Gaussian curvature of a given metric The exercise is from do Carmo, Differential Forms and Application, p.97. Consider $\mathbb{R}^2$ with the following inner product: if $p=(x,y)\in\mathbb{R}^2$ and $u,v\in T_p\mathbb{R}^2$, then $$\langle u,v\rangle_p=\frac{u\cdot v}{(g(p))^2},$$where $u\cdot v$ is the canonical inner product of $\mathbb{R}^2$ and $g:\mathbb{R}^2\rightarrow\mathbb{R}$ is a differentiable positive function. Prove that the Gaussian Curvature of this metric is $$K=g(g_{xx}+g_{yy})-(g_x^2+g_y^2)$$ I need some help to solve it. I guess all the details are given in the exercise.
 A: I'm not sure if there's a simpler way to do it, but here's a way that only involves the basic theory of surfaces.
Your first fundamental form is $E = G = \frac{1}{g^2}, F=0.$ Hence you can compute the Christoffel symbols by solving
\begin{equation}
\begin{pmatrix}
1/g^2 & 0\\
0 & 1/g^2
\end{pmatrix}
\begin{pmatrix}
\Gamma_{11}^1\\
\Gamma_{11}^2
\end{pmatrix}
= \begin{pmatrix}
\frac{1}{2}E_x\\
F_x-\frac{1}{2}E_y 
\end{pmatrix}
= \begin{pmatrix}
-g_x/g^3\\
g_y/g^3
\end{pmatrix}
\quad
\Rightarrow
\quad
\begin{pmatrix}
\Gamma_{11}^1\\
\Gamma_{11}^2
\end{pmatrix}
= \begin{pmatrix}
-g_x/g\\
g_y/g
\end{pmatrix}
\end{equation}
\begin{equation}
\begin{pmatrix}
1/g^2 & 0\\
0 & 1/g^2
\end{pmatrix}
\begin{pmatrix}
\Gamma_{12}^1\\
\Gamma_{12}^2
\end{pmatrix}
= \begin{pmatrix}
\frac{1}{2}E_y\\
\frac{1}{2}G_x 
\end{pmatrix}
= \begin{pmatrix}
-g_y/g^3\\
-g_x/g^3
\end{pmatrix}
\quad
\Rightarrow
\quad
\begin{pmatrix}
\Gamma_{12}^1\\
\Gamma_{12}^2
\end{pmatrix}
= \begin{pmatrix}
-g_y/g\\
-g_x/g
\end{pmatrix}
\end{equation}
\begin{equation}
\begin{pmatrix}
1/g^2 & 0\\
0 & 1/g^2
\end{pmatrix}
\begin{pmatrix}
\Gamma_{22}^1\\
\Gamma_{22}^2
\end{pmatrix}
= \begin{pmatrix}
F_y-\frac{1}{2}G_x\\
\frac{1}{2}G_y 
\end{pmatrix}
= \begin{pmatrix}
g_x/g^3\\
-g_y/g^3
\end{pmatrix}
\quad
\Rightarrow
\quad
\begin{pmatrix}
\Gamma_{22}^1\\
\Gamma_{22}^2
\end{pmatrix}
= \begin{pmatrix}
g_x/g\\
-g_y/g
\end{pmatrix}
\end{equation}
Furthermore, $(\Gamma_{12}^2)_x = -g_{xx}/g+g_x^2/g^2,$ and $(\Gamma_{11}^2)_y = g_{yy}/g-g_y^2/g^2$, so that the Gauss formula (see do Carmo's Differential Geometry of Curves and Surfaces, chapter 4-3, eq. 5) gives
\begin{align}
K &= -\frac{1}{E}\big( (\Gamma_{12}^2)_x - (\Gamma_{11}^2)_y  + \Gamma_{12}^1\Gamma_{11}^2-\Gamma_{11}^2\Gamma_{22}^2-\Gamma_{11}^1\Gamma_{12}^2\big)\\
&= -g^2 \big( -g_{xx}/g+g_x^2/g^2 - g_{yy}/g + g_y^2/g^2 +g_y^2/g^2 +g_x^2/g^2 -g_y^2/g^2 -g_x^2/g^2\big)\\
&= - \big( -g_{xx}g-g_{yy}g + g_x^2 + g_y^2 \big)\\
&= g(g_{xx}+g_{yy}) - (g_x^2+g_y^2) .
\end{align}
A: I've found a simpler way to answer this using moving frames. We first choose a moving frame compatible with the inner product. Let $e_1=ga_1$ and $e_2=ga_2$, where $a_1,a_2$ are the vectors of canonical base, such that $\langle e_i,e_j\rangle_p=\delta_{ij}$. We know the forms of coframe, $\omega_1,\omega_2$, are linear combinations of the elements of base of dual, that is, $dx$ and $dy$. From that, we get $\omega_1=\frac{1}{g}dx$ and $\omega_2=\frac{1}{g}dy$.
From structure equations $d\omega_{i}(e_1,e_2)=\omega_{12}(e_i)$. We calculate the exterior derivative of coframe forms and, once $\omega_{12}$ is also a 1-forma and, therefore, a linear combination of $dx$ and $dy$, we obtain that $\omega_{12}=\frac{g_y}{g}dx-\frac{g_x}{g}dy$.
From Gauss Equation, $d\omega_{12}=-K\omega_1\wedge\omega_2$. Calculating exterior derivative of $\omega_{12}$ and obtaining its value at $(e_1,e_2)$, we have $d\omega_{12}(e_1,e_2)=\frac{-g(g_{xx}+g_{yy})+(g_x^2+g_y^2)}{g^2}dx\wedge dy(e_1,e_2)=\frac{-g(g_{xx}+g_{yy})+(g_x^2+g_y^2)}{g^2}g^2$. Gauss Equation gives us the desired result.
