Teorema Gauss-Bonnet face evidentă legătura dintre geometrie şi topologie .
Forma locală [ ] Fie (U, h) o parametrizare semigeodezică, cu U omeomorfă cu un disc plan deschis, compatibilă cu orientarea suprafeţei orientate S .
Fie
R
⊂
h
(
U
)
{\displaystyle R \subset h(U) \!}
γ
:
[
0
,
l
]
→
S
{\displaystyle \gamma : [0, l] \rightarrow S \!}
∂
R
=
I
m
γ
.
{\displaystyle \partial R = Im \gamma. \!}
Fie
γ
(
s
i
)
{\displaystyle \gamma (s_i) \!}
γ
,
θ
i
{\displaystyle \gamma, \theta_i \!}
i
=
0
,
k
+
1
¯
.
{\displaystyle i=\overline { 0, k+1 }. \!}
Atunci are loc formula:
∑
i
=
0
k
∫
s
i
s
i
+
1
k
g
(
s
)
+
∫
∫
R
K
d
s
+
∑
i
=
0
k
θ
i
=
2
π
{\displaystyle \sum_{i=0}^k \int_{s_i}^{s_{i+1}} k_g (s) + \int \int_R K \; ds + \sum_{i=0}^k \theta_i = 2 \pi \!}
unde
k
g
{\displaystyle k_g \!}
γ
,
{\displaystyle \gamma, \!}
K este curbura gaussiană şi
d
σ
{\displaystyle d \sigma\!}
Demonstraţie [ ] Fie
X
=
γ
′
(
s
)
{\displaystyle X= \gamma'(s) \!}
[
∇
X
d
s
]
=
[
∇
γ
′
(
s
)
d
s
]
=
k
g
(
s
)
.
{\displaystyle \bigg [ \frac {\nabla X}{d \mathit s} \bigg ]= \bigg [ \frac {\nabla \gamma' (s)}{d \mathit s} \bigg ] = k_g (s). \!}
Utilizăm următoarele leme:
Lema 1 [ ]
[
∇
Y
d
t
]
−
[
∇
X
d
t
]
=
d
φ
d
t
{\displaystyle \bigg [ \frac {\nabla Y}{dt} \bigg ] - \bigg [ \frac {\nabla X}{dt} \bigg ] = \frac {d \varphi}{dt} \!}
Lema 2 [ ] Fie (U, h) o parametrizare ortogonală, X un câmp unitar pe
γ
{\displaystyle \gamma \!}
φ
{\displaystyle \varphi \!}
h
1
{\displaystyle h_1 \!}
X .
Atunci:
[
∇
X
d
t
]
=
1
2
g
11
g
22
{
∂
g
22
∂
u
1
d
u
2
d
t
−
∂
g
11
∂
u
2
d
u
1
d
t
}
+
d
φ
d
t
{\displaystyle \bigg [ \frac {\nabla X}{d \mathit t} \bigg ] = \frac {1}{2 \sqrt {g_{11} g_{22}}} \bigg \{ \frac {\partial g_{22}}{\partial u^1} \frac {du^2}{dt} - \frac {\partial g_{11}}{\partial u^2} \frac {du^1}{dt} \bigg \} + \frac {d \varphi}{dt} \!}
Demonstraţie .
Normăm câmpurile
h
1
{\displaystyle h_1 \!}
h
2
{\displaystyle h_2 \!}
e
1
=
h
1
g
11
,
e
2
=
h
1
g
22
.
{\displaystyle e_1 = \frac {h_1}{\sqrt {g_{11}}}, \; e_2 = \frac {h_1}{\sqrt {g_{22}}}. \!}
Atunci
e
1
×
e
2
=
N
{\displaystyle e_1 \times e_2 = N \!}
[
∇
X
d
t
]
=
[
∇
e
1
d
t
]
+
d
φ
d
t
{\displaystyle \bigg [ \frac{\nabla X}{dt} \bigg ] = \bigg [ \frac {\nabla e_1}{dt} \bigg ] + \frac {d \varphi}{dt} \!}
Resurse [ ] 
![{\displaystyle \gamma :[0,l]\rightarrow S\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/7401d0316c9b824a05f31451931983f52ab58238)
![{\displaystyle {\bigg [}{\frac {\nabla X}{d{\mathit {s}}}}{\bigg ]}={\bigg [}{\frac {\nabla \gamma '(s)}{d{\mathit {s}}}}{\bigg ]}=k_{g}(s).\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/aca6d4931bfa76fef0effad69dc582a30068943f)
![{\displaystyle {\bigg [}{\frac {\nabla Y}{dt}}{\bigg ]}-{\bigg [}{\frac {\nabla X}{dt}}{\bigg ]}={\frac {d\varphi }{dt}}\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/40102a47c7bad0b6b2b46242d4241c5654ea9897)
![{\displaystyle {\bigg [}{\frac {\nabla X}{d{\mathit {t}}}}{\bigg ]}={\frac {1}{2{\sqrt {g_{11}g_{22}}}}}{\bigg \{}{\frac {\partial g_{22}}{\partial u^{1}}}{\frac {du^{2}}{dt}}-{\frac {\partial g_{11}}{\partial u^{2}}}{\frac {du^{1}}{dt}}{\bigg \}}+{\frac {d\varphi }{dt}}\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/cc0d831abb3bcbe38d662eae3387b48cc0431e5c)
![{\displaystyle {\bigg [}{\frac {\nabla X}{dt}}{\bigg ]}={\bigg [}{\frac {\nabla e_{1}}{dt}}{\bigg ]}+{\frac {d\varphi }{dt}}\!}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/26c3ab9c57814ab671556adf2640dc7f62a9749c)