, we immediately see that F is a local isometry (note that dim(N) = n). The completeness of (N, h) follows from the completeness of M andM with Remark 3.13. Hence F is a surjective Riemannian covering. Moreover, if ? : [0, 1] ? N is a h-geodesic, it is tangent to D R and since it projects by F to a g-geodesic ?
,
, (Z)| q ? ker(G * | q ) from which g(X, Z) = h(L R (X)| q , L R (Z)| q ) = 0 for all Z ? ker A. This shows that X ? (ker A) ? . Therefore, for all
, We next prove (iii) ? (ii)
, The fact that q 0 = (x 0 ,x 0 ; A 0 ) ? Q can be seen as follows: if (iii) ? (1) holds, we havê g(A 0 X, A 0 Y ) =?(G * | z 0 ((F * | z 0 ) ?1 X), G * | z 0 ((F * | z 0 ) ?1 Y )) = h((F * | z 0 ) ?1 X, (F * | z 0 ) ?1 Y ) = g(X, Y ), where we used that G is a Riemannian immersion, G(z 0 ) ?M and A 0 := G * | z 0 ? (F * | z 0 ) ?1 : T x 0 M ? Tx 0M
, , vol.1
, A(t))
,
, ) and (2) mean, respectively, that G is a totally geodesic map, which is moreover a Riemannian (1) immersion, (2) submersion. By Corollary 1
, )), for all vector fields X, Y on N, we easily see that R
,
, A(t)Z), where X, Y , Z are any (local) F -lifts of X, Y, Z on N
, A(t)) is the unique rolling curve along ? starting at q 0 = (x 0 ,x 0 ; A 0 ) and defined on [0, 1] and therefore curves of Q formed in this manner fill up the orbit O D R (q 0 ). Moreover, by Eq. (24) we have shown also that
, Then, the rolling system ? (R) of Q(M,M ) is not completely controllable
, Q such thatx 0 ?N and im(A 0 ) ? Tx 0N . We proceed to prove that ? Q,M (O D R (q 0 )) ?N. To this end, we will first prove that for every geodesic curve ? on M starting at any point q = (x,x; A), with x ? M,x ?N and im(A) ? TxN, the resulting geodesic curve? D R :=? D R (?, q) = ? Q,M (q D R (?, q)) stays inN and that if q D R (?, q) = (?,?
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