Ivancevic_Applied-Diff-Geom

456 Applied Differential Geometry: A Modern IntroductionThis gives the following relation between the upper triangular parts of ˆLand ÂΓ αβ γ = 2(L αβ γ ) 2 ,(for α < β).Using the relation  = ˆΓ+ ˆB, where ˆB = gB, we find the following relationbetween the upper triangular components of ˆL and ˆBB αβ γ = L αβ γ − 2(L αβ γ ) 2 ,(for α < β).Finally expressing ˆL and ˆB in terms of ˆΓ we have for the upper triangularpartsL αβ γ =√12 Γ αβ γ , B αβ γ = −Γ αβ γ +√12 Γ αβ γ , (for α < β).Furthermore, the diagonal elements of  and ˆΓ are identical. This impliesthat ˆB is antisymmetric in agreement with our expectations.3.16.3 4D GeneralizationsRecall that we can get a 4D space–time simply by adding a time coordinateaccording to the prescription [Rosquist and Goliath (1997)](4) ds 2 = −(dq 0 ) 2 + ds 2 ,( )where ds 2 is a 3D positive–definite metric. It follows that (4) 0 0Γ = .0 ΓFor the cases obtained above this will lead to inequivalent space–times. Oneway to generalize the 3D Lax pair is( )( )i (4) p0 00 0L = ,(4) A = ,0 L0 Afor which(4) B =( 0 00 BThis Lax pair gives the geodesic Hamiltonian of the corresponding space–time metric as quadratic invariant.3.16.3.1 Case IAdding a time dimension to (3.266) we get the metric [Rosquist and Goliath(1997)]).ds 2 = −(dq 0 ) 2 + g 11 (dq 1 ) 2 + (dq 2 ) 2 + g 33 (dq 3 ) 2 ,where