如果你也在 怎样代写拓扑学topology这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。拓扑学topology在数学中，拓扑学（来自希腊语中的τόπος，”地方、位置”，和λόγος，”研究”）关注的是几何对象在连续变形下保持的属性，如拉伸、扭曲、皱缩和弯曲；也就是说，在不关闭孔、打开孔、撕裂、粘连或穿过自身的情况下。
拓扑学topology拓扑空间是一个被赋予了结构的集合，称为拓扑，它允许定义子空间的连续变形，以及更广泛地定义所有种类的连续性。欧几里得空间，以及更一般的，公制空间都是拓扑空间的例子，因为任何距离或公制都定义了一个拓扑。拓扑学中所考虑的变形是同构和同形。在这种变形下不变的属性是一种拓扑属性。拓扑属性的基本例子有：维度，它可以区分线和面；紧凑性，它可以区分线和圆；连接性，它可以区分一个圆和两个不相交的圆。
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数学代写|拓扑学作业代写topology代考|Tautological bundles
We are going to construct so-called tautological fibre bundles over the Grassmannian $\mathfrak{G}_{n, k}$.
- (The tautological vector bundle) Define
$$
\mathcal{V}\left(\mathfrak{G}{n, k}\right)=\left{(A, v) \in \mathfrak{G}{n, k} \times \mathbb{R}^{n} ; v \in V_{A}\right} ;
$$
i.e., $v$ belongs to the $k$-linear subspace $V$ of $\mathbb{R}^{n}$ such that $A=A_{V}$, via the usual bijection $G_{n, k} \cong \mathfrak{G}{n, k}$. The restriction of the projection to the first factor defines the smooth surjective map $$ \tau{n, k}: \mathcal{V}\left(\mathfrak{G}{n, k}\right) \rightarrow \mathfrak{G}{n, k} .
$$
数学代写|拓扑学作业代写topology代考|Pull-back
We introduce a fundamental construction on smooth fibred bundles. Here we state it in full generality; later we will apply it to the tautological bundles of Section 5.1.
Let us give a smooth fibre bundle
$$
\xi:=f: E \rightarrow X
$$
with fibres $E_{x}$ diffeomorphic to the manifold $F$ (recall Section 4.3.1).
Let $g \in \mathcal{E}(M, X)$. Then set
$$
\begin{gathered}
g^{} E={(p, y) \in M \times E ; g(p)=f(y)}, \ g^{}: g^{} E \rightarrow E, g^{}(p, y)=y \
g^{} f: g^{} E \rightarrow M, g^{} f(p, y)=p \end{gathered} $$ Obviously we have the commutative diagram of smooth maps, denoted by $\left[g, g^{}\right]$
$$
\begin{array}{lll}
g^{} E & \stackrel{g^{}}{\rightarrow} & E \
\downarrow g^{} f & & \downarrow f \ M & \stackrel{g}{\rightarrow} & X \end{array} . $$ Moreover, for every $p \in M, x=g(p)$, then $g^{} E_{p}:=\left(g^{} f\right)^{-1}(p)$ is equal to the fibre $E_{x}$. Hence, also every $g^{} E_{p}$ is diffeomorphic to $F$.
