Gauss and Codazzi-Mainardi Equations Codazzi-Mainardi Equations Theorem
L v − M u = L Γ 12 1 + M ( Γ 12 2 − Γ 11 1 ) − N Γ 11 2 L_v - M_u = L\Gamma_{12}^1 + M (\Gamma_{12}^2 - \Gamma_{11}^1) - N\Gamma_{11}^2 L v − M u = L Γ 12 1 + M ( Γ 12 2 − Γ 11 1 ) − N Γ 11 2 M v − N u = L Γ 22 1 + M ( Γ 22 2 − Γ 12 1 ) − N Γ 12 2 M_v - N_u = L\Gamma_{22}^1 + M (\Gamma_{22}^2 - \Gamma_{12}^1) - N\Gamma_{12}^2 M v − N u = L Γ 22 1 + M ( Γ 22 2 − Γ 12 1 ) − N Γ 12 2 proof . First, consider the first equation,
σ u u v = σ u v u \sigma_{uuv} = \sigma_{uvu} σ uuv = σ uvu Using Gauss equations,
Γ 11 1 σ u + Γ 11 2 σ v + L N ) v = ( Γ 12 1 σ u + Γ 12 2 σ v + M N ) u \Gamma_{11}^1\sigma_u + \Gamma_{11}^2\sigma_v + LN)_v = (\Gamma_{12}^1\sigma_u + \Gamma_{12}^2\sigma_v + MN)_u Γ 11 1 σ u + Γ 11 2 σ v + L N ) v = ( Γ 12 1 σ u + Γ 12 2 σ v + MN ) u ( Γ 11 1 ) v σ u + Γ 11 1 σ u v + ( Γ 11 2 ) v σ v + Γ 11 2 σ v v + L v N + L N v = ( Γ 12 1 ) u σ u + Γ 12 1 σ u u + ( Γ 12 2 ) u σ v + Γ 12 2 σ u v + M u N + M N u \begin{align*} &(\Gamma_{11}^1)_v \sigma_u + \Gamma_{11}^1\sigma_{uv} + (\Gamma_{11}^2)_v \sigma_v + \Gamma_{11}^2\sigma_{vv} + L_v\mathbf N + L\mathbf N_v\\ &= (\Gamma_{12}^1)_u \sigma_u + \Gamma_{12}^1\sigma_{uu} + (\Gamma_{12}^2)_u \sigma_v + \Gamma_{12}^2\sigma_{uv} + M_u \mathbf N + M\mathbf N_u \end{align*} ( Γ 11 1 ) v σ u + Γ 11 1 σ uv + ( Γ 11 2 ) v σ v + Γ 11 2 σ vv + L v N + L N v = ( Γ 12 1 ) u σ u + Γ 12 1 σ uu + ( Γ 12 2 ) u σ v + Γ 12 2 σ uv + M u N + M N u Then, take dot product N \mathbf N N N \mathbf N N σ u , σ v , N u , N v \sigma_u,\sigma_v, \mathbf N_u, \mathbf N_v σ u , σ v , N u , N v N ⋅ N = 1 \mathbf N\cdot \mathbf N = 1 N ⋅ N = 1
Γ 11 1 σ u v ⋅ N + Γ 11 2 σ v v ⋅ N + L v = Γ 12 1 σ u u ⋅ N + Γ 12 1 σ u v ⋅ N + M u \Gamma_{11}^1\sigma_{uv}\cdot \mathbf N + \Gamma_{11}^2\sigma_{vv}\cdot\mathbf N + L_v = \Gamma_{12}^1\sigma_{uu}\cdot\mathbf N + \Gamma_{12}^1\sigma_{uv}\cdot\mathbf N + M_u Γ 11 1 σ uv ⋅ N + Γ 11 2 σ vv ⋅ N + L v = Γ 12 1 σ uu ⋅ N + Γ 12 1 σ uv ⋅ N + M u Γ 11 1 M + Γ 11 2 N + L v = Γ 12 1 L + Γ 12 1 M + M u \Gamma_{11}^1 M + \Gamma_{11}^2N + L_v = \Gamma_{12}^1L + \Gamma_{12}^1M + M_u Γ 11 1 M + Γ 11 2 N + L v = Γ 12 1 L + Γ 12 1 M + M u reorder the equations, we have the first equation.
