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# 结构方程

以下令正则曲面 (S,n)(S,\boldsymbol n) 正向参数化 σ:DS\boldsymbol \sigma: D \to S
固定点 p=σ(u,v)S\boldsymbol p = \boldsymbol \sigma(u,v) \in S

对切空间 TpST_{\boldsymbol p}S 进行代数式分析,基于 Gram–Schmidt 正交化构造正交归一基底 A={ε1,ε2,ε3}\mathscr A = \{\boldsymbol \varepsilon_1, \boldsymbol \varepsilon_2, \boldsymbol \varepsilon_3\} \quad

ε1=σuσu,ε2=σv(σvε1)ε1σv(σvε1)ε1,ε3=n\boldsymbol \varepsilon_1 = \frac{\boldsymbol \sigma_u}{\left|\boldsymbol \sigma_u\right|}, \quad \boldsymbol \varepsilon_2 = \frac{\boldsymbol \sigma_v - (\boldsymbol \sigma_v \cdot \boldsymbol \varepsilon_1) \boldsymbol \varepsilon_1}{\left|\boldsymbol \sigma_v - (\boldsymbol \sigma_v \cdot \boldsymbol \varepsilon_1) \boldsymbol \varepsilon_1\right|}, \quad \boldsymbol \varepsilon_3 = \boldsymbol n

A\mathscr A 为伴随 σ\boldsymbol \sigma活动标架 (Moving Frame)「動標構」

Carton Darboux 在曲率计算研究中,引入活动标架用于将复杂的曲率计算转为简单的代数计算,此标架也称 Carton 标架


以下通过坐标转换,获取 Carton 标架下的坐标

由于 σu,σvn\boldsymbol \sigma_u, \boldsymbol \sigma_v \perp \boldsymbol n,其在 ε3\boldsymbol \varepsilon_3 方向的投影为 00,所以有

{σu=a11ε1+a12ε2+0ε3σv=a21ε1+a22ε2+0ε3[σu]A=(a11a120),[σv]A=(a21a220)\begin{cases} \boldsymbol \sigma_u = a_1^1 \boldsymbol \varepsilon_1 + a_1^2 \boldsymbol \varepsilon_2 + 0 \cdot \boldsymbol \varepsilon_3 \\ \boldsymbol \sigma_v = a_2^1 \boldsymbol \varepsilon_1 + a_2^2 \boldsymbol \varepsilon_2 + 0 \cdot \boldsymbol \varepsilon_3 \end{cases} \Longrightarrow [\boldsymbol \sigma_u]_{\mathscr A} = \begin{pmatrix} a_1^1 \\ a_1^2 \\ 0 \end{pmatrix}, \quad [\boldsymbol \sigma_v]_{\mathscr A} = \begin{pmatrix} a_2^1 \\ a_2^2 \\ 0 \end{pmatrix}

其中 a1j=σuεj,a2j=σvεj(j=1,2)a_1^j = \boldsymbol \sigma_u \cdot \boldsymbol \varepsilon_j, \quad a_2^j = \boldsymbol \sigma_v \cdot \boldsymbol \varepsilon_j \quad (j = 1,2)
aa 的下标指示偏导方向编号,上标指示基底方向编号

由于重点在于分析二维切空间中的表现,去掉垂直于切空间的方向,令

A=(a11a21a12a22)A = \begin{pmatrix} a_1^1 & a_2^1 \\ a_1^2 & a_2^2 \end{pmatrix}

AA 成为二维空间内两种基底的过渡矩阵

(σu,σv)=(ε1,ε2)A(\boldsymbol \sigma_u, \boldsymbol \sigma_v) = (\boldsymbol \varepsilon_1, \boldsymbol \varepsilon_2) A

并且此时

σu×σv=det(A)e1×e2=det(A)nσu×σv=det(A)=EGF2\boldsymbol \sigma_u \times \boldsymbol \sigma_v = \det(A) \boldsymbol e_1 \times \boldsymbol e_2 = \det(A) \boldsymbol n \\ \left|\boldsymbol \sigma_u \times \boldsymbol \sigma_v\right| = \det(A) = \sqrt{EG - F^2}

