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- Thread starter sit.think.solve
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cristo

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After reading this, do you ahave any other specific questions?

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garrett

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The covariant derivative uses a connection, while the Lie derivative doesn't.

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Just to spotlight one of these: The Lie derivative L_X(Y) is basically a dynamical invariant. It measures how compatible the flows of the vector fields are, i.e. how much they commute with each other. If X is generated by the flow [itex]\phi_t[/itex] and Y is generated by [itex]\psi_s[/itex], then the Lie derivative [X,Y] at point P is the tangent vector at time 0 of the curve given by: [itex]t \mapsto \psi_{-t} \circ \phi_{-t} \circ \psi_t \circ \phi_t (P)[/itex].

Berger describes this as moving P forward in time along the X-curves by t, then moving along the Y-curve by t, then moving backward in time along the X-curve and finally moving backward in time along the Y-curve. If you've ended up back at P, then [X,Y]=0 at P.

In particular, if [itex] \phi_s \circ \psi_t=\psi_t \circ \phi_s [/itex] for all s and t, then [X,Y]=0.

The covariant derivative acts similarly except instead of pushing Y along the X-curves via X's flow, we are pushing Y along X-curve via parallel transport. This explanation, though, is a bit of circular logic, since one usually uses the specific covariant derivative to generate the parallel transport.

Berger describes this as moving P forward in time along the X-curves by t, then moving along the Y-curve by t, then moving backward in time along the X-curve and finally moving backward in time along the Y-curve. If you've ended up back at P, then [X,Y]=0 at P.

In particular, if [itex] \phi_s \circ \psi_t=\psi_t \circ \phi_s [/itex] for all s and t, then [X,Y]=0.

The covariant derivative acts similarly except instead of pushing Y along the X-curves via X's flow, we are pushing Y along X-curve via parallel transport. This explanation, though, is a bit of circular logic, since one usually uses the specific covariant derivative to generate the parallel transport.

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