13. Connections and Curvature

13.1Connection

Let $E\to M$ be a smooth vector bundle of rank $k$. Then$$\Gamma (T^{\ast }M\otimes E)={\Omega }^1(E)$$which is adjunction $T^{\ast }M\otimes E\leftrightarrow \mathrm{Hom}(TM,E)$. If $\alpha \in {\Omega }^1(E)$, then $\alpha (X)\in \Gamma (E)$, $\forall X\in \mathfrak{X}(M)$.

定义 13.1.1. A connection $\nabla$ on $E$ is a $\mathbb{R}$-linear map $$\nabla :\Gamma (E)\to {\Omega }^1(E)$$satisfying the Leibniz rule $$\nabla (f\cdot s)=df\otimes s+f\nabla (s),\forall f\in {\mathcal{C}}^{\infty }(M),s\in \Gamma (E)$$

Properties:

(1)	$\nabla$ does not increase the support of a section, i.e. if $U\subset M$ open and $s\in \Gamma (E)$, $s{|}_U\equiv 0$, then $\nabla s{|}_U\equiv 0$. 证明. Take $p\in U$. Then there exists a smooth function $f\in {\mathcal{C}}^{\infty }(M)$, with $f(p)=1$ and $\mathrm{supp}f\subset U$. Then $f\cdot s\equiv 0$, so by $\mathbb{R}$-linearity: $$0=\nabla (f\cdot s)=df\otimes s+f\nabla (s)$$Evaluated at $p$ $$0=\underset{=0,\text{because}s(p)=0}{\underbrace{(df\otimes s)(p)}}+f(p)\nabla (s)(p)=\nabla (s)(p)$$This implies $\nabla (s)=0$ on $U$, because $p\in U$ was arbitrary. $\square$
(2)	The value of $\nabla (s)$ at any point $p\in M$ depends only on the restriction of $s$ to an arbitrarily small neighborhood of $p$. If $s,s^{\prime }\in \Gamma (E)$, s.t. $s{|}_U\equiv s^{\prime }{|}_U$ for some $U\ni p$, then $$\nabla (s){|}_U=\nabla (s^{\prime }){|}_U$$$\nabla (s)(p)$ depends only on the germ of $s$ at $p$. This is called differential operator.
(3)	If ${\nabla }^1$ and ${\nabla }^0$ are connections, so is $t{\nabla }^1+(1-t){\nabla }^0:={\nabla }^t$, $\forall t\in \mathbb{R}$. 证明. ${\nabla }^t$ is $\mathbb{R}$-linear. $$\begin{array}{rl}{\nabla }^t(fs)& =t{\nabla }^1(fs)+(1-t){\nabla }^0(fs)\\ & =t(df\otimes s+f{\nabla }^1(s))+(1-t)(df\otimes s+f{\nabla }^0(s))\\ & =df\otimes s+f{\nabla }^t(s)\end{array}$$ $\square$
(4)	If ${\nabla }^1$, ${\nabla }^0$ are connections, then ${\nabla }^1-{\nabla }^0\in {\Omega }^1(\mathrm{End}(E))$, $\mathrm{End}(E)=\mathrm{Hom}(E,E)=E^{\ast }\otimes E$. 证明. ${\nabla }^1-{\nabla }^0$ is $\mathbb{R}$-linear. The Leibniz rule gives $({\nabla }^1-{\nabla }^0)(fs)=f({\nabla }^1-{\nabla }^0)(s)$. $\Rightarrow$ $({\nabla }^1-{\nabla }^0)(s)(p)$ depends only on $s(p)$. $({\nabla }^1-{\nabla }^0{)}_p:E_p\to T_p^{\ast }M\otimes E_p$. ${\nabla }^1-{\nabla }^0\in \Gamma (E^{\ast }\otimes T^{\ast }M\otimes E)={\Omega }^1(E^{\ast }\otimes E)={\Omega }^1(\mathrm{End}(E))$. $\square$

命题 13.1.2. Every vector bundle $E$ admits connections. The space of connections is naturally an affine space whose vector space of translation is ${\Omega }^1(\mathrm{End}(E))$.

证明. Suppose $E$ has connections. Then the difference of two connections is an element in ${\Omega }^1(\mathrm{End}(E))$ by (4). Conversely, let $A\in {\Omega }^1(\mathrm{End}(E))$ and $\nabla$ a connection on $E$.

$\square$

Claim 13.1.3. $\begin{array}{rl}\nabla +A:\Gamma (E)& \to \Omega (E)\\ s& \mapsto \nabla (s)+A(s)\end{array}$ is a connection.

证明. $\nabla +A$ $\mathbb{R}$-linear is clear.
$(\nabla +A)(fs)=\nabla (fs)+A(fs)=df\otimes s+f\nabla (s)+fA(s)=df\otimes s+f(\nabla +A)(s)$.

$\square$

13.2Existence of Connections

Pick a system of local trivializations for $E$.$${\psi }_i:{\pi }^{-1}(U_i)\to U_i\times {\mathbb{R}}^k$$On $E{|}_{U_i}$, we define ${\nabla }^i$ as follows: Let $s_j\in \Gamma (E{|}_{U_i})$ be defined by $s_j(p)={\psi }_i^{-1}(p,e_j)$, where $e_1,\dots ,e_k$ is the standard basis of ${\mathbb{R}}^k$. Every section $s\in \Gamma (E{|}_{U_i})$ has the form $s=\sum _{j=1}^k{f}_j{s}_j$ for uniquely determined functions $f_j\in {\mathcal{C}}^{\infty }(U_i)$.$${\nabla }^i(s):=\sum _{j=1}^{k}d{f}_j\otimes s_j$$This is clearly $\mathbb{R}$-linear.$$\begin{array}{rl}{\nabla }^i(fs)& ={\nabla }^i\left(\sum _{j=1}^{k}f\cdot f_j{s}_j\right)\\ & =\sum _{j=1}^{k}d(f{f}_j)\otimes s_j\\ & =\sum _{j=1}^k(fd{f}_j+f_{j}df)\otimes s_j\\ & =df\otimes s+f\cdot {\nabla }^i(s)\end{array}$$so ${\nabla }^i$ is a connection on $E{|}_{U_i}$.

Let ${\rho }_i$ be a smooth partition of unity subordinate to the covering of $M$ by the $U_i$. Define $\nabla :=\sum _i{\rho }_i{\nabla }^i\cdot As$ in (3), we can show that $\nabla$ is a connection. ${\nabla }^i(s_j)=0$ by definition.

Terminology. $s_1,\dots ,s_k$ form a frame for $E{|}_{U_i}$.

If $s$ is a section, s.t. $\nabla (s)\equiv 0$ for some connection $\nabla$, then $s$ is called parallel or covariantly constant.

${\Omega }^l(E)=\Gamma ({\Lambda }^l{T}^{\ast }M\otimes )$ $l$-forms on $M$, with values in $E$.

引理 13.2.1. For every connection $\nabla$ on $E$, there is a unique $\mathbb{R}$-linear map $\overline{\nabla }:{\Omega }^l(E)\to {\Omega }^{l+1}(E)$ satisfying: $$\overline{\nabla }(\omega \otimes s)=d\omega \otimes s+(-1{)}^l\omega \wedge \nabla s,\forall \omega \in {\Omega }^l(M),s\in \Gamma (E)$$(13.1)Moreover, this $\overline{\nabla }$ satisfies $$\overline{\nabla }(f(\omega \otimes s))=(df\wedge \omega )\otimes s+f\overline{\nabla }(\omega \otimes s),f\in {\mathcal{C}}^{\infty }(M)$$

证明. Every element in ${\Omega }^l(E)$ is locally a sum of terms of the form $\omega \otimes s$. Define $\overline{\nabla }$ using a partition of unity and linearity, so $\overline{\nabla }$ is uniquely determined by (13.1). $$\begin{array}{rl}\overline{\nabla }(f(\omega \otimes s))& =\overline{\nabla }((f\omega )\otimes s)\\ & =d(f\omega )\otimes s+(-1{)}^{l}f(\omega \wedge \nabla s)\\ & =(df\wedge \omega )\otimes s+fd\omega \otimes s+f(-1{)}^l\omega \wedge \nabla s\\ & =(df\wedge \omega )\otimes s+f\nabla (\omega \otimes s)\end{array}$$

$\square$

13.3Curvature

Let $\nabla$ be a connection on $E$.

引理 13.3.1. The composition $\overline{\nabla }\circ \nabla :\Gamma (E)\to {\Omega }^2(E)$ is function-linear.

证明. $$\begin{array}{rl}(\overline{\nabla }\circ \nabla )(fs)& =\overline{\nabla }(df\otimes s+f\nabla (s))\\ & =\overline{\nabla }(df\otimes s)+\overline{\nabla }(f\nabla (s))\\ & =\underset{d^2\equiv 0}{\underbrace{dd}}f\otimes s-df\wedge \nabla s+df\wedge \nabla s+f\overline{\nabla }(\nabla s)\\ & =f(\overline{\nabla }\circ \nabla )(s)\end{array}$$

$\square$

The lemma shows that $\overline{\nabla }\nabla (s)=F^{\nabla }(s)$, where $F^{\nabla }\in {\Omega }^2(\mathrm{End}(E))$.

定义 13.3.2. $F^{\nabla }$ is called the curvature of $\nabla$.

Let $E\to M$ be smooth vector bundle with connection $\nabla$. Let $s_1,\dots ,s_k$ be frame. Then$$\nabla (s_i)=\sum _{j=1}^k{\omega }_{ij}\otimes s_j$$where ${\omega }_{ij}\in {\Omega }^1(M)$. We have $s=\sum _{i=1}^k{f}_i{s}_i$, $\nabla (s)=\sum _{i=1}^k(d{f}_i\otimes s_i+f_i\nabla (s_i))$ completely determined by $({\omega }_{ij})$. $F^{\nabla }\in {\Omega }^2(\mathrm{End}(E))$.$$F^{\nabla }(s_i)=\sum _{j=1}^k{\Omega }_{ij}\otimes s_j$$where ${\Omega }_{ij}\in {\Omega }^2(M)$.