L v − M u = L Γ 12 1 + M ( Γ 12 2 − Γ 11 1 ) − N Γ 11 2 L_v - M_u = L\Gamma_{12}^1 + M (\Gamma_{12}^2 - \Gamma_{11}^1) - N\Gamma_{11}^2 L v − M u = L Γ 12 1 + M ( Γ 12 2 − Γ 11 1 ) − N Γ 11 2 Similarly, the second equation can be obtained by σ v v u = σ u v v \sigma_{vvu} = \sigma_{uvv} σ vvu = σ uvv
Gauss Equations Theorem For K K K
E K = ( Γ 11 2 ) v − ( Γ 12 2 ) u + Γ 11 1 Γ 12 2 + Γ 11 2 Γ 22 2 − Γ 12 1 Γ 11 2 − ( Γ 12 2 ) 2 F K = ( Γ 12 1 ) u − ( Γ 11 1 ) v + Γ 12 2 Γ 12 1 + Γ 11 2 Γ 22 1 F K = ( Γ 12 2 ) v − ( Γ 22 2 ) u + Γ 12 1 Γ 12 2 + Γ 22 1 Γ 11 2 G K = ( Γ 22 1 ) u − ( Γ 12 1 ) v + Γ 22 1 Γ 11 1 + Γ 22 2 Γ 12 1 − ( Γ 12 2 ) 2 − Γ 12 2 Γ 22 1 \begin{align*} EK &= (\Gamma_{11}^2)_v - (\Gamma_{12}^2)_u + \Gamma_{11}^1 \Gamma_{12}^2 + \Gamma_{11}^2 \Gamma_{22}^2 - \Gamma_{12}^1\Gamma_{11}^2 - (\Gamma_{12}^2)^2\\ FK &= (\Gamma_{12}^1)_u - (\Gamma_{11}^1)_v + \Gamma_{12}^2 \Gamma_{12}^1 + \Gamma_{11}^2 \Gamma_{22}^1\\ FK &= (\Gamma_{12}^2)_v - (\Gamma_{22}^2)_u + \Gamma_{12}^1 \Gamma_{12}^2 + \Gamma_{22}^1 \Gamma_{11}^2\\ GK &= (\Gamma_{22}^1)_u - (\Gamma_{12}^1)_v + \Gamma_{22}^1 \Gamma_{11}^1 + \Gamma_{22}^2 \Gamma_{12}^1 - (\Gamma_{12}^2)^2 - \Gamma_{12}^2\Gamma_{22}^1 \end{align*} E K F K F K G K = ( Γ 11 2 ) v − ( Γ 12 2 ) u + Γ 11 1 Γ 12 2 + Γ 11 2 Γ 22 2 − Γ 12 1 Γ 11 2 − ( Γ 12 2 ) 2 = ( Γ 12 1 ) u − ( Γ 11 1 ) v + Γ 12 2 Γ 12 1 + Γ 11 2 Γ 22 1 = ( Γ 12 2 ) v − ( Γ 22 2 ) u + Γ 12 1 Γ 12 2 + Γ 22 1 Γ 11 2 = ( Γ 22 1 ) u − ( Γ 12 1 ) v + Γ 22 1 Γ 11 1 + Γ 22 2 Γ 12 1 − ( Γ 12 2 ) 2 − Γ 12 2 Γ 22 1 proof .