进一步,为了获取 Carton 标架下的坐标,在 TpST_{\boldsymbol p}S 上定义构造坐标函数 du,dvdu,dv 与微分形式,令 θj=a1jdu+a2jdv(j=1,2)\theta^j = a_1^j du + a_2^j dv \quad (j = 1,2),则

(θ1θ2)=A(dudv)\begin{pmatrix} \theta^1 \\ \theta^2 \end{pmatrix} = A \begin{pmatrix} du \\ dv \end{pmatrix}

这其实是坐标变换,由于 AA(ε1,ε2)(\boldsymbol \varepsilon_1, \boldsymbol \varepsilon_2)(σu,σv)(\boldsymbol \sigma_u, \boldsymbol \sigma_v) 的过渡矩阵
所以等式右边是 (σu,σv)(\boldsymbol \sigma_u, \boldsymbol \sigma_v) 下的坐标经过坐标函数映射后的结果
等式左边是 Carton 标架下的坐标

此时全微分

dσ=σudu+σvdv=(σu,σv)(dudv)=(ε1,ε2)A(dudv)=(ε1,ε2)(θ1θ2)=θ1e1+θ2e2d\boldsymbol \sigma = \boldsymbol \sigma_u \, du + \boldsymbol \sigma_v \, dv = (\boldsymbol \sigma_u, \boldsymbol \sigma_v) \begin{pmatrix} du \\ dv \end{pmatrix} = (\boldsymbol \varepsilon_1, \boldsymbol \varepsilon_2) A \begin{pmatrix} du \\ dv \end{pmatrix} = (\boldsymbol \varepsilon_1, \boldsymbol \varepsilon_2) \begin{pmatrix} \theta^1 \\ \theta^2 \end{pmatrix} = \theta^1 \boldsymbol e_1 + \theta^2 \boldsymbol e_2

θj\theta^j 具有以下计算性质

命题

  • θ1θ1=θ2θ2=0\theta_1 \wedge \theta_1 = \theta_2 \wedge \theta_2 = 0

  • θ1θ2=θ2θ1=det(A)dudv\theta_1 \wedge \theta_2 = \theta_2 \wedge \theta_1 = \det(A) \, du \wedge dv


活动标架的关键是 活动,对于曲面上的点 p\boldsymbol p,在其沿曲面移动时,活动标架会 不同程度的移动
依此,曲面研究的重点:曲率 转化为活动标架的变化情况

定义 联络ωij=dεiεj(i,j=1,2,3)\omega_i^j = d \boldsymbol \varepsilon_i \cdot \boldsymbol \varepsilon_j \quad (i,j = 1,2,3)
显然 ωij\omega_i^j 指示两个活动标架之间的变化关系,并且存在固定方向,非对称性继承于楔积

dεi=ωi1ε1+ωi2ε2+ωi3ε3d\boldsymbol \varepsilon_i = \omega_i^1 \boldsymbol \varepsilon_1 + \omega_i^2 \boldsymbol \varepsilon_2 + \omega_i^3 \boldsymbol \varepsilon_3

联络具有以下性质

命题

ωii=0,ωij=ωji\omega_i^i = 0 , \quad \omega_i^j = - \omega_j^i


借助 TpST_{\boldsymbol p}S 形算子 Σp:TpSTpS\Sigma_{\boldsymbol p}: T_{\boldsymbol p}S \to T_{\boldsymbol p}S,定义 bij=εiΣp(εj)(i,j=1,2)b_{ij} = \boldsymbol \varepsilon_i \cdot \Sigma_{\boldsymbol p}(\boldsymbol \varepsilon_j) \quad (i,j = 1,2),构造矩阵

