7. 移动标架与Gauss-Bonnet公式 7.1. Cartan结构方程# 我们经常发现使用移动（正交）标架的方法很方便，它在计算具有对称性的度量的联络和曲率时非常有效，例如旋转对称度量和齐次度量。该方法最初由Elie Cartan开发，随后由陈省身进一步完善。令 { e i } i = 1 n \left\{e_{i}\right\}_{i=1}^{n} { e i } i = 1 n U ⊂ M n U \subset M^{n} U ⊂ M n T ∗ M n T^{*} M^{n} T ∗ M n { ω i } i = 1 n \left\{\omega^{i}\right\}_{i=1}^{n} { ω i } i = 1 n ω i ( e j ) = δ j i \omega^{i}\left(e_{j}\right)=\delta_{j}^{i} ω i ( e j ) = δ j i i , j = 1 , … , n i, j=1, \ldots, n i , j = 1 , … , n
Rm i j ( X , Y ) = 1 2 ⟨ ∇ X , Y 2 e i − ∇ Y , X 2 e i , e j ⟩ = 1 2 ⟨ ( ∇ X ω i k ) ( Y ) e k + ω i k ( Y ) ∇ X e k − ( ∇ Y ω i k ) ( X ) e k − ω i k ( X ) ∇ Y e k , e j ⟩ \operatorname{Rm}_{i}^{j}(X, Y) = \frac{1}{2}\left\langle\nabla_{X, Y}^{2} e_{i}-\nabla_{Y, X}^{2} e_{i}, e_{j}\right\rangle = \frac{1}{2}\left\langle\left(\nabla_{X} \omega_{i}^{k}\right)(Y) e_{k}+\omega_{i}^{k}(Y) \nabla_{X} e_{k}-\left(\nabla_{Y} \omega_{i}^{k}\right)(X) e_{k}-\omega_{i}^{k}(X) \nabla_{Y} e_{k}, e_{j}\right\rangle Rm i j ( X , Y ) = 2 1 ⟨ ∇ X , Y 2 e i − ∇ Y , X 2 e i , e j ⟩ = 2 1 ⟨ ( ∇ X ω i k ) ( Y ) e k + ω i k ( Y ) ∇ X e k − ( ∇ Y ω i k ) ( X ) e k − ω i k ( X ) ∇ Y e k , e j ⟩ = d ω i k ( X , Y ) ⟨ e k , e j ⟩ + 1 2 ( ω i k ( Y ) ω k ℓ ( X ) − ω i k ( X ) ω k ℓ ( Y ) ) ⟨ e ℓ , e j ⟩ = d \omega_{i}^{k}(X, Y)\left\langle e_{k}, e_{j}\right\rangle+\frac{1}{2}\left(\omega_{i}^{k}(Y) \omega_{k}^{\ell}(X)-\omega_{i}^{k}(X) \omega_{k}^{\ell}(Y)\right)\left\langle e_{\ell}, e_{j}\right\rangle = d ω i k ( X , Y ) ⟨ e k , e j ⟩ + 2 1 ( ω i k ( Y ) ω k ℓ ( X ) − ω i k ( X ) ω k ℓ ( Y ) ) ⟨ e ℓ , e j ⟩ 并且（1.80）得证。
对于曲面 M 2 M^{2} M 2
d ω 1 = ω 2 ∧ ω 2 1 , d ω 2 = ω 1 ∧ ω 1 2 Rm 2 1 = d ω 2 1 \begin{gathered} d \omega^{1}=\omega^{2} \wedge \omega_{2}^{1}, \quad d \omega^{2}=\omega^{1} \wedge \omega_{1}^{2} \\ \operatorname{Rm}_{2}^{1}=d \omega_{2}^{1} \end{gathered} d ω 1 = ω 2 ∧ ω 2 1 , d ω 2 = ω 1 ∧ ω 1 2 Rm 2 1 = d ω 2 1 特别地，高斯曲率由以下公式给出：
K ≑ ⟨ Rm ( e 1 , e 2 ) e 2 , e 1 ⟩ = 2 Rm 2 1 ( e 1 , e 2 ) = 2 d ω 2 1 ( e 1 , e 2 ) K \doteqdot\left\langle\operatorname{Rm}\left(e_{1}, e_{2}\right) e_{2}, e_{1}\right\rangle=2 \operatorname{Rm}_{2}^{1}\left(e_{1}, e_{2}\right)=2 d \omega_{2}^{1}\left(e_{1}, e_{2}\right) K ≑ ⟨ Rm ( e 1 , e 2 ) e 2 , e 1 ⟩ = 2 Rm 2 1 ( e 1 , e 2 ) = 2 d ω 2 1 ( e 1 , e 2 ) 练习 1.70 （联络1-形式的公式） 。证明
d ω k ( e i , e j ) = 1 2 ω i k ( e j ) − 1 2 ω j k ( e i ) d \omega^{k}\left(e_{i}, e_{j}\right)=\frac{1}{2} \omega_{i}^{k}\left(e_{j}\right)-\frac{1}{2} \omega_{j}^{k}\left(e_{i}\right) d ω k ( e i , e j ) = 2 1 ω i k ( e j ) − 2 1 ω j k ( e i ) 利用此式及第一结构方程（1.79），推导出联络1-形式的公式：
ω i k ( e j ) = d ω i ( e j , e k ) + d ω j ( e i , e k ) − d ω k ( e j , e i ) (1.81) \omega_{i}^{k}\left(e_{j}\right)=d \omega^{i}\left(e_{j}, e_{k}\right)+d \omega^{j}\left(e_{i}, e_{k}\right)-d \omega^{k}\left(e_{j}, e_{i}\right) \tag{1.81} ω i k ( e j ) = d ω i ( e j , e k ) + d ω j ( e i , e k ) − d ω k ( e j , e i ) ( 1.81 ) 注意，这与Christoffel符号公式（1.5）有相似之处。度量可以写成：
