数学代写|黎曼几何代写Riemannian geometry代考|MATH3405

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数学代写|黎曼几何代写Riemannian geometry代考|Vector Bundles

Heuristically, a smooth vector bundle on a smooth manifold $M$ is a smooth family of vector spaces parametrized by points in $M$.

Definition 2.47 Let $M$ be an $n$-dimensional manifold. A smooth vector bundle of rank $k$ over $M$ is a smooth manifold $E$ with a surjective smooth map $\pi: E \rightarrow M$ such that:
(i) the set $E_{q}:=\pi^{-1}(q)$, the $f$ iber of $E$ at $q$, is a $k$-dimensional vector space;
(ii) for every $q \in M$ there exist a neighborhood $O_{q}$ of $q$ and a linear-on-fibers diffeomorphism (called a local trivialization) $\psi: \pi^{-1}\left(O_{q}\right) \rightarrow O_{q} \times \mathbb{R}^{k}$ such that the following diagram commutes:
The space $E$ is called total space and $M$ is the base of the vector bundle. We will refer to $\pi$ as the canonical projection, and rank $E$ will denote the rank of the bundle.
Remark $2.48$ A vector bundle $E$, as a smooth manifold, has dimension
$$\operatorname{dim} E=\operatorname{dim} M+\operatorname{rank} E=n+k .$$
In the case when there exists a global trivialization map (i.e., when one can choose a local trivialization with $O_{q}=M$ for all $q \in M$ ), then $E$ is diffeomorphic to $M \times \mathbb{R}^{k}$ and we say that $E$ is trivializable.

Example 2.49 For any smooth $n$-dimensional manifold $M$, the tangent bundle $T M$, defined as the disjoint union of the tangent spaces at all points of $M$,
$$T M=\bigcup_{q \in M} T_{q} M,$$
has the natural structure of a $2 n$-dimensional smooth manifold, equipped with the vector bundle structure (of rank $n$ ) induced by the canonical projection map
$$\pi: T M \rightarrow M, \quad \pi(v)=q \quad \text { if } \quad v \in T_{q} M .$$
In the same way one can consider the cotangent bundle $T^{} M$, defined as $$T^{} M=\bigcup_{q \in M} T_{q}^{*} M .$$

数学代写|黎曼几何代写Riemannian geometry代考|Submersions and Level Sets of Smooth Maps

If $\varphi: M \rightarrow N$ is a smooth map, we define the rank of $\varphi$ at $q \in M$ to be the rank of the linear map $\varphi_{*, q}: T_{q} M \rightarrow T_{\varphi(q)} N$. It is, of course, just the rank of the matrix of partial derivatives of $\varphi$ in any coordinate chart, or the dimension

of $\operatorname{im}\left(\varphi_{*, q}\right) \subset T_{\varphi(q)} N$. If $\varphi$ has the same rank $k$ at every point, we say $\varphi$ has constant rank and write rank $\varphi=k$.

An immersion is a smooth map $\varphi: M \rightarrow N$ with the property that $\varphi_{}$ is injective at each point (or equivalently $\operatorname{rank} \varphi=\operatorname{dim} M$ ). Similarly, a submersion is a smooth map $\varphi: M \rightarrow N$ such that $\varphi_{}$ is surjective at each point (equivalently, $\operatorname{rank} \varphi=\operatorname{dim} N$ ).

Theorem $2.56$ (Rank theorem) Suppose that $M$ and $N$ are smooth manifolds of dimensions $m$ and $n$ respectively and that $\varphi: M \rightarrow N$ is a smooth map with constant rank $k$ in a neighborhood of $q \in M$. Then there exist coordinates $\left(x_{1}, \ldots, x_{m}\right)$ centered at $q$ and $\left(y_{1}, \ldots, y_{n}\right)$ centered at $\varphi(q)$ in which $\varphi$ has the following coordinate representation:
$$\varphi\left(x_{1}, \ldots, x_{m}\right)=\left(x_{1}, \ldots, x_{k}, 0, \ldots, 0\right) .$$
Remark $2.57$ The previous theorem can be rephrased in the following way.
Let $\varphi: M \rightarrow N$ be a smooth map between two smooth manifolds. Then the following are equivalent:
(i) $\varphi$ has constant rank in a neighborhood of $q \in M$;
(ii) there exist coordinates near $q \in M$ and $\varphi(q) \in N$ in which the coordinate representation of $\varphi$ is linear.