Reconsider the equation
( Γ 11 1 ) v σ u + Γ 11 1 σ u v + ( Γ 11 2 ) v σ v + Γ 11 2 σ v v + L v N + L N v = ( Γ 12 1 ) u σ u + Γ 12 1 σ u u + ( Γ 12 2 ) u σ v + Γ 12 2 σ u v + M u N + M N u \begin{align*} &(\Gamma_{11}^1)_v \sigma_u + \Gamma_{11}^1\sigma_{uv} + (\Gamma_{11}^2)_v \sigma_v + \Gamma_{11}^2\sigma_{vv} + L_v\mathbf N + L\mathbf N_v\\ &= (\Gamma_{12}^1)_u \sigma_u + \Gamma_{12}^1\sigma_{uu} + (\Gamma_{12}^2)_u \sigma_v + \Gamma_{12}^2\sigma_{uv} + M_u \mathbf N + M\mathbf N_u \end{align*} ( Γ 11 1 ) v σ u + Γ 11 1 σ uv + ( Γ 11 2 ) v σ v + Γ 11 2 σ vv + L v N + L N v = ( Γ 12 1 ) u σ u + Γ 12 1 σ uu + ( Γ 12 2 ) u σ v + Γ 12 2 σ uv + M u N + M N u Now, take dot product σ u \sigma_u σ u σ u ⋅ σ u = 1 , σ v ⋅ σ u = N ⋅ σ u = 0 \sigma_u\cdot\sigma_u = 1, \sigma_v\cdot \sigma_u = \mathbf N\cdot\sigma_u = 0 σ u ⋅ σ u = 1 , σ v ⋅ σ u = N ⋅ σ u = 0
( Γ 11 1 ) v + Γ 11 1 σ u v σ u + Γ 11 2 σ v v σ u + L N v σ u = ( Γ 12 1 ) u + Γ 12 1 σ u u σ u + Γ 12 2 σ u v σ u + M N u σ u \begin{align*} &(\Gamma_{11}^1)_v + \Gamma_{11}^1\sigma_{uv}\sigma_u + \Gamma_{11}^2\sigma_{vv}\sigma_u + L\mathbf N_v\sigma_u\\ &= (\Gamma_{12}^1)_u + \Gamma_{12}^1\sigma_{uu}\sigma_u + \Gamma_{12}^2\sigma_{uv}\sigma_u + M\mathbf N_u\sigma_u \end{align*} ( Γ 11 1 ) v + Γ 11 1 σ uv σ u + Γ 11 2 σ vv σ u + L N v σ u = ( Γ 12 1 ) u + Γ 12 1 σ uu σ u + Γ 12 2 σ uv σ u + M N u σ u Then, note that Gauss equations represents σ u u , σ u v , σ v v \sigma_{uu},\sigma_{uv}, \sigma_{vv} σ uu , σ uv , σ vv σ u , σ v , N \sigma_u, \sigma_v, \mathbf N σ u , σ v , N σ u \sigma_u σ u
σ u u σ u = Γ 11 1 , σ u v σ u = Γ 12 1 , σ v v σ u = Γ 22 1 \sigma_{uu}\sigma_u = \Gamma_{11}^1, \sigma_{uv}\sigma_u = \Gamma_{12}^1, \sigma_{vv}\sigma_u = \Gamma_{22}^1 σ uu σ u = Γ 11 1 , σ uv σ u = Γ 12 1 , σ vv σ u = Γ 22 1 Replacing them back to the equation,
( Γ 11 1 ) v + Γ 11 1 Γ 12 1 + Γ 11 2 Γ 22 1 + L N v σ u = ( Γ 12 1 ) u + Γ 12 1 Γ 11 1 + Γ 12 2 Γ 12 1 + M N u σ u (\Gamma_{11}^1)_v + \Gamma_{11}^1\Gamma_{12}^1 + \Gamma_{11}^2\Gamma_{22}^1 + L\mathbf N_v\sigma_u = (\Gamma_{12}^1)_u + \Gamma_{12}^1\Gamma_{11}^1 + \Gamma_{12}^2\Gamma_{12}^1 + M\mathbf N_u\sigma_u ( Γ 11 1 ) v + Γ 11 1 Γ 12 1 + Γ 11 2 Γ 22 1 + L N v σ u = ( Γ 12 1 ) u + Γ 12 1 Γ 11 1 + Γ 12 2 Γ 12 1 + M N u σ u Finally, consider N u σ u , N v σ u \mathbf N_u\sigma_u, \mathbf N_v\sigma_u N u σ u , N v σ u
N u σ u = L G − M F E G − F 2 , N v σ u = M G − N F E G − F 2 \mathbf N_u\sigma_u = \frac{LG-MF}{EG-F^2}, \mathbf N_v\sigma_u = \frac{MG-NF}{EG-F^2} N u σ u = EG − F 2 L G − MF , N v σ u = EG − F 2 MG − NF Therefore,
L N v σ u − M N u σ u = − F K L\mathbf N_v\sigma_u - M\mathbf N_u\sigma_u = -FK L N v σ u − M N u σ u = − F K and re-order the equation above, we obtain the second equation in the claim.
Similarly, we can obtain 4 equations from σ u u v = σ u v u , σ u v u = σ v v u \sigma_{uuv} = \sigma_{uvu}, \sigma_{uvu} = \sigma_{vvu} σ uuv = σ uvu , σ uvu = σ vvu σ u , σ v \sigma_u, \sigma_v σ u , σ v
Theorem If two surface patches has the same first and second fundamental forms, then exists a direct isometry between them.
