Tag: differential

I am trying to understand the proof of the first and second variation of the Arclength formulas for Riemannian manifests. I want a check that the following covariant derivatives switch. I find it intuitive but I also want to have a formal test.

Some notation: Let $ gamma (t, s): overbrace {(a, b)} ^ {t} times overbrace {(- epsilon, epsilon)} ^ {s} rightarrow M $ be a variation of $ gamma_0 (t) $ with $ | dot { gamma_ {0} |} = lambda forall t in (a, b) $ Y $ V = gamma _ {*} left ( frac { partial} { partial s} right) $ The variational vector field.

So I would like to demonstrate that $ frac {D} {ds} ( dot { gamma_t}) = frac {D} {dt} V $ or in other words that $ frac {D} {ds}, frac {D} {dt} $ commute. Note that $ frac {D} {ds} $ is the covariant derivative throughout the map $ gamma $ therefore we will use the property for connections across maps that $ nabla ^ { gamma} _X (Z circ gamma (p)) = nabla _ { gamma _ * (X)} Z | _p $.

Indeed $ frac {D} {ds} left ( dot { gamma_t} right) = frac {D} {ds} ( gamma _ * ( frac {d} {dt}) circ ( gamma (s, t))) = D _ { gamma _ * left ( frac {d} {dt} right)} gamma _ * ( frac {d} {ds}) = D _ { gamma _ * left ( frac {d} {ds} right)} gamma _ * ( frac {d} {dt}) $.

The last equality follows from $ D _ { gamma _ * left ( frac {d} {dt} right)} gamma _ * ( frac {d} {ds}) – D _ { gamma _ * left ( frac {d} {ds} right)} gamma _ * ( frac {d} {dt}) = left ( gamma _ * ( frac {d} {dt}), gamma _ * ( frac {d} {ds}) right) = gamma _ * (( frac {d} {ds}, frac {d} {dt})) = 0 $ as $ frac {d} {ds}, frac {d} {dt} $ they are vector coordinate fields.