The Christoffel Symbols are Intrinsic
📂Geometry The Christoffel Symbols are Intrinsic Theorem The Christoffel symbols Γ i j k \Gamma_{ij}^{k} Γ ij k intrinsic .
Γ i j k = 1 2 ∑ l = 1 2 g l k ( ∂ g l j ∂ u i − ∂ g i j ∂ u l + ∂ g i l ∂ u j )
\Gamma_{ij}^{k} = \dfrac{1}{2} \sum \limits_{l=1}^{2} g^{lk} \left( \dfrac{\partial g_{lj}}{\partial u_{i}} - \dfrac{\partial g_{ij}}{\partial u_{l}} + \dfrac{\partial g_{il}}{\partial u_{j}} \right)
Γ ij k = 2 1 l = 1 ∑ 2 g l k ( ∂ u i ∂ g l j − ∂ u l ∂ g ij + ∂ u j ∂ g i l )
Explanation Gauss proved it.
The Christoffel symbols depend only on the Riemann metric and are independent of the normal vector . Therefore, by using Christoffel symbols, one can understand the structure of a surface without leaving the surface.
Proof First, the partial derivatives of each index of the Riemann metric coefficients are as follows.
∂ g i l ∂ u j = ∂ ∂ u j ⟨ x i , x l ⟩ = ⟨ x i j , x l ⟩ + ⟨ x i , x l j ⟩
\dfrac{\partial g_{il}}{\partial u_{j}} = \dfrac{\partial}{\partial u_{j}} \left\langle \mathbf{x}_{i} , \mathbf{x}_{l} \right\rangle
= \langle \mathbf{x}_{ij} , \mathbf{x}_{l} \rangle + \langle \mathbf{x}_{i}, \mathbf{x}_{lj} \rangle
∂ u j ∂ g i l = ∂ u j ∂ ⟨ x i , x l ⟩ = ⟨ x ij , x l ⟩ + ⟨ x i , x l j ⟩
∂ g i j ∂ u l = ∂ ∂ u l ⟨ x i , x j ⟩ = ⟨ x i l , x j ⟩ + ⟨ x i , x j l ⟩
\dfrac{\partial g_{ij}}{\partial u_{l}} = \dfrac{\partial}{\partial u_{l}} \left\langle \mathbf{x}_{i} , \mathbf{x}_{j} \right\rangle
= \langle \mathbf{x}_{il} , \mathbf{x}_{j} \rangle + \langle \mathbf{x}_{i}, \mathbf{x}_{jl} \rangle
∂ u l ∂ g ij = ∂ u l ∂ ⟨ x i , x j ⟩ = ⟨ x i l , x j ⟩ + ⟨ x i , x j l ⟩
∂ g l j ∂ u i = ∂ ∂ u i ⟨ x l , x j ⟩ = ⟨ x l i , x j ⟩ + ⟨ x l , x j i ⟩
\dfrac{\partial g_{lj}}{\partial u_{i}} = \dfrac{\partial}{\partial u_{i}} \left\langle \mathbf{x}_{l} , \mathbf{x}_{j} \right\rangle
= \langle \mathbf{x}_{li} , \mathbf{x}_{j} \rangle + \langle \mathbf{x}_{l}, \mathbf{x}_{ji} \rangle
∂ u i ∂ g l j = ∂ u i ∂ ⟨ x l , x j ⟩ = ⟨ x l i , x j ⟩ + ⟨ x l , x ji ⟩