Question: How is ${\Omega }_{ij}$ determined by ${\omega }_{ij}$?$$\begin{array}{rl}{F}^{\nabla }(s_i)& =\overline{\nabla }\circ \nabla (s_i)\\ & =\overline{\nabla }\left(\sum _j{\omega }_{ij}\otimes s_j\right)\\ & =\sum _{j=1}^k(d{\omega }_{ij}\otimes s_j-{\omega }_{ij}\wedge \nabla (s_j))\\ & =\sum _{j=1}^k\left[d{\omega }_{ij}\otimes s_j-{\omega }_{ij}\wedge \sum _{l=1}^k{\omega }_{lj}\otimes s_l\right]\\ & =\sum _{j=1}^k\left[d{\omega }_{ij}-\sum _{l=1}^k{\omega }_{il}\wedge {\omega }_{lj}\right]\otimes s_j.\end{array}$$So$${\Omega }_{ij}=d{\omega }_{ij}-\sum _{l=1}^k{\omega }_{il}\wedge {\omega }_{lj}$$This can be denoted as$$\boxed{\Omega =d\omega -\omega \wedge \omega }$$Then$$\begin{array}{rl}d{\Omega }_{ij}& =-\sum _{l}d{\omega }_{il}\wedge {\omega }_{lj}+\sum _l{\omega }_{il}\wedge d{\omega }_{lj}\\ & =-\sum _l\left[{\Omega }_{il}+\sum _m{\omega }_{im}\wedge {\omega }_{ml}\right]\wedge {\omega }_{lj}+\sum _l\left[{\omega }_{il}\wedge \left({\Omega }_{lj}+\sum _m{\omega }_{lm}\wedge {\omega }_{mj}\right)\right]\\ & =\sum _l({\omega }_{il}\wedge {\Omega }_{lj}-{\Omega }_{il}\wedge {\omega }_{lj})+\underset{=0}{\underbrace{\sum _{l,m}({\omega }_{il}\wedge {\omega }_{lm}\wedge {\omega }_{mj}-{\omega }_{im}\wedge {\omega }_{ml}\wedge {\omega }_{lj})}}\end{array}$$

13.4Bianchi Identity

$$d{\Omega }_{ij}=\sum _l\left[{\omega }_{il}\wedge {\Omega }_{lj}-{\Omega }_{il}\wedge {\omega }_{lj}\right]$$which can be denoted as$$\boxed{d\Omega =\omega \wedge \Omega -\Omega \wedge \omega }$$Let $s_1^{\prime },\dots ,s_k^{\prime }$ be another frame.$$s_i^{\prime }=\sum g_{ij}{s}_j$$where $g=(g_{ij})$ invertible. Then$$\nabla (s_i^{\prime })=\sum _{j=1}^k{\omega }_{ij}^{\prime }{s}_j^{\prime }$$where $({\omega }_{ij}{)}^{\prime }$ is the connection matrix of $\nabla$ with respect to $s_1^{\prime },\dots ,s_k^{\prime }$.$$\begin{array}{rl}\nabla (s_i^{\prime })& =\nabla \left(\sum _j{g}_{ij}{s}_j\right)\\ & =\sum _j(d{g}_{ij}\otimes s_j+g_{ij}\nabla (s_j))\\ & =\sum _j\left(d{g}_{ij}\otimes s_j+g_{ij}\sum _{l=1}^k{\omega }_{jk}\otimes s_l\right)\\ & =\sum _{l=1}^k\left(d{g}_{ij}+\sum _{j=1}^k{g}_{ij}{\omega }_{jl}\right)\otimes s_l\\ & =\sum _{l=1}^k\left(d{g}_{il}+\sum _j{g}_{ij}{\omega }_{jl}\right)\otimes \sum _{m=1}^k{g}_{lm}^{-1}{s}_m^{\prime }\\ & =\sum _{m=1}^k\left(\left(\sum _{l}d{g}_{il}+\sum _j{g}_{ij}{\omega }_{jl}\right)g_{lm}^{-1}\right)\otimes s_m^{\prime }\end{array}$$Then$${\omega }_{im}^{\prime }=\sum _l\left(d{g}_{il}+\sum _j{g}_{ij}{\omega }_{jl}\right)g_{lm}^{-1}$$This can be denoted as$$\boxed{{\omega }^{\prime }=dg{g}^{-1}+g\omega g^{-1}}$$The following terms are the same:$$\begin{array}{rl}\text{a choice of local trivialization}& \iff \text{a choice of a frame}\\ & \iff \text{a choice of gauge}\end{array}$$A change of frame is called a gauge transformation $g$.$$\begin{array}{rl}{\Omega }^{\prime }& =d{\omega }^{\prime }-{\omega }^{\prime }\wedge {\omega }^{\prime }\\ & =d(dg{g}^{-1}+g\omega g^{-1})-(dg{g}^{-1}+g\omega g^{-1})\wedge (dg{g}^{-1}+g\omega g^{-1})\\ & =d^{2}g{g}^{-1}-dg\wedge d{g}^{-1}+dg\wedge \omega g^{-1}+gd\omega g^{-1}-g\omega \wedge d{g}^{-1}\\ & -dg{g}^{-1}\wedge dg{g}^{-1}-g\omega g^{-1}\wedge dg{g}^{-1}-dg{g}^{-1}\wedge g\omega g^{-1}-g\omega g^{-1}\wedge g\omega g^{-1}\\ & =g\Omega g^{-1}\end{array}$$Compare the following two equation:$$\begin{array}{r}{\omega }^{\prime }=dg{g}^{-1}+g\omega g^{-1}\\ {\Omega }^{\prime }=g\Omega g^{-1}\end{array}$$The first one shows that connection is not a “tensor”, while the second one shows that curvature is a “tensor”.

Let $\nabla :\Gamma (E)\to \Omega (E)$.

定义 13.4.1. ${\nabla }_{X}s=⟨\nabla s,X⟩\in \Gamma (E)$ for every $X\in \mathfrak{X}(M)$ is the covariant derivative of $s$ (in the direction of $X$).

Let $(U,\phi )$ be a chart for $M$.$$\begin{array}{rl}{x}_1,\dots ,x_n:U& \to \mathbb{R}\\ p& \mapsto {\phi }_i(p)\end{array}$$where $d{x}_1,\dots ,d{x}_n$ are the dual frame to ${\mathcal{O}}_1,\dots ,{\mathcal{O}}_n$.

Let $s_1,\dots ,s_k$ be a frame for $E{|}_U$. $\nabla$ is represented by $\omega =({\omega }_{ij})$ with respect to $s_1,\dots ,s_k$. Then ${\omega }_{ij}=\sum _{\alpha }^n{\omega }_{ij}^{\alpha }d{x}_{\alpha }$ for unique ${\omega }_{ij}^{\alpha }\in {\mathcal{C}}^{\infty }(U)$, where ${\omega }_{ij}^{\alpha }=⟨{\omega }_{ij},{\partial }_{\alpha }⟩$.$$\begin{array}{rl}{\nabla }_{{\partial }_{\alpha }}{s}_i& =⟨\nabla s_i,{\partial }_{\alpha }⟩=⟨\sum _j{\omega }_{ij}\otimes s_j,{\partial }_{\alpha }⟩=\sum _j⟨{\omega }_{ij},{\partial }_{\alpha }⟩\otimes s_j=\sum _j{\omega }_{ij}^{\alpha }{s}_j\\ s& =\sum _{i=1}^k{f}_i{s}_i\\ {\nabla }_{{\partial }_{\alpha }}s& =⟨\nabla s,{\partial }_{\alpha }⟩=⟨\sum _{i=1}^{k}d{f}_i\otimes s_i+f_i\nabla s_i,{\partial }_{\alpha }⟩=\sum _{j=1}^k{\partial }_{\alpha }{f}_j{s}_j+\sum _{i,j}{f}_i{\omega }_{ij}^{\alpha }{s}_j\\ & =\sum _{j=1}^k\left({\partial }_{\alpha }{f}_j+\sum _{i=1}^k{f}_i{\omega }_{ij}^{\alpha }\right)s_j\end{array}$$$A^{\alpha }:=({\omega }_{ij}^{\alpha })$ is $k\times k$ matrix of ${\mathcal{C}}^{\infty }$-functions on $U$.$${\nabla }_{{\partial }_{\alpha }}s={\partial }_{\alpha }s+A^{\alpha }(s)$$We define$${\nabla }_{\alpha }:={\partial }_{\alpha }+A^{\alpha }$$

13.5Parallel Transport

Let $M\subset {\mathbb{R}}^n$ open. Let $E\stackrel{\pi }{\to }M$ smooth vector bundle of rank $k$, with a connection $\nabla$. Let $y_1,\dots ,y_n$ be linear coordinates on ${\mathbb{R}}^n$. Let $c:[0,1]\to M$ be smooth curve, then write it is in terms of coordinates$$\begin{array}{rl}c(t)& =(y_1(t),\dots ,y_n(t))\\ \dot{c}(t)& =\sum _{\alpha =1}^n\frac{d{y}_{\alpha }}{dt}\frac{\partial }{\partial y_{\alpha }}=\sum _{\alpha =1}^n{\dot{y}}_{\alpha }{\partial }_{\alpha }\end{array}$$Assume $E$ is trivial. $E\cong M\times {\mathbb{R}}^k$. Let $s_1,\dots ,s_k$ be the frame with $s_i(p)=(p,e_i)$. $s\in \Gamma (E)$ is written uniquely as $s=\sum _{i=1}^k{x}_i{s}_i$.$$\begin{array}{rl}{\nabla }_{\dot{c}}(s)& =⟨\nabla s,\dot{c}⟩\\ & =⟨\sum _{i=1}^{k}d{x}_i\otimes s_i+x_i\nabla s_i,\sum _{\alpha =1}^n{\dot{y}}_{\alpha }{\partial }_{\alpha }⟩\\ & =\sum _{i,\alpha }⟨\sum _{j=1}^n\frac{\partial x_i}{\partial y_j}d{y}_j\otimes s_i+x_i\sum _{j=1}^k{\omega }_{ij}\otimes s_j,{\dot{y}}_{\alpha }{\partial }_{\alpha }⟩\\ & =\sum _{i,\alpha }{\dot{y}}_{\alpha }\frac{\partial x_i}{\partial y_{\alpha }}{s}_i+x_i\sum _{j=1}^n{\omega }_{ij}^{\alpha }{\dot{y}}_{\alpha }{s}_j\\ & =\sum _{j,\alpha }{\dot{y}}_{\alpha }\frac{\partial x_j}{\partial y_{\alpha }}{s}_j+\sum _{i,j,\alpha }{x}_i{\omega }_{ij}^{\alpha }{\dot{y}}_{\alpha }{s}_j\\ & =\sum _{j=1}^k\left(\sum _{\alpha =1}^n\left(\frac{\partial x_j}{\partial y_{\alpha }}{\dot{y}}_{\alpha }+\sum _{i=1}^k{x}_i{\omega }_{ij}^{\alpha }\right){\dot{y}}_{\alpha }\right)s_j\end{array}$$