数学代写|拓扑学作业代写TOPOLOGY代考|Categories of vector bundles
Let $M$ be a smooth manifold (possibly with boundary). Let
$$
f: M \rightarrow \mathfrak{G}{n, k} $$ be a smooth map. Then we can consider the pull-back vector bundle $f^{} \tau{n, k}$; that is,
$$
\begin{array}{ccc}
f^{} \mathcal{V}\left(\mathfrak{G}{n, k}\right) & \stackrel{f^{}}{\rightarrow} & \mathcal{V}\left(\mathfrak{G}{n, k}\right) \
\downarrow f^{} \tau_{n, k} & & \downarrow_{n, k} \
M & \stackrel{f}{\rightarrow} & \mathfrak{G}{n, k} \end{array} . $$ By the strict definition, the total space of $\operatorname{id}{\mathfrak{G}{n, k}}^{} \tau{n, k}$ is a submanifold of $\mathfrak{G}{n, k} \times\left(\mathfrak{G}{n, k} \times \mathbb{R}^{n}\right)$; however, the projection to the product of the first and third factors gives a canonical fibred diffeomorphism to the total space of $\tau_{n, k}$. Modulo this normalized embedding, we can stipulate that
$$
\mathrm{id}{\mathrm{G}{n, k}}^{} \tau_{n, k}=\tau_{n, k} \text {. }
$$
Similarly, for every $f: M \rightarrow \mathfrak{G}{n, k}$ as above, the total space of $f^{} \tau{n, k}$ has a canonical embedding in $M \times \mathbb{R}^{n}$; modulo this normalization we can state that
$$
\mathrm{id}{M}^{}\left(f^{} \tau{n, k}\right)=f^{} \tau_{n, k} \text {. }
$$
Such a normalization is performed by default. Note also that the composition of $f^{}$ with the natural projection of $\mathcal{V}\left(\mathfrak{G}{n, k}\right)$ to $\mathbb{R}^{n}$ gives a map which is linear and injective at every fibre of $f^{}\left(\mathcal{V}\left(\mathfrak{G}{n, k}\right)\right)$, from which we can tautologically reconstruct the map $f$.
Denote $\mathcal{N}={(n, k) \in \mathbb{N} \times \mathbb{N} ; 0 \leq k \leq n}$. For every $(n, k) \in \mathcal{N}$, set
$$
\mathcal{V}{n, k}(M):=\left{f^{} \tau{n, k} ; f \in \mathcal{E}\left(M, \mathfrak{G}{n, k}\right)\right} $$ and $$ \mathcal{V}(M)=\bigcup{(n, k) \in \mathcal{N}} \mathcal{V}{n, k}(M) $$ We see immediately that $$ M \Rightarrow \mathcal{V}(M) \text {, } $$ $$ g: N \rightarrow M \Rightarrow g^{\bullet}: \mathcal{V}(M) \rightarrow \mathcal{V}(N), g^{\bullet}\left(f^{} \tau{n, k}\right)=(f \circ g)^{} \tau_{n, k} $$ so that $$ (g \circ h)^{\bullet}=h^{\bullet} \circ g^{\bullet} $$ defines a contravariant functor from the category of smooth manifolds (with boundary) to this category of smooth vector bundles. Moreover, for every $f$ and every $g$ as above there is the natural vector bundle map $$ \left[g, g^{}\right]: g^{\bullet}\left(f^{} \tau_{n, k}\right) \rightarrow f^{} \tau_{n, k} .
$$
拓扑学代写
数学代写|拓扑学作业代写TOPOLOGY代考|TAUTOLOGICAL BUNDLES
我们将在 Grassmannian 上构建所谓的重言式纤维束Gn,ķ.
- (The tautological vector bundle) Define
- $$
- \mathcal{V}\left(\mathfrak{G}{n, k}\right)=\left{(A, v) \in \mathfrak{G}{n, k} \times \mathbb{R}^{n} ; v \in V_{A}\right} ;
- $$
- i.e., $v$ belongs to the $k$-linear subspace $V$ of $\mathbb{R}^{n}$ such that $A=A_{V}$, via the usual bijection $G_{n, k} \cong \mathfrak{G}{n, k}$. The restriction of the projection to the first factor defines the smooth surjective map $$ \tau{n, k}: \mathcal{V}\left(\mathfrak{G}{n, k}\right) \rightarrow \mathfrak{G}{n, k} .
- $$
数学代写|拓扑学作业代写TOPOLOGY代考|PULL-BACK
我们介绍了光滑纤维束的基本结构。在这里，我们完全笼统地陈述它；稍后我们将把它应用到第 5.1 节的重言式丛中。
让我们给出一个光滑的纤维束
$$
\xi:=f: E \rightarrow X
$$
with fibres $E_{x}$ diffeomorphic to the manifold $F$ (recall Section 4.3.1).
