Rappelons que
![{\displaystyle T_{v}\exp _{p}:T_{p}M\cong T_{v}T_{p}M\supset T_{v}B_{\epsilon }(0)\longrightarrow T_{q}M.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4299f50363b242c4140ac37b86a973ccedcc96c6)
Nous procédons en trois étapes :
: construisons une courbe
telle que
,
et
. Comme
, on peut poser
. Alors,
![{\displaystyle T_{v}\exp _{p}(v)={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl (}\exp _{p}\circ \alpha (t){\Bigr )}{\Big \vert }_{t=0}={\frac {\mathrm {d} }{\mathrm {d} t}}\gamma (t,p,v){\Big \vert }_{t=0}=v.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e11523fb13984bfdd79631ee780e1017c9ca952)
Calculons maintenant le produit scalaire
.
Décomposons
en une composante
tangente à
et une composante
normale à
. En particulier, posons
,
.
L'étape précédente implique alors directement :
![{\displaystyle \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w)\rangle =\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{T})\rangle +\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\alpha \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(v)\rangle +\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\langle v,w_{T}\rangle +\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6dcb5a4dda210c73167c34f5e3ae0040a50975ce)
Il faut donc montrer que le second terme est nul, car, selon le lemme de Gauss, on devrait avoir
![{\displaystyle \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\langle v,w_{N}\rangle =0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a2751367f00f4afa68999d7388262c7ad619f52)
:
définissons la courbe
![{\displaystyle \alpha :]-\epsilon ,\epsilon [\times [0,1]\longrightarrow T_{p}M,\qquad (s,t)\longmapsto t\cdot v(s),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23733997c8720a12eadb854dcdab7b28c419fb21)
avec
et
. On remarque au passage que
![{\displaystyle \alpha (0,1)=v(0)=v,\qquad {\frac {\partial \alpha }{\partial t}}(0,t)=v(0)=v,\qquad {\frac {\partial \alpha }{\partial s}}(0,t)=tw_{N}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fa24153a1d1f6e7b2a3c26d6af99ff45721f7d3)
Posons alors
![{\displaystyle f:]-\epsilon ,\epsilon [\times [0,1]\longrightarrow M,\qquad (s,t)\longmapsto \exp _{p}(t\cdot v(s)),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/178afdec23dea9a592ca4f8db3e4d1b04793716b)
et calculons :
![{\displaystyle T_{v}\exp _{p}(v)=T_{\alpha (0,1)}\exp _{p}\left({\frac {\partial \alpha }{\partial t}}(0,1)\right)={\frac {\partial }{\partial t}}{\Bigl (}\exp _{p}\circ \alpha (s,t){\Bigr )}{\Big \vert }_{t=1,s=0}={\frac {\partial f}{\partial t}}(0,1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8e0f4cbbb0bb884fbdd65673c68cb7a242bf26b)
et
![{\displaystyle T_{v}\exp _{p}(w_{N})=T_{\alpha (0,1)}\exp _{p}\left({\frac {\partial \alpha }{\partial s}}(0,1)\right)={\frac {\partial }{\partial s}}{\Bigl (}\exp _{p}\circ \alpha (s,t){\Bigr )}{\Big \vert }_{t=1,s=0}={\frac {\partial f}{\partial s}}(0,1).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/395e02c24c0ce5dc3bf65169d5f015515cee62bd)
Donc,
![{\displaystyle \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\rangle (0,1).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/488306c099abc8962703e60cd76b899910fe20bc)
On va vérifier maintenant que ce produit scalaire est en fait indépendant de la variable t, et donc que, par exemple,
![{\displaystyle \langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\rangle (0,1)=\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\rangle (0,0)=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42bc2c5f238d894d9c0678815b66dde08ec0311c)
car, selon ce qui a été donné plus haut,
![{\displaystyle \lim _{t\rightarrow 0}{\frac {\partial f}{\partial s}}(t,0)=\lim _{t\rightarrow 0}T_{tv}\exp _{p}(tw_{N})=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/768a3de0156331951d34550c51762b690a7ddd9f)
étant donné que la différentielle est une application linéaire. Ceci prouverait alors le lemme.
- On vérifie que
: c'est du calcul direct. En effet, on prend d'abord conscience du fait que les applications
sont des géodésiques, i.e.
. Alors,
![{\displaystyle {\frac {\partial }{\partial t}}\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\rangle =\langle \underbrace {{\frac {D}{\partial t}}{\frac {\partial f}{\partial t}}} _{=0},{\frac {\partial f}{\partial s}}\rangle +\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial t}}{\frac {\partial f}{\partial s}}\rangle =\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial s}}{\frac {\partial f}{\partial t}}\rangle ={\frac {\partial }{\partial s}}\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial t}}\rangle -\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial s}}{\frac {\partial f}{\partial t}}\rangle .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c9c5480fd6dd4d5b94a94971b86a5cc0441204c)
Donc, en particulier,
![{\displaystyle 0={\frac {1}{2}}{\frac {\partial }{\partial s}}\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial t}}\rangle =\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial s}}{\frac {\partial f}{\partial t}}\rangle ={\frac {\partial }{\partial t}}\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\rangle ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3fa85db817352e3c28e5ce248d1f7c536b166b5)
car on a
.