Cartan's first structure equation proof

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this is my first post on this site but it looks like the sort of ppl that i would like to associate myself with.

Unfourtanately, I have not had any formal schooling for any mathematics above calculus but i have read a few books and papers and am trying to make due.

I was studying about the cartan's first structure equation and was looking at this proof :

http://www.pzgnet.cc/images/cartan/eq1.png

where [tex]\nabla_x[/tex] is a koszul connection, [tex]e_i[/tex] is a basis and [tex]\partial_j A^j_i[/tex] is a change of basis from e and [tex]\omega[/tex] is a standard connection in the actual equation :

http://www.pzgnet.cc/images/cartan/eq2.png

In step 3 why can the exterior derivitave be applied to [tex]A^j_i[/tex]?

I am no impact no idea on this step and it seems quite important so i don't want to skip it. Any ideas what I am missing?

**nb in equation 2 [tex]\omega^i_j[/tex] should be [tex]\omega^j_i[/tex]
 
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If [itex]f[/itex] is a function, how is the exterior derivative [itex]df[/itex] defined?

It may prove useful to think about vector fields [itex]X[/itex] and/or a coodinate basis

[tex]\left\{\frac{\partial}{\partial x_i} \right\}[/itex]<br /> <br /> to answer this.<br /> <br /> Regards,<br /> George[/tex]
 
that may be my biggest problem. I am a visual learner, but I don't have a good physical picture of 'd'.

Algebraically, d is defined a [tex]d = \frac{\partial}{\partial x_i} dx^i[/tex].
d applied to the 0-form [tex]f[/tex] gives a 1-form [tex]df[/tex].

Then it would follow that [tex]X(f) = df (X)[/tex] as in the definition of a 1-form.

However is [tex]A^j_i[/tex] equivalent to a scalar because the contravariant and covariant parts of the [tex][\frac{1}{1}][/tex] valent tensor cancel out?

This is opposed to [tex]\omega^k_i[/tex], a [tex][\frac{1}{1}][/tex] tensor as well, which acts as a 1-form on X. Is there a standard of how a tensor acts or is it based on the definition of the tensor?
 
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For each [itex]i[/itex] and [itex]j[/itex], [itex]A^{i}_{j}[/itex] is a component, i.e., a function, while each [itex]\omega^{i}_{j}[/itex] is a 1-form, and, consequently, they are very different animals.

Consider a couple of examples. A vector field [itex]X[/itex] can be expressed in terms a set of basis fields as [itex]X = X^{i} e_{i}[/itex]. Each [itex]X^i[/itex] is is a component, i.e., a scalar-valued function of the base space, while each [itex]e_i[/itex] is a vector field.

Similary, if [itex]\omega^{i}_{j} = \omega^{i}_{jk} dx^k[/itex], each [itex]\omega^{i}_{jk}[/itex] is a component, while each [itex]dx^k[/itex] is a 1-form.

Bottom line: sometime indices label comonents, and sometimes they label other objects.

Welcome to Physics Forums, and, if I haven't answered all your questions, or if my explanation is not very clear, keep asking questions.

Regards,
George
 
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thank you very much sir, that clears it up perfectly.

However i do have one more inquery. What is a good physical/geometrical description of an exterior derivitave?
 
r16 said:
What is a good physical/geometrical description of an exterior derivitave?

Unfortunately, I don't know a good physical/geometrical description of an exterior derivative. Maybe someone else does.

I just work abstractly with its properties.

Regards,
George