命题 13.5.1. Let $E\to M$ be any smooth vector bundle, with a connection $\nabla$. Let $c:[0,1]\to M$ be a smooth curve and $v\in E_{c(0)}$. Then there exists a unique lift $\tilde{c}:[0,1]\to E$ with $\pi \circ \tilde{c}=c$, with $\tilde{c}(0)=v$ and ${\nabla }_{\dot{c}}s\equiv 0$ if $s$ is a section of $E{|}_{\mathrm{im}c}$ given by $\tilde{c}$.

证明. By compactness of $[0,1]$, we can choose a finite subdivision $t_0=0<t_1<\cdots <t_l=1$, s.t. $c{|}_{[t_i,t_{i+1}]}$ has image in an open set in $M$, which is the domain of a chart and over which $E$ is trivial.

Without loss of generality, we only need to prove the proposition for $c$ with image in a chart where $E$ is trivial.

We write $c(t)=(y_1(t),\dots ,y_n(t))$ and use the frame $s_1,\dots ,s_k$ given by the trivialization.

$\tilde{c}(t)={\sum }_{i=1}^k{x}_i(t)s_i(c(t))$ with $v=\tilde{c}(0)=\sum _{i=1}^k{x}_i(0)s_i(c(0))$.

The equation ${\nabla }_{\dot{c}}s\equiv 0$ is equivalent to $$\sum _{\alpha =1}^n\left(\frac{\partial x_j}{\partial y_{\alpha }}+\sum _{i=1}^k{x}_i{\omega }_{ij}^{\alpha }\right){\dot{y}}_{\alpha }\equiv 0,\forall j\in \{1,\dots ,k\}$$$${\dot{x}}_j+\sum _{i,\alpha }{x}_i{\omega }_{ij}^{\alpha }\equiv 0$$This is a linear system of ODE for the function $x_1,\dots ,x_k$.

For every initial condition $x_1(0),\dots ,x_k(0)$, there is a unique smooth solution, which, moreover, depends linearly on the initial condition.

$\square$

推论 13.5.2. Let $E\stackrel{\pi }{\to }M$ and $\nabla$ be as in the proposition. Then every smooth curve $c:[0,1]\to M$ defines a unique linear map $$\begin{array}{rl}{E}_{c(0)}& \to E_{c(1)}\\ \tilde{c}(0)=v& \mapsto \tilde{c}(1)\end{array}$$This linear map is an isomorphism

Let $E\stackrel{\pi }{\to }M$ be any smooth vector bundle over $M$. $p,q\in M$, $E_p$, $E_q$ are $k$-dimensional vector spaces.

(1)	If $p,q\in U\subset M$ and $\psi :E{|}_U\to U\times {\mathbb{R}}^k$ trivialization, then $\psi$ identifies $E_p$ with $\{p\}\times {\mathbb{R}}^k$ and $E_q$ with $\{q\}\times {\mathbb{R}}^k$, so those are identified using $\psi$.
(2)	If a connection $\nabla$ on $E$ is given and there exists a smooth path $c(0)=p$, $c(1)=q$, then $P_c=$parallel transport along $c$ defines an isomorphism between $E_p$ and $E_q$. $P_c$ depends not just on $\nabla$ but also on $c$.

In a trivialization, every $\nabla$ is given by ${\nabla }_{\alpha }={\partial }_{\alpha }+A^{\alpha }$.$${\omega }_{ij}=\sum _{\alpha }{\omega }_{ij}^{\alpha }d{y}_{\alpha },A^{\alpha }={\omega }_{ij}^{\alpha }$$

命题 13.5.3. Let $E\stackrel{\pi }{\to }M$ be a smooth vector bundle of rank $k$ and $s_1,\dots ,s_k$ be frame. Let $\nabla$ be a connection on $E$, ${\omega }_{ij}$ its connection matrix with respect to $s_1,\dots ,s_k$ and ${\Omega }_{ij}$ its curvature matrix. If we pick a chart with coordinate functions $y_1,\dots ,y_n$, then $$[{\nabla }_{\alpha },{\nabla }_{\beta }]s_i=\sum _{j=1}^k{\Omega }_{ij}({\partial }_{\alpha },{\partial }_{\beta })s_j$$

推论 13.5.4. $F^{\nabla }\equiv 0$ $\iff$ ${\Omega }_{ij}$ for every local frame $\iff$ $[{\nabla }_{\alpha },{\nabla }_{\beta }]=0$.

证明. $$\begin{array}{rl}{\Omega }_{ij}({\partial }_{\alpha },{\partial }_{\beta })& =\left(d{\omega }_{ij}-\sum _{l=1}^k{\omega }_{il}\wedge {\omega }_{lj}\right)({\partial }_{\alpha },{\partial }_{\beta })\\ & =L_{{\partial }_{\alpha }}({\omega }_{ij}^{\beta })-L_{{\partial }_{\beta }}({\omega }_{ij}^{\alpha })-\sum _{l=1}^k{\omega }_{il}^{\alpha }{\omega }_{lj}^{\beta }-{\omega }_{il}^{\beta }{\omega }_{lj}^{\alpha }\\ & ={\partial }_{\alpha }{\omega }_{ij}^{\beta }-{\partial }_{\beta }{\omega }_{ij}^{\alpha }-\sum _{l=1}^k({\omega }_{il}^{\alpha }{\omega }_{lj}^{\beta }-{\omega }_{il}^{\beta }{\omega }_{lj}^{\alpha })\end{array}$$$$\begin{array}{rl}{\nabla }_{\alpha }{\nabla }_{\beta }{s}_i& ={\nabla }_{\alpha }(⟨\nabla s_i,{\partial }_{\beta }⟩)={\nabla }_{\alpha }\left(⟨\sum _{j=1}^k{\omega }_{ij}\otimes s_j,{\partial }_{\beta }⟩\right)\\ & ={\nabla }_{\alpha }\left(\sum _{j=1}^k{\omega }_{ij}^{\beta }{s}_j\right)=⟨\nabla \left(\sum _{j=1}^k{\omega }_{ij}{s}_j\right),{\partial }_{\alpha }⟩\\ & =\sum _{j=1}^k⟨d{\omega }_{ij}^{\beta }\otimes s_j+{\omega }_{ij}^{\beta }\nabla s_j,{\partial }_{\alpha }⟩\\ & =\sum _{j=1}^k({\partial }_{\alpha }{\omega }_{ij}^{\beta })s_j+\sum _{j=1}^k{\omega }_{ij}^{\beta }\sum _{l=1}^k{\omega }_{jl}^{\alpha }{s}_l\\ & =\sum _{j=1}^k\left({\partial }_{\alpha }{\omega }_{ij}^{\beta }+\sum _{l=1}^k{\omega }_{il}^{\beta }{\omega }_{lj}^{\alpha }\right)s_j\end{array}$$$$\begin{array}{rlrlrl}\Rightarrow & & & ({\nabla }_{\alpha }{\nabla }_{\beta }-{\nabla }_{\beta }{\nabla }_{\alpha })s_i& & \\ & & & =\sum _{j=1}^k\left({\partial }_{\alpha }{\omega }_{ij}^{\beta }+\sum _{l=1}^k{\omega }_{il}^{\beta }{\omega }_{lj}^{\alpha }-{\partial }_{\beta }{\omega }_{ij}^{\alpha }-\sum _{l=1}^k{\omega }_{il}^{\alpha }{\omega }_{lj}^{\beta }\right)s_j& & \\ & & & =\sum _{j=1}^k\left({\partial }_{\alpha }{\omega }_{ij}^{\beta }-{\partial }_{\beta }{\omega }_{ij}^{\alpha }-\sum _{l=1}^k\left({\omega }_{il}^{\alpha }{\omega }_{lj}^{\beta }-{\omega }_{il}^{\beta }{\omega }_{lj}^{\alpha }\right)\right)s_j& & \\ & & & =\sum _{j=1}^k{\Omega }_{ij}({\partial }_{\alpha },{\partial }_{\beta })s_j& & \end{array}$$

$\square$

Over a curve, every $\nabla$ admits local trivializations by parallel sections, i.e. a parallel frame.

定理 13.5.5. $E$ admits a system of local trivializations by $\nabla$-parallel frames if and only if $F^{\nabla }\equiv =0$.

定义 13.5.6. $s\in \Gamma (E)$ is $\nabla$-parallel if $\nabla s\equiv 0$. A frame $s_1,\dots ,s_k$ is $\nabla$-parallel if $\nabla s_i\equiv 0$, $\forall i$.

证明. Suppose $s_1,\dots ,s_k$ is a $\nabla$-parallel local frame. Then $0=\nabla s_i=\sum _j{\omega }_{ij}\otimes s_j$ $\Rightarrow$ ${\omega }_{ij}=0$, $\forall i,j$. $\Rightarrow$ ${\Omega }_{ij}=d{\omega }_{ij}-\sum _l{\omega }_{il}\wedge {\omega }_{lj}=0$, so $F^{\nabla }\equiv 0$.

Conversely, if $F^{\nabla }\equiv 0$, we want to find local $\nabla$-parallel frames. Since the statement is local, we work on $M={\mathbb{R}}^n$.

