Given $X_{u} • X_{v} = 0$ and an adapted frame $X_{u} = p \vec{e_{1}}$ and $X_{v} = q \vec{e_{2}}$ where $p$ and $q$ are positive functions on $U$.
- Find $\theta_{1}$, $\theta_{2}$, and then find an expression for $K$ in terms of $p$, $q$, and their derivatives.
Given $X_{u}$ and $X_{v}$ we can find the first fundamental form.
$$I = p^2 du^2 + q^2 dv^2$$
We have that $I = \theta_{1}^2 + \theta_{2}^2$
From $I$ we see that $\theta_{1} = p du$ and $\theta_{2} = qdv$
From here we can see that $d\theta_{1} = -p_{v} du \wedge dv$
Likewise, $d\theta_{2} = q_{u} du \wedge dv$
To find $K$ we first need to find $w_{12}$ and this is where I am running into trouble.
I have the following equations
$$d\theta_{1} + w_{12} \wedge \theta_{2} = 0$$ $$d\theta_{2} - w_{12} \wedge \theta_{1} = 0$$
Looking at the first equation we get
$$d\theta_{1} = -w_{12} \wedge \theta_{2}$$ $$-p_{v} du \wedge dv = -w_{12} \wedge q dv$$
This makes it look like $w_{12}$ has a $du$ piece.
Doing the same with the second equation
$$d\theta_{2} = w_{12} \wedge \theta_{1}$$ $$q_{u} du \wedge dv = w_{12} \wedge p du$$
This makes it look like $w_{12}$ has a $dv$ piece.
But $w_{12}$ cannot have both a $du$ and $dv$ piece. Then those wedges would be 0.
But if you actually try to figure out was $w_{12}$ is for each equation you get the following.
For equation 1: $w_{12} = \frac{p_{v}}{q} du$
For equation 2: $w_{12} = \frac{q_{u}}{p} dv$
So, they give different $w_{12}$. What does $w_{12} look like if it has both a du and dv piece? Any advice on how to fix this? Thanks!