g = ∑ i = 1 n ω i ⊗ ω i g=\sum_{i=1}^{n} \omega^{i} \otimes \omega^{i} g = i = 1 ∑ n ω i ⊗ ω i 联络1-形式 ω i j \omega_{i}^{j} ω i j { e i } i = 1 n \left\{e_{i}\right\}_{i=1}^{n} { e i } i = 1 n
∇ X e i ≑ ∑ j = 1 n ω i j ( X ) e j (1.76) \nabla_{X} e_{i} \doteqdot \sum_{j=1}^{n} \omega_{i}^{j}(X) e_{j} \tag{1.76} ∇ X e i ≑ j = 1 ∑ n ω i j ( X ) e j ( 1.76 ) 对所有 i , j = 1 , … , n i, j=1, \ldots, n i , j = 1 , … , n U U U X X X
ω i j = − ω j i \omega_{i}^{j}=-\omega_{j}^{i} ω i j = − ω j i 因为对所有 X X X
0 ≡ X ⟨ e i , e j ⟩ = ⟨ ∇ X e i , e j ⟩ + ⟨ e i , ∇ X e j ⟩ 0 \equiv X\left\langle e_{i}, e_{j}\right\rangle=\left\langle\nabla_{X} e_{i}, e_{j}\right\rangle+\left\langle e_{i}, \nabla_{X} e_{j}\right\rangle 0 ≡ X ⟨ e i , e j ⟩ = ⟨ ∇ X e i , e j ⟩ + ⟨ e i , ∇ X e j ⟩ 由 ω i ( e j ) = δ j i \omega^{i}\left(e_{j}\right)=\delta_{j}^{i} ω i ( e j ) = δ j i
∇ X ω i = − ω j i ( X ) ω j (1.77) \nabla_{X} \omega^{i}=-\omega_{j}^{i}(X) \omega^{j} \tag{1.77} ∇ X ω i = − ω j i ( X ) ω j ( 1.77 ) 在 U U U R m i j \mathrm{Rm}_{i}^{j} Rm i j
Rm i j ( X , Y ) e j ≑ 1 2 Rm ( X , Y ) e i (1.78) \operatorname{Rm}_{i}^{j}(X, Y) e_{j} \doteqdot \frac{1}{2} \operatorname{Rm}(X, Y) e_{i} \tag{1.78} Rm i j ( X , Y ) e j ≑ 2 1 Rm ( X , Y ) e i ( 1.78 ) 因此 Rm i j ( X , Y ) = 1 2 ⟨ Rm ( X , Y ) e i , e j ⟩ \operatorname{Rm}_{i}^{j}(X, Y)=\frac{1}{2}\left\langle\operatorname{Rm}(X, Y) e_{i}, e_{j}\right\rangle Rm i j ( X , Y ) = 2 1 ⟨ Rm ( X , Y ) e i , e j ⟩
定理 1.69 （Cartan结构方程） 。第一和第二Cartan结构方程为：
d ω i = ω j ∧ ω j i Rm i j = d ω i j − ω i k ∧ ω k j \begin{align*} d \omega^{i} & =\omega^{j} \wedge \omega_{j}^{i} \tag{1.79}\\ \operatorname{Rm}_{i}^{j} & =d \omega_{i}^{j}-\omega_{i}^{k} \wedge \omega_{k}^{j} \tag{1.80} \end{align*} d ω i Rm i j = ω j ∧ ω j i = d ω i j − ω i k ∧ ω k j ( 1.79 ) ( 1.80 ) 证明。我们计算：
d ω i ( X , Y ) = 1 2 ( ∇ X ω i ) ( Y ) − 1 2 ( ∇ Y ω i ) ( X ) = − 1 2 ω j i ( X ) ω j ( Y ) + 1 2 ω j i ( Y ) ω j ( X ) d \omega^{i}(X, Y) =\frac{1}{2}\left(\nabla_{X} \omega^{i}\right)(Y)-\frac{1}{2}\left(\nabla_{Y} \omega^{i}\right)(X) = -\frac{1}{2} \omega_{j}^{i}(X) \omega^{j}(Y) + \frac{1}{2} \omega_{j}^{i}(Y) \omega^{j}(X) d ω i ( X , Y ) = 2 1 ( ∇ X ω i ) ( Y ) − 2 1 ( ∇ Y ω i ) ( X ) = − 2 1 ω j i ( X ) ω j ( Y ) + 2 1 ω j i ( Y ) ω j ( X ) 从而（1.79）得证。由（1.8）和
∇ 2 e i = ( ∇ ω i k ) e k + ω i k ∇ e k \nabla^{2} e_{i}=\left(\nabla \omega_{i}^{k}\right) e_{k}+\omega_{i}^{k} \nabla e_{k} ∇ 2 e i = ( ∇ ω i k ) e k + ω i k ∇ e k **练习 1.71 （第二Bianchi恒等式）。**证明
( d ∇ R m ) i j ≑ d R m i j − ω i k ∧ R m k j + ω k j ∧ R m i k = 0 \left(d_{\nabla} \mathrm{Rm}\right)_{i}^{j} \doteqdot d \mathrm{Rm}_{i}^{j}-\omega_{i}^{k} \wedge \mathrm{Rm}_{k}^{j}+\omega_{k}^{j} \wedge \mathrm{Rm}_{i}^{k}=0 ( d ∇ Rm ) i j ≑ d Rm i j − ω i k ∧ Rm k j + ω k j ∧ Rm i k = 0 这里 d ∇ R m i j d_{\nabla} \mathrm{Rm}_{i}^{j} d ∇ Rm i j R m \mathrm{Rm} Rm T ∗ M n ⊗ T M n T^{*} M^{n} \otimes T M^{n} T ∗ M n ⊗ T M n
练习 1.72 。证明：若 ( M 2 , g ) \left(M^{2}, g\right) ( M 2 , g ) u : M 2 → R u: M^{2} \rightarrow \mathbb{R} u : M 2 → R
R ( e u g ) = e − u ( R ( g ) − Δ g u ) R\left(e^{u} g\right)=e^{-u}\left(R(g)-\Delta_{g} u\right) R ( e u g ) = e − u ( R ( g ) − Δ g u ) 解答见例如 [163] 中引理 5.3。注意，可以使用（1.81）来进行计算，而不是采用 [163] 中的证明方法。
