Differential Geometry 1-forms.

Question

I'm really having trouble understanding this and I'm not sure if this is the right way to prove this.

For a 1-form Ø = $\sum f_i \theta_i$ , prove that

$$dØ =\sum_j(df_i + \sum_{i} f_i w^{ij}) ˄ \theta_j$$

This is my approach:

$dØ= d(\sum f_i\theta_i)$ = $\sum (df_i\theta_i + f_id\theta_i)$

By theorem 2.8.3 (Cartan Structural Equation) $d\theta_i=\sum_{j}w^{ij} ˄ \theta_j$

So, this implies that dØ= $\sum (df_i\theta_i + f_i\sum_{j}w^{ij} )˄ \theta_j$ = $\sum (df_i\theta_i + \sum_{j}f_iw^{ij} )˄ \theta_j$.

However, I'm confused as to what happens to the $\theta_i$ on the first half. Any help here will be great! Thanks.

Answer 1

The formula is $d\phi=\sum df_i\wedge\theta_i+f_i\wedge d\theta_i$

Answer 2

You have

$$d\phi= d\left(\sum f_i\theta_i\right) = \sum_i (df_i\wedge\theta_i + f_i\color{blue}{d\theta_i}) \tag{1}$$

and then

$$\color{blue}{d\theta_i}=\color{blue}{\sum_{j}w_{ij} \wedge \theta_j} \tag{2}$$

Everything is ok so far. But then you say

So, this implies that $$d\phi = \sum_i \left(df_i\wedge\theta_i + f_i\sum_{j}w_{ij}\right)\wedge\theta_j$$

which is not right. If you substitute $(2)$ into $(1)$ you get $$d\phi= d\left(\sum_i f_i\theta_i\right) = \sum_i\left(df_i\wedge\theta_i + f_i\color{blue}{\left(\sum_{j}w_{ij} \wedge \theta_j\right)}\right)$$

See if you can you continue from here.