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Jan23-10, 11:49 AM
P: 1,402
Quote Quote by Fredrik View Post
Suppose [itex]\phi:\mathbb R^n\rightarrow\mathbb R^n[/itex], and that v is a tangent vector at x. Let the I be the identity map on [itex]\mathbb R^n[/itex]. (Note that this function satisfies the definition of a coordinate system). Show that the components of [itex]\phi_*v[/itex] in I are the same as the components of the matrix [itex]J_\phi (x)v'[/itex], where [itex]J_\phi(x)[/itex] is the Jacobian matrix of [itex]\phi[/itex] at x, and v' is the matrix of components of v in I.
Let M stand for the domain of [itex]\phi[/itex], and N the codomain. I'll call the (in this case global) coordinates of M with the identity map [itex]m^i[/itex], and the (in this case global) coordinates of N with the identity map [itex]n^i[/itex]. The components of v, in the tangent space of some point in M, with the identity map, will be some linear combination of basis tangent vectors, which have been defined as partial derivative operators, thus:

[tex]v = v^i \frac{\partial }{\partial m^i}.[/tex]

In this case, the Jacobian for components is the matrix of partial derivatives:

[tex]\left [ \frac{\partial n^j}{\partial m^i} \right ]_{ji}.[/tex]


[tex]J v' = J v = \frac{\partial n^j}{\partial m^i} v^i \frac{\partial \left f}{\partial n^j}.[/tex]


[tex]\phi_* v\left ( f \right ) = v\left ( f \circ \phi \right ),[/tex]

which, in component form, becomes

[tex]v^i \frac{\partial }{\partial m^i} \left ( f \circ \phi \right ) = v^i \frac{\partial f}{\partial n^j} \frac{\partial n^j}{\partial m^i} = Jv'\left ( f \right )[/tex]

by the chain rule.