7.2. 度量的共形变化下的曲率# 拉普拉斯算子和Hessian算子在计算共形相关度量的曲率关系时出现。
练习 1.73 。令 g ~ = e 2 u g \tilde{g}=e^{2u} g g ~ = e 2 u g { ω i } i = 1 n \left\{\omega^{i}\right\}_{i=1}^{n} { ω i } i = 1 n g g g { ω ~ i } i = 1 n \left\{\tilde{\omega}^{i}\right\}_{i=1}^{n} { ω ~ i } i = 1 n ω ~ i = e u ω i \tilde{\omega}^{i}=e^{u} \omega^{i} ω ~ i = e u ω i g ~ \tilde{g} g ~ { e i } i = 1 n \left\{e_{i}\right\}_{i=1}^{n} { e i } i = 1 n { e ~ i } i = 1 n \left\{\tilde{e}_{i}\right\}_{i=1}^{n} { e ~ i } i = 1 n { ω i } \left\{\omega^{i}\right\} { ω i } { ω ~ i } \left\{\tilde{\omega}^{i}\right\} { ω ~ i } e ~ i = e − u e i \tilde{e}_{i}=e^{-u} e_{i} e ~ i = e − u e i R m i j \mathrm{Rm}_{i}^{j} Rm i j R m i j ~ \widetilde{\mathrm{Rm}_{i}^{j}} Rm i j
Rm i j ~ = Rm i j + ∇ e k ∇ e i u ⋅ ω k ∧ ω j − ∇ e k ∇ e j u ⋅ ω k ∧ ω i + ∣ ∇ u ∣ 2 ω i ∧ ω j + d u ∧ ( e j ( u ) ω i − e i ( u ) ω j ) \begin{align*} \widetilde{\operatorname{Rm}_{i}^{j}} & =\operatorname{Rm}_{i}^{j}+\nabla_{e_{k}} \nabla_{e_{i}} u \cdot \omega^{k} \wedge \omega^{j}-\nabla_{e_{k}} \nabla_{e_{j}} u \cdot \omega^{k} \wedge \omega^{i} \\ & +|\nabla u|^{2} \omega^{i} \wedge \omega^{j}+d u \wedge\left(e_{j}(u) \omega^{i}-e_{i}(u) \omega^{j}\right) \tag{1.82} \end{align*} Rm i j = Rm i j + ∇ e k ∇ e i u ⋅ ω k ∧ ω j − ∇ e k ∇ e j u ⋅ ω k ∧ ω i + ∣∇ u ∣ 2 ω i ∧ ω j + d u ∧ ( e j ( u ) ω i − e i ( u ) ω j ) ( 1.82 ) 其中 ∇ \nabla ∇ g g g
注释 1.74 。我们采用的约定是，对于1-形式 α \alpha α β \beta β α ∧ β = 1 2 ( α ⊗ β − β ⊗ α ) \alpha \wedge \beta=\frac{1}{2}(\alpha \otimes \beta-\beta \otimes \alpha) α ∧ β = 2 1 ( α ⊗ β − β ⊗ α )
应用（1.82）到（1.78），我们计算出Ricci曲率的关系为
Rc ~ ( e ~ ℓ , e ~ i ) = ∑ m = 1 n ⟨ Rm ~ ( e ~ m , e ~ ℓ ) e ~ i , e ~ m ⟩ = 2 ∑ m = 1 n ⟨ Rm i j ~ ( e ~ m , e ~ ℓ ) ⋅ e ~ j , e ~ m ⟩ = 2 ∑ m = 1 n Rm i m ~ ( e ~ m , e ~ ℓ ) = e − 2 u ( Rc ( e ℓ , e i ) + ( 2 − n ) ∇ e ℓ ∇ e i u − δ ℓ i Δ u + ∣ ∇ u ∣ 2 ( 2 − n ) δ i ℓ − ( 2 − n ) e ℓ ( u ) e i ( u ) ) . \begin{aligned} \widetilde{\operatorname{Rc}}\left(\tilde{e}_{\ell}, \tilde{e}_{i}\right) & =\sum_{m=1}^{n}\left\langle\widetilde{\operatorname{Rm}}\left(\tilde{e}_{m}, \tilde{e}_{\ell}\right) \tilde{e}_{i}, \tilde{e}_{m}\right\rangle=2 \sum_{m=1}^{n}\left\langle\widetilde{\operatorname{Rm}_{i}^{j}}\left(\tilde{e}_{m}, \tilde{e}_{\ell}\right) \cdot \tilde{e}_{j}, \tilde{e}_{m}\right\rangle \\ & =2 \sum_{m=1}^{n} \widetilde{\operatorname{Rm}_{i}^{m}}\left(\tilde{e}_{m}, \tilde{e}_{\ell}\right) \\ & =e^{-2 u}\binom{\operatorname{Rc}\left(e_{\ell}, e_{i}\right)+(2-n) \nabla_{e_{\ell}} \nabla_{e_{i}} u-\delta_{\ell i} \Delta u}{+|\nabla u|^{2}(2-n) \delta_{i \ell}-(2-n) e_{\ell}(u) e_{i}(u)} . \end{aligned} Rc ( e ~ ℓ , e ~ i ) = m = 1 ∑ n ⟨ Rm ( e ~ m , e ~ ℓ ) e ~ i , e ~ m ⟩ = 2 m = 1 ∑ n ⟨ Rm i j ( e ~ m , e ~ ℓ ) ⋅ e ~ j , e ~ m ⟩ = 2 m = 1 ∑ n Rm i m ( e ~ m , e ~ ℓ ) = e − 2 u ( + ∣∇ u ∣ 2 ( 2 − n ) δ i ℓ − ( 2 − n ) e ℓ ( u ) e i ( u ) Rc ( e ℓ , e i ) + ( 2 − n ) ∇ e ℓ ∇ e i u − δ ℓ i Δ u ) . 因此，如果 g ~ = e 2 u g \tilde{g}=e^{2 u} g g ~ = e 2 u g g g g g ~ \tilde{g} g ~