Since x i j = x j i \mathbf{\mathbf{x}}_{ij} = \mathbf{\mathbf{x}}_{ji} x ij = x ji
∂ g i l ∂ u j − ∂ g i j ∂ u l + ∂ g l j ∂ u i = ⟨ x i j , x l ⟩ + ⟨ x i , x l j ⟩ − ⟨ x i l , x j ⟩ − ⟨ x i , x j l ⟩ + ⟨ x l i , x j ⟩ + ⟨ x l , x j i ⟩ = ⟨ x i j , x l ⟩ + ⟨ x l , x j i ⟩ = 2 ⟨ x i j , x l ⟩
\begin{align*}
\dfrac{\partial g_{il}}{\partial u_{j}} - \dfrac{\partial g_{ij}}{\partial u_{l}} + \dfrac{\partial g_{lj}}{\partial u_{i}}
=&\ \langle \mathbf{x}_{ij} , \mathbf{x}_{l} \rangle + \langle \mathbf{x}_{i}, \mathbf{x}_{lj} \rangle - \langle \mathbf{x}_{il} , \mathbf{x}_{j} \rangle - \langle \mathbf{x}_{i}, \mathbf{x}_{jl} \rangle + \langle \mathbf{x}_{li} , \mathbf{x}_{j} \rangle + \langle \mathbf{x}_{l}, \mathbf{x}_{ji} \rangle
\\ =&\ \langle \mathbf{x}_{ij} , \mathbf{x}_{l} \rangle + \langle \mathbf{x}_{l}, \mathbf{x}_{ji} \rangle
\\ =&\ 2 \langle \mathbf{x}_{ij} , \mathbf{x}_{l} \rangle
\end{align*}
∂ u j ∂ g i l − ∂ u l ∂ g ij + ∂ u i ∂ g l j = = = ⟨ x ij , x l ⟩ + ⟨ x i , x l j ⟩ − ⟨ x i l , x j ⟩ − ⟨ x i , x j l ⟩ + ⟨ x l i , x j ⟩ + ⟨ x l , x ji ⟩ ⟨ x ij , x l ⟩ + ⟨ x l , x ji ⟩ 2 ⟨ x ij , x l ⟩
Therefore, the Christoffel symbols are
Γ i j k = ∑ l = 1 2 ⟨ x i j , x l ⟩ g l k = 1 2 ∑ l = 1 2 ( ∂ g l j ∂ u i − ∂ g i j ∂ u l + ∂ g i l ∂ u j ) g l k
\Gamma_{ij}^{k} = \sum \limits_{l=1}^{2} \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle g^{lk} = \dfrac{1}{2} \sum \limits_{l=1}^{2} \left( \dfrac{\partial g_{lj}}{\partial u_{i}} - \dfrac{\partial g_{ij}}{\partial u_{l}} + \dfrac{\partial g_{il}}{\partial u_{j}} \right) g^{lk}
Γ ij k = l = 1 ∑ 2 ⟨ x ij , x l ⟩ g l k = 2 1 l = 1 ∑ 2 ( ∂ u i ∂ g l j − ∂ u l ∂ g ij + ∂ u j ∂ g i l ) g l k
Example Suppose we are given a Monge patch as in x ( u 1 , u 2 ) = ( u 1 , u 2 , f ( u 1 , u 2 ) ) \mathbf{x}(u_{1}, u_{2}) = \left( u_{1}, u_{2}, f(u_{1}, u_{2}) \right) x ( u 1 , u 2 ) = ( u 1 , u 2 , f ( u 1 , u 2 ) )
x 1 = ∂ u 1 x = ( 1 , 0 , f 1 ) and x 2 = ( 0 , 1 , f 2 )
\mathbf{x}_{1} = \partial_{u_{1}}\mathbf{x} = (1,0,f_{1}) \quad \text{and} \quad \mathbf{x}_{2} = (0,1,f_{2})