Moveover, for V ⊂ R 2 V\subset\mathbb R^2 V ⊂ R 2 E F G L M N EFGLMN EFG L MN V V V E > 0 , G > 0 , E G − F 2 > 0 E>0,G>0,EG-F^2>0 E > 0 , G > 0 , EG − F 2 > 0 p p p V V V U ⊂ V , p ∈ U U\subset V, p\in U U ⊂ V , p ∈ U σ : U → R 3 \sigma:U\rightarrow\mathbb R^3 σ : U → R 3 σ \sigma σ E F G L M N EFGLMN EFG L MN
In other words, given appropriate functions for first and second fundamental forms, a surface will always exist locally, and being unique up to a direct isometry.
Example: Cylinder Let the E = G = 1 , F = 0 , L = − 1 , M = N = 0 E=G=1, F=0, L=-1, M=N=0 E = G = 1 , F = 0 , L = − 1 , M = N = 0 E > 0 , G > 0 , E G − F 2 > 0 E>0,G>0, EG-F^2 >0 E > 0 , G > 0 , EG − F 2 > 0 Γ \Gamma Γ
Using Gauss equations and plug in the numbers
σ u u = Γ 11 1 σ u + Γ 11 2 σ v + L N = − N , σ u v = ⋯ = 0 , σ v v = ⋯ = 0 \sigma_{uu} = \Gamma_{11}^1\sigma_u + \Gamma_{11}^2\sigma_v + L\mathbf N = -\mathbf N, \sigma_{uv} = \cdots = 0, \sigma_{vv} =\cdots = 0 σ uu = Γ 11 1 σ u + Γ 11 2 σ v + L N = − N , σ uv = ⋯ = 0 , σ vv = ⋯ = 0 Since σ v \sigma_v σ v σ v u = σ v v = 0 \sigma_{vu} = \sigma_{vv} = 0 σ vu = σ vv = 0 σ v = a \sigma_v = \mathbf a σ v = a σ u v = 0 \sigma_{uv} = 0 σ uv = 0
σ ( u , v ) = b ( u ) + a v \sigma(u,v) = \mathbf b(u) + \mathbf a v σ ( u , v ) = b ( u ) + a v where N = − b ′ ′ \mathbf N = -\mathbf b'' N = − b ′′ N u , N v \mathbf N_u,\mathbf N_v N u , N v − N u = a σ u + b σ v , − N v = c σ u + d σ v -\mathbf N_u = a\sigma_u + b\sigma_v, -\mathbf N_v = c\sigma_u + d\sigma_v − N u = a σ u + b σ v , − N v = c σ u + d σ v
[ a c b d ] = F I − 1 F I I = [ 1 0 0 1 ] − 1 [ − 1 0 0 0 ] = [ − 1 0 0 0 ] \begin{bmatrix}a&c\\b&d\end{bmatrix} = F_I^{-1}F_{II} = \begin{bmatrix}1&0\\0&1\end{bmatrix}^{-1} \begin{bmatrix}-1&0\\0&0\end{bmatrix} = \begin{bmatrix}-1&0\\0&0\end{bmatrix} [ a b c d ] = F I − 1 F II = [ 1 0 0 1 ] − 1 [ − 1 0 0 0 ] = [ − 1 0 0 0 ] so that N u = σ u , N v = 0 \mathbf N_u = \sigma_u,\mathbf N_v = 0 N u = σ u , N v = 0
N u = d d u ( − b ′ ′ ) = − b ′ ′ ′ = b ′ = σ u \mathbf N_u = \frac{d}{du}(-\mathbf b'') = -\mathbf b''' =\mathbf b' = \sigma_u N u = d u d ( − b ′′ ) = − b ′′′ = b ′ = σ u Since b ′ ′ ′ + b ′ = 0 \mathbf b''' + \mathbf b' = 0 b ′′′ + b ′ = 0 b ′ ′ + b = − N + b \mathbf b'' + \mathbf b = -\mathbf N + \mathbf b b ′′ + b = − N + b b ( u ) = c sin u + d cos u \mathbf b(u) = \mathbf c \sin u + \mathbf d \cos u b ( u ) = c sin u + d cos u b \mathbf b b b ′ ′ \mathbf b'' b ′′ c = e 1 , d = e 2 \mathbf c = e_1,\mathbf d = e_2 c = e 1 , d = e 2
