用户: Eee/Problem set on Geometry and Topology

1留言

2Milnor and Stasheff

Problem 1-C	For any smooth manifold $M$ show that the collection $F=C^{\infty }(M,\mathbb{R})$ of smooth real valued functions on $M$ can be made into a ring, and that every point $x\in M$ determines a ring homomorphism $r_x:F\to \mathbb{R}$ and hence a maximal ideal in $F$.   If $M$ is compact, show that every maximal ideal in $F$ arises this way from a point in $M$.   More generally, if there is a countable basis for the topology of $M$, show that every ring homomorphism $F\to \mathbb{R}$ is obtained in this way. Thus the smooth manifold $M$ is completely determined by the ring $F$.
Solution.	First, the collection $F=C^{\infty }(M,\mathbb{R})$ is naturally a ring with the regular addition and multiplication. Now we consider the ring homomorphism $r_x:F\to \mathbb{R}$ more specifically. For any $f\in F$, we set $r_x(f)=f(x)\in \mathbb{R}$ and then $r_x$ becomes a ring homomorphism naturally. Then we need to show $\ker r_x$ is a maximal ideal of ring $F$.   We remark that $\ker r_x=\{f\in F\mid f(x)=0\}$, and for any $g\in F$, we have $g\cdot \ker r_x\subset \ker r_x$, which shows $\ker r_x$ is an ideal. For the maximality, for any $g\in F\backslash \ker r_x$, we show that $Spa{n}_F(\{g\}\cup \ker r_x)=F$.   Since $g\ne \ker r_x$, we see $c=g(x)\ne 0$. Consider the constant map $c(y)=c$, $\forall y\in M$, then $(g-c)(x)=g(x)-c(x)=c-c=0$ which implies $g-c\in \ker r_x$. Then $c\in Spa{n}_F(\{g\}\cup \ker r_x)$. Specially, the multiplicative identity $1=\frac{1}{c}c\in Spa{n}_F(\{g\}\cup \ker r_x)$, then we see $Spa{n}_F(\{g\}\cup \ker r_x)=F$. (Actually, we just used a algebraic property: the kernel of a ring homomorphism is a maximal ideal, if the image ring is a field.)   Then we move on to the second question, let $I$ be any maximal ideal in $F$. First we state that there is a point $x_0\in M$ such that $f(x_0)=0$ for any $f\in I$. Otherwise, for any point $x\in M$, there exists a function $f_x\in I$ s.t. $f_x(0)\ne 0$. By the smoothness of $f$, there exists a neighborhood $U_x$ of $x$ such that $f$ is nowhere $0$ in $U_x$. Then we get an open covering $\{U_x{\}}_{x\in M}$ of $M$, by the compactness of $M$, it has a sub-covering $\{U_{x_i}{\}}_{i=1}^n$ of $M$. Hence we have $\{f_{x_i}\}$ satisfying $f_{x_i}$ is nowhere $0$ in $U_{x_i}$. We set $f=f_{x_1}^2+\cdots +f_{x_n}^2$, then $f\in I$ and $f>0$ in the entire $M$. Then $\frac{1}{f}\in F$ exists, and the multiplicative identity $1=\frac{1}{f}\cdot f\in I$, then $I=F$, a contradiction! Thus there is a point $x_0\in M$ such that $f(x_0)=0$ for any $f\in I$, from the last question, we have a maximal ideal $\ker r_{x_0}$ and apparently $I\subset \ker r_{x_0}$. By the maximality of these two ideals, they are identical.   The ring homomorphism $F\to \mathbb{R}$ defines a maximal ideal, the kernel ring. Actually, the compactness of $M$ is not necessary. For a second countable space, i.e. there is a countable topology basis, any open covering has a countable sub-covering. Back to the last paragraph, the open covering $\{U_x{\}}_{x\in M}$ (additionally, we choose $U_x$ small enough s.t. $\overline{U_x}$ is compact) has a countable sub-covering $\{U_i{\}}_{i=1}^{\infty }$. Then we have a function set $\{f_i{\}}_{i=1}^{\infty }$ satisfying $f_i$ is nowhere $0$ in $U_i$.   Since $\overline{U_i}$ is compact, we consider a greater open set $V_i\supset \overline{U_i}$ satisfying $\overline{V_i}$ is still compact. Then $|f_i|$ is bounded in $V_i$ due to the compactness of $V_i$. Consider a special smooth function $g_i$ s.t. $g_i{|}_{U_i}\equiv 1$ and $g_i{|}_{M\backslash V_i}\equiv 0$. Then we define ${\tilde{f}}_i=c_i\cdot g_i\cdot f_i$, where $c_i$ is a constant small enough s.t. $|{\tilde{f}}_i|\le 1$. Thus, we could define the desired function $\tilde{f}=\sum _{n=1}^{\infty }\frac{1}{n^2}{\tilde{f}}_n^2$. We could verify that $\tilde{f}$ is convergent and positive at every point of $M$. Similar to the last question, we could show that this maximal ideal arises from a point of $M$ and the point is unique.   Thus, every ring homomorphism defines a maximal ideal, which is proved to be identical to a maximal ideal defined by a unique point.
