- #1

"Don't panic!"

- 601

- 8

Does this follow from a more general case? That is, if one has two manifolds [itex]M,N[/itex] then, given a smooth function, [itex]f:M\rightarrow N[/itex] that maps between the two, one can define a

*differential map*[tex]df_{p}:T_{p}M\rightarrow T_{f(p)}N[/tex] that maps vectors in the tangent space [itex]T_{p}M[/itex] to a point [itex]p\in M[/itex] to tangent vectors in the tangent space [itex]T_{f(p)}N[/itex] to the point [itex]f(p)\in N[/itex]. Given this we can define such a map by the way a "new" vector [itex]df_{p}(\mathbf{v})\in T_{f(p)}N[/itex] (where [itex]\mathbf{v}\in T_{p}M[/itex]) acts on a function [itex]g:N\rightarrow\mathbb{R}[/itex] defined on [itex]N[/itex], i.e. [tex]\left(df_{p}(\mathbf{v})\right)(g)=\mathbf{v}_{p}(g\circ f)[/tex] In other words, the action of [itex]df_{p}(\mathbf{v})[/itex] on [itex]g[/itex] in [itex]N[/itex] should be equal to the action of [itex]\mathbf{v}[/itex] on [itex]g\circ f[/itex] in [itex]M[/itex].

Given this, then the first definition I gave is just a special case, in which [itex]N=\mathbb{R}[/itex] and [itex]df_{p}:T_{p}M\rightarrow T_{f(p)}\mathbb{R}\cong\mathbb{R}[/itex] becomes a one-form in the dual space [itex]T^{\ast}_{p}M[/itex] to the tangent space [itex]T_{p}M[/itex] at [itex]p\in M[/itex]. Also, we find that [itex]g:\mathbb{R}\rightarrow\mathbb{R}[/itex] and so [itex]g=\text{id}[/itex] is simply the identity map. Hence we find, [tex]\left(df_{p}(\mathbf{v})\right)(\text{id})\equiv df_{p}(\mathbf{v})=\mathbf{v}_{p}(\text{id}\circ f)=\mathbf{v}_{p}(f)[/tex]

Would this be correct, or am I missing something?