### 数学代写|黎曼几何代写Riemannian geometry代考|MAST90029

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## 数学代写|黎曼几何代写Riemannian geometry代考|Nonautonomous Vector Fields

Definition $2.13$ A nonautonomous vector field is family of vector fields $\left{X_{t}\right}_{t \in \mathbb{R}}$ such that the $\operatorname{map} X(t, q)=X_{t}(q)$ satisfies the following properties:
(C1) the map $t \mapsto X(t, q)$ is measurable for every fixed $q \in M$;
(C2) the map $q \mapsto X(t, q)$ is smooth for every fixed $t \in \mathbb{R}$;
(C3) for every system of coordinates defined in an open set $\Omega \subset M$ and every compact $K \subset \Omega$ and compact interval $I \subset \mathbb{R}$ there exist two functions $c(t), k(t)$ in $L^{\infty}(I)$ such that, for all $(t, x),(t, y) \in I \times K$,
$$|X(t, x)| \leq c(t), \quad|X(t, x)-X(t, y)| \leq k(t)|x-y|$$
Conditions (C1) and (C2) are equivalent to requiring that for every smooth function $a \in C^{\infty}(M)$ the scalar function $\left.(t, q) \mapsto X_{t} a\right|_{q}$ defined on $\mathbb{R} \times M$ is measurable in $t$ and smooth in $q$.

Remark $2.14$ In what follows we are mainly interested in nonautonomous vector fields of the following form:
$$X_{t}(q)=\sum_{i=1}^{m} u_{i}(t) f_{i}(q)$$
where the $u_{i}$ are $L^{\infty}$ functions and the $f_{i}$ are smooth vector fields on $M$. For this class of nonautonomous vector fields, assumptions (C1)-(C2) are trivially satisfied. Regarding $(\mathrm{C} 3)$, thanks to the smoothness of $f_{i}$, for every compact set $K \subset \Omega$ we can find two positive constants $C_{K}, L_{K}$ such that, for all $i=1, \ldots, m$, and $j=1, \ldots, n$, we have

$$\left|f_{i}(x)\right| \leq C_{K}, \quad\left|\frac{\partial f_{i}}{\partial x_{j}}(x)\right| \leq L_{K}, \quad \forall x \in K,$$
and we obtain, for all $(t, x),(t, y) \in I \times K$,
$$|X(t, x)| \leq C_{K} \sum_{i=1}^{m}\left|u_{i}(t)\right|, \quad|X(t, x)-X(t, y)| \leq L_{K} \sum_{i=1}^{m}\left|u_{i}(t)\right| \cdot|x-y| .$$
The existence and uniqueness of integral curves of a nonautonomous vector field are guaranteed by the following theorem (see [BP07]).

## 数学代写|黎曼几何代写Riemannian geometry代考|Differential of a Smooth Map

A smooth map between manifolds induces a map between the corresponding tangent spaces.

Definition $2.17$ Let $\varphi: M \rightarrow N$ be a smooth map between smooth manifolds and let $q \in M$. The differential of $\varphi$ at the point $q$ is the linear map
$$\varphi_{, q}: T_{q} M \rightarrow T_{\varphi(q)} N$$ defined as follows: $$\varphi_{, q}(v)=\left.\frac{d}{d t}\right|{t=0} \varphi(\gamma(t)) \quad \text { if } \quad v=\left.\frac{d}{d t}\right|{t=0} \gamma(t), \quad q=\gamma(0) .$$
It is easily checked that this definition depends only on the equivalence class of $\gamma$.

The differential $\varphi_{: q}$ of a smooth map $\varphi: M \rightarrow N$ (see Figure 2.1), also called its pushforward, is sometimes denoted by the symbols $D_{q} \varphi$ or $d_{q} \varphi$. Exercise 2.18 Let $\varphi: M \rightarrow N, \psi: N \rightarrow Q$ be smooth maps between manifolds. Prove that the differential of the composition $\psi \circ \varphi: M \rightarrow Q$ satisfies $(\psi \circ \varphi){}=\psi{} \circ \varphi_{}$.

As we said, a smooth map induces a transformation of tangent vectors. If we deal with diffeomorphisms, we can also obtain a pushforward for a vector field.

## 数学代写|黎曼几何代写Riemannian geometry代考|Lie Brackets

In this section we introduce a fundamental notion for sub-Riemannian geometry, the Lie bracket of two vector fields $X$ and $Y$. Geometrically it is defined as an infinitesimal version of the pushforward of the second vector field along the flow of the first. As explained below, it measures how much $Y$ is modified by the flow of $X$.

