M-\bd U) \ar[ur]^{excision:\cong}\ar[llllll]& &} }\] $\Rightarrow $ commutes(다른 면이 모두 commute하므로)\\ $\therefore$ we have a well-defined map $\bd :\Gamma M\rightarrow \Gamma(\bd M)\\ \therefore M:R$-orientable $\Rightarrow \bd M : R$-orientable.\\ {\bf 2. Double of $M$} Consider the double of $M$, $DM=M_{1} \coprod \bd M \times [0,1] \coprod M_{2}/\sim$, where $M_{1}=M_{2}=M$ and $x \in \bd M_{1} \Rightarrow x \sim (x,0) \in \bd M \times [0,1], x \in \bd M_{2} \Rightarrow x \sim (x,1) \in \bd M \times [0,1]$. %그림 1% \psset{unit=0.7cm}{ \begin{center} \begin{pspicture}(-1,-1)(11,4)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \psellipse(4,2)(0.4,1) \pspolygon[fillstyle=solid, linecolor=white, fillcolor=white](4,1)(4,3)(3,3)(3,1) \psellipse[linestyle=dashed](4,2)(0.4,1) \psellipse(6,2)(0.4,1) \pspolygon[fillstyle=solid, linecolor=white, fillcolor=white](6,1)(6,3)(5.5,3)(5.5,1) \psellipse[linestyle=dashed](6,2)(0.4,1) \psellipse[fillstyle=vlines,hatchangle=45](5,2)(2.3,0.4) \pscurve(4,1)(2,0.7)(0,1.5)(0,3)(2,3.3)(4,3) \pscurve(6,1)(8,0.7)(10,1.5)(10,3)(8,3.3)(6,3) \psline[linewidth=0.04](4,1)(6,1) \psline[linewidth=0.04](4,3)(6,3) \rput(2.4,3.4){\Large{$M$}} \rput(7.6,3.4){\Large{$M$}} \psline{->}(2.5,0.3)(2.5,1.2) \rput(2.5,0){\Large{$M_{1}$}} \psline{->}(7.5,0.3)(7.5,1.2) \rput(7.5,0){\Large{$M_{2}$}} \psline{->}(5,0.3)(5,1.2) \rput(5,0){$\bd M \times [0,1]$} \psline{->}(5.5,3.8)(5.5,2) \rput(5.5,4){\Large{$U$}} \end{pspicture} \end{center}} {\bf Note} If $M$ is $R$-orientable, $DM$ is $R$-orientable. (unique orientation compatible with that of $M$) \begin{pf2} $U$: "coordinate neighborhood across the boundary" can be obtained as follow: Given $x \in V \subset M$, %그림 2% \psset{unit=0.6cm}{ \begin{center} \begin{pspicture}(-2,-1)(15,5.5)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \pspolygon[fillstyle=vlines,hatchangle=45](0,0)(5,0)(5,2)(0,2) \pspolygon[fillstyle=vlines,hatchangle=90](0,2)(5,2)(5,3)(0,3) \pspolygon[fillstyle=vlines,hatchangle=45](0,3)(5,3)(5,5)(0,5) \psline[linecolor=white](0,0)(5,0) \psline[linecolor=white](5,0)(5,5) \psline[linecolor=white](5,5)(0,5) \psline[linecolor=white](0,5)(0,0) \pspolygon[fillstyle=vlines,hatchangle=45](9,3)(14,3)(14,5)(9,5) \psline[linecolor=white](14,3)(14,5) \psline[linecolor=white](14,5)(9,5) \psline[linecolor=white](9,5)(14,3) \pswedge[fillstyle=solid](11.5,3){1}{0}{180} \pswedge[fillstyle=solid](2.5,2){1}{180}{360} \pswedge[fillstyle=solid](2.5,3){1}{0}{180} \psline(1.5,2)(1.5,3) \psline(3.5,2)(3.5,3) \psdots(2.5,1.5) \psdots(2.5,3.5) \psdots(11.5,3.5) \psline{->}(11.5,2.5)(5.5,1.5) \rput(9.5,1.5){$r$:reflection} \rput(2.7,1.5){$x^{'}$} \rput(2.7,3.5){$x$} \rput(11.7,3.5){$x$} \psline{->}(2.5,0)(2.5,1.2) \rput(2.5,-0.2){$V^{'}=r(V_{2})$} \psline{->}(2.5,5.2)(2.5,3.8) \rput(2.5,5.3){$V_{1}$} \psline{->}(11.5,5.2)(11.5,3.8) \rput(11.5,5.3){$V_{2}$} \rput(-0.5,1){$r(\rb^{n}_{+})$} \rput(-1.5,2.5){$\rb^{n-1} \times [0,1]$} \rput(-0.5,4){$M_{1}$} \rput(14.5,4){$M_{2}$} \end{pspicture} \end{center}} $\Rightarrow \xymatrix @M=1ex @C=1em @R=1em @*[c]{% H_{n}(M_{2},M_{2}-x)= H_{n}(V_{2},V_{2}-x) \ar[r]^-{r_{*}}_-{\times (-1)} \ar[d]_{\cong} & H_{n}(V^{'},V^{'}-x^{'}) \ar[d]_{\cong} \\ H_{n-1}(S^{n-1}) \ar[r]^{r_{*}}_{\textrm{deg}\,r=-1} & H_{n-1}(S^{n-1})}$ $\Rightarrow \xymatrix @M=1ex @C=1em @R=1em @*[c]{% H_{n}(M_{1},M_{1}-x) \ar[r]^{id.} & H_{n}(DM,DM-x) \ar[dr]^{id.} & \\ &&H_{n}(DM,DM-U) = R \\ H_{n}(M_{2},M_{2}-x) \ar[r]_{ \times (-1)} & H_{n}(DM,DM-x^{'}) \ar[ur]_{id.} & }$ $\Rightarrow (M_{1},s)\,\, \textrm{and}\,\, (M_{2},-s)$ can be spliced together to give an orientation of $DM$. \end{pf2} {\bf 3. (Fundamental orientation class for $\bd M$)} Let $M(\bd M \neq \emptyset)$ be compact, $R$-orientable. (i.e.,$\overset{\circ}{M}$ is $R$-orientable.) \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% H_{n}(M, M-\overset{\circ}{M}) \ar[r]^-{j} \ar[d]_{\cong}^{\textrm{excision}} \ar@{}[dr]|<<<{\displaystyle{\circlearrowleft}} & \Gamma \overset{\circ}{M} \\ H_{n}(DM,DM-\overset{\circ}{M}) \ar[ur]_-{j_{\overset{\circ}{M}}} \ar@{}[dr]|{\displaystyle{\circlearrowleft}} & \\ H_{n}(DM,DM-M) \ar[u]^{i_{*}}_{\cong} \ar[r]^-{j_{M}}_-{\cong} & \Gamma M \ar[uu]_{r:\textrm{restriction}}} \] Here $i_{*}$ is an isomorphism since $DM-\overset{\circ}{M}$ and $DM-M$ are homotopically equivalent to $M_{2}$, and $j_{M}$ is isomorphism since $M$ is compact. Note that $r$ is an isomorphism since an orientation on $\overset{\circ}{M}$ can be extended uniquely on $DM$.({\bf by 2.}) Hence $j$ is an isomorphism. $\therefore$ An orientation of $\overset{\circ}{M}, s \in \Gamma \overset{\circ}{M}$ determines a unique class $\zeta \in H_{n}(M,\bd M)$ via $j(\cong)$, so that $\zeta|_{x \in \overset{\circ}{M}} = s(x) \in H_{n}(\overset{\circ}{M},\overset{\circ}{M}-x)$.\\ Now consider %Diagram 그리기% \[ \xymatrix @M=1ex @C=1ex @R=3ex @*[c] { % % 윗줄 & H_{n}(M,M-\overset{\circ}{V}) \ar[rr]_{\cong} \ar[dd]_<{\cong}^<{\textrm{exc}^{-1} \circ \bd} & & \Gamma\overset{\circ}{V} \ar[dd]_{\cong}^{\bd_{V}}\\ % 가운뎃 줄1 \zeta \in H_{n}(M,\bd M) \ar[ur] \ar[rr]_<<<<<{\cong}^<<<<<{j} \ar[dd]^{\bd} & & \Gamma M \ar[ur]_{r} \ar[dd]^{\bd} & \\ % 가운뎃 줄 2 & H_{n-1}(\bd M, \bd M- \bd V) \ar[rr]^{\cong} & & \Gamma(\bd V) \\ % 아래줄 \bd \zeta = \zeta_{\bd M} \in H_{n-1}(\bd M) \ar[ur] \ar[rr]^{j}_{\cong} & & \Gamma(\bd M) \ar[ur]_{r}& } \] diagram에서 옆면과 위아래면은 commutativity는 자명하고, 뒷면의 commutativity는 1의 내용으로 부터 성립한다. 