We write ∂ 1 g(t, x) for ∂ s g(s, x) evaluated at t. Let us express the metric g(t) ina coordinate system; without loss of generality we can differentiate at time 0. Let(x 1 , ..., x n ) be a coordinate system at the point x(0), in which we have:V (t) = v i (t)∂ x iIn these local coordinates we get:C(t) = c i (t)∂ x ig(t, x(t)) = g i,j (t, x(t))dx i ⊗ dx jddt | t=0〈V (t), C(t)〉 g(t,x(t))= d dt | t=0g i,j (t, x(t))v i (t)c j (t)= (∂ 1 g i,j (0, x)v i (0)c j (0) + d (g i,j (0, x(t))v i (t)c j (t))dt | t=0 〉= ∂ 1 g i,j (0, x)v i (0)c j (0) +〈∇ 0 ẋ(0)〉V (t), C(0) g(0,x(0))+〈V (0), ∇ 0 ẋ(0) C(0) g(0,x(0)) 〈〉= 〈V (0), C(0)〉 ∂1 g(0,x(0) + ∇ 0 ẋ(0)〉V (0), C(0) g(0,x(0))+〈V (0), ∇ 0 ẋ(0) C(0) .g(0,x(0))In order to compute the g(t) norm of a tangent valued process we will use whatMalliavin calls “the transfer principle”, as explained in [13],[12].Recall the equivalence between a given connection on a manifold M and asplitting on T T M, i.e. T T M = H ∇ T T M ⊕ V T T M [19]. We have a bijection:For X, Y ∈ Γ(T M) we have:V v : T M π(v) −→ V v T T Mu ↦−→ d (v + tu)| dt t=0.∇ X Y (x) = V −1X(x) ((dY (x)(X(x)))v ),where (.) v is the projection of a vector in T T M onto the vertical subspace V T T Mparallely to H ∇ T T M.For a T (M)-valued process T t , we define:D S,t T t = (V Tt ) −1 ((∗dT t ) v,t ), (1.2)where (.) v,t is defined as before but for the connection ∇ t . The above generalizationmakes sense for a tangent valued process coming from a Stratonovich equation likeU t e i , where U t is a solution of the Stratonovich differential equation (1.1).383
For the solution U t of (1.1) we get()d 〈U t e i , U t e j 〉 g(t,π(Ut))= 〈U t e i , U t e j 〉 ∂1 g(t,π(U t)))dt (1.3)+ 〈 〉D S,t U t e i , U t e j + 〈 g(t,π(U t))U t e i , D S,t U t e j (1.4)〉g(t,π(U t))We would like to find a symmetric A such that the left hand side of the aboveequation vanishes for all time (i.e. U t ∈ (O(M), g(t))). Denote by ev ei : F(M) →T M the ordinary evaluation, and d ev ei : T F(M) → T T M its differential.It is easy to see that d ev ei sends V T F(M) to V T T M and sends H ∇h T F(M) toH ∇ T T M. We obtain:n∑D S,t U t e i = A α,i (t, U t )U t e α dt. (1.5)For simplicity, we take for notation: (∂ 1 G(t, U)) i,j = 〈Ue i , Ue j 〉 ∂tg(t) andα=1(G(t, U)) i,j = 〈Ue i , Ue j 〉 g(t).It is now easy to find the condition for A:(G(t, U t )A(t, U t )) j,i + (G(t, U t )A(t, U t )) i,j = −(∂ 1 G(t, U t )) i,j (1.6)Given orthogonality G(t, U t ) = Id and so by (1.6) A differs from − 1 2 ∂ 1G by skewsymmetric matrice, therefore will be equal to it if we demand symmetry. Converselyif A = − 1 2 ∂ 1G then (1.3) and equation (1.2) we see G(t, U t ) = Id.Remark : The SDE in proposition 1.2 does not explode because on anycompact time interval all coefficients and their derivatives up to order 2 in spaceand order 1 in time are bounded.Remark : The condition of symmetry is linked to a good definition of paralleltransport with moving metrics in some sense.To see where the condition of symmetry comes from we may observe whathappens in the constant metric case. It is easy to see that the usual definitionof parallel transport along a semi-martingale which depends on the vanishing ofthe Stratonovich integral of connection form, is equivalent to isometry and thesymmetry condition for the drift in the following SDE in F(M):⎧⎪⎨⎪⎩dŨt = ∑ di=1 L i(Ũt) ∗ dW i + A(Ũt) α,β V α,β (Ũt) dtŨ 0 ∈ (O(M), g)Ũ t ∈ (O(M), g) (isometry)A(., .) α,β ∈ S(n) (vertical evolution).394
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