B=(b11b12b21b22)B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}

依据形算子的自伴性有:bij=bji,tB=Bb_{ij} = b_{ji}, \quad {}^tB = B
由于 Σp(εi)\Sigma_{\boldsymbol p}(\boldsymbol \varepsilon_i) 也处于基底 (ε1,ε2)(\boldsymbol \varepsilon_1, \boldsymbol \varepsilon_2) 所在的切空间内,取内积得到坐标 Σp(εi)=b1jε1+b2jε2\Sigma_{\boldsymbol p}(\boldsymbol \varepsilon_i) = b_{1j} \boldsymbol \varepsilon_1 + b_{2j} \boldsymbol \varepsilon_2

进一步

(Σp(e1),Σp(e2))=(e1,e2)B\left( \Sigma_{\boldsymbol p}(\boldsymbol e_1), \Sigma_{\boldsymbol p}(\boldsymbol e_2) \right) = (\boldsymbol e_1, \boldsymbol e_2) B

所以 BB 是线性映射 Σp\Sigma_{\boldsymbol p} 关于基底 (e1,e2)(\boldsymbol e_1, \boldsymbol e_2) 的矩阵表示,BB 的特征值 λ1,2\lambda_{1,2} 即为主曲率

K=det(B),H=12tr(B),κ1,2=λ1,2K = \det(B), \quad H = \frac{1}{2} \mathrm{tr}(B), \quad \kappa_{1,2} = \lambda_{1,2}

至此,曲率计算已完全转变为矩阵计算


特别地,联络的计算性质如下

命题

ω13=b11θ1+b12θ2,ω23=b21θ1+b22θ2\omega_1^3 = b_{11} \theta^1 + b_{12} \theta^2, \quad \omega_2^3 = b_{21} \theta^1 + b_{22} \theta^2

并且得到以下结构方程

定理 结构方程

  • dθ1=ω21θ2,dθ2=ω12θ1d \theta^1 = -\omega_2^1 \wedge \theta^2, \quad d \theta^2 = -\omega_1^2 \wedge \theta^1

  • dω21=Kθ1θ2d \omega_2^1 = K \cdot \theta^1 \wedge \theta^2

其中 KK 为 Gauss 曲率

# Riemannian 度规

回到常见的 R2\mathbb R^2

以下令开集 DR2D \subset \mathbb R^2
定点 q=(uv)D\boldsymbol q = \binom{u}{v} \in D
切空间 TqD=R2T_{\boldsymbol q}D = \mathbb R^2
正交归一标准基底 e1=(10),e2=(01)\boldsymbol e_1 = \binom{1}{0}, \quad \boldsymbol e_2 = \binom{0}{1} \quad

定义
对于 ξ,ηTqD\boldsymbol \xi, \boldsymbol \eta \in T_{\boldsymbol q}D
若函数 gg 定义为

g(q,ξ,η)=gq(ξ,η)R2g(\boldsymbol q, \boldsymbol \xi, \boldsymbol \eta) = g_{\boldsymbol q}(\boldsymbol \xi, \boldsymbol \eta) \in \mathbb R^2

满足

  • 对任意 qD\boldsymbol q \in Dgqg_{\boldsymbol q} 成为 TqDT_{\boldsymbol q}D 上的内积,即满足:正定,对称,线性
  • gg 关于 q\boldsymbol q 成为 DD 上的 CC^\infty 光滑函数

则称 ggRiemannian 度规 (Riemannian Metric)「リーマン計量」

gg 为 Riemannian 度规

gq(ξ,η)=gq(ξ1e1+ξ2e2,η1e1+η2e2)=i,j=12ξiηjgq(ei,ej)g_{\boldsymbol q}(\boldsymbol \xi, \boldsymbol \eta) = g_{\boldsymbol q}(\xi_1 \boldsymbol e_1 + \xi_2 \boldsymbol e_2, \eta_1 \boldsymbol e_1 + \eta_2 \boldsymbol e_2) = \sum_{i,j=1}^2 \xi_i \eta_j g_{\boldsymbol q}(\boldsymbol e_i, \boldsymbol e_j)

其中

ξ=(ξ1ξ2),η=(η1η2)\boldsymbol \xi = \binom{\xi_1}{\xi_2}, \quad \boldsymbol \eta = \binom{\eta_1}{\eta_2}