x 1 = ∂ u 1 x = ( 1 , 0 , f 1 ) and x 2 = ( 0 , 1 , f 2 )
At this time, f i = ∂ u i f f_{i} = \partial_{u_{i}}f f i = ∂ u i f Γ 11 1 \Gamma_{11}^{1} Γ 11 1
Extrinsically computing x 1 × x 2 = ( − f 1 , − f 2 , 1 )
\mathbf{x}_{1} \times \mathbf{x}_{2} = (-f_{1}, -f_{2}, 1)
x 1 × x 2 = ( − f 1 , − f 2 , 1 )
The unit normal is
n = ( − f 1 , − f 2 , 1 ) ( f 1 ) 2 + ( f 2 ) 2 + 1
\mathbf{n} = \dfrac{(-f_{1}, -f_{2}, 1)}{\sqrt{(f_{1})^{2} + (f_{2})^{2} + 1}}
n = ( f 1 ) 2 + ( f 2 ) 2 + 1 ( − f 1 , − f 2 , 1 )
Gauss formula
x i j = L i j n + ∑ k = 1 2 Γ i j k x k
\mathbf{x}_{ij} = L_{ij} \mathbf{n} + \sum \limits_{k=1}^{2} \Gamma_{ij}^{k} \mathbf{x}_{k}
x ij = L ij n + k = 1 ∑ 2 Γ ij k x k
The second derivative of x \mathbf{x} x
x 11 = ( 0 , 0 , f 11 ) = L 11 n + Γ 11 1 x 1 + Γ 11 2 x 2
\mathbf{x}_{11} = (0, 0, f_{11}) = L_{11}\mathbf{n} + \Gamma_{11}^{1}\mathbf{x}_{1} + \Gamma_{11}^{2}\mathbf{x}_{2}
x 11 = ( 0 , 0 , f 11 ) = L 11 n + Γ 11 1 x 1 + Γ 11 2 x 2
Therefore, the coefficients of the second fundamental form are L i j = ⟨ x i j , n ⟩ L_{ij} = \langle \mathbf{x}_{ij}, \mathbf{n} \rangle L ij = ⟨ x ij , n ⟩
L 11 = ⟨ ( 0 , 0 , f 11 ) , ( − f 1 , − f 2 , 1 ) ( f 1 ) 2 + ( f 2 ) 2 + 1 ⟩ = f 11 ( f 1 ) 2 + ( f 2 ) 2 + 1
L_{11} = \left\langle (0,0,f_{11}), \dfrac{(-f_{1}, -f_{2}, 1)}{\sqrt{(f_{1})^{2} + (f_{2})^{2} + 1}}\right\rangle = \dfrac{f_{11}}{\sqrt{(f_{1})^{2} + (f_{2})^{2} + 1}}
L 11 = ⟨ ( 0 , 0 , f 11 ) , ( f 1 ) 2 + ( f 2 ) 2 + 1 ( − f 1 , − f 2 , 1 ) ⟩ = ( f 1 ) 2 + ( f 2 ) 2 + 1 f 11
If we break down x 11 \mathbf{x}_{11} x 11
x 11 = ( 0 , 0 , f 11 ) = L 11 ( f 1 ) 2 + ( f 2 ) 2 + 1 ( − f 1 , − f 2 , 1 ) + Γ 11 1 ( 1 , 0 , f 1 ) + Γ 11 2 ( 0 , 1 , f 2 )
\mathbf{x}_{11} = (0, 0, f_{11}) = \dfrac{L_{11}}{\sqrt{(f_{1})^{2} + (f_{2})^{2} + 1}}(-f_{1}, -f_{2}, 1) + \Gamma_{11}^{1}(1,0,f_{1}) + \Gamma_{11}^{2}(0,1,f_{2})
x 11 = ( 0 , 0 , f 11 ) = ( f 1 ) 2 + ( f 2 ) 2 + 1 L 11 ( − f 1 , − f 2 , 1 ) + Γ 11 1 ( 1 , 0 , f 1 ) + Γ 11 2 ( 0 , 1 , f 2 )
Looking at the first component, we obtain the following equation.