따라서 앞면도 commute하다.($r$을 작용했을 때 일치하면 section의 uniqueness에 의해 일치한다.)\\ {\bf 4. Lefschetz Duality} Let $M$ be compact, $R$-orientable manifold with $\bd M \neq \emptyset$. Consider \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & H^{q-1}(M) \ar[r] \ar[d]_{\zeta_{\cap}} \ar@{}[dr]|{\Large{(1)}} & H^{q-1}(\bd M) \ar[r]^-{\del} \ar[d]_{\cong ,\textrm{P.D} }^{\bd \zeta_{\cap}} \ar@{}[dr]|{\Large{(2)}} & H^{q}(M,\bd M) \ar[r] \ar[d]^{\zeta \cap} \ar@{}[dr]|{\Large{(3)}} & H^{q}(M) \ar[r] \ar[d]^{\zeta_{\cap}} & \cdots \\ \cdots \ar[r] & H_{n-q+1}(M,\bd M) \ar[r]_-{\bd} & H_{n-q}(\bd M) \ar[r] & H_{n-q}(M) \ar[r] & H_{n-q}(M,\bd M) \ar[r] & \cdots} \] ,where $\zeta \in H_{n}(M,\bd M)$ is the fundamental orientation class. (1) is commutative up to sign $(-1)^{q-1}$ : Note that we have the following on the chain level. $\bd(\zeta \cap a) = (-1)^{q-1}(\bd \zeta \cap a - \zeta \cap \del a). \qquad \qquad \qquad (*)$ (2) $\bd \zeta \cap a = \zeta \cap \del a : \qquad (*)\,\, \textrm{on the chain level}\, \Rightarrow !$. (3) Clear.\\ Show $\zeta_{\cap} : H^{q}(M) \rightarrow H_{n-q}(M, \bd M)$ is an isomorphism : \begin{center} $H^{q}(M) = \check{H}^{q}(M) \overset{\cong, \textrm{A.D.}}{\rightarrow} H_{n-q}(DM,DM-M) \overset{\textrm{excision, cf.3}}{\cong} H_{n-q}(M,\bd M)$ \end{center} and note $\zeta \in H_{n}(M, \bd M) = \Gamma\overset{\circ}{M}$ is a restriction of $\zeta_{M} \in H_{n}(DM,DM-M)$ and A.D. map is essentially $\zeta_{M \cap}$. Now 5-lemma $\Rightarrow \zeta_{\cap} : H^{q}(M,\bd M) \rightarrow H_{n-q}(M)$ is also an isomorphism.\\ {\bf Application}\\ {\bf 5.} Let $M$ be a compact manifold with $\bd M$. (1) $\chi(DM) = 2 \chi(M) - \chi(\bd M)$. (2) $(M : R-\textrm{orientable} \Rightarrow) \chi(\bd M)=\textrm{even}$. \begin{pf2} %그림% \psset{unit=0.5cm}{ \begin{center} \begin{pspicture}(-1,-1)(11,5)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \psellipse(4,2)(0.4,1) \pspolygon[fillstyle=solid, linecolor=white, fillcolor=white](4,1)(4,3)(3,3)(3,1) \psellipse[linestyle=dashed](4,2)(0.4,1) \psellipse(6,2)(0.4,1) \pspolygon[fillstyle=solid, linecolor=white, fillcolor=white](6,1)(6,3)(5.5,3)(5.5,1) \psellipse[linestyle=dashed](6,2)(0.4,1) \pscurve(4,1)(2,0.7)(0,1.5)(0,3)(2,3.3)(4,3) \pscurve(6,1)(8,0.7)(10,1.5)(10,3)(8,3.3)(6,3) \psline[linewidth=0.04](4,1)(6,1) \psline[linewidth=0.04](4,3)(6,3) \psline(0,0.7)(3,0.4) \psline(3,0.4)(6,0.7) \psline(4,3.5)(7,3.8) \psline(7,3.8)(10,3.5) \rput(3,0.2){\Large{$N_{1}$}} \rput(7,4.0){\Large{$N_{2}$}} \end{pspicture} \end{center}} Use