由此有基于 Riemannian 度规的基本量(注意 q=(uv)\boldsymbol q = \binom{u}{v},与内积的对称性)

{E(q)=gq(e1,e1)F(q)=gq(e1,e2)=gq(e2,e1)G(q)=gq(e2,e2)\begin{cases} E(\boldsymbol q) = g_{\boldsymbol q}(\boldsymbol e_1, \boldsymbol e_1) \\ F(\boldsymbol q) = g_{\boldsymbol q}(\boldsymbol e_1, \boldsymbol e_2) = g_{\boldsymbol q}(\boldsymbol e_2, \boldsymbol e_1) \\ G(\boldsymbol q) = g_{\boldsymbol q}(\boldsymbol e_2, \boldsymbol e_2) \end{cases}

那么

gq(ξ,η)=(ξ1,ξ2)(EFFG)(η1η2)=tξ(EFFG)ηg_{\boldsymbol q}(\boldsymbol \xi, \boldsymbol \eta) = (\xi_1, \xi_2) \begin{pmatrix} E & F \\ F & G \end{pmatrix} \begin{pmatrix} \eta_1 \\ \eta_2 \end{pmatrix} = {}^t\boldsymbol \xi \begin{pmatrix} E & F \\ F & G \end{pmatrix} \boldsymbol \eta

因为 E,F,GE,F,GDD 上光滑,所以由内积的正定性得到:对于任意 qD\boldsymbol q \in D

(E(q)F(q)F(q)G(q))\begin{pmatrix} E(\boldsymbol q) & F(\boldsymbol q) \\ F(\boldsymbol q) & G(\boldsymbol q) \end{pmatrix}

都为正定对称矩阵EGF2>0,E>0EG - F^2 \gt 0, E \gt 0

反过来,如果 DD 上的函数 E,F,GE,F,G 满足上述正定条件且光滑,那么

gq(ξ,η)=tξ(E(q)F(q)F(q)G(q))ηg_{\boldsymbol q}(\boldsymbol \xi, \boldsymbol \eta) = {}^t\boldsymbol \xi \begin{pmatrix} E(\boldsymbol q) & F(\boldsymbol q) \\ F(\boldsymbol q) & G(\boldsymbol q) \end{pmatrix} \boldsymbol \eta

成为 DD 上的 Riemannian 度规

示例
简单地令 gq(ξ,η)=ξηg_{\boldsymbol q}(\boldsymbol \xi, \boldsymbol \eta) = \boldsymbol \xi \cdot \boldsymbol \eta 为 Euclidean 内积
那么 gg 成为 DD 上的 Riemannian 度规,称为 Euclidean 度量

示例

H={(uv)R2v>0},gq(ξ,η)=ξηv2\mathbf H = \left\{ \binom{u}{v} \in \mathbb R^2 \mid v > 0 \right\},\quad g_{\boldsymbol q}(\boldsymbol \xi, \boldsymbol \eta) = \frac{\boldsymbol \xi \cdot \boldsymbol \eta}{v^2}

那么 gg 成为 H\mathbf H 上的 Riemannian 度规,称为 Poincaré 度量

证明

计算即得

gq(ξ,η)=tξ(1v2001v2)ηg_{\boldsymbol q}(\boldsymbol \xi, \boldsymbol \eta) = {}^t\boldsymbol \xi \begin{pmatrix} \frac{1}{v^2} & 0 \\ 0 & \frac{1}{v^2} \end{pmatrix} \boldsymbol \eta

EGF2=1v4>0,E=1v2>0EG - F^2 = \frac{1}{v^4} \gt 0, E = \frac{1}{v^2} \gt 0


实际应用中,往往需要从曲面的参数化中导出 Riemannian 度规

SSR3\mathbb R^3 中的正则曲面,参数化 σ:DS\boldsymbol \sigma: D \to S,定义

gqσ(ξ,η)=(dσ)q(ξ)(dσ)q(η)=(σuξ1+σvξ2)(σuη1+σvη2)g_{\boldsymbol q}^{\boldsymbol \sigma}(\boldsymbol \xi, \boldsymbol \eta) = (d\boldsymbol \sigma)_{\boldsymbol q}(\boldsymbol \xi) \cdot (d\boldsymbol \sigma)_{\boldsymbol q}(\boldsymbol \eta) = (\boldsymbol \sigma_u \xi_1 + \boldsymbol \sigma_v \xi_2) \cdot (\boldsymbol \sigma_u \eta_1 + \boldsymbol \sigma_v \eta_2)