0 = L 11 ( − f 1 ) ( f 1 ) 2 + ( f 2 ) 2 + 1 + Γ 11 1 ⟹ Γ 11 1 = L 11 ( f 1 ) ( f 1 ) 2 + ( f 2 ) 2 + 1 ⟹ Γ 11 1 = f 1 f 11 ( f 1 ) 2 + ( f 2 ) 2 + 1
\begin{align*}
&& 0 =&\ \dfrac{L_{11}(-f_{1})}{\sqrt{(f_{1})^{2} + (f_{2})^{2} + 1}} + \Gamma_{11}^{1}
\\[1em] \implies&& \Gamma_{11}^{1} =&\ \dfrac{L_{11}(f_{1})}{\sqrt{(f_{1})^{2} + (f_{2})^{2} + 1}}
\\[1em] \implies&& \Gamma_{11}^{1} =&\ \dfrac{f_{1} f_{11}}{(f_{1})^{2} + (f_{2})^{2} + 1}
\end{align*}
⟹ ⟹ 0 = Γ 11 1 = Γ 11 1 = ( f 1 ) 2 + ( f 2 ) 2 + 1 L 11 ( − f 1 ) + Γ 11 1 ( f 1 ) 2 + ( f 2 ) 2 + 1 L 11 ( f 1 ) ( f 1 ) 2 + ( f 2 ) 2 + 1 f 1 f 11
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Intrinsically computing According to the theorem above, Γ 11 1 \Gamma_{11}^{1} Γ 11 1
Γ 11 1 = 1 2 ∑ l = 1 2 g l 1 ( ∂ g l 1 ∂ u 1 − ∂ g 11 ∂ u l + ∂ g 1 l ∂ u 1 ) = 1 2 [ g 11 ( ∂ g 11 ∂ u 1 − ∂ g 11 ∂ u 1 + ∂ g 11 ∂ u 1 ) + g 21 ( ∂ g 21 ∂ u 1 − ∂ g 11 ∂ u 2 + ∂ g 12 ∂ u 1 ) ] = 1 2 [ g 11 ∂ g 11 ∂ u 1 + 2 g 21 ∂ g 21 ∂ u 1 − g 21 ∂ g 11 ∂ u 2 ]
\begin{align*}
\Gamma_{11}^{1} =&\ \dfrac{1}{2} \sum \limits_{l=1}^{2} g^{l1} \left( \dfrac{\partial g_{l1}}{\partial u_{1}} - \dfrac{\partial g_{11}}{\partial u_{l}} + \dfrac{\partial g_{1l}}{\partial u_{1}} \right)
\\ =&\ \dfrac{1}{2} \left[ g^{11} \left( \dfrac{\partial g_{11}}{\partial u_{1}} - \dfrac{\partial g_{11}}{\partial u_{1}} + \dfrac{\partial g_{11}}{\partial u_{1}} \right) + g^{21} \left( \dfrac{\partial g_{21}}{\partial u_{1}} - \dfrac{\partial g_{11}}{\partial u_{2}} + \dfrac{\partial g_{12}}{\partial u_{1}} \right) \right]
\\ =&\ \dfrac{1}{2} \left[ g^{11} \dfrac{\partial g_{11}}{\partial u_{1}} + 2g^{21}\dfrac{\partial g_{21}}{\partial u_{1}} - g^{21}\dfrac{\partial g_{11}}{\partial u_{2}} \right]
\end{align*}
Γ 11 1 = = = 2 1 l = 1 ∑ 2 g l 1 ( ∂ u 1 ∂ g l 1 − ∂ u l ∂ g 11 + ∂ u 1 ∂ g 1 l ) 2 1 [ g 11 ( ∂ u 1 ∂ g 11 − ∂ u 1 ∂ g 11 + ∂ u 1 ∂ g 11 ) + g 21 ( ∂ u 1 ∂ g 21 − ∂ u 2 ∂ g 11 + ∂ u 1 ∂ g 12 ) ] 2 1 [ g 11 ∂ u 1 ∂ g 11 + 2 g 21 ∂ u 1 ∂ g 21 − g 21 ∂ u 2 ∂ g 11 ]
The coefficients of the first fundamental form are g i j = ⟨ x 1 , x 2 ⟩ g_{ij} = \left\langle \mathbf{x}_{1}, \mathbf{x}_{2} \right\rangle g ij = ⟨ x 1 , x 2 ⟩
[ g i j ] = [ 1 + ( f 1 ) 2 f 1 f 2 f 1 f 2 1 + ( f 2 ) 2 ]
\left[ g_{ij} \right] = \begin{bmatrix} 1+(f_{1})^{2} & f_{1}f_{2} \\[1em] f_{1}f_{2} & 1 + (f_{2})^{2} \end{bmatrix}
[ g ij ] = 1 + ( f 1 ) 2 f 1 f 2 f 1 f 2 1 + ( f 2 ) 2
The inverse matrix is
[ g i j ] − 1 = [ g l k ] = 1 ( f 1 ) 2 + ( f 2 ) 2 + 1 [ 1 + ( f 2 ) 2 − f 1 f 2 − f 1 f 2 1 + ( f 1 ) 2 ]
\left[ g_{ij} \right]^{-1} = \left[ g^{lk} \right] = \dfrac{1}{(f_{1})^{2} + (f_{2})^{2} + 1}\begin{bmatrix} 1+(f_{2})^{2} & -f_{1}f_{2} \\[1em] -f_{1}f_{2} & 1 + (f_{1})^{2} \end{bmatrix}
[ g ij ] − 1 = [ g l k ] = ( f 1 ) 2 + ( f 2 ) 2 + 1 1 1 + ( f 2 ) 2 − f 1 f 2 − f 1 f 2 1 + ( f 1 ) 2
So, when we gather everything needed, we get the following.