For $n=1$, we find a $\nabla$-parallel frame over $\mathbb{R}$ by parallel transport.

For $n>1$, we prove the statement by induction.

Let $p⩾1$ and assume we have a $\nabla$-parallel frame over ${\mathbb{R}}^p\times \{0\}\subset {\mathbb{R}}^{p+1}$. By construction, $E$ is trivial on ${\mathbb{R}}^n$, so we may pick an arbitrary frame for $E$. We need to find a gauge transformation $g$, so that $s_i^{\prime }=\sum _j{g}_{ij}{s}_j$ gives a $\nabla$-parallel frame $s_1^{\prime },\dots ,s_k^{\prime }$. We want to solve$$0={\omega }_{ij}^{\prime }=(dg\cdot g^{-1}+g\omega g^{-1}{)}_{ij}$$$$\begin{array}{rlrlr}\iff & & {\omega }_{ij}^{\prime \alpha }=0,\forall \alpha & & \\ \iff & & (dg+g\omega {)}_{ij}^{\alpha }=0,\forall \alpha & & \\ \iff & & {\partial }_{\alpha }{g}_{ij}+\sum _l{g}_{il}{\omega }_{lj}^{\alpha }=0,\forall \alpha & & \end{array}\text{\hspace{1em}(*)}$$For the inductive step, we assume, we have a $g$ s.t. (*) holds for $\alpha ⩽p$.

In the inductive step, we assume the statement has been proved for ${\mathbb{R}}^p$. This means ${\omega }_{ij}^{\alpha }=0$ for $\alpha ⩽p$.

To obtain the statement over ${\mathbb{R}}^{p+1}$, we need to solve$${\partial }_{\alpha }{g}_{ij}=0\text{for}\alpha ⩽p\text{and}{\partial }_{p+1}{g}_{ij}+\sum _l{g}_{il}{\omega }_{lj}^{p+1}=0\forall i,j\text{\hspace{1em}(**)}$$Fix all $y_{\beta }$ except $y_{p+1}$. We treat the second equation in (**) as an ODE in $y_{p+1}$. With initial condition $g(0)=\mathbb{I}$, this ODE has a unique solution.

Varying the starting point (the $y$-coordinates other than $y_{p+1}$), the solutions of the ODE vary smoothly.

The assumption that $F^{\nabla }\equiv 0$, means $[{\nabla }_{\alpha },{\nabla }_{\beta }]=0$, $\forall \alpha ,\beta$. Take $\alpha ⩽p$, $\beta =p+1$. Then$${\partial }_{\alpha }{\omega }_{il}^{p+1}-{\partial }_{p+1}\cancel{{\omega }_{il}^{\alpha }}+\sum _j({\omega }_{ij}^{p+1}\cancel{{\omega }_{jl}^{\alpha }}-\cancel{{\omega }_{ij}^{\alpha }}{\omega }_{jl}^{p+1})=0$$$$\begin{array}{rlrlr}\Rightarrow & & {\partial }_{\alpha }{\omega }_{il}^{p+1}=0& & \\ \Rightarrow & & {\omega }_{ij}^{\prime p+1}=(dg\cdot g^{-1}+g\omega g^{-1}{)}_{ij}^{p+1}=0& & \end{array}$$because $g$ solves the second equation in (**).

$\square$

推论 13.5.7. A vector bundle $E\stackrel{\pi }{\to }M$ admits a flat connection $\nabla$ if and only if it admits a system of trivializations with constant transition maps.

证明. If $E$ admits $\nabla$ with $F^{\nabla }\equiv 0$, then we can find local trivialization given by $\nabla$-parallel frames.

$$\begin{array}{rl}{\psi }_U:{\pi }^{-1}(U)& \to U\times {\mathbb{R}}^k\\ v=\sum _i{\lambda }_i{s}_i& \mapsto (\pi (v),({\lambda }_1,\dots ,{\lambda }_k))\end{array}\begin{array}{rl}{\psi }_V:{\pi }^{-1}(V)& \to V\times {\mathbb{R}}^k\\ v=\sum _i{\mu }_i{s}_i^{\prime }& \mapsto (\pi (v),({\mu }_1,\dots ,{\mu }_k))\end{array}$$$$\begin{array}{rl}{\psi }_V\circ {\psi }_U^{-1}:(U\cap V)\times {\mathbb{R}}^k& \to (U\cap V)\times {\mathbb{R}}^k\\ (p,\omega )& \mapsto (p,g(p)\omega )\end{array}$$where $g:U\cap V\to G{L}_k(\mathbb{R})$ is smooth. $$\cancel{{\omega }^{\prime }}=dg{g}^{-1}+g\cancel{\omega }{g}^{-1}\iff dg\equiv 0,\text{so}g\text{is constant.}$$Conversely suppose we have $(U_i,{\psi }_i)$ a system of local trivialization for $E$, s.t. each ${\psi }_j\circ {\psi }_i^{-1}$ has the form $(p,\omega )\mapsto (p,g(p)\omega )$ with $g$ constant.

On $E{|}_{U_i}$, we define a connection $\nabla$ by making the constant sections in the trivial bundle parallel, i.e. $s_i(p)={\psi }^{-1}(p,e_i)$. $$\nabla \left(\sum _j{f}_j{s}_j\right)=\sum _{j}d{f}_j\otimes s_j$$

$\square$

Claim 13.5.8. If $U_i\cap U_j\ne \varnothing$, then ${\nabla }^i={\nabla }^j$ on ${\pi }^{-1}(U_i\cap U_j)$.

证明. $s_i^{\prime }=\sum _j{g}_{ij}{s}_j$, with $d{g}_{ij}\equiv 0$.
$\Rightarrow$ $s_i^{\prime }$ which are ${\nabla }^j$ parallel are also ${\nabla }^i$ parallel.
$\Rightarrow$ ${\nabla }^i={\nabla }^j$
$\Rightarrow$ The ${\nabla }^i$ fit together to a global connection $\nabla$. Since ${\nabla }^i$ is flat, so is $\nabla$.

$\square$

注 13.5.9. To prove existence of connections on arbitrary $E$, we also took local trivializations $(U_i,{\psi }_i)$ and the corresponding flat connection ${\nabla }^i$. If the transition functions are not constant, the ${\nabla }^i$ do not agree on the overlaps of their domains.

$\nabla =\sum _i{\rho }_i{\nabla }^i$, ${\rho }_i$ a smooth partition of unity, is not flat.

13.6Compatible

$E\stackrel{\pi }{\to }M$ admits a positive definite metric $⟨,⟩:\Gamma (E)\times \Gamma (E)\to {\mathcal{C}}^{\infty }(M)$.

定义 13.6.1. A connection $\nabla$ on $E$ is compatible with $⟨,⟩$, if $$d⟨s_1,s_2⟩=⟨\nabla s_1,s_2⟩+⟨s_1,\nabla s_2⟩,\forall s_1,s_2\in \Gamma (E)$$(13.2)

引理 13.6.2. $\nabla$ is compatible with $⟨,⟩$ if and only if for every orthonormal local frame $s_1,\dots ,s_k$, the connection matrix $\omega$ representing $\nabla$ is skew-symmetric, i.e. ${\omega }_{ij}=-{\omega }_{ji}$, $\forall i,j$.

证明. Let $s_1,\dots ,s_k$ be orthonormal frame with respect to $⟨,⟩$. Then $$⟨s_i,s_j⟩=\text{const.},\forall i,j$$If $\nabla$ is compatible with $⟨,⟩$, then $$\begin{array}{rl}0& =d⟨s_i,s_j⟩\\ & =⟨\nabla s_i,s_j⟩+⟨s_i,\nabla s_j⟩\\ & =⟨\sum _l{\omega }_{il}\otimes s_l,s_j⟩+⟨s_j,\sum _l{\omega }_{jl}\otimes s_l⟩\\ & =\sum _l({\omega }_{il}⟨s_l,s_j⟩+{\omega }_{jl}⟨s_i,s_l⟩)\\ & ={\omega }_{ij}+{\omega }_{ji}\end{array}$$$$\begin{array}{rlrlr}\iff & & {\omega }_{ij}=-{\omega }_{ji}& & \end{array}$$

Conversely, assume $\omega$ is skew-symmetric$$⟨\nabla s_i,s_j⟩+⟨s_i,\nabla s_j⟩={\omega }_{ij}+{\omega }_{ji}=0$$$$⟨s_i,s_j⟩=\text{const.}\Rightarrow d⟨s_i,s_j⟩=0$$$\Rightarrow$ (13.2) holds for the basis sections.

Let $s=\sum _i{f}_i{s}_i$ and $s^{\prime }=\sum _j{g}_j{s}_j$. Then$$⟨s,s^{\prime }⟩=\sum _i{f}_i{g}_i\Rightarrow d⟨s,s^{\prime }⟩=\sum _i{f}_{i}d{g}_i+\sum _i{g}_{i}d{f}_i$$$$\begin{array}{rl}⟨\nabla s,s^{\prime }⟩+⟨s,\nabla s^{\prime }⟩& =\sum _{i,j}⟨d{f}_i\otimes s_i+f_i\nabla s_i,g_j{s}_j⟩+⟨f_i{s}_i,d{g}_j\otimes s_j+g_j\nabla s_j⟩\\ & =\sum _i{g}_{i}d{f}_i+\sum _i{f}_{i}d{g}_i+\sum _{i,j}{f}_i{g}_j\underset{=0\text{by above}}{\underbrace{(⟨\nabla s_i,s_j⟩+⟨s_i,\nabla s_j⟩)}}\end{array}$$$$\begin{array}{rlrlr}\Rightarrow & & d⟨s,s^{\prime }⟩=⟨\nabla s,s^{\prime }⟩+⟨s,\nabla s^{\prime }⟩& & \end{array}$$

$\square$

引理 13.6.3. If $\nabla$ is compatible with $⟨,⟩$, then $\Omega$ is skew-symmetric for every orthonormal frame.