利用第一基本量,也可以写作

gqσ(ξ,η)=E(q)ξ1η1+F(q)(ξ1η2+ξ2η1)+G(q)ξ2η2=tξ(E(q)F(q)F(q)G(q))ηg_{\boldsymbol q}^{\boldsymbol \sigma}(\boldsymbol \xi, \boldsymbol \eta) = E(\boldsymbol q) \xi_1 \eta_1 + F(\boldsymbol q)(\xi_1 \eta_2 + \xi_2 \eta_1) + G(\boldsymbol q) \xi_2 \eta_2 = {}^t\boldsymbol \xi \begin{pmatrix} E(\boldsymbol q) & F(\boldsymbol q) \\ F(\boldsymbol q) & G(\boldsymbol q) \end{pmatrix} \boldsymbol \eta

基于第一基本量的性质,可以得到 EGF2>0,E>0EG - F^2 \gt 0, E \gt 0
所以 gg 成为 DD 上的 Riemannian 度规


以下令 ggDD 上的 Riemannian 度规,定义内积与范数

ξ,ηq:=gq(ξ,η),ξq:=gq(ξ,ξ)\langle \boldsymbol \xi, \boldsymbol \eta \rangle _{\boldsymbol q} := g_{\boldsymbol q}(\boldsymbol \xi, \boldsymbol \eta), \quad \|\boldsymbol \xi\|_{\boldsymbol q} := \sqrt{g_{\boldsymbol q}(\boldsymbol \xi, \boldsymbol \xi)}

此时,将 1,2D\partial_1, \partial_2 \in D 关于内积 ()q(\cdot)_{\boldsymbol q} 正交归一化,得到 TqDT_{\boldsymbol q}D 上的正交归一基底,实为 活动标架 的二维版本

ε1(q)=e1e1q=1E(q)e1,ε2(q)=e2(e2,ε1)qε1(q)e2(e2,ε1)qε1(q)q=1EGF2(FEe1+Ee2)\boldsymbol \varepsilon_1(\boldsymbol q) = \frac{\boldsymbol e_1}{\|\boldsymbol e_1\|_{\boldsymbol q}} = \frac{1}{\sqrt{E(\boldsymbol q)}} \boldsymbol e_1, \quad \boldsymbol \varepsilon_2(\boldsymbol q) = \frac{\boldsymbol e_2 - (\boldsymbol e_2, \boldsymbol \varepsilon_1)_{\boldsymbol q} \boldsymbol \varepsilon_1(\boldsymbol q)}{\left\|\boldsymbol e_2 - (\boldsymbol e_2, \boldsymbol \varepsilon_1)_{\boldsymbol q} \boldsymbol \varepsilon_1(\boldsymbol q)\right\|_{\boldsymbol q}} = \frac{1}{\sqrt{EG - F^2}} \left(-\frac{F}{\sqrt{E}} \boldsymbol e_1 + \sqrt{E} \boldsymbol e_2\right)

因此,对于任意 ξ,ηTqD\boldsymbol \xi, \boldsymbol \eta \in T_{\boldsymbol q}D,有唯一表示

{ξ=ξ1ε1(q)+ξ2ε2(q)η=η1ε1(q)+η2ε2(q)\begin{cases} \boldsymbol \xi = \xi_1 \boldsymbol \varepsilon_1(\boldsymbol q) + \xi_2 \boldsymbol \varepsilon_2(\boldsymbol q)\\ \boldsymbol \eta = \eta_1 \boldsymbol \varepsilon_1(\boldsymbol q) + \eta_2 \boldsymbol \varepsilon_2(\boldsymbol q) \end{cases}