∂ g 11 ∂ u 1 = ∂ ∂ u 1 ( 1 + ( f 1 ) 2 ) = 2 f 1 f 11 ∂ g 21 ∂ u 1 = ∂ ∂ u 1 ( f 1 f 2 ) = f 11 f 2 + f 1 f 21 ∂ g 11 ∂ u 2 = ∂ ∂ u 2 ( 1 + ( f 1 ) 2 ) = 2 f 1 f 12
\begin{align*}
\dfrac{\partial g_{11}}{\partial u_{1}} =&\ \dfrac{\partial }{\partial u_{1}}\left( 1+ (f_{1})^{2} \right) = 2f_{1}f_{11}
\\ \dfrac{\partial g_{21}}{\partial u_{1}} =&\ \dfrac{\partial }{\partial u_{1}}\left( f_{1}f_{2} \right) = f_{11}f_{2} + f_{1}f_{21}
\\ \dfrac{\partial g_{11}}{\partial u_{2}} =&\ \dfrac{\partial }{\partial u_{2}}\left( 1+ (f_{1})^{2} \right) = 2f_{1}f_{12}
\end{align*}
∂ u 1 ∂ g 11 = ∂ u 1 ∂ g 21 = ∂ u 2 ∂ g 11 = ∂ u 1 ∂ ( 1 + ( f 1 ) 2 ) = 2 f 1 f 11 ∂ u 1 ∂ ( f 1 f 2 ) = f 11 f 2 + f 1 f 21 ∂ u 2 ∂ ( 1 + ( f 1 ) 2 ) = 2 f 1 f 12
And
g 11 = 1 + ( f 2 ) 2 ( f 1 ) 2 + ( f 2 ) 2 + 1 g 21 = − f 1 f 2 ( f 1 ) 2 + ( f 2 ) 2 + 1
\begin{align*}
g^{11} =&\ \dfrac{1+(f_{2})^{2}}{(f_{1})^{2} + (f_{2})^{2} + 1}
\\ g^{21} =&\ \dfrac{-f_{1}f_{2}}{(f_{1})^{2} + (f_{2})^{2} + 1}
\end{align*}
g 11 = g 21 = ( f 1 ) 2 + ( f 2 ) 2 + 1 1 + ( f 2 ) 2 ( f 1 ) 2 + ( f 2 ) 2 + 1 − f 1 f 2
Now, if we substitute, we get as follows.
