Undergrad Pushforward/Pullback of Vector Field

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The discussion revolves around the pushforward and pullback of vector fields, specifically reconciling two different formulations of the pushforward. The user seeks clarification on the formula f*(X) = Tf o X o f-1 and its relationship to f*(X)(g) = X(g o f). They emphasize the importance of evaluating at a specific point in the context of mappings between manifolds. Additionally, they explore the implications of a left action Lg on a curve γ(t) and confirm the correctness of their calculations regarding the derivative of Lg at t=0. Overall, the conversation highlights the complexities of vector field transformations in differential geometry.
knowwhatyoudontknow
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Pushforward of vector field.
I am looking at the following document. In section 2.3 they have the formula for the pushforward:

f*(X) := Tf o X o f-1

I am having trouble trying to reconcile this with the more familiar equation:

f*(X)(g ) = X(g o f)

Any help would be appreciated.
 
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I read this as an unfortunate way to reference the point of evaluation: Given ##f\, : \,M\longrightarrow N## we need a point ##p\in M## to evaluate at: ##p=f^{-1}(q)## for ##q\in N.##

Have a look at eq. 38 at the end in https://www.physicsforums.com/insights/pantheon-derivatives-part-iii/
I also collected a few of the many ways to write it.
 
OK. Thanks. If I have γ(t) = exp(tv) so that γ(0) = p and γ'(0) = v and Lg is a left action then is the following correct?

Lg*γ'(0) = (Lg o γ(t))'(0)

= d/dt|t=0Lg(γ(t))

= dLg/dt|t=0.v

= gv

In other words dLg/dt|t=0 ≡ g
 
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knowwhatyoudontknow said:
OK. Thanks. If I have γ(t) = exp(tv) so that γ(0) = p and γ'(0) = v and Lg is a left action then is the following correct?

Lg*γ'(0) = (Lg o γ(t))'(0)

= d/dt|t=0Lg(γ(t))

= dLg/dt|t=0.v

= gv

In other words dLg/dt|t=0 ≡ g
Looks ok to me.
 
Great. Thanks.
 

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