THÈSE Koléhè Abdoulaye COULIBALY-PASQUIER

where Ut 3 is the horizontal lift of Zt 3 , to be correct we have to express a systemof stochastic differential equations as in Kendall [19], Ut1 is the horizontal liftof Zt 1 , and dWti are Brownian motion that drives Zt 1 , the mirror map m t x,yconsists in transporting a vector along the unique minimal g(t)-geodesic thatjoins x to y and then reflecting it in the hyperplane of (T y M, g(t)) which isperpendicular to the incoming geodesic.By isometry property of the horizontal lift of the g(t)-BM (see [4]),is an R n -valued Brownian motion.(U 3 t ) −1 m t Z 1 t ,Z3 t U 1 t dW it ,Let T2 N = (T1 N + 1) ∧ inf{t > T N 2 1 , ρ t (Zt 1 , Zt 3 √) > πcst+ɛ∧ Cn(d,K,cst−ɛ)2} ∧ T ∧ C2 N .• if ρ T N1(Z 1 T N 1, Z 3 T N 1√) > πcst+ɛ∧ Cn(d,K,cst−ɛ)2then T N 4 2 = T1 N .Iterate step 1 and 2 successively (changing T0 N by T2 N and T1 N by T3 N in step 1,changing T1 N by T3 N and T2 N by T4 N in step 2 ..., after time T if we have no coupling,we let Z 3 evolve independently of Zt1 until the end), we build by induction theprocess Zt3 and a sequence of stopping times. We sketch it as:• if C N < TT N 0independent−→ T N 1coupling−→ T N 2independent−→ T N 3coupling−→ T N 4 ... C NZ 3 t =Z1 t−→ 0• if C N > TT N 0independent−→ T N 1coupling−→ T N 2independent−→ T N 3coupling−→ T N 4 ... T independent−→ 0Proposition 3.5 The two processes Z 3 and Z 2 are equal in law.proof : It is clear that before N the two processes are equal so equal in law.After:ZN 3 = Z2 N .⎧⎨⎩∗dZt 3 = ∑ i U t 3 e i ∗ dB i , when t ∈ [T2k N, T 2k+1 N ] and T 2k+1 N ≤ C N∗dZt 3 = ∑ i U t 3 ∗ d((Ut 3 ) −1 m t UZt 1,Z3 t 1 )e i dWt i , when t ∈ [T2k+1 N , T 2k+2 N ], and T 2k+2 N ≤ C NtZt 3 = Zt 1 , C N ≤ tWe write:Z 3 t = ∑ ∞k=01 [T Nk ,T N k+1 ] (t) ∗ dZ 3 t= ∑ k:even .... + ∑ k:odd1883