证明. $$\begin{array}{rl}{\Omega }_{ij}& =d{\omega }_{ij}-\sum _l{\omega }_{il}\wedge {\omega }_{lj}\\ & =-d{\omega }_{ji}-\sum _l{\omega }_{li}\wedge {\omega }_{jl}\\ & =-(d{\omega }_{ij}-\sum _l{\omega }_{jl}\wedge {\omega }_{li})\\ & =-{\Omega }_{ji}\end{array}$$

$\square$

定义 13.6.4. $A\in \Gamma (\mathrm{End}E)$ is skew-symmetric with respect to $⟨,⟩$ if $$⟨As,s^{\prime }⟩=-⟨s,A{s}^{\prime }⟩,\forall s,s^{\prime }\in \Gamma (E)$$$\mathrm{End}E=\text{Skew}-\mathrm{End}E\oplus \text{Sym}-\mathrm{End}(E)$.

命题 13.6.5. For every metric $⟨,⟩$, there exist compatible connections $\nabla$. All such connections is naturally an affine space for ${\Omega }^1(\text{Skew}-\mathrm{End}(E))$.

证明. Let $\{U_i\mid i\in I\}$ be an open cover of $M$, s.t. $E{|}_{U_i}$ is trivial. Then on every $U_i$, we have an orthonormal local frame for $E$ with respect to $⟨,⟩$. Let $s_1,\dots ,s_k$ be such an orthonormal frame over $U_i$. Define ${\nabla }^i$ on $E{|}_{U_i}$ by ${\nabla }^i(s_j)\equiv 0$.

Claim 13.6.6. ${\nabla }^i$ is compatible with $⟨,⟩$.

证明. With respect to orthonormal frame $s_1,\dots ,s_k$, ${\omega }_{ij}\equiv 0$. So ${\omega }_{ij}$ is skew-symmetric. Let ${\rho }_i$ be a smooth partition of unity subordinate to $U_i$.

$\square$

Define $\nabla :=\sum _i{\rho }_i{\nabla }^i$. This is a connection on $E$ compatible with $⟨,⟩$, because each ${\nabla }^i$ is.

Suppose $\nabla$, ${\nabla }^{\prime }$ are both compatible with $⟨,⟩$. Set $\nabla -{\nabla }^{\prime }=A\in {\Omega }^1(\mathrm{End}E)$. Then$$\begin{array}{rl}⟨As,s^{\prime }⟩& =⟨(\nabla -{\nabla }^{\prime })s,s^{\prime }⟩\\ & =⟨\nabla s,s^{\prime }⟩-⟨{\nabla }^{\prime }s,s⟩\\ & =d(⟨s,s^{\prime }⟩)-⟨s,\nabla s^{\prime }⟩-d(⟨s,s^{\prime }⟩)+⟨s,{\nabla }^{\prime }s⟩\\ & =-⟨s,(\nabla -{\nabla }^{\prime })s^{\prime }⟩\\ & =-⟨s,A{s}^{\prime }⟩\end{array}$$So $A\in {\Omega }^1(\text{Skew}-\mathrm{End}(E))$.

If $\nabla$ is compatible with the metric and $A\in {\Omega }^1(\text{Skew}-\mathrm{End}E)$, then $\nabla +A$ is also compatible.

$\square$

例 13.6.7. $k=1$: Let $s$ be a (local) section of $E$, $s$ nowhere zero. $$\nabla s=\alpha \otimes s={\omega }_{11}\otimes s$$$${\Omega }_{11}=d{\omega }_{11}-\sum _l{\omega }_{1l}\wedge {\omega }_{l1}=d{\omega }_{11}-\cancel{{\omega }_{11}\wedge {\omega }_{11}}$$$$\begin{array}{rlrlr}\Rightarrow & & d{\Omega }_{11}=0& & \end{array}$$Suppose we have a metric $⟨,⟩$ and $⟨s,s⟩\equiv 1$. If $\nabla$ is compatible with $⟨,⟩$, then $\nabla s\equiv 0$. $$⟨s,s⟩=\text{const.}\Rightarrow 0=2⟨s,\nabla s⟩\Rightarrow \nabla s\equiv 0,\text{by 1-dimension}$$Conclusion: Every compatible connection $\nabla$ on a rank $1$ bundle is flat. $\Rightarrow$ Every rank $1$ bundle admits a flat connection.

13.7Affine Connection

定义 13.7.1. A connection on $E=TM$ is called an affine connection on $M$.$$\begin{array}{rcccl}\Gamma (E)& \to & {\Omega }^1(E)& \stackrel{i_X}{\to }& \Gamma (E)\\ s& \mapsto & \nabla s& \mapsto & {\nabla }_{X}s\end{array}$$where $X\in \mathfrak{X}(M)$. If $E=TM$, then $s\in \mathfrak{X}(M)$.

例 13.7.2. There is no affine connection $\nabla$ satisfying ${\nabla }_{X}Y={\nabla }_{Y}X$, $\forall X,Y\in \mathfrak{X}(M)$.

13.8Torsion

定义 13.8.1. If $\nabla$ is an affine connection on $M$, then $$T^{\nabla }(X,Y):={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y],\text{for}X,Y\in \mathfrak{X}(M)$$$T^{\nabla }$ is the torsion of $\nabla$.

定义 13.8.2. $\nabla$ is symmetric if it is torsion-free, i.e. $T^{\nabla }\equiv 0$.

命题 13.8.3 (Properties of $T^{\nabla }$).  

(1)	$T^{\nabla }$ is skew-symmetric in $X$, $Y$.
(2)	$T^{\nabla }$ is ${\mathcal{C}}^{\infty }(M)$-linear in $X$ and $Y$.

证明. $$\begin{array}{rl}{T}^{\nabla }(fX,Y)& ={\nabla }_{fX}Y-{\nabla }_{Y}fX-[fX,Y]\\ & =f{\nabla }_{X}Y-⟨df\otimes X+f\nabla X,Y⟩-f[X,Y]+\underset{=df(Y)}{\underbrace{L_{Y}f}}X\\ & =f\cdot T^{\nabla }(X,Y)\end{array}$$

$\square$

Let $x_1,\dots ,x_n$ be local coordinates on $M$ given by some charts $(U,\phi )$. Then ${\partial }_1,\dots ,{\partial }_n$ form a local frame for $TM{|}_U=TU$. If $\nabla$ is any affine connection on $M$, we can write$$\begin{array}{rl}\nabla {\partial }_i& =\sum _{j=1}^n{\omega }_{ij}\otimes {\partial }_j{\omega }_{ij}=\sum _{l=1}^n{\omega }_{ij}^{l}d{x}_l\\ {\nabla }_{{\partial }_l}{\partial }_i& =\sum _{j=1}^n⟨{\omega }_{ij},{\partial }_l⟩{\partial }_j=\sum _{j=1}^n{\omega }_{ij}^l{\partial }_j=\sum _{j=1}^n{\Gamma }_{li}^j{\partial }_j\end{array}$$The ${\Gamma }_{li}^j$ are called the Christoffel symbols of $\nabla$ with respect to the local coordinates $x_1,\dots ,x_n$$$T^{\nabla }({\partial }_{\alpha },{\partial }_{\beta })={\nabla }_{{\partial }_{\alpha }}{\partial }_{\beta }-{\nabla }_{{\partial }_{\beta }}{\partial }_{\alpha }-\cancel{[{\partial }_{\alpha },{\partial }_{\beta }]}=\sum _{j=1}^n({\Gamma }_{\alpha \beta }^j-{\Gamma }_{\beta \alpha }^j){\partial }_j$$

引理 13.8.4. $T^{\nabla }\equiv 0$ $\iff$ ${\Gamma }_{\alpha \beta }^j={\Gamma }_{\beta \alpha }^j$, $\forall \alpha ,\beta ,j\in \{1,\dots ,n\}$ and all local coordinate systems on $M$.

定义 13.8.5. ${\nabla }^{\ast }$ on $E^{\ast }$ by$$\begin{array}{rl}d\lambda (s)& =\lambda (\nabla s)+({\nabla }^{\ast }\lambda )(s)\\ d⟨\lambda ,s⟩& =⟨\lambda ,\nabla s⟩+⟨{\nabla }^{\ast }\lambda ,s⟩\end{array}$$where $\forall \lambda \in \Gamma (E^{\ast }),s\in \Gamma (E)$.

Claim 13.8.6. ${\nabla }^{\ast }$ is a connection on $E^{\ast }$.

证明. $$({\nabla }^{\ast }\lambda )(s)=d\lambda (s)-\lambda (\nabla (s))$$$$\begin{array}{rl}({\nabla }^{\ast }(f\lambda ))(s)& =d(f\lambda )(s)-(f\lambda )(\nabla s)\\ & =d(f\cdot \lambda (s))-(f\cdot \lambda )(\nabla s)\\ & =\lambda (s)df+fd\lambda (s)-f\cdot \lambda (\nabla (s))\\ & =\lambda (s)df+f({\nabla }^{\ast }\lambda )(s)\\ & =(df\cdot \lambda +f{\nabla }^{\ast }\lambda )(s)\end{array}$$Let $s_1,\dots ,s_k$ be a local frame for $E$, and ${\lambda }_1,\dots ,{\lambda }_k$ the dual frame for $E^{\ast }$, i.e.$$\begin{array}{rl}{\lambda }_i(s_j)& ={\delta }_{ij}\\ 0& ={\lambda }_i(\nabla s_j)+({\nabla }^{\ast }{\lambda }_i)(s_j)\\ & ={\lambda }_i\left(\sum _{m=1}^k{\omega }_{jm}\otimes s_m\right)+\left(\sum _{m=1}^k{\omega }_{il}^{\ast }\otimes {\lambda }_l\right)(s_j)\\ & ={\omega }_{ji}+{\omega }_{ij}^{\ast }\end{array}$$$$\begin{array}{rlrlr}\Rightarrow & & {\omega }_{ij}^{\ast }=-{\omega }_{ji},{\omega }^{\ast }=-{\omega }^t& & \end{array}$$If $\nabla$ is an affine connection of $M$, then ${\nabla }^{\ast }$ is a connection on $T^{\ast }M$.

