Connection form uniquely determined by linearly independent $\theta_1,\theta_2$?

Question

I'm working through a tutorial for a differential geometry class. The question is:

Consider the structure equations for a map $\bar x:\mathbb R^2\to\mathbb E^2$. Suppose that $\theta_1,\theta_2$ are everywhere linearly independent. Show that given $\theta_1,\theta_2$, the connection form $w_{12}$ is uniquely determined by the first structure equation.

I really don't know how to go about it. But I've had a go:

(1) Using the second structure equation first: $$\begin{align}0&=dw+w\land w \\&=dw+\left(\begin{matrix}0&w_{12}\\-w_{12}&0\end{matrix}\right)\land\left(\begin{matrix}0&w_{12}\\-w_{12}&0\end{matrix}\right) \\&= dw \end{align}$$

So, we know that $w$ is an exact form.

(2) Now, using the first structure equation: $$\begin{cases} \\d\theta_1+w_{12}\land\theta_2 =0 \\d\theta_2-w_{12}\land\theta_1 =0 \end{cases}$$ I'm not sure to where go from here. I know this is system of equations and I can do row operations on them but I also know that $\theta_1,\theta_2$ are linearly independent, so no amount of row operations can negate either of the $\theta s$.

The notation is new to me so it's likely that I'm missing something trivial.

Any help would be grand, cheers.

Answer 1

Let $\omega_{12}=a_1\theta_1+a_2\theta_2$, $d\theta_1=b_1\theta_1\wedge\theta_2$ and $d\theta_2=b_2\theta_1\wedge\theta_2$ for some functions $a_1,a_2, b_1, b_2$ on $\mathbb R^2$. Plug into the 1st structure equations and get $a_1=-b_1, a_2=b_2.$