R ~ = ∑ i = 1 n Rc ~ ( e ~ i , e ~ i ) = e − 2 u ( R − 2 ( n − 1 ) Δ u − ( n − 2 ) ( n − 1 ) ∣ ∇ u ∣ 2 ) \begin{align*} \tilde{R} & =\sum_{i=1}^{n} \widetilde{\operatorname{Rc}}\left(\tilde{e}_{i}, \tilde{e}_{i}\right) \\ & =e^{-2 u}\left(R-2(n-1) \Delta u-(n-2)(n-1)|\nabla u|^{2}\right) \tag{1.83} \end{align*} R ~ = i = 1 ∑ n Rc ( e ~ i , e ~ i ) = e − 2 u ( R − 2 ( n − 1 ) Δ u − ( n − 2 ) ( n − 1 ) ∣∇ u ∣ 2 ) ( 1.83 ) 注释 1.75 。如果我们令 u ≑ − 1 n f u \doteqdot-\frac{1}{n} f u ≑ − n 1 f f : M n → R f: M^{n} \rightarrow \mathbb{R} f : M n → R
R ~ = e 2 n f ( R + 2 ( 1 − 1 n ) Δ f − ( 1 − 2 n ) ( 1 − 1 n ) ∣ ∇ f ∣ 2 ) \tilde{R}=e^{\frac{2}{n} f}\left(R+2\left(1-\frac{1}{n}\right) \Delta f-\left(1-\frac{2}{n}\right)\left(1-\frac{1}{n}\right)|\nabla f|^{2}\right) R ~ = e n 2 f ( R + 2 ( 1 − n 1 ) Δ f − ( 1 − n 2 ) ( 1 − n 1 ) ∣∇ f ∣ 2 ) 形式上，如果我们取 n → ∞ n \rightarrow \infty n → ∞
R ~ → R + 2 Δ f − ∣ ∇ f ∣ 2 \begin{equation*} \tilde{R} \rightarrow R+2 \Delta f-|\nabla f|^{2} \tag{1.84} \end{equation*} R ~ → R + 2Δ f − ∣∇ f ∣ 2 ( 1.84 ) 7.3. 高斯-博内公式# 我们使用移动框架法来证明黎曼几何中最基本的结果之一，即高斯-博内公式。它指出，对于一个封闭的黎曼曲面 ( M 2 , g ) \left(M^{2}, g\right) ( M 2 , g ) 2 π 2\pi 2 π M 2 M^{2} M 2
定理 1.76 (高斯-博内) 。如果 ( M 2 , g ) \left(M^{2}, g\right) ( M 2 , g )
∫ M K d A = 2 π ⋅ χ ( M ) \int_{M} K dA = 2\pi \cdot \chi(M) ∫ M K d A = 2 π ⋅ χ ( M ) 该公式的证明将占据本小节的剩余部分。令 { e 1 , e 2 } \left\{e_1, e_2\right\} { e 1 , e 2 } U ⊂ M U \subset M U ⊂ M
d A = ω 1 ∧ ω 2 dA = \omega^1 \wedge \omega^2 d A = ω 1 ∧ ω 2 高斯-博内积分的被积项在局部是联络1-形式 − ω 1 2 -\omega_1^2 − ω 1 2
K d A = − d ω 1 2 (1.85) K dA = -d\omega_1^2 \tag{1.85} K d A = − d ω 1 2 ( 1.85 ) 我们希望应用Stokes定理。由于 ω 1 2 \omega_1^2 ω 1 2 M M M V ∈ C ∞ ( M ) V \in C^{\infty}(M) V ∈ C ∞ ( M ) p 1 , … , p k p_1, \dots, p_k p 1 , … , p k V V V M M M M 2 − { p 1 , … , p k } M^2 - \{p_1, \dots, p_k\} M 2 − { p 1 , … , p k }
e 1 = V ∣ V ∣ e_1 = \frac{V}{|V|} e 1 = ∣ V ∣ V 然后 e 2 e_2 e 2 { e 1 , e 2 } \{e_1, e_2\} { e 1 , e 2 }
引理 1.77 。对于 ε < inj ( M 2 , g ) \varepsilon < \operatorname{inj}(M^2, g) ε < inj ( M 2 , g ) B ( p i , ε ) ≑ { x ∈ M : d ( x , p i ) < ε } B(p_i, \varepsilon) \doteqdot \{x \in M: d(x, p_i) < \varepsilon\} B ( p i , ε ) ≑ { x ∈ M : d ( x , p i ) < ε } i = 1 , … , k i=1, \dots, k i = 1 , … , k
∫ M K d A = ∑ i = 1 k lim ε → 0 ∫ ∂ B ( p i , ε ) ω 1 2 \int_M K dA = \sum_{i=1}^k \lim_{\varepsilon \rightarrow 0} \int_{\partial B(p_i, \varepsilon)} \omega_1^2 ∫ M K d A = i = 1 ∑ k ε → 0 lim ∫ ∂ B ( p i , ε ) ω 1 2 证明。根据(1.85)、Stokes定理，并考虑方向，我们有
∫ M K d A = − lim ε → 0 ∫ M − ⋃ i = 1 k B ( p i , ε ) d ω 1 2 = lim ε → 0 ∫ ⋃ i = 1 k ∂ B ( p i , ε ) ω 1 2 (1.86) \int_M K dA = -\lim_{\varepsilon \rightarrow 0} \int_{M - \bigcup_{i=1}^k B(p_i, \varepsilon)} d\omega_1^2 = \lim_{\varepsilon \rightarrow 0} \int_{\bigcup_{i=1}^k \partial B(p_i, \varepsilon)} \omega_1^2 \tag{1.86} ∫ M K d A = − ε → 0 lim ∫ M − ⋃ i = 1 k B ( p i , ε ) d ω 1 2 = ε → 0 lim ∫ ⋃ i = 1 k ∂ B ( p i , ε ) ω 1 2 ( 1.86 ) 我们将证明(1.86)右侧的边界积分趋近于向量场 V V V 2 π 2\pi 2 π
引理 1.78.
