I think the horizontal connection vector fields are just the horizontal distribution θⁱʲ =0, which is 4-dimensional and thus can be spanned by 4 linear independent vector fields Dᵢ. And the Lie algebra of the Lorentz group acts on the principal bundle K, probably like fA, A being the Lie algebra of the Lorentz group, f being a section of K, giving 6 vertical vector fields Dᵢᵣ on K. I think there is freedom to choose the 6 Dᵢᵣ. Is vertical in contrast to horizontal in the next line? I think so because K is 10 dimensional, the 4-dimensional base manifold plus the 6-dimensional fiber. Thus vertical means tangent to the fiber and horizontal means tangent to the base manifold. θ ʲᵏ are Lie-algebra-valued 1-forms and θ ͥ are torsion-free 1-forms. I think < θⁱʲ ,Dᵣ>=0 is the defining equation for the horizontal connection vector fields. Then after Dᵣ are chosen, imposing θ ͥ to be the dual 1-forms of Dᵣ, =δͥᵣ, the torsion-free condition Dθ ͥ=0 and =0 are arrived (In a Riemannian manifold, the connection coefficients are symmetric, so DDᵣ=0 leads to θ ʲᵏ(Dᵣ)=0.). Similarly, imposing θ ʲᵏ to be the dual 1-forms of Dᵢᵣ (after Dᵢᵣ are chosen, of course), =1/2(δʲᵢδᵏᵣ-δʲᵣδᵏᵢ), then the defining equation for the curvature dθ ʲᵏ+θ ʲᵣ^θ ͬᵏ=Rʲᵏ is arrived. I got a bonus: If DDᵣ=0, then the torsion in the frame Dᵣ vanishes. As for why [Dᵢ, Dᵣ]=RᵢᵣᵘᵛDᵤᵥ, I still have no idea. Its novel to me that translations in different directions dont commute in a manifold whose curvature doesnt vanish.
Posted on: Sat, 08 Mar 2014 13:35:35 +0000
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