Finally, σ u × σ v = λ N ⟹ b ′ × a = λ b ⟹ a = ( 0 , 0 , λ ) \sigma_u\times \sigma_v = \lambda \mathbf N\implies \mathbf b' \times \mathbf a = \lambda \mathbf b\implies \mathbf a = (0, 0, \lambda) σ u × σ v = λ N ⟹ b ′ × a = λ b ⟹ a = ( 0 , 0 , λ ) λ ≠ 0 \lambda \neq 0 λ = 0 λ = 1 \lambda =1 λ = 1 σ ( u , v ) = ( cos u , sin u , v ) \sigma(u,v) = (\cos u, \sin u, v) σ ( u , v ) = ( cos u , sin u , v )
Example: Sphere Let E = cos 2 v , F = 0 , G = 1 , L = − cos 2 v , M = 0 , N = − 1 E = \cos^2 v, F = 0, G = 1, L =-\cos^2 v, M = 0, N = -1 E = cos 2 v , F = 0 , G = 1 , L = − cos 2 v , M = 0 , N = − 1
First, using the Weingarten map we have that
[ a c b d ] = F I − 1 F I I = [ cos 2 v 0 0 1 ] − 1 [ − cos 2 v 0 0 − 1 ] = [ − 1 0 0 − 1 ] \begin{bmatrix}a&c\\b&d\end{bmatrix} = F_I^{-1}F_{II} = \begin{bmatrix}\cos^2 v&0\\0&1\end{bmatrix}^{-1} \begin{bmatrix}-\cos^2v&0\\0&-1\end{bmatrix} = \begin{bmatrix}-1&0\\0&-1\end{bmatrix} [ a b c d ] = F I − 1 F II = [ cos 2 v 0 0 1 ] − 1 [ − cos 2 v 0 0 − 1 ] = [ − 1 0 0 − 1 ] so that σ u = N u , σ v = N v \sigma_u = \mathbf N_u, \sigma_v = \mathbf N_v σ u = N u , σ v = N v N = σ + a \mathbf N = \sigma + \mathbf a N = σ + a a \mathbf a a ∥ σ + a ∥ = 1 \|\sigma+\mathbf a\| = 1 ∥ σ + a ∥ = 1
The parameterization is given by
σ ( u , v ) = ( cos v cos u , cos v sin u , sin v ) \sigma(u, v) = (\cos v\cos u, \cos v\sin u, \sin v) σ ( u , v ) = ( cos v cos u , cos v sin u , sin v ) Example: Not a surface patch E = 1 , F = 0 , G = cos 2 u , L = cos 2 u , M = 0 , N = 1 E = 1, F = 0, G = \cos^2 u, L = \cos^2 u, M = 0, N = 1 E = 1 , F = 0 , G = cos 2 u , L = cos 2 u , M = 0 , N = 1 G G G F F F
Γ 12 2 = − 2 cos u sin u 2 cos 2 u = − 2 tan u , Γ 22 1 = G G u 2 G = cos u sin u \Gamma_{12}^2 = \frac{-2\cos u\sin u}{2\cos^2 u} = -2\tan u, \Gamma_{22}^1 = \frac{GG_u}{2G} = \cos u\sin u Γ 12 2 = 2 cos 2 u − 2 cos u sin u = − 2 tan u , Γ 22 1 = 2 G G G u = cos u sin u Therefore, the RHS of the second equation is
cos 2 u cos u sin u + 2 tan u ≠ 0 \cos^2 u \cos u \sin u + 2\tan u\neq 0 cos 2 u cos u sin u + 2 tan u = 0 But the LHS is 0 0 0 M , N M,N M , N
Therefore, the CM equation is not satisifed and there is no surface patch.