$\square$

命题 13.8.7. $\nabla$ is torsion-free if and only if

证明. Let $x_1,\dots ,x_n$ be local coordinates, given by a chart $(U,\phi )$. Then ${\partial }_1,\dots ,{\partial }_n$ is a local frame for $TM$ and $d{x}_1,\dots ,d{x}_n$ is the dual frame for $T^{\ast }M$.

Every $1$-form $\alpha$ on $U$ is of the form$$\beta =\sum _{i=1}^n{f}_{i}d{x}_i$$$$\begin{array}{rlrlr}\Rightarrow & & d\beta =\sum _{i=1}^{n}d{f}_i\wedge d{x}_i=\sum _i\sum _j\frac{\partial f_i}{\partial x_j}d{x}_j\wedge d{x}_i=\sum _{i<j}\left(\frac{\partial f_i}{\partial x_j}-\frac{f_j}{\partial x_i}\right)d{x}_j\wedge d{x}_i& & \end{array}$$$$\begin{array}{rl}{\nabla }^{\ast }\beta & =\sum _i{\nabla }^{\ast }(f_{i}d{x}_i)\\ & =\sum _{i}d{f}_i\otimes d{x}_i+f_i{\nabla }^{\ast }d{x}_i\\ & =\sum _{i}d{f}_i\otimes d{x}_i+f_i\sum _j{\omega }_{ij}^{\ast }\otimes d{x}_j\\ & =\sum _i\left(d{f}_i\otimes d{x}_i-f_i\sum _j{\omega }_{ji}\otimes d{x}_j\right)\\ & =\sum _i\left(\sum _j\frac{\partial f_i}{\partial x_j}d{x}_j\otimes d{x}_i-f_i\sum _{j,\alpha }{\omega }_{ji}^{\alpha }d{x}_{\alpha }\otimes d{x}_j\right)\\ & =\sum _{i,j}\frac{\partial f_i}{\partial x_j}d{x}_j\otimes d{x}_i-\sum _{i,j,\alpha }{f}_i{\omega }_{ji}^{\alpha }d{x}_{\alpha }\otimes d{x}_j\\ & =\sum _{j,\alpha }\frac{\partial f_j}{\partial x_{\alpha }}d{x}_{\alpha }\otimes d{x}_j-\sum _{i,j,\alpha }{f}_i{\omega }_{ji}^{\alpha }d{x}_{\alpha }\otimes d{x}_j\\ & =\sum _{j,\alpha }\left(\frac{\partial f_j}{\partial x_{\alpha }}-\sum _i{f}_i{\omega }_{ji}^{\alpha }\right)d{x}_{\alpha }\otimes d{x}_j\end{array}$$$$\wedge ({\nabla }^{\ast }\beta )=\sum _{j<\alpha }\left(\frac{\partial f_j}{\partial x_{\alpha }}-\frac{\partial f_{\alpha }}{\partial x_j}-\sum _i{f}_i({\omega }_{ji}^{\alpha }-{\omega }_{\alpha i}^j)\right)d{x}_{\alpha }\wedge d{x}_j$$$$\wedge ({\nabla }^{\ast }\beta )=d\beta ,\forall \beta \iff {\omega }_{ji}^{\alpha }-{\omega }_{\alpha i}^j=0,\forall j,\alpha \iff {\Gamma }_{\alpha j}^i={\Gamma }_{j\alpha }^i,\forall \alpha ,j\iff T^{\nabla }\equiv 0$$$\beta =df$:$$\begin{array}{rl}{\nabla }^{\ast }\beta & ={\nabla }^{\ast }\left(\sum _{i=1}^n\frac{\partial f}{\partial x_i}d{x}_i\right)\\ & =\sum _{i=1}^{n}d\left(\frac{\partial f}{\partial x_i}\right)d{x}_i+\frac{\partial f}{\partial x_i}{\nabla }^{\ast }(d{x}_i)\\ & =\sum _{i,j}\frac{{\partial }^{2}f}{\partial x_i\partial x_j}d{x}_j\otimes d{x}_i-\frac{\partial f}{\partial x_i}{\omega }_{ji}\otimes d{x}_j\\ & =\sum _{i,j}\left(\frac{{\partial }^{2}f}{\partial x_i\partial x_j}d{x}_j\otimes d{x}_i-\sum _{\alpha }\frac{\partial f}{\partial x_i}{\omega }_{ji}^{\alpha }d{x}_{\alpha }\otimes d{x}_j\right)\\ & =\sum _{\alpha ,j}\left(\frac{{\partial }^{2}f}{\partial x_j\partial x_{\alpha }}-\sum _i\frac{\partial f}{\partial x_i}{\Gamma }_{\alpha j}^i\right)d{x}_{\alpha }\otimes d{x}_j\end{array}$$$T^{\nabla }\equiv 0$ $\iff$ $\underset{\in \Gamma (T^{\ast }M\otimes T^{\ast }M)}{\underbrace{{\nabla }^{\ast }df}}$ is symmetric $\forall f\in {\mathcal{C}}^{\infty }(M)$.$$\begin{array}{rl}{T}^{\nabla +A}(X,Y)& =(\nabla +A{)}_{X}Y-(\nabla +A{)}_{Y}X-[X,Y]\\ & =\underset{=T^{\nabla }}{\underbrace{{\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]}}+A_X(Y)-A_Y(X)\end{array}$$where $A\in {\Omega }_1(\mathrm{End}(TM))$, $A_X\in \Gamma (\mathrm{End}(TM))$, $A_{X}Y$ is evaluation of the endomorphism $A_X$ on $Y$.

$\square$

13.9Riemannian Geometry

定理 13.9.1. Let $⟨,⟩$ be a metric on $TM$ (a Riemannian metric on $M$). For every ${\mathcal{C}}^{\infty }(M)$-bilinear skew-symmetric$$T:\mathfrak{X}(M)\times \mathfrak{X}(M)\to \mathfrak{X}(M)$$There exists a unique affine connection $\nabla$ compatible with $⟨,⟩$ and $T^{\nabla }=T$.

证明. Uniqueness: Suppose $\nabla$ is compatible, with $⟨,⟩$ and $T^{\nabla }=T$.$$d⟨X,Y⟩=⟨\nabla X,Y⟩+⟨X,\nabla Y⟩,\forall X,Y\in TM$$$$\begin{array}{rl}{L}_Z⟨X,Y⟩& =⟨{\nabla }_{Z}X,Y⟩+⟨X,{\nabla }_{Z}Y⟩,\forall X,Y,Z\in TM\\ T(Z,Y)& ={\nabla }_{Z}Y-{\nabla }_{Y}Z-[Z,Y]\end{array}$$$$\begin{array}{rl}⟨{\nabla }_{Z}X,Y⟩& =L_Z⟨X,Y⟩-⟨X,{\nabla }_{Z}Y⟩\\ & =L_Z⟨X,Y⟩-⟨X,T(Z,Y)⟩-⟨X,{\nabla }_{Y}Z⟩-⟨X,[Z,Y]⟩\\ & =L_Z⟨X,Y⟩-⟨X,T(Z,Y)⟩-L_Y⟨X,Z⟩+⟨{\nabla }_{Y}X,Z⟩-⟨X,[Z,Y]⟩\\ & =L_Z⟨X,Y⟩-⟨X,T(Z,Y)⟩-L_Y⟨X,Z⟩+⟨T(Y,X),Z⟩+⟨{\nabla }_{X}Y,Z⟩\\ & +⟨[Y,X],Z⟩-⟨X,[Z,Y]⟩\\ & =L_Z⟨X,Y⟩-⟨X,T(Z,Y)⟩-L_Y⟨X,Z⟩+⟨T(Y,X),Z⟩+L_X⟨Y,Z⟩\\ & -⟨Y,{\nabla }_{X}Z⟩+⟨[Y,X],Z⟩-⟨X,[Z,Y]⟩\\ & =L_Z⟨X,Y⟩-⟨X,T(Z,Y)⟩-L_Y⟨X,Z⟩+⟨T(Y,X),Z⟩+L_X⟨Y,Z⟩\\ & -⟨Y,T(X,Z)⟩-⟨Y,{\nabla }_{Z}X⟩-⟨Y,[X,Z]⟩+⟨[Y,X],Z⟩-⟨X,[Z,Y]⟩\end{array}$$Therefore, we have the so called Koszul formula.$$\boxed{\begin{array}{rl}⟨{\nabla }_{Z}X,Y⟩& =\frac{1}{2}(L_Z⟨X,Y⟩-L_Y⟨X,Z⟩+L_X⟨Y,Z⟩-⟨X,T(Z,Y)⟩+⟨Z,T(Y,X)⟩\\ & -⟨Y,T(X,Z)⟩-⟨Y,[X,Z]⟩+⟨[Y,X],Z⟩-⟨X,[Z,Y]⟩)\end{array}}$$This shows ${\nabla }_{Z}X$ is uniquely determined $\forall Z,X\in \mathfrak{X}(M)$.

Existence: Define ${\nabla }_{Z}X$ by the Koszul formula. Fix $M$ and $⟨,⟩$ on $TM$. Let $\nabla$ be the Levi-Civita connection of $⟨,⟩$. Let $x_1,\dots ,x_n$ be local coordinates given by a chart ${\partial }_1,\dots ,{\partial }_n$ the coordinate vector fields.$$\begin{array}{rl}{\gamma }_{ij}& =⟨{\partial }_i,{\partial }_j⟩\\ ⟨{\nabla }_{{\partial }_i}{\partial }_j,{\partial }_k⟩& =\frac{1}{2}(L_{{\partial }_i}{\gamma }_{jk}+L_{{\partial }_j}{\gamma }_{ki}-L_{{\partial }_k}{\gamma }_{ij})=\frac{1}{2}({\partial }_i{\gamma }_{jk}+{\partial }_j{\gamma }_{ki}-{\partial }_k{\gamma }_{ij})\\ {\nabla }_{{\partial }_i}{\partial }_j& =\sum _{k=1}^n{\omega }_{jk}^i{\partial }_k=\sum _{k=1}^n{\Gamma }_{ij}^k{\partial }_k\\ ⟨{\nabla }_{{\partial }_i}{\partial }_j,{\partial }_l⟩& =\sum _{k=1}^n{\Gamma }_{ij}^k{\gamma }_{kl}=\frac{1}{2}({\partial }_i{\gamma }_{jk}+{\partial }_j{\gamma }_{ki}-{\partial }_k{\gamma }_{ij})\end{array}$$Formula of ${\Gamma }_{ij}^k$ in terms of ${\gamma }_{ij}$.