Γ 11 1 = 1 2 [ g 11 ∂ g 11 ∂ u 1 + 2 g 21 ∂ g 21 ∂ u 1 − g 21 ∂ g 11 ∂ u 2 ] = 1 2 [ 1 + ( f 2 ) 2 ( f 1 ) 2 + ( f 2 ) 2 + 1 2 f 1 f 11 + 2 − f 1 f 2 ( f 1 ) 2 + ( f 2 ) 2 + 1 ( f 11 f 2 + f 1 f 21 ) − − f 1 f 2 ( f 1 ) 2 + ( f 2 ) 2 + 1 2 f 1 f 12 ] = 1 ( f 1 ) 2 + ( f 2 ) 2 + 1 [ ( 1 + ( f 2 ) 2 ) f 1 f 11 + ( − f 1 f 2 ) ( f 11 f 2 + f 1 f 21 ) − ( − f 1 f 2 ) f 1 f 12 ] = 1 ( f 1 ) 2 + ( f 2 ) 2 + 1 [ f 1 f 11 + f 1 ( f 2 ) 2 f 11 − f 1 ( f 2 ) 2 f 11 − ( f 1 ) 2 f 2 f 21 + ( f 1 ) 2 f 2 f 12 ] = 1 ( f 1 ) 2 + ( f 2 ) 2 + 1 [ f 1 f 11 ] = f 1 f 11 ( f 1 ) 2 + ( f 2 ) 2 + 1
\begin{align*}
\Gamma_{11}^{1} =&\ \dfrac{1}{2} \left[ g^{11} \dfrac{\partial g_{11}}{\partial u_{1}} + 2g^{21}\dfrac{\partial g_{21}}{\partial u_{1}} - g^{21}\dfrac{\partial g_{11}}{\partial u_{2}} \right]
\\ =&\ \dfrac{1}{2} \left[ \dfrac{1+(f_{2})^{2}}{(f_{1})^{2} + (f_{2})^{2} + 1} 2f_{1}f_{11} + 2\dfrac{-f_{1}f_{2}}{(f_{1})^{2} + (f_{2})^{2} + 1} \left( f_{11}f_{2} + f_{1}f_{21} \right)- \dfrac{-f_{1}f_{2}}{(f_{1})^{2} + (f_{2})^{2} + 1}2f_{1}f_{12} \right]
\\ =&\ \dfrac{1}{(f_{1})^{2} + (f_{2})^{2} + 1}\left[ \left( 1+(f_{2})^{2} \right)f_{1}f_{11} + \left( -f_{1}f_{2} \right) \left( f_{11}f_{2} + f_{1}f_{21} \right)- (-f_{1}f_{2})f_{1}f_{12} \right]
\\ =&\ \dfrac{1}{(f_{1})^{2} + (f_{2})^{2} + 1}\left[ f_{1}f_{11} + f_{1}(f_{2})^{2}f_{11} - f_{1}(f_{2})^{2}f_{11} -(f_{1})^{2}f_{2}f_{21} + (f_{1})^{2}f_{2}f_{12} \right]
\\ =&\ \dfrac{1}{(f_{1})^{2} + (f_{2})^{2} + 1}\left[ f_{1}f_{11} \right]
\\ =&\ \dfrac{f_{1}f_{11}}{(f_{1})^{2} + (f_{2})^{2} + 1}
\end{align*}
Γ 11 1 = = = = = = 2 1 [ g 11 ∂ u 1 ∂ g 11 + 2 g 21 ∂ u 1 ∂ g 21 − g 21 ∂ u 2 ∂ g 11 ] 2 1 [ ( f 1 ) 2 + ( f 2 ) 2 + 1 1 + ( f 2 ) 2 2 f 1 f 11 + 2 ( f 1 ) 2 + ( f 2 ) 2 + 1 − f 1 f 2 ( f 11 f 2 + f 1 f 21 ) − ( f 1 ) 2 + ( f 2 ) 2 + 1 − f 1 f 2 2 f 1 f 12 ] ( f 1 ) 2 + ( f 2 ) 2 + 1 1 [ ( 1 + ( f 2 ) 2 ) f 1 f 11 + ( − f 1 f 2 ) ( f 11 f 2 + f 1 f 21 ) − ( − f 1 f 2 ) f 1 f 12 ] ( f 1 ) 2 + ( f 2 ) 2 + 1 1 [ f 1 f 11 + f 1 ( f 2 ) 2 f 11 − f 1 ( f 2 ) 2 f 11 − ( f 1 ) 2 f 2 f 21 + ( f 1 ) 2 f 2 f 12 ] ( f 1 ) 2 + ( f 2 ) 2 + 1 1 [ f 1 f 11 ] ( f 1 ) 2 + ( f 2 ) 2 + 1 f 1 f 11
We can see that computing Christoffel symbols intrinsically is more complicated compared to when not doing so.
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