Applying Gauss Equations Theorem Suppose that a surface patch σ ( u , v ) \sigma(u,v) σ ( u , v ) E = G = v − 2 , v > 0 E=G=v^{-2}, v > 0 E = G = v − 2 , v > 0 L , N L,N L , N u u u
proof . First, the non-zero FFF derivatives are E v = G v = − 2 v − 3 E_v = G_v = -2v^{-3} E v = G v = − 2 v − 3
Γ 11 2 = − 2 v − 5 2 v − 4 = 1 / v , Γ 12 1 = − 1 / v , Γ 22 2 = − 1 / v \Gamma_{11}^2 = \frac{-2v^{-5} }{2v^{-4} } = 1/v, \Gamma_{12}^1 = -1/v, \Gamma_{22}^2 = -1/v Γ 11 2 = 2 v − 4 − 2 v − 5 = 1/ v , Γ 12 1 = − 1/ v , Γ 22 2 = − 1/ v The CM equations then gives
L v = − L v − 1 − N v − 1 = − v − 1 ( L + N ) , N u = 0 L_v = -Lv^{-1} - Nv^{-1} = -v^{-1}(L+N), N_u = 0 L v = − L v − 1 − N v − 1 = − v − 1 ( L + N ) , N u = 0 N u = 0 N_u=0 N u = 0 N N N u u u
Then, consider Gauss equations, plug in the equations for E K EK E K G K GK G K
v − 2 K = − v − 2 − 0 + 0 + 0 − 0 − 0 ⟹ K = − 1 v^{-2}K= -v^{-2} - 0 + 0 + 0 - 0- 0 \implies K = -1 v − 2 K = − v − 2 − 0 + 0 + 0 − 0 − 0 ⟹ K = − 1 where K K K
K = L N − M 2 E G − F 2 = v 4 L N = − 1 ⟹ L N = − v − 4 K = \frac{LN-M^2}{EG-F^2} = v^4LN = -1\implies LN = -v^{-4} K = EG − F 2 L N − M 2 = v 4 L N = − 1 ⟹ L N = − v − 4 does not depend on u u u
Going back to the first CM equation where L v = − v − 1 ( L + N ) L_v = -v^{-1}(L+N) L v = − v − 1 ( L + N )
L v = − v − 1 ( L + L − 1 v − 4 ) ⟹ L v 5 L v = 1 − L 2 v 4 L_v = -v^{-1}(L + L^{-1}v^{-4})\implies Lv^5 L_v = 1 - L^2 v^4 L v = − v − 1 ( L + L − 1 v − 4 ) ⟹ L v 5 L v = 1 − L 2 v 4 If F = 0 , M = 0 F=0, M=0 F = 0 , M = 0
Γ 11 1 = E u 2 E , Γ 11 2 = − E v 2 G \Gamma_{11}^1 = \frac{E_u}{2E}, \Gamma_{11}^2 = \frac{-E_v}{2G} Γ 11 1 = 2 E E u , Γ 11 2 = 2 G − E v Γ 12 1 = E v 2 E , Γ 12 2 = G u 2 G \Gamma_{12}^1 = \frac{E_v}{2E}, \Gamma_{12}^2 = \frac{G_u}{2G} Γ 12 1 = 2 E E v , Γ 12 2 = 2 G G u Γ 22 1 = − G u 2 E , Γ 22 2 = G v 2 G \Gamma_{22}^1 = \frac{-G_u}{2E}, \Gamma_{22}^2 = \frac{G_v}{2G} Γ 22 1 = 2 E − G u , Γ 22 2 = 2 G G v Therefore, the CM equations become
L v = 1 2 ( L E v E + N E v G ) = 1 2 E v ( L E + N G ) L_v = \frac{1}{2}\big(\frac{LE_v}{E} + \frac{NE_v}{G}\big) = \frac{1}{2}E_v(\frac{L}{E}+\frac{N}{G}) L v = 2 1 ( E L E v + G N E v ) = 2 1 E v ( E L + G N ) N u = − 1 2 ( − L G u E + − N G u G ) = 1 2 G u ( L E + N G ) N_u = -\frac{1}{2}\big(\frac{-LG_u}{E} + \frac{-NG_u}{G}) =\frac{1}{2}G_u(\frac{L}{E}+\frac{N}{G}) N u = − 2 1 ( E − L G u + G − N G u ) = 2 1 G u ( E L + G N ) and then note that the principal curvatures are simply the roots of ( L − κ E ) ( N − κ G ) = 0 (L-\kappa E)(N-\kappa G) = 0 ( L − κ E ) ( N − κ G ) = 0 M = F = 0 M=F=0 M = F = 0 κ 1 = L E , κ 2 = N G \kappa_1 = \frac{L}{E}, \kappa_2 = \frac{N}{G} κ 1 = E L , κ 2 = G N