且内积与范数为

ξ,ηq=ξ1ε1(q)+ξ2ε2(q),η1ε1(q)+η2ε2(q)q=ξ1η1+ξ2η2\langle \boldsymbol \xi, \boldsymbol \eta \rangle_ {\boldsymbol q} = \langle \xi_1 \boldsymbol \varepsilon_1(\boldsymbol q) + \xi_2 \boldsymbol \varepsilon_2(\boldsymbol q),\ \eta_1 \boldsymbol \varepsilon_1(\boldsymbol q) + \eta_2 \boldsymbol \varepsilon_2(\boldsymbol q) \rangle_{\boldsymbol q} = \xi_1 \eta_1 + \xi_2 \eta_2

ξq=ξ,ξq=ξ12+ξ22\|\boldsymbol \xi\|_{\boldsymbol q} = \sqrt{\langle \boldsymbol \xi, \boldsymbol \xi \rangle_{\boldsymbol q}} = \sqrt{\xi_1^2 + \xi_2^2}

即在活动标架下,内积与欧氏内积相同
注意:在 g=gσg = g^{\boldsymbol \sigma} 时,dσq(εi)=eid \boldsymbol \sigma_{\boldsymbol q}(\boldsymbol \varepsilon_i) = \boldsymbol e_i

取活动标架下的坐标 (θ1θ2),θj=a1jdu+a2jdv\binom{\theta^1}{\theta^2},\ \theta^j = a_1^j du + a_2^j dv,对于过度矩阵 AA

(e1e2)=(ε1ε2)AA=(ε1ε2)1(\boldsymbol e_1 \ \boldsymbol e_2) = (\boldsymbol \varepsilon_1 \ \boldsymbol \varepsilon_2) A \Longrightarrow A = (\boldsymbol \varepsilon_1 \ \boldsymbol \varepsilon_2)^{-1}

(θ1θ2)=A(dudv)\binom{\theta^1}{\theta^2} = A \binom{du}{dv}

所以,活动标架下的线性结合得以转为微分形式

θ1ε1+θ2ε2=(ε1ε2)(θ1θ2)=A1A(dudv)=(dudv)\theta^1 \boldsymbol \varepsilon_1 + \theta^2 \boldsymbol \varepsilon_2 = (\boldsymbol \varepsilon_1 \ \boldsymbol \varepsilon_2) \binom{\theta^1}{\theta^2} = A^{-1} A \binom{du}{dv} = \binom{du}{dv}


特别地,由于分析对象是二维空间,仅考虑前两个活动标架的联络 ω21\omega_2^1

命题
!ω21\exists! \omega_2^1DD 上的微分 1 形式,使得

dθ1=ω21θ2,dθ2=ω12θ1d \theta^1 = -\omega_2^1 \wedge \theta^2,\quad d \theta^2 = \omega_1^2 \wedge \theta^1

证明

ω21=fdu+gdv\omega_2^1 = f du + g dv,计算验证

ω21=(fg)(dudv)=(fg)(ε1ε2)(θ1θ2)=f~θ1+g~θ2\omega_2^1 = (f \ g) \binom{du}{dv} = (f \ g)(\boldsymbol \varepsilon_1 \ \boldsymbol \varepsilon_2) \binom{\theta^1}{\theta^2} = \tilde{f} \theta^1 + \tilde{g} \theta^2

其中 (f~g~):=(fg)(ε1ε2)(\tilde{f} \ \tilde{g}) := (f \ g)(\boldsymbol \varepsilon_1 \ \boldsymbol \varepsilon_2)
所以

ω21θ2=f~θ1θ2=f~detAdudv\omega_2^1 \wedge \theta^2 = \tilde{f} \theta^1 \wedge \theta^2 = \tilde{f} \mathrm{det}A \, du \wedge dv

