# Homework Help: Cartan's first structure equation proof

1. May 4, 2006

### r16

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 $$\nabla_x$$ is a koszul connection, $$e_i$$ is a basis and $$\partial_j A^j_i$$ is a change of basis from e and $$\omega$$ 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 $$A^j_i$$?

I am no impact no idea on this step and it seems quite important so i dont want to skip it. Any ideas what im missing?

**nb in equation 2 $$\omega^i_j$$ should be $$\omega^j_i$$

Last edited: May 4, 2006
2. May 4, 2006

### George Jones

Staff Emeritus
If $f$ is a function, how is the exterior derivative $df$ defined?

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

$$\left\{\frac{\partial}{\partial x_i} \right\}[/itex] to answer this. Regards, George 3. May 4, 2006 ### r16 that may be my biggest problem. I am a visual learner, but I dont have a good physical picture of 'd'. Algebraically, d is defined a [tex] d = \frac{\partial}{\partial x_i} dx^i$$.
d applied to the 0-form $$f$$ gives a 1-form $$df$$.

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

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

This is opposed to $$\omega^k_i$$, a $$[\frac{1}{1}]$$ 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?

Last edited: May 4, 2006
4. May 5, 2006

### George Jones

Staff Emeritus
For each $i$ and $j$, $A^{i}_{j}$ is a component, i.e., a function, while each $\omega^{i}_{j}$ is a 1-form, and, consequently, they are very different animals.

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

Similary, if $\omega^{i}_{j} = \omega^{i}_{jk} dx^k$, each $\omega^{i}_{jk}$ is a component, while each $dx^k$ 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

Last edited: May 5, 2006
5. May 5, 2006

### r16

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?

6. May 5, 2006

### George Jones

Staff Emeritus
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