lim ϵ → 0 ∫ ∂ B ( p i , ε ) ω 1 2 = 2 π ⋅ index p i ( V ) \begin{equation*} \lim _{\epsilon \rightarrow 0} \int_{\partial B\left(p_{i}, \varepsilon\right)} \omega_{1}^{2}=2 \pi \cdot \operatorname{index}_{p_{i}}(V) \tag{1.87} \end{equation*} ϵ → 0 lim ∫ ∂ B ( p i , ε ) ω 1 2 = 2 π ⋅ index p i ( V ) ( 1.87 ) 假设该引理成立，我们可以完成 高斯-博内公式的证明。使用 (1.86)、(1.87) 和庞加莱-霍普夫定理，我们有
∫ M K d A = 2 π ∑ i = 1 k index p i ( V ) = 2 π ⋅ index M ( V ) = 2 π ⋅ χ ( M ) \int_{M} K d A=2 \pi \sum_{i=1}^{k} \operatorname{index}_{p_{i}}(V)=2 \pi \cdot \operatorname{index}_{M}(V)=2 \pi \cdot \chi(M) ∫ M K d A = 2 π i = 1 ∑ k index p i ( V ) = 2 π ⋅ index M ( V ) = 2 π ⋅ χ ( M ) 我们以 引理 1.78 的证明结束本节。给定 1 ≤ i ≤ k 1 \leq i \leq k 1 ≤ i ≤ k ν \nu ν ∂ B ( p i , ε ) \partial B\left(p_{i}, \varepsilon\right) ∂ B ( p i , ε ) T T T ∂ B ( p i , ε ) \partial B\left(p_{i}, \varepsilon\right) ∂ B ( p i , ε ) { ν , T } \{\nu, T\} { ν , T } ∂ B ( p i , ε ) \partial B\left(p_{i}, \varepsilon\right) ∂ B ( p i , ε )
γ i : [ 0 , L i ] → ∂ B ( p i , ε ) ⊂ M \gamma_{i}:\left[0, L_{i}\right] \rightarrow \partial B\left(p_{i}, \varepsilon\right) \subset M γ i : [ 0 , L i ] → ∂ B ( p i , ε ) ⊂ M 使得 d γ i / d s = T d \gamma_{i} / d s=T d γ i / d s = T
θ : [ 0 , L i ] → R \theta:\left[0, L_{i}\right] \rightarrow \mathbb{R} θ : [ 0 , L i ] → R 如下所示
T ( γ i ( s ) ) ≑ cos θ ( s ) ⋅ e 1 ( γ i ( s ) ) + sin θ ( s ) ⋅ e 2 ( γ i ( s ) ) T\left(\gamma_{i}(s)\right) \doteqdot \cos \theta(s) \cdot e_{1}\left(\gamma_{i}(s)\right)+\sin \theta(s) \cdot e_{2}\left(\gamma_{i}(s)\right) T ( γ i ( s ) ) ≑ cos θ ( s ) ⋅ e 1 ( γ i ( s ) ) + sin θ ( s ) ⋅ e 2 ( γ i ( s ) ) 我们要求 θ \theta θ 2 π 2 \pi 2 π ν \nu ν
ν ( γ i ( s ) ) = sin θ ( s ) ⋅ e 1 ( γ i ( s ) ) − cos θ ( s ) ⋅ e 2 ( γ i ( s ) ) \nu\left(\gamma_{i}(s)\right)=\sin \theta(s) \cdot e_{1}\left(\gamma_{i}(s)\right)-\cos \theta(s) \cdot e_{2}\left(\gamma_{i}(s)\right) ν ( γ i ( s ) ) = sin θ ( s ) ⋅ e 1 ( γ i ( s ) ) − cos θ ( s ) ⋅ e 2 ( γ i ( s ) ) 围绕 γ i \gamma_{i} γ i θ \theta θ
θ ( L i ) − θ ( 0 ) = − 2 π ( index p i ( V ) − 1 ) \theta\left(L_{i}\right)-\theta(0)=-2 \pi\left(\operatorname{index}_{p_{i}}(V)-1\right) θ ( L i ) − θ ( 0 ) = − 2 π ( index p i ( V ) − 1 ) 上述关于角度 θ \theta θ
子引理.