Let $g \in \mathcal{E}(M, X)$. Then set
$$
\begin{gathered}
g^{} E={(p, y) \in M \times E ; g(p)=f(y)}, \ g^{}: g^{} E \rightarrow E, g^{}(p, y)=y \
g^{} f: g^{} E \rightarrow M, g^{} f(p, y)=p \end{gathered} $$ Obviously we have the commutative diagram of smooth maps, denoted by $\left[g, g^{}\right]$
$$
\begin{array}{lll}
g^{} E & \stackrel{g^{}}{\rightarrow} & E \
\downarrow g^{} f & & \downarrow f \ M & \stackrel{g}{\rightarrow} & X \end{array} . $$ Moreover, for every $p \in M, x=g(p)$, then $g^{} E_{p}:=\left(g^{} f\right)^{-1}(p)$ is equal to the fibre $E_{x}$. Hence, also every $g^{} E_{p}$ is diffeomorphic to $F$.
数学代写|拓扑学作业代写TOPOLOGY代考|CATEGORIES OF VECTOR BUNDLES
Let $M$ be a smooth manifold (possibly with boundary). Let
$$
f: M \rightarrow \mathfrak{G}{n, k} $$ be a smooth map. Then we can consider the pull-back vector bundle $f^{} \tau{n, k}$; that is,
$$
\begin{array}{ccc}
f^{} \mathcal{V}\left(\mathfrak{G}{n, k}\right) & \stackrel{f^{}}{\rightarrow} & \mathcal{V}\left(\mathfrak{G}{n, k}\right) \
\downarrow f^{} \tau_{n, k} & & \downarrow_{n, k} \
M & \stackrel{f}{\rightarrow} & \mathfrak{G}{n, k} \end{array} . $$ By the strict definition, the total space of $\operatorname{id}{\mathfrak{G}{n, k}}^{} \tau{n, k}$ is a submanifold of $\mathfrak{G}{n, k} \times\left(\mathfrak{G}{n, k} \times \mathbb{R}^{n}\right)$; however, the projection to the product of the first and third factors gives a canonical fibred diffeomorphism to the total space of $\tau_{n, k}$. Modulo this normalized embedding, we can stipulate that
$$
\mathrm{id}{\mathrm{G}{n, k}}^{} \tau_{n, k}=\tau_{n, k} \text {. }
$$
Similarly, for every $f: M \rightarrow \mathfrak{G}{n, k}$ as above, the total space of $f^{} \tau{n, k}$ has a canonical embedding in $M \times \mathbb{R}^{n}$; modulo this normalization we can state that
$$
\mathrm{id}{M}^{}\left(f^{} \tau{n, k}\right)=f^{} \tau_{n, k} \text {. }
$$
Such a normalization is performed by default. Note also that the composition of $f^{}$ with the natural projection of $\mathcal{V}\left(\mathfrak{G}{n, k}\right)$ to $\mathbb{R}^{n}$ gives a map which is linear and injective at every fibre of $f^{}\left(\mathcal{V}\left(\mathfrak{G}{n, k}\right)\right)$, from which we can tautologically reconstruct the map $f$.
Denote $\mathcal{N}={(n, k) \in \mathbb{N} \times \mathbb{N} ; 0 \leq k \leq n}$. For every $(n, k) \in \mathcal{N}$, set
$$
\mathcal{V}{n, k}(M):=\left{f^{} \tau{n, k} ; f \in \mathcal{E}\left(M, \mathfrak{G}{n, k}\right)\right} $$ and $$ \mathcal{V}(M)=\bigcup{(n, k) \in \mathcal{N}} \mathcal{V}{n, k}(M) $$ We see immediately that $$ M \Rightarrow \mathcal{V}(M) \text {, } $$ $$ g: N \rightarrow M \Rightarrow g^{\bullet}: \mathcal{V}(M) \rightarrow \mathcal{V}(N), g^{\bullet}\left(f^{} \tau{n, k}\right)=(f \circ g)^{} \tau_{n, k} $$ so that $$ (g \circ h)^{\bullet}=h^{\bullet} \circ g^{\bullet} $$ defines a contravariant functor from the category of smooth manifolds (with boundary) to this category of smooth vector bundles. Moreover, for every $f$ and every $g$ as above there is the natural vector bundle map $$ \left[g, g^{}\right]: g^{\bullet}\left(f^{} \tau_{n, k}\right) \rightarrow f^{} \tau_{n, k} .
$$
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