$\square$

Setting $T=0$, we get

推论 13.9.2 (Fundamental Lemma of Riemannian Geometry). For every metric on $TM$, there exists a unique, compatible, torsion-free connection.

定义 13.9.3. This connection $\nabla$ as in the corollary is called the Levi-Civita connection of $(M;⟨,⟩)$.

定义 13.9.4. If $\nabla$ is the Levi-Civita connection, then$$R(X,Y)Z:=(F^{\nabla }(X,Y))Z$$is called the Riemann curvature tensor of the metric $⟨,⟩$.

This is trilinear over ${\mathcal{C}}^{\infty }(M)$.$$\begin{array}{rl}R:\mathfrak{X}(M)\times \mathfrak{X}(M)\times \mathfrak{X}(M)& \to \mathfrak{X}(M)\\ (X,Y,Z)& \mapsto R(X,Y)Z\end{array}$$Equivalently, we can consider $R$ as follows:$$\begin{array}{rl}R:\mathfrak{X}(M)\times \mathfrak{X}(M)\times \mathfrak{X}(M)\times \mathfrak{X}(M)& \to {\mathcal{C}}^{\infty }(M)\\ (X,Y,Z,W)& \mapsto ⟨R(X,Y)Z,W⟩\end{array}$$

命题 13.9.5 (Symmetries of $R$).  

(1)	$R(X,Y)Z=-R(Y,X)Z$, because $F^{\nabla }$ is a $2$-form.
(2)	$R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0$, $\forall X,Y,Z$. Sometimes it is called the first Bianchi Identity.
(3)	$⟨R(X,Y)Z,W⟩=-⟨R(X,Y)W,Z⟩$, $\forall X,Y,Z,W$.
(4)	$⟨R(X,Y)Z,W⟩=⟨R(Z,W)X,Y⟩$, $\forall X,Y,Z,W$

证明. (2) It is enough to prove (2) for $X,Y,Z$ with pairwise vanishing brackets. $$F^{\nabla }(X,Y)s={\nabla }_X{\nabla }_{Y}s-{\nabla }_Y{\nabla }_{X}s-\cancel{{\nabla }_{[X,Y]}s}$$In this case, left hand side of (2) $$\begin{array}{rl}& {\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z+{\nabla }_Y{\nabla }_{Z}X-{\nabla }_Z{\nabla }_{Y}X+{\nabla }_Z{\nabla }_{X}Y-{\nabla }_X{\nabla }_{Z}Y\\ & ={\nabla }_X\underset{=0}{\underbrace{({\nabla }_{Y}Z-{\nabla }_{Z}Y)}}+{\nabla }_Y\underset{=0}{\underbrace{({\nabla }_{Z}X-{\nabla }_{X}Z)}}+{\nabla }_Z\underset{=0\text{since}{T}^{\nabla }=0}{\underbrace{({\nabla }_{X}Y-{\nabla }_{Y}X)}}\\ & =0\end{array}$$

(3): We need to prove $⟨R(X,Y)Z,Z⟩=0$, $\forall X,Y,Z$. We may assume that $X,Y,Z$ have vanishing brackets. $$⟨R(X,Y)Z,Z⟩=⟨{\nabla }_X{\nabla }_{Y}Z,Z⟩-⟨{\nabla }_Y{\nabla }_{X}Z,Z⟩$$Consider $$\begin{array}{rl}{L}_X⟨Z,Z⟩& =⟨{\nabla }_{X}Z,Z⟩+⟨Z,{\nabla }_{X}Z⟩=Z⟨{\nabla }_{X}Z,Z⟩\\ L_Y{L}_X⟨Z,Z⟩& =2L_Y⟨{\nabla }_{X}Z,Z⟩=2(⟨{\nabla }_Y{\nabla }_{X}Z,Z⟩+⟨{\nabla }_{X}Z,{\nabla }_{Y}Z⟩)\end{array}$$$L_Y{L}_X⟨Z,Z⟩$ is symmetric in $X,Y$, since $⟨X,Y⟩=0$ and $⟨{\nabla }_{X}Z,{\nabla }_{Y}Z⟩$ is symmetric in $X,Y$. Therefore, $⟨{\nabla }_Y{\nabla }_{X}Z,Z⟩$ is symmetric in $X,Y$. Thus$$\begin{array}{rlrlr}\Rightarrow & & ⟨R(X,Y)Z,Z⟩=0& & \end{array}$$

(4):

Sum for upper left-hand face is $⟨R(Y,X)W,Z⟩+⟨R(W,Y)X,Z⟩+⟨R(X,W)Y,Z⟩$. Sum of labels is $=0$ by (1)+(2)+(3) for top left and right and bottom front and back faces.

Sum the top left and right and subtract the bottom front and back faces: $\Rightarrow$ $$\begin{array}{rlrlr}\Rightarrow & & \text{The middle nodes cancel}& & \\ \Rightarrow & & 0=⟨R(X,Y)Z,W⟩-⟨R(Z,W)X,Y⟩& & \end{array}$$

$\square$

Let $M$ with metric and $R$ its Riemann tensor.

定义 13.9.6. Take $p\in M$, $X,Y\in T_{p}M$ linearly independent $$K(X,Y):=\frac{⟨R(X,Y)Y,X⟩}{⟨X,X⟩⟨Y,Y⟩-⟨X,Y{⟩}^2}$$This is called the sectional curvature of $(M,⟨,⟩)$ with respect to the plane $\sigma$ spanned by $X,Y$ in $T_{p}M$.

Claim 13.9.7. $K(X,Y)$ depends only on $\sigma =\mathrm{span}\{X,Y\}$.

证明. $K(\lambda X,Y)=\frac{{\lambda }^2⟨R(X,Y)Y,X⟩}{{\lambda }^2\cdot (⟨X,X⟩⟨Y,Y⟩-⟨X,Y{⟩}^2)}=K(X,Y)\ne 0$.

Since $K(X,Y)=K(Y,X)$, we also get $K(X,\lambda Y)=K(X,Y)$.$$\begin{array}{rl}K(X,Y+\lambda X)& =\frac{⟨R(X,Y)(Y+\lambda X),X⟩+⟨R(X,\lambda X)(Y+\lambda X),X⟩}{⟨X,X⟩(⟨Y,Y⟩+{\lambda }^2⟨X,X⟩+2\lambda ⟨X,Y⟩)-⟨X,Y+\lambda X{⟩}^2}\\ & =K(X,Y)\end{array}$$This shows $K(X,Y)$ is the same $\forall X,Y\in \sigma$.

$\square$

命题 13.9.8. The collection of all sectional curvatures determines $R$.

证明. Let $V$ be a vector space with positive definite $⟨,⟩$.

Let $R,R^{\prime }:V\times V\times V\to V$ be two trilinear maps satisfying the symmetry of the curvature tensor. Then if $K(X,Y)=\frac{⟨R(X,Y)Y,X⟩}{⟨X,X⟩⟨Y,Y⟩-⟨X,Y{⟩}^2}$ equals $K^{\prime }$ computed in the same way from $R^{\prime }$ for all linear independent $X$, $Y$, $R=R^{\prime }$.

$R(X,Y)Z=0=R^{\prime }(X,Y)Z$, if $X$, $Y$ are linear independent.

Assume $X$, $Y$ linearly independent, then $K(X,Y)=K^{\prime }(X,Y)$ implies $$⟨R(X,Y)Y,X⟩=⟨R^{\prime }(X,Y)Y,X⟩,\forall X,Y\text{linearly independent}$$$$\begin{array}{rlrlrl}\Rightarrow & & & ⟨R(X+Z,Y)Y,X+Z⟩=⟨R^{\prime }(X+Z,Y)Y,X+Z⟩& & \\ \iff & & & \cancel{⟨R(X,Y)Y,X⟩}+⟨R(X,Y)Y,Z⟩+\underset{=⟨R(Y,Z)X,Y⟩=⟨R(X,Y)Y,Z⟩}{\underbrace{⟨R(Z,Y)Y,Z⟩}}+\cancel{⟨R(Z,Y)Y,Z⟩}=(R\leftrightarrow R^{\prime })& & \\ \iff & & & 2⟨R(X,Y)Y,Z⟩=2⟨R^{\prime }(X,Y)Y,Z⟩,\forall Z& & \end{array}$$After one more polarization $Y\mapsto Y+W$, we conclude$$⟨R(X,Y)Z,W⟩=⟨R^{\prime }(X,Y)Z,W⟩,\forall X,Y,Z,W$$$$\begin{array}{rlrlr}\Rightarrow & & R=R^{\prime }& & \end{array}$$

$\square$

例 13.9.9. Let $M={\mathbb{R}}^n$, and $⟨,⟩$ constant, standard. $\nabla \frac{\partial }{\partial x_n}=0$ gives Levi-Civita $\Rightarrow$ $R\equiv 0$, so $K\equiv 0$.

例 13.9.10. Let $M\subset {\mathbb{R}}^{n+1}$ be smooth hypersurface. $⟨,⟩$ on ${\mathbb{R}}^{n+1}$ as in 13.9.9. $\nabla$ the Levi-Civita connection of ${\mathbb{R}}^{n+1}$. We restrict the constant scalars product on ${\mathbb{R}}^{n+1}$ to the tangent space of $M$ to get a metric $⟨,⟩$ on $TM$. $${\mathbb{R}}^{n+1}\times {\mathbb{R}}^{n+1}{|}_M=T{\mathbb{R}}^{n+1}{|}_M=TM\oplus T{M}^{\perp }$$where $T{M}^{\perp }$ is the normal bundle of $M$.

If $M$ is orientable, then there is a uniquely defined unit normal vector field to $M$, so that the orientation of $M$ together with the positive or of $\mathbb{R}$ defines the standard orientation of ${\mathbb{R}}^{n+1}$.

