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T. Hangan ELASTIC STRIPS AND DIFFERENTIAL GEOMETRY

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184 T. <strong>Hangan</strong><br />

5. The connection associated to the system (5).<br />

In componentwise form the system (5) on an n−dimensional manifold M writes<br />

(6) x i ′′′ = A i jk x j ′′ x k ′ + B i jkl x j ′ x k ′ x l ′ + C i j x j ′ , i = 1, 2,..., n.<br />

where one has to sum up when repeated indices appear. The invariance of the system<br />

(6) with respect to coordinate transformations implies the following:<br />

REMARK 3. The functions<br />

Ŵ i jk = 1 3 Ai jk<br />

transform like the coefficients of a linear connection Ŵ, see [8].<br />

REMARK 4. The connection Ŵ decomposes as a sum Ŵ = S + T where S is a<br />

symmetric connection ( without torsion ) and T is the torsion of Ŵ. Thus<br />

S i jk = 1 2 (Ŵi jk + Ŵi kj ) , T i jk = 1 2 (Ŵi jk − Ŵi kj ).<br />

REMARK 5. Given a smooth curve, s ∈ [0, L] → c(s) ∈ M, its third order jet<br />

j 3 c(s) (c) = (c(s), c′ (s), c ′′ (s), c ′′′ (s)) at c(s) defines a vector a 3 (c(s)) ∈ T c(s) M with<br />

components<br />

a i 3 (c) = ci′′′ − 3Ŵ i rs cr′′ c s′ − (Ŵ i rs,t − (Ŵi pr + 2T i pr )Ŵ p st)c r′ c s′ c t′ .<br />

It will be called acceleration of third order of c at c(s). In terms of the covariant<br />

derivation ∇ associated to Ŵ and of the torsion T of Ŵ, a 3 (c) writes<br />

REMARK 6. System (5) can be written<br />

a 3 (c) = ∇ c ′(∇ c ′c ′ ) + 2T(∇ c ′c ′ , c ′ ).<br />

a 3 (c(s)) = 3 (c ′ (s)) + C 1 (c ′ ))<br />

where 3 is a tensor field on M of type (1,3) symmetric in its three covariant indices<br />

and C 1 is a tensor field of type (1,1) i.e. a field of endomorphisms of the tangent bundle<br />

T M .In components , the field expresses with the functions A i jk and Bi jkl<br />

from (6).<br />

6. Differential geometric objects associated to system (3), (4).<br />

In accord with Remark 3 , the linear conection Ŵ associated to system (3), (4) has<br />

components<br />

Ŵκκ κ = 1 − 3ω2<br />

3κ(1 + ω 2 ) , Ŵκ κω = − 8ω(1 + 3ω2 )<br />

3(1 + ω 2 ) 2 ,<br />

Ŵωκ κ = − 4ω(1 + 3ω2 )<br />

3(1 + ω 2 ) 2 , Ŵκ ωω = − 4κ(2 + 3ω2 + 9ω 4 )<br />

3(1 + ω 2 ) 3

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