ω21θ1=g~θ1θ2=g~detAdudv\omega_2^1 \wedge \theta^1 = -\tilde{g} \theta^1 \wedge \theta^2 = -\tilde{g} \mathrm{det}A \, du \wedge dv

dθj=hjdudvd \theta^j = h_j \, du \wedge dv,则

dθ1=ω21θ2h1=f~detAd \theta^1 = -\omega_2^1 \wedge \theta^2 \Leftrightarrow h_1 = -\tilde{f} \mathrm{det}A

dθ2=ω12θ1h2=g~detAd \theta^2 = \omega_1^2 \wedge \theta^1 \Leftrightarrow h_2 = \tilde{g} \mathrm{det}A

解得

f~=h1detA,g~=h2detA\tilde{f} = -\frac{h_1}{\mathrm{det}A},\quad \tilde{g} = \frac{h_2}{\mathrm{det}A}

(f~g~)=(fg)(ε1ε2)(\tilde{f} \ \tilde{g}) = (f \ g)(\boldsymbol \varepsilon_1 \ \boldsymbol \varepsilon_2) 可唯一确定 f,gf,g,从而唯一确定 ω21\omega_2^1

利用 ω21\omega_2^1,可以得到 Riemannian 度规下的 Gauss 曲率定义
dω21=Kθ1θ2d \omega_2^1 = K \theta^1 \wedge \theta^2 导出的 KK 称为 gg Gauss 曲率
g=gσg = g^{\boldsymbol \sigma} 时,KK 即为曲面 SS 在点 σ(q)\boldsymbol \sigma(\boldsymbol q) 处的 Gauss 曲率

示例
在 Poincaré 度量下

展开

ε1=1gq(1,1)=vE=(v0),ε2=2gq(2,2)=vG=(0v)\boldsymbol \varepsilon_1 = \frac{\partial_1}{\sqrt{g_{\boldsymbol q}(\partial_1, \partial_1)}} = \frac{v}{\sqrt{E}} = \binom{v}{0}, \quad \boldsymbol \varepsilon_2 = \frac{\partial_2}{\sqrt{g_{\boldsymbol q}(\partial_2, \partial_2)}} = \frac{v}{\sqrt{G}} = \binom{0}{v}

A=(ε1ε2)1=(1v001v)A = (\boldsymbol \varepsilon_1 \ \boldsymbol \varepsilon_2)^{-1} = \begin{pmatrix} \frac{1}{v} & 0 \\ 0 & \frac{1}{v} \end{pmatrix}

(θ1θ2)=A(dudv)=(1v001v)(dudv)=(duvdvv)\binom{\theta^1}{\theta^2} = A \binom{du}{dv} = \begin{pmatrix} \frac{1}{v} & 0 \\ 0 & \frac{1}{v} \end{pmatrix} \binom{du}{dv} = \binom{\frac{du}{v}}{\frac{dv}{v}}

ω21=fdu+gdv\omega_2^1 = f du + g dv

{dθ1=1v2dudv=ω21θ2=(fdu+gdv)dvv=fvdudvdθ2=1v2dvdu=ω21θ1=(fdu+gdv)duv=gvdvdu{f=1vg=0\begin{cases} d \theta^1 = -\frac{1}{v^2} du \wedge dv = -\omega_2^1 \wedge \theta^2 = -\left(f du + g dv\right) \wedge \frac{dv}{v} = -\frac{f}{v} du \wedge dv \\ d \theta^2 = -\frac{1}{v^2} dv \wedge du = \omega_2^1 \wedge \theta^1 = \left(f du + g dv\right) \wedge \frac{du}{v} = \frac{g}{v} dv \wedge du \end{cases} \Longrightarrow \begin{cases} f = -\frac{1}{v} \\ g = 0 \end{cases}

所以 ω21=1vdu\omega_2^1 = -\frac{1}{v} du \quad,求得全微分

dω21=d(1v)du=v2dvdu=1v2dudvd \omega_2^1 = -d\left(\frac{1}{v}\right) \wedge du = v^{-2} dv \wedge du = -\frac{1}{v^2} du \wedge dv

另一边由楔积求得

θ1θ2=1v2dudv\theta^1 \wedge \theta^2 = \frac{1}{v^2} du \wedge dv

因此 K=1K = -1