In the case of a submersion, from Theorem $2.56$ one can deduce the following result.

数学代写|黎曼几何代写Riemannian geometry代考|Basic Definitions

We start by introducing a bracket-generating family of vector fields.
Definition $3.1$ Let $M$ be a smouth manifold and let $\mathcal{F} \subset \operatorname{Vec}(M)$ be a family of smooth vector fields. The Lie algebra generated by $\mathcal{F}$ is the smallest subalgebra of $\operatorname{Vec}(M)$ containing $\mathcal{F}$, namely
$$\operatorname{Lie} \mathcal{F}:=\operatorname{span}\left{\left[X_{1}, \ldots,\left[X_{j-1}, X_{j}\right]\right], X_{i} \in \mathcal{F}, j \in \mathbb{N}\right}$$
We will say that $\mathcal{F}$ is bracket-generating (or that it satisfies the Hörmander condition) if
$$\operatorname{Lie}{q} \mathcal{F}:={X(q), X \in \text { Lie } \mathcal{F}}=T{q} M, \quad \forall q \in M$$

Moreover, for $s \in \mathbb{N}$, we define
$$\operatorname{Lie}^{s} \mathcal{F}:=\operatorname{span}\left{\left[X_{1}, \ldots,\left[X_{j-1}, X_{j}\right]\right], X_{i} \in \mathcal{F}, j \leq s\right}$$
We say that the family $\mathcal{F}$ has step s at $q$ if $s \in \mathbb{N}$ is the minimal integer satisfying
$$\operatorname{Lie}{q}^{s} \mathcal{F}:=\left{X(q) X \in \operatorname{Lie}^{s} \mathcal{F}\right}=T{q} M$$
Notice that, in general, the step $s$ may depend on the point on $M$ and that $s=s(q)$ can be unbounded on $M$ even for bracket-generating families.

数学代写|黎曼几何代写Riemannian geometry代考|Vector Bundles

(i) 集合和q:=圆周率−1(q)， 这F伊伯尔和在q, 是一个ķ维向量空间；
(ii) 对于每个q∈米有一个社区○q的q和纤维上的线性微分同胚（称为局部平凡化）ψ:圆周率−1(○q)→○q×Rķ使得下图通勤：

数学代写|黎曼几何代写Riemannian geometry代考|Submersions and Level Sets of Smooth Maps

(i)披在附近有恒定的排名q∈米;
(ii) 附近有坐标q∈米和披(q)∈ñ其中的坐标表示披是线性的。

数学代写|黎曼几何代写Riemannian geometry代考|Basic Definitions

\operatorname{Lie} \mathcal{F}:=\operatorname{span}\left{\left[X_{1}, \ldots,\left[X_{j-1}, X_{j}\right]\right ], X_{i} \in \mathcal{F}, j \in \mathbb{N}\right}\operatorname{Lie} \mathcal{F}:=\operatorname{span}\left{\left[X_{1}, \ldots,\left[X_{j-1}, X_{j}\right]\right ], X_{i} \in \mathcal{F}, j \in \mathbb{N}\right}

\operatorname{Lie}^{s} \mathcal{F}:=\operatorname{span}\left{\left[X_{1}, \ldots,\left[X_{j-1}, X_{j}\对]\right], X_{i} \in \mathcal{F}, j \leq s\right}\operatorname{Lie}^{s} \mathcal{F}:=\operatorname{span}\left{\left[X_{1}, \ldots,\left[X_{j-1}, X_{j}\对]\right], X_{i} \in \mathcal{F}, j \leq s\right}

\operatorname{Lie}{q}^{s} \mathcal{F}:=\left{X(q) X \in \operatorname{Lie}^{s} \mathcal{F}\right}=T{q } 米\operatorname{Lie}{q}^{s} \mathcal{F}:=\left{X(q) X \in \operatorname{Lie}^{s} \mathcal{F}\right}=T{q } 米

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