Lemme de Gauss (géométrie riemannienne)

Rappelons que

${\displaystyle T_{v}\exp _{p}:T_{p}M\cong T_{v}T_{p}M\supset T_{v}B_{\epsilon }(0)\longrightarrow T_{q}M.}$

Nous procédons en trois étapes :

• ${\displaystyle T_{v}\exp _{p}(v)=v\ }$ : construisons une courbe ${\displaystyle \alpha$ :\mathbb {R} \supset I\rightarrow T_{p}M} telle que ${\displaystyle \alpha (0):=v\in T_{p}M}$, ${\displaystyle \alpha '(0):=v\in T_{v}T_{p}M\cong T_{p}M}$ et ${\displaystyle |v|=cste}$. Comme ${\displaystyle T_{v}T_{p}M\cong T_{p}M\cong \mathbb {R} ^{n}}$, on peut poser ${\displaystyle \alpha (t):=e^{t}v\ }$. Alors,

${\displaystyle T_{v}\exp _{p}(v)={\frac {\mathrm {d} }{\mathrm {d} t}}{\Bigl (}\exp _{p}\circ \alpha (t){\Bigr )}{\Big \vert }_{t=0}={\frac {\mathrm {d} }{\mathrm {d} t}}\gamma (t,p,v){\Big \vert }_{t=0}=v.}$

Calculons maintenant le produit scalaire ${\displaystyle \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w)\rangle }$.

Décomposons ${\displaystyle w\ }$ en une composante ${\displaystyle w_{T}\ }$ tangente à ${\displaystyle v\ }$ et une composante ${\displaystyle w_{N}\ }$ normale à ${\displaystyle v\ }$. En particulier, posons ${\displaystyle w_{T}:=\alpha v\ }$, ${\displaystyle \alpha \in \mathbb {R} }$.

L'étape précédente implique alors directement :

${\displaystyle \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w)\rangle =\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{T})\rangle +\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\alpha \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(v)\rangle +\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\langle v,w_{T}\rangle +\langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle .}$

Il faut donc montrer que le second terme est nul, car, selon le lemme de Gauss, on devrait avoir

${\displaystyle \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\langle v,w_{N}\rangle =0.}$

• ${\displaystyle \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =0}$ :

définissons la courbe

${\displaystyle \alpha$ :]-\epsilon ,\epsilon [\times [0,1]\longrightarrow T_{p}M,\qquad (s,t)\longmapsto t\cdot v(s),}

avec ${\displaystyle v(0):=v\ }$ et ${\displaystyle v'(0):=w_{N}\ }$. On remarque au passage que

${\displaystyle \alpha (0,1)=v(0)=v,\qquad {\frac {\partial \alpha }{\partial t}}(0,t)=v(0)=v,\qquad {\frac {\partial \alpha }{\partial s}}(0,t)=tw_{N}.}$

Posons alors

${\displaystyle f:]-\epsilon ,\epsilon [\times [0,1]\longrightarrow M,\qquad (s,t)\longmapsto \exp _{p}(t\cdot v(s)),}$

et calculons :

${\displaystyle T_{v}\exp _{p}(v)=T_{\alpha (0,1)}\exp _{p}\left({\frac {\partial \alpha }{\partial t}}(0,1)\right)={\frac {\partial }{\partial t}}{\Bigl (}\exp _{p}\circ \alpha (s,t){\Bigr )}{\Big \vert }_{t=1,s=0}={\frac {\partial f}{\partial t}}(0,1)}$

et

${\displaystyle T_{v}\exp _{p}(w_{N})=T_{\alpha (0,1)}\exp _{p}\left({\frac {\partial \alpha }{\partial s}}(0,1)\right)={\frac {\partial }{\partial s}}{\Bigl (}\exp _{p}\circ \alpha (s,t){\Bigr )}{\Big \vert }_{t=1,s=0}={\frac {\partial f}{\partial s}}(0,1).}$

Donc,

${\displaystyle \langle T_{v}\exp _{p}(v),T_{v}\exp _{p}(w_{N})\rangle =\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\rangle (0,1).}$

On va vérifier maintenant que ce produit scalaire est en fait indépendant de la variable t, et donc que, par exemple,

${\displaystyle \langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\rangle (0,1)=\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\rangle (0,0)=0,}$

car, selon ce qui a été donné plus haut,

${\displaystyle \lim _{t\rightarrow 0}{\frac {\partial f}{\partial s}}(t,0)=\lim _{t\rightarrow 0}T_{tv}\exp _{p}(tw_{N})=0}$

étant donné que la différentielle est une application linéaire. Ceci prouverait alors le lemme.

• On vérifie que ${\displaystyle {\frac {\partial }{\partial t}}\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\rangle =0}$ : c'est du calcul direct. En effet, on prend d'abord conscience du fait que les applications ${\displaystyle t\mapsto f(s,t)}$ sont des géodésiques, i.e. ${\displaystyle {\frac {D}{\partial t}}{\frac {\partial f}{\partial t}}=0}$. Alors,

${\displaystyle {\frac {\partial }{\partial t}}\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\rangle =\langle \underbrace {{\frac {D}{\partial t}}{\frac {\partial f}{\partial t}}} _{=0},{\frac {\partial f}{\partial s}}\rangle +\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial t}}{\frac {\partial f}{\partial s}}\rangle =\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial s}}{\frac {\partial f}{\partial t}}\rangle ={\frac {\partial }{\partial s}}\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial t}}\rangle -\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial s}}{\frac {\partial f}{\partial t}}\rangle .}$

Donc, en particulier,

${\displaystyle 0={\frac {1}{2}}{\frac {\partial }{\partial s}}\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial t}}\rangle =\langle {\frac {\partial f}{\partial t}},{\frac {D}{\partial s}}{\frac {\partial f}{\partial t}}\rangle ={\frac {\partial }{\partial t}}\langle {\frac {\partial f}{\partial t}},{\frac {\partial f}{\partial s}}\rangle ,}$

car on a ${\displaystyle |v|=cste}$.