ω 1 2 ( T ) = κ − T θ \begin{equation*} \omega_{1}^{2}(T)=\kappa-T \theta \tag{1.88} \end{equation*} ω 1 2 ( T ) = κ − Tθ ( 1.88 ) 其中 κ ≑ − ⟨ ∇ T T , ν ⟩ \kappa \doteqdot-\left\langle\nabla_{T} T, \nu\right\rangle κ ≑ − ⟨ ∇ T T , ν ⟩ ∂ B ( p i , ε ) \partial B\left(p_{i}, \varepsilon\right) ∂ B ( p i , ε )
子引理的证明。我们计算
∇ T T = ∇ T ( cos θ ⋅ e 1 + sin θ ⋅ e 2 ) = T ( cos θ ) ⋅ e 1 + T ( sin θ ) ⋅ e 2 + cos θ ⋅ ∇ T e 1 + sin θ ⋅ ∇ T e 2 = − sin θ ⋅ T θ ⋅ e 1 + cos θ ⋅ T θ ⋅ e 2 + cos θ ⋅ ω 1 2 ( T ) e 2 + sin θ ⋅ ω 2 1 ( T ) e 1 = − ν ⋅ ( T θ + ω 1 2 ( T ) ) \begin{aligned} \nabla_{T} T & =\nabla_{T}\left(\cos \theta \cdot e_{1}+\sin \theta \cdot e_{2}\right) \\ & =T(\cos \theta) \cdot e_{1}+T(\sin \theta) \cdot e_{2}+\cos \theta \cdot \nabla_{T} e_{1}+\sin \theta \cdot \nabla_{T} e_{2} \\ & =-\sin \theta \cdot T \theta \cdot e_{1}+\cos \theta \cdot T \theta \cdot e_{2}+\cos \theta \cdot \omega_{1}^{2}(T) e_{2}+\sin \theta \cdot \omega_{2}^{1}(T) e_{1} \\ & =-\nu \cdot\left(T \theta+\omega_{1}^{2}(T)\right) \end{aligned} ∇ T T = ∇ T ( cos θ ⋅ e 1 + sin θ ⋅ e 2 ) = T ( cos θ ) ⋅ e 1 + T ( sin θ ) ⋅ e 2 + cos θ ⋅ ∇ T e 1 + sin θ ⋅ ∇ T e 2 = − sin θ ⋅ Tθ ⋅ e 1 + cos θ ⋅ Tθ ⋅ e 2 + cos θ ⋅ ω 1 2 ( T ) e 2 + sin θ ⋅ ω 2 1 ( T ) e 1 = − ν ⋅ ( Tθ + ω 1 2 ( T ) ) 这意味着
κ = − ⟨ ∇ T T , ν ⟩ = T θ + ω 1 2 ( T ) \kappa=-\left\langle\nabla_{T} T, \nu\right\rangle=T \theta+\omega_{1}^{2}(T) κ = − ⟨ ∇ T T , ν ⟩ = Tθ + ω 1 2 ( T ) 设 d s d s d s ∂ B ( p i , ε ) \partial B\left(p_{i}, \varepsilon\right) ∂ B ( p i , ε )
∫ ∂ B ( p i , ε ) ω 1 2 = ∫ ∂ B ( p i , ε ) ω 1 2 ( T ) d s = ∫ ∂ B ( p i , ε ) ( κ − T θ ) d s = ∫ ∂ B ( p i , ε ) κ d s − θ ( L i ) + θ ( 0 ) = ∫ ∂ B ( p i , ε ) κ d s + 2 π ( index p i ( V ) − 1 ) \begin{aligned} \int_{\partial B\left(p_{i}, \varepsilon\right)} \omega_{1}^{2} & =\int_{\partial B\left(p_{i}, \varepsilon\right)} \omega_{1}^{2}(T) d s=\int_{\partial B\left(p_{i}, \varepsilon\right)}(\kappa-T \theta) d s \\ & =\int_{\partial B\left(p_{i}, \varepsilon\right)} \kappa d s-\theta\left(L_{i}\right)+\theta(0) \\ & =\int_{\partial B\left(p_{i}, \varepsilon\right)} \kappa d s+2 \pi\left(\operatorname{index}_{p_{i}}(V)-1\right) \end{aligned} ∫ ∂ B ( p i , ε ) ω 1 2 = ∫ ∂ B ( p i , ε ) ω 1 2 ( T ) d s = ∫ ∂ B ( p i , ε ) ( κ − Tθ ) d s = ∫ ∂ B ( p i , ε ) κ d s − θ ( L i ) + θ ( 0 ) = ∫ ∂ B ( p i , ε ) κ d s + 2 π ( index p i ( V ) − 1 ) 因为
lim ϵ → 0 ∫ ∂ B ( p i , ε ) κ d s = 2 π \lim _{\epsilon \rightarrow 0} \int_{\partial B\left(p_{i}, \varepsilon\right)} \kappa d s=2 \pi ϵ → 0 lim ∫ ∂ B ( p i , ε ) κ d s = 2 π 我们得出
∑ i = 1 k lim ϵ → 0 ∫ ∂ B ( p i , ε ) ω 1 2 = 2 π ⋅ ∑ i = 1 k index p i ( V ) \sum_{i=1}^{k} \lim _{\epsilon \rightarrow 0} \int_{\partial B\left(p_{i}, \varepsilon\right)} \omega_{1}^{2}=2 \pi \cdot \sum_{i=1}^{k} \operatorname{index}_{p_{i}}(V) i = 1 ∑ k ϵ → 0 lim ∫ ∂ B ( p i , ε ) ω 1 2 = 2 π ⋅ i = 1 ∑ k index p i ( V ) 至此，引理和高斯-博内公式均已证明。