定义 13.9.11. $\begin{array}{rl}G:M& \to S^n\subset {\mathbb{R}}^{n+1}\\ p& \mapsto n(p)\end{array}$ is the Gauss map of $M$.

定义 13.9.12. $\begin{array}{rl}L:T_{p}M& \to T_{p}M\\ v& \mapsto ({\tilde{\nabla }}_{v}n)(p)\end{array}$ is the Weingarten map of $M$ at $p$.

引理 13.9.13. $L$ is self adjoint with respect to $⟨,⟩$.

证明. Let $X,Y\in \mathfrak{X}(M)$.$$\begin{array}{rl}⟨L(X),Y⟩& =⟨{\tilde{\nabla }}_{X}n,Y⟩\\ & =L_X\cancel{⟨n,Y⟩}-⟨n,{\tilde{\nabla }}_{X}Y⟩\\ & =-⟨n,{\tilde{\nabla }}_{Y}X+\cancel{[X,Y]}⟩\\ & =-⟨n,{\tilde{\nabla }}_{Y}X⟩\\ & =-L_Y\cancel{⟨n,X⟩}+⟨{\tilde{\nabla }}_{Y}n,X⟩\\ & =⟨L(Y),X⟩\\ & =⟨X,L(Y)⟩\end{array}$$

$\square$

引理 13.9.14. $D_{p}G=L$.

证明. $D_{p}G:T_{p}M\to T_{G(p)}{S}^n=T_{p}M$, since both are orthogonal complement of $n$.

Let $c:(-\epsilon ,\epsilon )\to M$ be a smooth curve, with $c(0)=p$ and $\dot{c}(0)=v$. Then $$\begin{array}{rl}{D}_{p}G(v)& =(D_{c(0)}G)(\dot{c}(0))\\ & =D_0(G\circ c)\left(\frac{\partial }{\partial t}\right)\\ & =\frac{d}{dt}n(c(t)){|}_{t=0}\\ & ={\tilde{\nabla }}_{\dot{c}(0)}n\\ & =L(v)\end{array}$$

$\square$

Let $X$, $Y\in \mathfrak{X}(M)$.$${\tilde{\nabla }}_{X}Y=({\tilde{\nabla }}_{X}Y{)}_t+({\tilde{\nabla }}_{X}Y{)}_n\text{with respect to}{\mathbb{R}}^{n+1}=T_{p}M\oplus \mathbb{R}n(p)$$

定义 13.9.15. Define ${\nabla }_{X}Y=\pi ({\tilde{\nabla }}_{X}Y)$, $\pi :{\mathbb{R}}^{n+1}\to T_{p}M$ is the projection with kernel $\mathbb{R}n(p)$.

引理 13.9.16. $\nabla$ is the Levi Civita connection of $M$.

证明. Step 1: $\nabla$ is a connection on $TM$. ${\nabla }_{X}Y$ is $\mathbb{R}$-linear in $X$, $Y$ and it is ${\mathcal{C}}^{\infty }(M)$-linear in $X$. $${\nabla }_X(fY)=\pi ({\tilde{\nabla }}_X(fY))=\pi (L_{X}f\cdot Y+f{\tilde{\nabla }}_{X}Y)=L_{X}f\cdot Y+f\cdot {\nabla }_{X}Y$$Leibniz rule for $\nabla$.

Step 2: $\nabla$ on $TM$ is compatible with $⟨,⟩$. $$⟨{\nabla }_{X}Y,Z⟩+⟨Y,{\nabla }_{X}Z⟩=⟨{\tilde{\nabla }}_{X}Y,Z⟩+⟨Y,{\tilde{\nabla }}_{X}Z⟩=L_X⟨Y,Z⟩,X,Y,Z\in \mathfrak{X}(M)$$

Step 3: $$0=T^{\tilde{\nabla }}(X,Y)={\tilde{\nabla }}_{X}Y-{\tilde{\nabla }}_{Y}X-[X,Y],X,Y\in \mathfrak{X}(M)$$(13.3)projecting to $TM$ gives $$0={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]=T^{\nabla }(X,Y)$$In (13.3), take $⟨-,n⟩$ $$0=⟨{\tilde{\nabla }}_{X}Y,n⟩-⟨{\tilde{\nabla }}_{Y}X,n⟩\stackrel{\text{Lemma 1 Proof}}{=}-⟨L(X),Y⟩+⟨X,L(Y)⟩$$$$\begin{array}{rlrlr}\iff & & L\text{is self adjoint with respect to}⟨,⟩.& & \end{array}$$

$\square$

$X,Y\in \mathfrak{X}(M)$, ${\tilde{\nabla }}_{X}Y={\nabla }_{X}Y+⟨{\tilde{\nabla }}_{X}Y,n⟩n={\nabla }_{X}Y-⟨L(X),Y⟩n$.

Take $X,Y,Z\in \mathfrak{X}(M)$.$$\begin{array}{rl}{\tilde{\nabla }}_X{\tilde{\nabla }}_{Y}Z& ={\tilde{\nabla }}_X({\nabla }_{Y}Z-⟨L(Y),Z⟩n)\\ & ={\tilde{\nabla }}_X{\nabla }_{Y}Z-L_X⟨L(Y),Z⟩\cdot n-⟨L(Y),Z⟩{\tilde{\nabla }}_{X}n\\ & ={\nabla }_X{\nabla }_{Y}Z-⟨L(X),{\nabla }_{Y}Z⟩\cdot n-⟨{\nabla }_{X}L(Y),Z⟩\cdot n-⟨L(Y),{\nabla }_{X}Z⟩\cdot n\\ & -⟨L({\nabla }_{X}Y),Z⟩\cdot n-⟨L(Y),Z⟩L(X)\end{array}$$Similarly for ${\tilde{\nabla }}_Y{\tilde{\nabla }}_{X}Z$$${\tilde{\nabla }}_{[X,Y]}Z={\nabla }_{[X,Y]}Z-⟨L([X,Y],Z)⟩n$$$$\begin{array}{rl}0& =\tilde{R}(X,Y)Z={\tilde{\nabla }}_X{\tilde{\nabla }}_{Y}Z-{\tilde{\nabla }}_Y{\tilde{\nabla }}_{X}Z-{\tilde{\nabla }}_{[X,Y]}Z\\ & ={\nabla }_X{\nabla }_{Y}Z-⟨L(Y),Z⟩L(X)-(⟨L(X),{\nabla }_{Y}Z⟩+⟨{\nabla }_{X}L(Y),Z⟩+⟨L(Y),{\nabla }_{X}Z⟩)n\\ & -{\nabla }_Y{\nabla }_{X}Z+⟨L(X),Z⟩L(Y)+(⟨L(Y),{\nabla }_{X}Z⟩+⟨{\nabla }_{Y}L(X),Z⟩+⟨L(X),{\nabla }_{Y}Z⟩)n\\ & -{\nabla }_{[X,Y]}Z+⟨L([X,Y]),Z⟩n\end{array}$$Projecting to $TM$, we get the Gauss equation$$\begin{array}{rlrlr}\Rightarrow & & R(X,Y)Z=⟨L(Y),Z⟩L(X)-⟨L(X),Z⟩L(Y)& & \end{array}$$Projecting to $n$$$\begin{array}{rlrlrl}\Rightarrow & & 0=\cancel{-⟨L(X),{\nabla }_{Y}Z⟩}& -⟨{\nabla }_{X}L(Y),t⟩\cancel{-⟨L(Y),{\nabla }_{X}Z⟩}+\cancel{⟨L(Y),{\nabla }_{X}Z⟩}& & \\ & & & +⟨{\nabla }_{Y}L(X),Z⟩+\cancel{⟨L(X),{\nabla }_{Y}Z⟩}+⟨L([X,Y]),Z⟩& & \\ \Rightarrow & & ⟨L([X,Y]),Z⟩& =⟨{\nabla }_{X}L(Y),Z⟩-⟨{\nabla }_{Y}L(X),Z⟩\forall X,Y,Z\in \mathfrak{X}(M)& & \\ \Rightarrow & & L([X,Y])& ={\nabla }_{X}L(Y)-{\nabla }_{Y}L(X)\forall X,Y\in \mathfrak{X}(M)& & \end{array}$$This is called the Codazzi-Mainardi equation. We can apply the Gauss equation to any smooth hypersurface $M\subset {\mathbb{R}}^{n+1}$. If $M$ is an affine hyperplane, $G$ is constant, so $L\equiv DG\equiv 0$ $\Rightarrow$ $R(X,Y)Z=0$.

If $M\subset {\mathbb{R}}^{n+1}$ is the unit sphere $S^n$, then $G=\mathrm{Id}$ $\Rightarrow$ $L=DG=\mathrm{Id}$. By the Gauss equation$$R(X,Y)Z=⟨Y,Z⟩X-⟨X,Z⟩Y$$$X,Y\in T_p{S}^n$, linear independent:$$K(X,Y)=\frac{⟨R(X,Y)Y,X⟩}{⟨X,X⟩⟨Y,Y⟩-⟨X,Y{⟩}^2}=\frac{⟨Y,Y⟩⟨X,X⟩-⟨X,Y⟩⟨Y,X⟩}{⟨X,X⟩⟨Y,Y⟩-⟨X,Y{⟩}^2}=1$$If $M=S^n(r)$ is the sphere of Radius $r$ in ${\mathbb{R}}^{n+1}$, then$$G=\frac{1}{r}\Rightarrow L=\frac{1}{r}\Rightarrow R(X,Y)Z=\frac{1}{r^2}(⟨Y,Z⟩X-⟨X,Z⟩Y)\Rightarrow K(X,Y)=\frac{1}{r^2}$$

注 13.9.17. $(M,⟨,⟩)$ is any Riemannian manifold. Consider $(M,\underset{⟨,{⟩}_{\lambda }}{\underbrace{\lambda ⟨,⟩}})$ for $\lambda >0$. Then $K(X,Y{)}_{⟨,{⟩}_{\lambda }}=\frac{1}{\lambda }K(X,Y{)}_{⟨,⟩}$.