By Bennett Chow; Peng Lu; and Lei Ni

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**Example text**

Therefore dαi (q) = d(k (Xi ))(q), which implies d d ci dαi (q) = i=1 ci d(k (Xi ))(q) i=1 d = dk ( ci Xi )(q) = d(k (cQ ))(q). (49) i=1 In other words, with respect to the basis {((Xi (q), 0), (0, ei ))}1≤i≤d of the „ « B At . Here A = A(q) is space (HQ×Rd )(q,c) , the matrix of Q×Rd (q, c) is −A 0 the d × d positive definite symmetric matrix6 (Ajk ) = k(q)(Xj (q), Xk (q)) and B = B(q) is the d × d antisymmetric matrix (B ,m ), where d B m ci dαi (q)(X (q), Xm (q)) = LX k(cQ , Xm ) (q) = i=1 − LXm k(cQ , X ) (q) − k(q) cQ (q), [X , Xm ](q) .

Pulling back both sides of XµZ (k )∗ XµZ ω = (k )∗ XµZ ωQ = dµZ by k gives (k )∗ ωQ = k (XµZ ∗ ωQ ) ∗ = (k ) dµZ = d((k ) µZ ) = dPZ . Therefore XPZ = (k )∗ XµZ . From T πQ ◦ k = T τQ it follows that T τQ ◦ XPZ = T πQ ◦ k ◦ (k )∗ XµZ = T πQ ◦ XµZ ◦ k = Z ◦ πQ ◦ k = Z ◦ τQ . This proves (89). Since XPZ = (k )∗ XµZ and XµZ = ZT ∗ Q , it follows that XPZ = ZT Q if and only if ZT Q = (k )∗ ZT ∗ Q . This equality holds if and only if T ϕs = (k )−1 ◦ (T ϕ−s )t ◦ k , that is, for every q ∈ Q and every vq ∈ Tq Q we have k (ϕs (q))(Tq ϕs vq ) = k (q)(vq )Tϕ−s (q) ϕs .

Then the evolution of Yh in Lc is given by the distributional Hamiltonian vector field YhLc of hLc relative to (HLc , Lc ). s. H Lc , hLc ). (107) c∈Rd We can refine the above decomposition further by considering accessible sets of HLc . Let M be an accessible set HLc . 42, M is an immersed submanifold of Lc . The restriction HM of H to points in M coincides with HLc ∩ T M . Let M be the restriction of Lc 46 Nonholonomically constrained motions to HM and hM the restriction of hLc to M . s. H c∈Rd M , hM ).