7.4 移动标架适应于超曲面# 研究黎曼流形中的超曲面对探究这些流形的几何具有重要作用。例如，在三维流形中，极小曲面有助于理解周围三维流形的拓扑结构。设( P n , g P ) \left(P^{n}, g_{P}\right) ( P n , g P ) D D D M n − 1 ⊂ P n M^{n-1} \subset P^{n} M n − 1 ⊂ P n { e i } i = 1 n \left\{e_{i}\right\}_{i=1}^{n} { e i } i = 1 n M n − 1 M^{n-1} M n − 1 U ⊂ P n U \subset P^{n} U ⊂ P n ( P n , g P ) \left(P^{n}, g_{P}\right) ( P n , g P ) { ω i j } i , j = 1 n \left\{\omega_{i}^{j}\right\}_{i, j=1}^{n} { ω i j } i , j = 1 n
D X e i = ∑ j = 1 n ω i j ( X ) e j . D_{X} e_{i}=\sum_{j=1}^{n} \omega_{i}^{j}(X) e_{j} . D X e i = j = 1 ∑ n ω i j ( X ) e j . 现在假设标架适应于M n − 1 M^{n-1} M n − 1 e n ≑ ν e_{n} \doteqdot \nu e n ≑ ν M n − 1 M^{n-1} M n − 1
g M ( X , Y ) ≑ g P ( X , Y ) \begin{equation*} g_{M}(X, Y) \doteqdot g_{P}(X, Y) \tag{1.89} \end{equation*} g M ( X , Y ) ≑ g P ( X , Y ) ( 1.89 ) 对于X , Y ∈ T M n − 1 X, Y \in T M^{n-1} X , Y ∈ T M n − 1
h ( X , Y ) ≑ ⟨ D X ν , Y ⟩ = ω n j ( X ) ⟨ Y , e j ⟩ \begin{equation*} h(X, Y) \doteqdot\left\langle D_{X} \nu, Y\right\rangle=\omega_{n}^{j}(X)\left\langle Y, e_{j}\right\rangle \tag{1.90} \end{equation*} h ( X , Y ) ≑ ⟨ D X ν , Y ⟩ = ω n j ( X ) ⟨ Y , e j ⟩ ( 1.90 ) 对于切于M n − 1 M^{n-1} M n − 1 X X X Y Y Y h i j ≑ h ( e i , e j ) = ω n j ( e i ) h_{i j} \doteqdot h\left(e_{i}, e_{j}\right)=\omega_{n}^{j}\left(e_{i}\right) h ij ≑ h ( e i , e j ) = ω n j ( e i )
ω n j = ∑ i = 1 n − 1 h i j ω i \omega_{n}^{j}=\sum_{i=1}^{n-1} h_{i j} \omega^{i} ω n j = i = 1 ∑ n − 1 h ij ω i 平均曲率是第二基本形式的迹：
H ≑ ∑ i = 1 n h ( e i , e i ) H \doteqdot \sum_{i=1}^{n} h\left(e_{i}, e_{i}\right) H ≑ i = 1 ∑ n h ( e i , e i ) 诱导的g M g_{M} g M ∇ \nabla ∇
∇ X e i ≑ ( D X e i ) T = ∑ j = 1 n − 1 ω i j ( X ) e j \nabla_{X} e_{i} \doteqdot\left(D_{X} e_{i}\right)^{T}=\sum_{j=1}^{n-1} \omega_{i}^{j}(X) e_{j} ∇ X e i ≑ ( D X e i ) T = j = 1 ∑ n − 1 ω i j ( X ) e j 这里T { }^{T} T { ω i j } i , j = 1 n − 1 \left\{\omega_{i}^{j}\right\}_{i, j=1}^{n-1} { ω i j } i , j = 1 n − 1 ( M n − 1 , g M ) \left(M^{n-1}, g_{M}\right) ( M n − 1 , g M )
( Rm P ) i j = d ω i j − ∑ k = 1 n ω i k ∧ ω k j , i , j = 1 , … , n ( Rm M ) i j = d ω i j − ∑ k = 1 n − 1 ω i k ∧ ω k j , i , j = 1 , … , n − 1 \begin{array}{ll} \left(\operatorname{Rm}_{P}\right)_{i}^{j}=d \omega_{i}^{j}-\sum_{k=1}^{n} \omega_{i}^{k} \wedge \omega_{k}^{j}, & i, j=1, \ldots, n \\ \left(\operatorname{Rm}_{M}\right)_{i}^{j}=d \omega_{i}^{j}-\sum_{k=1}^{n-1} \omega_{i}^{k} \wedge \omega_{k}^{j}, & i, j=1, \ldots, n-1 \end{array} ( Rm P ) i j = d ω i j − ∑ k = 1 n ω i k ∧ ω k j , ( Rm M ) i j = d ω i j − ∑ k = 1 n − 1 ω i k ∧ ω k j , i , j = 1 , … , n i , j = 1 , … , n − 1 因此，对于i , j = 1 , … , n − 1 i, j=1, \ldots, n-1 i , j = 1 , … , n − 1
( Rm M ) i j = ( Rm P ) i j + ω i n ∧ ω n j = ( Rm P ) i j − h i k h j ℓ ω k ∧ ω ℓ . \begin{aligned} \left(\operatorname{Rm}_{M}\right)_{i}^{j} & =\left(\operatorname{Rm}_{P}\right)_{i}^{j}+\omega_{i}^{n} \wedge \omega_{n}^{j} \\ & =\left(\operatorname{Rm}_{P}\right)_{i}^{j}-h_{i k} h_{j \ell} \omega^{k} \wedge \omega^{\ell} . \end{aligned} ( Rm M ) i j = ( Rm P ) i j + ω i n ∧ ω n j = ( Rm P ) i j − h ik h j ℓ ω k ∧ ω ℓ . 应用于( e k , e ℓ ) \left(e_{k}, e_{\ell}\right) ( e k , e ℓ )