Problem 3-E	Show that the set of isomorphism classes of $1$-dimensional vector bundles over $B$ forms an abelian group with respect to the tensor product operation. Show that a given $\mathbb{R}$-bundle $\xi$ possesses a Euclidean metric if and only if $\xi$ represents an element of order $\le 2$ in this group.
Proof.	The last problem is interesting. We assert that if the bundle $\xi$ is orientable and possesses a Euclidean metric, then $\xi$ is trivial.   If $\xi$ possesses a Euclidean metric $g：E\to \mathbb{R}$, then $g$ assigns every fiber $E_b$ two vectors $v_b^1$ and $v_b^2$ satisfying $g(v_b^i)=1$, where $b\in B$.   If $\xi$ is orientable, then we could choose a vector $v_b^i$ on every $b$, which then becomes a global section $v(b)$. The continuity of $v$ is given by the continuity of $g$. Since $\xi$ has a non-where zero global section $v$, then $\xi$ is trivial.   Note that $\xi \otimes \xi$ is always orientable, and we could endow a Euclidean metric induced by $\xi$, thus $\xi \otimes \xi$ is trivial.
Problem 3-F	Let $B$ be a Tychonoff space and let $\mathbb{R}(B)$ denote the ring of continuous real valued functions on $B$. For any vector bundle $\xi$ over $B$, let $S(\xi )$ denote the $\mathbb{R}$-module consisting of all cross-sections of $\xi$. 1. Show that $S(\xi \oplus \eta )\cong S(\xi )\oplus S(\eta )$.(This is trivial.) Show that $\xi$ is trivial if and only if $S(\eta )$ is free. (If $S(\eta )$ is free, then it possesses a basis which consisits of $n$ non-where dependent cross-sections making $\xi$ trivial.) 2. If $\xi \oplus \eta$ is trivial, show that $S(\xi )$ is a finitely generated projective module. Conversely if $Q$ is a finitely generated projective module over $\mathbb{R}(B)$, show that $Q\cong S(\xi )$ for some $\xi$. Proof   Since $S(\xi )\oplus S(\eta )=S(\xi \oplus \eta )$ which is a free module, then $S(\xi )$ is projective. If $Q$ is a finitely generated projective module, then there exists another module $P$ such that $Q\oplus P=\mathbb{R}(B{)}^n$.   Then we construct a trivial bundle $B\times \mathbb{R}(B{)}^n$ on $B$. Let $\{e_1,\cdots ,e_n\}$ be a basis of the $\mathbb{R}(B{)}^n$, and then we could denote $Q$ by $\{(f_1,\cdots ,f_n)\in \mathbb{R}(B{)}^n\}$. Then we consider the subbundle $\xi =\{(b,f_1{e}_1+\cdots f_n{e}_n)\in B\times \mathbb{R}(B{)}^n\mid (f_1,\cdots ,f_n)\in Q\}$, then we could show that this is a bundle on $B$ such that $S(\xi )\cong Q$. 3. Show that $\xi \cong \eta$ if and only if $S(\xi )=S(\eta )$.(Trivial)
5.	$\mathbb{R}{\mathbb{P}}^2$ 不能作为紧致三维流形的边界. 证明一   我们证明一个一般性结论: 若流形 $M$ 是紧致流形 $B$ 的边界, 则 $\chi (M)$ 是偶数.   将 $B$ 与 $B$ 沿边界 $M$ 粘在一起得到闭流形 $\tilde{B}$, 取 $U,V$ 分别是包含第一、二个 $B$ 的开集, 且满足 $U,V$ 同伦等价于 $B$, 及 $U\cap V$ 同伦等价于 $M$. 有 M-V 序列$$\cdots \to H_k(U\cap V)=H_k(M)\to H_k(U)\oplus H_k(V)=H_k(B)\oplus H_k(B)\to H_k(\tilde{B})\to \cdots$$故 $2\chi (B)=\chi (M)+\chi (\tilde{B})$.   若 $\dim M$ 是奇数, 则 $\chi (M)=0$. 若 $\dim M$ 是偶数, 则 $\dim \tilde{B}$ 是奇数, 故 $\chi (\tilde{B})=0$, 推出 $\chi (M)=2\chi (B)$ 是偶数.   回到此题, ${\mathbb{P}}^2=e_0\cup e_1\cup e_2$ 推出 $\chi ({\mathbb{P}}^2)=1$, 故 ${\mathbb{P}}^2$ 不能作为三维紧致流形的边界. 证明二   我们用配边理论中的 Stiefel-Whitney class 来解决.   设 $w_k(T{\mathbb{P}}^2)\in H^k({\mathbb{P}}^2,{\mathbb{Z}}_2)$ 是 $T{\mathbb{P}}^2$ 的 Stiefel-Whitney class, $a\in H_1({\mathbb{P}}^2,{\mathbb{Z}}_2)$ 是非平凡元, 有$$w(T{\mathbb{P}}^2)=(1+a{)}^3=1+3a+3a^2+a^3=1+a+a^2.$$故 $w_2(T{\mathbb{P}}^2)=a^2\ne 0$.   取 $\eta \in H_2({\mathbb{P}}^2,{\mathbb{Z}}_2)$ 是生成元, 则 Stiefel-Whitney 数 $⟨w_2(T{\mathbb{P}}^2),\eta ⟩=⟨a^2,\eta ⟩\ne 0.$   由 Pontrjagin 定理: 若紧致流形 $B$ 有边界 $M$, 则 $M$ 的所有 Stiefel-Whitney 数均为 $0$. 故 ${\mathbb{P}}^2$ 不能作为三维紧致流形的边界.

