lavinia Science Advisor Messages 3,385 Reaction score 760 Thread starter Oct 22, 2011 #1 On a 2 dimensional Riemannian manifold how does one derive the covariant derivative from the connection 1 form on the tangent unit circle bundle?
On a 2 dimensional Riemannian manifold how does one derive the covariant derivative from the connection 1 form on the tangent unit circle bundle?
Ben Niehoff Science Advisor Gold Member Messages 1,891 Reaction score 170 Oct 22, 2011 #2 Its the same formula in any dimension. Let [itex]X = X^a e_a[/itex] be a vector field, where [itex]e_a[/itex] is an orthonormal frame. Then [tex]\nabla X = (D X^a) \otimes e_a = (d X^a + \omega^a{}_b X^b) \otimes e_a[/tex]
Its the same formula in any dimension. Let [itex]X = X^a e_a[/itex] be a vector field, where [itex]e_a[/itex] is an orthonormal frame. Then [tex]\nabla X = (D X^a) \otimes e_a = (d X^a + \omega^a{}_b X^b) \otimes e_a[/tex]