( Rm M ) i j k ℓ = ( Rm P ) i j k ℓ + h i ℓ h j k − h i k h j ℓ \begin{equation*} \left(\operatorname{Rm}_{M}\right)_{i j k \ell}=\left(\operatorname{Rm}_{P}\right)_{i j k \ell}+h_{i \ell} h_{j k}-h_{i k} h_{j \ell} \tag{1.91} \end{equation*} ( Rm M ) ijk ℓ = ( Rm P ) ijk ℓ + h i ℓ h jk − h ik h j ℓ ( 1.91 ) 练习 1.79 证明
( Rc M ) j k = ( Rc P ) j k − ( R P ) n j k n + H h j k − h j ℓ h ℓ k \left(\operatorname{Rc}_{M}\right)_{j k}=\left(\operatorname{Rc}_{P}\right)_{j k}-\left(R_{P}\right)_{n j k n}+H h_{j k}-h_{j \ell} h_{\ell k} ( Rc M ) jk = ( Rc P ) jk − ( R P ) njkn + H h jk − h j ℓ h ℓ k 以及
R M = R P − 2 ( Rc P ) n n + H 2 − ∣ h ∣ 2 \begin{equation*} R_{M}=R_{P}-2\left(\operatorname{Rc}_{P}\right)_{n n}+H^{2}-|h|^{2} \tag{1.92} \end{equation*} R M = R P − 2 ( Rc P ) nn + H 2 − ∣ h ∣ 2 ( 1.92 ) 对于j = 1 , … , n − 1 j=1, \ldots, n-1 j = 1 , … , n − 1
( Rm P ) n j = d ω n j − ∑ k = 1 n − 1 ω n k ∧ ω k j \left(\operatorname{Rm}_{P}\right)_{n}^{j}=d \omega_{n}^{j}-\sum_{k=1}^{n-1} \omega_{n}^{k} \wedge \omega_{k}^{j} ( Rm P ) n j = d ω n j − k = 1 ∑ n − 1 ω n k ∧ ω k j (1,1)-张量W ≑ ∑ j = 1 n − 1 ω n j e j W \doteqdot \sum_{j=1}^{n-1} \omega_{n}^{j} e_{j} W ≑ ∑ j = 1 n − 1 ω n j e j W W W T M n − 1 T M^{n-1} T M n − 1
d ∇ W = ∑ j = 1 n − 1 ( R m P ) n j e j d_{\nabla} W=\sum_{j=1}^{n-1}\left(\mathrm{Rm}_{P}\right)_{n}^{j} e_{j} d ∇ W = j = 1 ∑ n − 1 ( Rm P ) n j e j 这是一个取值于T M n − 1 T M^{n-1} T M n − 1
练习 1.80 (Codazzi 方程) 证明对于切于M n − 1 M^{n-1} M n − 1 X , Y , Z X, Y, Z X , Y , Z
( ∇ X h ) ( Y , Z ) − ( ∇ Y h ) ( X , Z ) = − ⟨ Rm P ( X , Y ) Z , ν ⟩ \left(\nabla_{X} h\right)(Y, Z)-\left(\nabla_{Y} h\right)(X, Z)=-\left\langle\operatorname{Rm}_{P}(X, Y) Z, \nu\right\rangle ( ∇ X h ) ( Y , Z ) − ( ∇ Y h ) ( X , Z ) = − ⟨ Rm P ( X , Y ) Z , ν ⟩ 在第4章和第9章中关于梯度Ricci孤立子的研究中，我们将特别有用地考虑由光滑函数f : M n → R f: M^{n} \rightarrow \mathbb{R} f : M n → R f f f c ∈ R c \in \mathbb{R} c ∈ R f ( x ) = c f(x)=c f ( x ) = c x ∈ M x \in M x ∈ M ∇ f ( x ) ≠ 0 \nabla f(x) \neq 0 ∇ f ( x ) = 0 f − 1 ( c ) f^{-1}(c) f − 1 ( c ) f − 1 ( c ) f^{-1}(c) f − 1 ( c )
h ( V , W ) = Hess ( f ) ( V , W ) ∣ ∇ f ∣ \begin{equation*} h(V, W)=\frac{\operatorname{Hess}(f)(V, W)}{|\nabla f|} \tag{1.93} \end{equation*} h ( V , W ) = ∣∇ f ∣ Hess ( f ) ( V , W ) ( 1.93 ) 事实上，ν = ∇ f ∣ ∇ f ∣ \nu=\frac{\nabla f}{|\nabla f|} ν = ∣∇ f ∣ ∇ f f − 1 ( c ) f^{-1}(c) f − 1 ( c ) V , W ∈ T f − 1 ( c ) V, W \in T f^{-1}(c) V , W ∈ T f − 1 ( c )
h ( V , W ) = ⟨ ∇ V ν , W ⟩ = ⟨ ∇ V ∇ f ∣ ∇ f ∣ , W ⟩ = ⟨ ∇ V ∇ f ∣ ∇ f ∣ , W ⟩ = 1 ∣ ∇ f ∣ Hess ( f ) ( V , W ) \begin{aligned} h(V, W) & =\left\langle\nabla_{V} \nu, W\right\rangle=\left\langle\nabla_{V} \frac{\nabla f}{|\nabla f|}, W\right\rangle \\ & =\left\langle\frac{\nabla_{V} \nabla f}{|\nabla f|}, W\right\rangle=\frac{1}{|\nabla f|} \operatorname{Hess}(f)(V, W) \end{aligned} h ( V , W ) = ⟨ ∇ V ν , W ⟩ = ⟨ ∇ V ∣∇ f ∣ ∇ f , W ⟩ = ⟨ ∣∇ f ∣ ∇ V ∇ f , W ⟩ = ∣∇ f ∣ 1 Hess ( f ) ( V , W ) 因为⟨ ∇ f , W ⟩ = 0 \langle\nabla f, W\rangle=0 ⟨ ∇ f , W ⟩ = 0 f f f ∇ ∇ f ≥ 0 \nabla \nabla f \geq 0 ∇∇ f ≥ 0 ∇ ∇ f > 0 \nabla \nabla f>0 ∇∇ f > 0 f − 1 ( c ) f^{-1}(c) f − 1 ( c ) h ≥ 0 h \geq 0 h ≥ 0 h > 0 h>0 h > 0