MV-sequence : "$N_{1} \cap N_{2} \rightarrow N_{1} \bigoplus N_{2} \rightarrow DM$". $\Rightarrow \chi(N_{1}) + \chi(N_{2}) = \chi(N_{1} \cap N_{2}) + \chi(DM) \Rightarrow (1)$ (2) $n$: even $\Rightarrow$ dim($\bd M$) = odd $\Rightarrow \chi (\bd M) =0$. $n$ : odd $\Rightarrow \chi(DM) = 0 \Rightarrow \chi(\bd M) = 2 \chi(M).$ \end{pf2} {\bf Example} \,\,\, $\rb P^{2n} (\chi = 1),\,\, \mathbb{C}P^{2n} (\chi = 2n+1)\,\,\, \textrm{and}\,\,\, \mathbb{H}P^{2n} (\chi =2n+1)$\,\, can not be a boundary of some manifolds.\\ {\bf 6. (Signature of $M^{4k}$)} Let $M^{4k}$ be $R$-orientable, closed (and connected). Consider $I : H^{2k}(M) \bigotimes H^{2k}(M) \rightarrow R \\ \Rightarrow I$ is a symmetric bilinear pairing. Let $R = \rb \Rightarrow I$ is a non-degenerated symmetric bilinear form. Signature of $M = \sigma(M) :=$ signature of $I$.\\ {\bf Linear Algebra} Let $V^{n}$ be a vector space over $\rb$. $ b : V \bigotimes V \rightarrow \rb$, a symmetric bilinear form. Choose $e_{1} \in V$ such that $b(e_{1},e_{1}) \neq 0$. (If not, $b \equiv 0$). May assume $b(e_{1},e_{1}) = \pm 1$. Let $\eps^{1} : V \rightarrow \rb , \eps^{1}(x) = b(x,e_{1})$. Then $\eps^{1}$ is onto. $\Rightarrow V_{1} =\ker \eps^{1} = \left< e_{1} \right>^{\perp}$. Apply the same argument to $V_{1}$ to get $e_{2}$ with $b(e_{2},e_{2}) = \pm 1$.\\ %행렬 $\Rightarrow b = \begin{pmatrix} 1 &&&&&&&& \\ & \ddots &&&&&&& \\ && 1&&&\textrm{{\huge 0}}&&& \\ &&& -1 &&&&& \\ &&&& \ddots &&&& \\ &&&&& -1 &&& \\ &&&\textrm{{\huge 0}}&&& 0 && \\ &&&&&&& \ddots & \\ &&&&&&&&0 \end{pmatrix} \qquad b_{ij} = b(e_{i},e_{j})$\\ sign $b$ := $\sharp$ of positive eigenvalues of $b - \sharp$ of negative eigenvalues of $b = r-s$ and rk $b = r+s$.\\ {\bf Witt index}\,\, Let $b$ be a non-degenerated symmetric bilinear form on $V$. $\nu(b):=$ Witt index of $b$ = dim(maximal totally isotropic subspace $U$ of $V$ i.e., $\forall x \in U \Rightarrow b(x,x) = 0 ( b|_{U} \equiv 0 )$). Then $\nu(b) = \frac{1}{2}(n-|\textrm{sign} b|) = \sharp$ of (1,-1) pairs in $\begin{pmatrix} 1 &&&&& \\ & \ddots &&&\textrm{{\huge 0}}& \\ &&1&&& \\ &&& -1 && \\ & \textrm{{\huge 0}}&&& \ddots & \\ &&&&&-1 \end{pmatrix}$. \begin{pf2} Let $U$ be a maximal totally isotropic subspace and $\{x_{1}, \cdots, x_{\nu}\}$ be a basis for $U$. Let $\alp_{1} : U \rightarrow \rb$ be the dual of $x_{1}$, i.e., $\alp_{1}(x_{1}) =1, \alp_{1}(x_{i}) = 0 , i \geq 2$ and extend $\alp_{1} : V \rightarrow \rb$ trivially ($ V = U \bigoplus U^{'} \,\,\textrm{and}\,\, \alp_{1}(U^{'}) = 0$) $b$: non-degenerate $\Rightarrow \exists y_{1} \in V$ such that $\alp_{1} = b( \,\, , y_{1})$ so that $b(x_{1}, y_{1}) =1, \,\,b(x_{i},y_{1})=0, i \geq 2$ and may assume $b(y_{1},y_{1})=0$. $(b(y_{1}+ax_{1},y_{1}+ax_{1}) = b(y_{1},y_{1}) + 2ab(x_{1},y_{1}) +a^{2}b(x_{1},x_{1}) = b(y_{1},y_{1})+2a \,\,\textrm{and let}\,\, a=-\frac{1}{2}b(y_{1},y_{1}))$.\\ Let $H_{1}= \textrm{span} \{x_{1},y_{1}\}$ and consider $\theta_{1} : V \rightarrow \rb^{2}$ given by $\theta_{1}(v) = (b(v,x_{1}),b(v,y_{1}))$. Then $\theta_{1}(x_{1}) = (0,1), \,\,\theta_{1}(y_{1}) = (1,0)$ , so $\ker \theta_{1} = H_{1}^{\perp}$ and $\{x_{2}, \cdots, x_{\nu}\} \subset \ker \theta_{1}$.\\ Apply the same argument to $x_{2}$ in $H_{1}^{\perp}$, etc., finally to get $V = H_{1} \bigoplus \cdots \bigoplus H_{\nu} \bigoplus W$(orthogonal direct sum) for some $W$. Now $W$ does not have an isotropic vector by the maximality of $U$. $\Rightarrow b|_{W}$ is definite and $b|_{H_{i}} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ with respect to $\{\frac{x_{i}+y_{i}}{\sqrt{2}} ,\frac{x_{i}-y_{i}}{\sqrt{2}}\}$. \end{pf2} \begin{thm} Let $M^{4k+1}$ be compact and $\rb$-orientable. Then (1) dim $H_{2k}(\bd M)$ = 2 dim(im\,$j_{*}$) = 2dim($\ker\, j_{*}$) = 2dim(im\, $j^{*}$), where $j_{*} : H_{2k}(\bd M) \rightarrow H_{2k}(M)$. (2) $\sigma(\bd M) = 0 $. \end{thm} \begin{pf2} \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% H_{2k+1}(M,\bd M) \ar[d]_{\bd} \ar@{}[dr]|{\displaystyle{\circlearrowleft}} & H^{2k}(M) \ar@{}[dr]|{\displaystyle{\circlearrowleft}} \ar[l]^-{\textrm{L.D}}_-{\cong} \ar[r]^-{\cong} \ar[d]^{j^{*}} & \htx(H_{2k}(M),\rb) \ar[d]^{\widetilde{j_{*}}} \\ H_{2k}(\bd M) \ar[d]^{j_{*}} & H^{2k}(\bd M) \ar[l]^{\textrm{P.D}}_{\cong} \ar[r]^-{\cong} & \htx(H_{2k}(\bd M),\rb) \\ H_{2k}(M) &&} \] dim $H_{2k}(\bd M)$ = dim $H^{2k}(\bd M)$ dim(im $j^{*}$) = dim (im $\bd$) =dim ($\ker\, j_{*}$) and dim(im $j^{*}$) = dim(im $\widetilde{j}_{*}$) = dim(im $j_{*}$) $\therefore$ dim $H_{2k}(\bd M)$ = dim(im $j_{*}$) + dim ($\ker j_{*}$) = 2dim(im $j_{*}$) = 2dim(im $j^{*}$) Now note $I \equiv 0$ on im $j^{*} \subset H^{2k}(\bd M)$ since $I(j^{*}a,j^{*}b) = \left<\zeta_{\bd M}, j^{*}a \cup j^{*}b \right> = \left< \zeta_{\bd M}, j^{*}(a \cup b)\right> = \left = \left = 0$ Hence $\nu(I) = \frac{1}{2}\textrm{dim}H^{2k} \Rightarrow \sigma(\bd M) = \sigma(I) = 0$. \end{pf2} {\bf Cobordism theory Reference} Milnor, "Characteristic classes" Stong, "Cobordism theory" \end{document}