Let $h$ be a $C^\infty$ function on $\mathbb{R}$. At each $(x,y) \in \mathbb{R}^2$ define $$\langle,\rangle_{(x,y)} = dx\otimes dx + h(x)^2dy\otimes dy.$$ Compute the Gaussian curvature of $\mathbb{R}^2$ with this metric.
The Gaussian curvature is given by $K=\Omega_2^1(e_1,e_2)$ in this case I think. In order to fin $K$ I need to find the connection 1-form $\omega_2^1$ and to find this I can use the first structural equation, but for that I need an orthonormal frame $(e_1,e_2)$ and a coframe $(\theta_1,\theta_2)$.
So given the metric how can I find an orthonormal frame with respect to it? If I set $e_1=a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y}$ and $e_2 = c\frac{\partial}{\partial x} + d\frac{\partial}{\partial y}$, then $$\langle e_1,e_2\rangle_p=ac+h^2(x)bd= \delta_{12}$$ and $$\langle e_1,e_1\rangle_p=a^2+h^2(x)b^2=1, \langle e_2,e_2\rangle_p=c^2+h^2(x)d^2=1$$ given these how do I find the coefficients? The two latters suggest that $a=1, b=0$ and $c=1$ and $d=0$, but this wont satisfy the first equation since it's always $1$ then.
Best Answer
Your approach is fine. It's not hard to see that the orthonormal frame you are looking for is given by $e_1=\frac{\partial}{\partial x}$ and $e_2=\frac{1}{h(x)}\frac{\partial}{\partial y}$ (instead of trying to solve the equations think about how to force the conditions you want).
The dual frame $(\theta^1,\theta^2)$ is then given by $\theta^1=dx$ and $\theta^2=h(x)dy$. In this case the strucutral equations then relate these and the connection $1$-forms by \begin{align*} d\theta^1&=-\omega_2^1 \wedge \theta^2 \\ d\theta^2 &= \omega_2^1 \wedge \theta^1. \end{align*}
On $\mathbb{R}^2$, the $1-$form $\omega_2^1$ is a linear combination of $dx$ and $dy$ so $\omega_{2}^1=adx + bdy$.
The first equation gives $0 = -(adx+bdy) \wedge h(x)dy = -ah(x)dx \wedge dy$ and since $h(x) > 0$, we have that $a = 0$.
The second equation gives that $h'(x)dy \wedge dz = (adx+bdy) \wedge dx = bdy \wedge dx = -bdx \wedge dy $ so $b = -h'(x)$ and $$\omega_2^1 = -h'(x)dy.$$
Now $\Omega_2^1 = d\omega_2^1 = -h''(x) dx \wedge dy$ and so $$K=\Omega_2^1(e_1,e_2)=-h''(x)dx \wedge dy\left( \frac{\partial}{\partial x}, \frac{1}{h(x)}\frac{\partial}{\partial y}\right) = -\frac{h''(x)}{h(x)}.$$