3Poincaré Hopf theorem

Definition 3.1. The index of vector field in ${\mathbb{R}}^n$.

  Consider first a vector field $X$ on an open set $U\subset {\mathbb{R}}^n$ with an isolated zero at $0\in U$. We can define a function $f_X:U\setminus \{0\}\to S^{n-1}$ by $f_X(p)=X(p)/\Vert X(p)\Vert$. If $i:S^{n-1}\to U$ is $i(p)=\epsilon p$, mapping $S^{n-1}$ into $U$, then the map $f_X\circ i:S^{n-1}\to S^{n-1}$ has a certain map degree.

  It is independent of $i$, for small $\epsilon$, since the maps $i_1,i_2:S^{n-1}\to U$ corresponding to ${\epsilon }_1$ and ${\epsilon }_2$ will be smoothly homotopic.

  The degree is called the index of $X$ at $0$.

Lemma 3.2. If $h:U\subset {\mathbb{R}}^n\to V\subset {\mathbb{R}}^n$ is a diffeomorphism with $h(0)=0$ and $X$ has an isolated zero at $0$, then the index of $h_{\ast }X$ at $0$ equals the index of $X$ at $0$.

Proof. Suppose first that $h$ is orientation preserving. Define $$H:{\mathbb{R}}^n\times [0,1]\to {\mathbb{R}}^n$$by $$H(x,t)=\{\begin{array}{ll}\frac{h(tx)}{t},& 0<t\le 1\\ d{h}_0(x),& t=0.\end{array}$$This is a smooth homotopy.

  Each map $H_t:x\mapsto H(x,t)$ is clearly a diffeomorphism, $0\le t\le 1$. Note that $H_0=d{h}_0\in G{L}^{+}(n)$, since $h$ is orientation preserving. There is also a smooth homotopy $\{H_t\}$, $-1\le t\le 0$ with each $H_t\in G{L}^{+}(n)$ and $H_{-1}=id$, since $G{L}^{+}(n)$ is connected. So, the map $h$ is smoothly homotopic to the identity, via maps which are diffeomorphisms.