Definition 2.22 Let $X, Y \in \operatorname{Vec}(M)$. We define their Lie bracket as the vector field
$$[X, Y]:=\left.\frac{\partial}{\partial t}\right|{t=0} e{}^{-t X} Y .$$ Remark $2.23$ The geometric meaning of the Lie bracket can be understood by writing explicitly \begin{aligned} {\left.[X, Y]\right|{q} } &=\left.\left.\frac{\partial}{\partial t}\right|{t=0} e_{}^{-t X} Y\right|{q}=\left.\frac{\partial}{\partial t}\right|{t=0} e_{}^{-t X}\left(\left.Y\right|{e^{t} X}(q)\right.\ &=\left.\frac{\partial}{\partial s \partial t}\right|{t=s=0} e^{-t X} \circ e^{s Y} \circ e^{t X}(q) \end{aligned}
Proposition 2.24 As derivations on functions, one has the identity
$$\lfloor X, Y \mid=X Y-Y X$$
Proof By definition of the Lie bracket we have $[X, Y] a=\left.(\partial / \partial t)\right|{t=0}$ $\left(e{}^{-t X} Y\right) a$. Hence we need to compute the first-order term in the expansion, with respect to $t$, of the map

$t \mapsto\left(e_{}^{-t X} Y\right) a .$ Using formula (2.28), we have $$\left(e_{}^{-t X} Y\right) a=Y\left(a \circ e^{-t X}\right) \circ e^{t X} .$$
By Remark 2.9, we have $a \circ e^{-t X}=a-t X a+O\left(t^{2}\right)$, hence
\begin{aligned} \left(e_{}^{-t X} Y\right) a &=Y\left(a-t X a+O\left(t^{2}\right)\right) \circ e^{t X} \ &=\left(Y a-t Y X a+\bar{O}\left(t^{2}\right)\right) \circ e^{t X} . \end{aligned} Denoting $b=Y a-t Y X a+O\left(t^{2}\right), b_{t}=b \circ e^{t X}$, and using again the above expansion, we get \begin{aligned} \left(e_{}^{-t X} Y\right) a &=\left(Y a-t Y X a+O\left(t^{2}\right)\right)+t X\left(Y a-t Y X a+O\left(t^{2}\right)\right)+O\left(t^{2}\right) \ &=Y a+t(X Y-Y X) a+O\left(t^{2}\right) \end{aligned}
which proves that the first-order term with respect to $t$ in the expansion is $(X Y-Y X) a$.
Proposition $2.24$ shows that $(\operatorname{Vec}(M),[\cdot, \cdot])$ is a Lie algebra.

## 数学代写|黎曼几何代写Riemannian geometry代考|Nonautonomous Vector Fields

（C1）地图吨↦X(吨,q)是可测量的每个固定的q∈米;
(C2) 地图q↦X(吨,q)对于每个固定的都是平滑的吨∈R;
(C3) 对于在开放集中定义的每个坐标系统Ω⊂米和每一个契约ķ⊂Ω和紧区间我⊂R存在两个功能C(吨),ķ(吨)在大号∞(我)这样，对于所有人(吨,X),(吨,是)∈我×ķ,

|X(吨,X)|≤C(吨),|X(吨,X)−X(吨,是)|≤ķ(吨)|X−是|

X吨(q)=∑一世=1米在一世(吨)F一世(q)

|F一世(X)|≤Cķ,|∂F一世∂Xj(X)|≤大号ķ,∀X∈ķ,

|X(吨,X)|≤Cķ∑一世=1米|在一世(吨)|,|X(吨,X)−X(吨,是)|≤大号ķ∑一世=1米|在一世(吨)|⋅|X−是|.

## 数学代写|黎曼几何代写Riemannian geometry代考|Lie Brackets

[X,是]:=∂∂吨|吨=0和−吨X是.评论2.23李括号的几何意义可以通过显式书写来理解

[X,是]|q=∂∂吨|吨=0和−吨X是|q=∂∂吨|吨=0和−吨X(是|和吨X(q) =∂∂s∂吨|吨=s=0和−吨X∘和s是∘和吨X(q)

⌊X,是∣=X是−是X

(和−吨X是)一个=是(一个∘和−吨X)∘和吨X.

(和−吨X是)一个=是(一个−吨X一个+○(吨2))∘和吨X =(是一个−吨是X一个+○¯(吨2))∘和吨X.表示b=是一个−吨是X一个+○(吨2),b吨=b∘和吨X，并再次使用上述展开式，我们得到

(和−吨X是)一个=(是一个−吨是X一个+○(吨2))+吨X(是一个−吨是X一个+○(吨2))+○(吨2) =是一个+吨(X是−是X)一个+○(吨2)

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