( κ 1 ) v = L v E − L E v E 2 = 1 2 E v ( κ 1 + κ 2 ) E − L E v E 2 = E 2 E v ( κ 1 + κ 2 ) − E 2 2 L E E v 2 E 2 = E v 2 E ( κ 1 + κ 2 − 2 κ 1 ) = E v 2 E ( κ 2 − κ 1 ) ( κ 2 ) u = N u G − N G u G 2 = G u 2 G ( κ 1 − κ 2 ) \begin{align*} (\kappa_1)_v &= \frac{L_vE - LE_v}{E^2}\\ &= \frac{\frac{1}{2}E_v(\kappa_1+\kappa_2)E - LE_v}{E^2}\\ &=\frac{E}{2}\frac{E_v(\kappa_1+\kappa_2) - \frac{E}{2}\frac{2L}{E}E_v}{2E^2}\\ &= \frac{E_v}{2E}(\kappa_1 + \kappa_2 - 2\kappa_1)\\ &= \frac{E_v}{2E}(\kappa_2 - \kappa_1)\\ (\kappa_2)_u &= \frac{N_uG - NG_u}{G^2} \\ &= \frac{G_u}{2G}(\kappa_1-\kappa_2) \end{align*} ( κ 1 ) v ( κ 2 ) u = E 2 L v E − L E v = E 2 2 1 E v ( κ 1 + κ 2 ) E − L E v = 2 E 2 E 2 E v ( κ 1 + κ 2 ) − 2 E E 2 L E v = 2 E E v ( κ 1 + κ 2 − 2 κ 1 ) = 2 E E v ( κ 2 − κ 1 ) = G 2 N u G − N G u = 2 G G u ( κ 1 − κ 2 ) Theorem The Gaussian curvature of a surface is perserved by local isometries.
Theorem The Gaussian curvature is given by
K = 1 ( E G − F 2 ) 2 ( det [ E v v 2 + F u v − G u u 2 E u 2 F u − E v 2 F v − G u 2 E F G v 2 F G ] − det [ 0 E v 2 G u 2 E v 2 E F G u 2 F G ] ) K = \frac{1}{(EG-F^2)^2}\big(\det\begin{bmatrix} \frac{E_{vv} }{2} + F_{uv} - \frac{G_{uu} }{2} &\frac{E_u}{2}&F_u-\frac{E_v}{2}\\ F_v-\frac{G_u}{2} &E&F\\ \frac{G_v}{2}&F&G \end{bmatrix} - \det\begin{bmatrix} 0&\frac{E_v}{2}&\frac{G_u}{2}\\ \frac{E_v}{2}&E&F\\ \frac{G_u}{2}&F&G \end{bmatrix}\big) K = ( EG − F 2 ) 2 1 ( det 2 E vv + F uv − 2 G uu F v − 2 G u 2 G v 2 E u E F F u − 2 E v F G − det 0 2 E v 2 G u 2 E v E F 2 G u F G ) Corollary If F = 0 F=0 F = 0
K = − 1 2 E G ( ∂ u ( G u E G ) + ∂ v ( E v E G ) ) K = -\frac{1}{2\sqrt{EG} } \big(\partial_u (\frac{G_u}{\sqrt{EG} }) +\partial_v (\frac{E_v}{\sqrt{EG} })\big) K = − 2 EG 1 ( ∂ u ( EG G u ) + ∂ v ( EG E v ) ) proof . Using the first formula of Gauss equations, and substitute each Chrsitoffel symbol
E K = ( Γ 11 2 ) v − ( Γ 12 2 ) u + Γ 11 1 Γ 12 2 + Γ 11 2 Γ 22 2 − Γ 12 1 Γ 11 2 − ( Γ 12 2 ) 2 = ( − E v 2 G ) v − ( G u 2 G ) u + E u 2 E G u 2 G + − E v 2 G G v 2 G − E v 2 E ( − E v 2 G ) − ( G u 2 G ) 2 = E v G v − E v v G 2 G 2 − G u u G − G u 2 2 G 2 + E u G u 4 E G − E v G v 2 G 2 + E v 2 4 E G − G u 2 4 G 2 ∂ u ( G u E G ) = G u u E G − G u ( E u G + E G u ) 2 ( E G ) 3 / 2 ∂ v ( E v E G ) = E v v E G − E v ( E v G + E G v ) 2 ( E G ) 3 / 2 \begin{align*} EK &= (\Gamma_{11}^2)_v - (\Gamma_{12}^2)_u + \Gamma_{11}^1 \Gamma_{12}^2 + \Gamma_{11}^2 \Gamma_{22}^2 - \Gamma_{12}^1\Gamma_{11}^2 - (\Gamma_{12}^2)^2\\ &= (-\frac{E_v}{2G})_v - (\frac{G_u}{2G})_u + \frac{E_u}{2E} \frac{G_u}{2G} + \frac{-E_v}{2G}\frac{G_v}{2G} - \frac{E_v}{2E}(-\frac{E_v}{2G}) - (\frac{G_u}{2G})^2\\ &= \frac{E_vG_v - E_{vv} G}{2G^2} - \frac{G_{uu} G - G_u^2}{2G^2}+ \frac{E_uG_u}{4EG} - \frac{E_vG_v}{2G^2} + \frac{E_v^2}{4EG} - \frac{G_u^2}{4G^2}\\ \partial_u (\frac{G_u}{\sqrt{EG} }) &= \frac{G_{uu} }{\sqrt{EG} } - \frac{G_u(E_uG + E G_u)}{2(EG)^{3/2} }\\ \partial_v (\frac{E_v}{\sqrt{EG} }) &= \frac{E_{vv} }{\sqrt{EG} } - \frac{E_v(E_vG + E G_v)}{2(EG)^{3/2} } \end{align*} E K ∂ u ( EG G u ) ∂ v ( EG E v ) = ( Γ 11 2 ) v − ( Γ 12 2 ) u + Γ 11 1 Γ 12 2 + Γ 11 2 Γ 22 2 − Γ 12 1 Γ 11 2 − ( Γ 12 2 ) 2 = ( − 2 G E v ) v − ( 2 G G u ) u + 2 E E u 2 G G u + 2 G − E v 2 G G v − 2 E E v ( − 2 G E v ) − ( 2 G G u ) 2 = 2 G 2 E v G v − E vv G − 2 G 2 G uu G − G u 2 + 4 EG E u G u − 2 G 2 E v G v + 4 EG E v 2 − 4 G 2 G u 2 = EG G uu − 2 ( EG ) 3/2 G u ( E u G + E G u ) = EG E vv − 2 ( EG ) 3/2 E v ( E v G + E G v ) Take 2 E G 2\sqrt{EG} 2 EG
Corollary If E = 1 , F = 0 E=1,F=0 E = 1 , F = 0 K = − 1 G ∂ 2 G ∂ u 2 K = -\frac{1}{\sqrt G}\frac{\partial^2\sqrt G}{\partial u^2} K = − G 1 ∂ u 2 ∂ 2 G
Example: Sphere Claim any plane map of any region of a sphere must distort distances.
proof . The claim is equivalent to that there is no isometry between open subset of a plane to a sphere. Known that the Gaussian curvature of plane is 0 0 0 K K K
Example Claim If E = e λ , F = 0 , G = e λ E=e^\lambda, F=0,G=e^\lambda E = e λ , F = 0 , G = e λ λ \lambda λ u , v u,v u , v
Δ λ + 2 K e λ = 0 \Delta \lambda + 2Ke^\lambda = 0 Δ λ + 2 K e λ = 0 Δ \Delta Δ
proof . Note that F = 0 F=0 F = 0
K = − 1 2 e λ ( ∂ u ( G u e λ ) + ∂ v ( E v e λ ) ) = − 1 2 e λ ( ∂ u ( e λ λ u e λ ) + ∂ v ( e λ λ v e λ ) ) = − 1 2 e λ ( λ u u + λ v v ) K = − 1 2 e λ Δ λ \begin{align*} K &= -\frac{1}{2e^\lambda } \big(\partial_u (\frac{G_u}{e^\lambda }) +\partial_v (\frac{E_v}{e^\lambda })\big)\\ &= -\frac{1}{2e^\lambda } \big(\partial_u (\frac{e^\lambda\lambda_u}{e^\lambda }) +\partial_v (\frac{e^\lambda\lambda_v}{e^\lambda })\big)\\ &= -\frac{1}{2e^\lambda }(\lambda_{uu} + \lambda_{vv})\\ K &= -\frac{1}{2e^\lambda}\Delta \lambda \end{align*} K K = − 2 e λ 1 ( ∂ u ( e λ G u ) + ∂ v ( e λ E v ) ) = − 2 e λ 1 ( ∂ u ( e λ e λ λ u ) + ∂ v ( e λ e λ λ v ) ) = − 2 e λ 1 ( λ uu + λ vv ) = − 2 e λ 1 Δ λ January 11, 2025 January 9, 2023