  This shows that $f_{h_{\ast }X}$ is smoothly homotopic to $f_X$ on a sufficiently small region of ${\mathbb{R}}^n\setminus \{0\}$. Hence the degree of $f_{h_{\ast }X}\circ i$ is the same as the degree of $f_X\circ i$.

  To deal with non-orientation preserving $h$, it obviously suffices to check the theorem for $h(x)=(x^1\cdots ,x^{n-1},x^n)$. In this case $$f_{h_{\ast }X}=h\circ f_X\circ h^{-1},$$which shows that degree of $f_{h_{\ast }X}\circ i$ equals the degree of $f_X\circ i$.

$\square$

  As a consequence of the last lemma, we can now define the index of a vector field on a manifold.

Definition 3.3. The index of vector field on manifold.

  If $X$ is a vector field on a manifold $M$, with an isolated zero at $p\in M$, we choose a coordinate system $(\phi ,U)$ with $\phi (p)=0$, and define the index of $X$ at $p$ to be the index of ${\phi }_{\ast }X$ at $0$.

There is an amazing theorem, and I haven’t read the proof.

Theorem 3.4. A closed and connected manifold has a non vanishing vector field if and only if $\chi (M)=0$.

4Topology

1.	Show that $SO(3)$ is a smooth manfiold, and find its fundamental group.
证明:	将 $SO(3)$ 视作空间 ${\mathbb{R}}^9$ 的子集, 由 $A^{t}A=I$ 给出了 $6$ 个不同的多项式函数, 再辅以 $\det A=1$ 可证明其上点都是光滑的, 故该簇的光滑点全体即为 $SO(3)$, 进而 $SO(3)$ 是光滑流形.   我们再证明 $SO(3)$ 拓扑同胚于 $\mathbb{R}{P}^3$. $SO(3)$ 表示 ${\mathbb{R}}^3$ 中的旋转全体, 给定一个点 $p\in S^2$, 则 $p$ 确定了一个定向的旋转轴, 按该定向确定的旋转方向旋转 $\phi$, $\phi \in [0,\pi ]$. 注意到 $(p,0)=(q,0),\forall p,q\in S^2$, $(p,\pi )=(-p,\pi )$. 因此 $SO(3)=S^2\times [0,\pi ]/\sim$, 其中 $\sim$ 由前面给出.   定义映射 $f:S^2\times [0,\pi ]\to {\mathbb{R}}^3$, $(p,\phi )\to \frac{\phi }{\pi }\cdot p$. 当我们将 $(p,0)$ 和 $(q,0)$ 等同时 ($\forall p,q\in S^2$), 得到双射 $\tilde{f}:S^2\times [0,\pi ]/(p,0)\sim (q,0)$ 到三维欧式空间中单位球的双射. 再粘合 $(p,\pi )$ 和 $(-p,\pi )$, 得到 $SO(3)$ 到 $\mathbb{R}{P}^3$ 的双射, 容易验证这是同胚. 故 $SO(3)\cong \mathbb{R}{P}^3$, 进而 ${\pi }_1(SO(3))={\pi }_1(\mathbb{R}{P}^3)={\pi }_1(\mathbb{R}{P}^2)={\mathbb{Z}}_2$.
2.	The unit tangent bundle of $S^2$ is the subset $$T^1(S^2)=\{p,v\in {\mathbb{R}}^3\mid \Vert p\Vert =1,(p,v)=0,\Vert v\Vert =1\}.$$Show that $T^1(S^2)$ is a smooth submanifold of the tangent bundle $T(S^2)$ and $T^1(S^2)$ is diffeomorphic to $\mathbb{R}{P}^3$.
证明:	We only prove the second question. From the last problem, it suffices to show that $T^1(S^2)\cong SO(3)$.   Firstly we choose an orthonomal frame $e_1,e_2,e_3$ in ${\mathbb{R}}^3$ and for any point $(p,v)$ in $T^1(S^2)$, there is a unique element in $SO(3)$, rotating $e_1$ to $p$ and $e_2$ to $v$. Then we only need to show the map $T^1(S^2)\to SO(3)$ is a diffeomorhism.