How to find Gaussian curvature here?

Question

$\textbf{Problem}$ Let $M=\mathbb{R}^{2}$ with the standard coordinates $(x,y)$. Consider the following Riemannian metric on $M$: $$dx^{2}+2\cos(\alpha(x,y))dxdy+dy^{2},$$ where $\alpha$ is a smooth function on $M$ such that $\alpha(x,y)\neq\pi k,k\in\mathbb{Z}$ for all $(x,y)\in\mathbb{R}.$

(a) Prove that the vector fields $$e_{1}=\frac{\partial}{\partial x},\quad e_{2}=\frac{1}{\sin(\alpha(x,y))}\bigg(\frac{\partial}{\partial y}-\cos(\alpha(x,y))\frac{\partial}{\partial x}\bigg)$$ constitute an orthonormal frame with respect to this Riemannian metric.

(b) Find the dual coframe to the frame $(e_{1},e_{2})$.

(c) Prove that the Gaussian curvature of this Riemannian metric is equal to $-\frac{\alpha_{xy}}{\sin\alpha}$.

I am having trouble with (c). I think the dual coframes are given by $$e_{1}^{*}=dx+\cos(\alpha(x,y))dy,\quad e_{2}^{*}=\sin(\alpha(x,y))dy.$$ I tried to use the equation $$d\omega_{12}=-Ke_{1}^{*}\wedge e_{2}^{*}$$ where $$0=de_{1}=\omega_{11}e_{1}+\omega_{12}e_{2}$$ and so, $\omega_{11}=\omega_{12}=0$.

But since $$e_{1}^{*}\wedge e_{2}^{*}=\sin(\alpha(x,y))dx\wedge dy,$$ it looks like $K=0$. What did I do wrong?

$\large\textbf{The following was added to the original question following levap's answer.}$

We have $$\omega_{12}=\langle\nabla_{e_{1}}e_{1},e_{2}\rangle e^{1}+\langle\nabla_{e_{2}}e_{1},e_{2}\rangle e^{2}$$ and $$\nabla_{e_{1}}e_{1}=\Gamma_{11}^{1}e_{1}+\Gamma_{11}^{2}e_{2},\quad\nabla_{e_{2}}e_{1}=\Gamma_{12}^{1}e_{1}+\Gamma_{12}^{2}e_{2}.$$ We have $$\begin{gather*}\Gamma_{11}^{1}=\frac{1}{2}g^{11}\bigg(\frac{\partial g_{11}}{\partial x^{1}}+\frac{\partial g_{11}}{\partial x^{1}}-\frac{\partial g_{11}}{\partial x^{1}}\bigg)+\frac{1}{2}g^{12}\bigg(\frac{\partial g_{12}}{\partial x^{1}}+\frac{\partial g_{21}}{\partial x^{1}}-\frac{\partial g_{11}}{\partial x^{2}}\bigg)\\ \Gamma_{11}^{2}=\frac{1}{2}g^{21}\bigg(\frac{\partial g_{11}}{\partial x^{1}}+\frac{\partial g_{11}}{\partial x^{1}}-\frac{\partial g_{11}}{\partial x^{1}}\bigg)+\frac{1}{2}g^{22}\bigg(\frac{\partial g_{12}}{\partial x^{1}}+\frac{\partial g_{21}}{\partial x^{1}}-\frac{\partial g_{11}}{\partial x^{2}}\bigg)\\ \Gamma_{12}^{1}=\frac{1}{2}g^{11}\bigg(\frac{\partial g_{11}}{\partial x^{2}}+\frac{\partial g_{12}}{\partial x^{1}}-\frac{\partial g_{12}}{\partial x^{1}}\bigg)+\frac{1}{2}g^{12}\bigg(\frac{\partial g_{12}}{\partial x^{2}}+\frac{\partial g_{22}}{\partial x^{1}}-\frac{\partial g_{12}}{\partial x^{2}}\bigg)\\ \Gamma_{12}^{2}=\frac{1}{2}g^{21}\bigg(\frac{\partial g_{11}}{\partial x^{2}}+\frac{\partial g_{12}}{\partial x^{1}}-\frac{\partial g_{12}}{\partial x^{1}}\bigg)+\frac{1}{2}g^{22}\bigg(\frac{\partial g_{12}}{\partial x^{2}}+\frac{\partial g_{22}}{\partial x^{1}}-\frac{\partial g_{12}}{\partial x^{2}}\bigg)\end{gather*}$$ and $$g=\begin{pmatrix}1&\cos(\alpha(x,y))\\ \cos(\alpha(x,y))&1\end{pmatrix},\quad g^{-1}=\begin{pmatrix}\frac{1}{1-\cos^{2}(\alpha(x,y))}&\frac{-\cos(\alpha(x,y))}{1-\cos^{2}(\alpha(x,y))}\\\frac{-\cos(\alpha(x,y))}{1-\cos^{2}(\alpha(x,y))}&\frac{1}{1-\cos^{2}(\alpha(x,y))} \end{pmatrix}.$$ The Christoffel symbols turn out to be $$\Gamma_{11}^{1}=\frac{\cos(\alpha(x,y))\alpha_{x}}{\sin(\alpha(x,y))},\quad\Gamma_{11}^{2}=\frac{-\alpha_{x}}{\sin(\alpha(x,y))},\quad\Gamma_{12}^{1}=0,\quad\Gamma_{12}^{2}=0.$$ So I got $$\nabla_{e_{1}}e_{1}=\begin{pmatrix}\bigg(\frac{\cos(\alpha(x,y))}{\sin(\alpha(x,y))}+\frac{\cos(\alpha(x,y))}{\sin^{2}(\alpha(x,y))}\bigg)\alpha_{x}\\\frac{-\alpha_{x}}{\sin^{2}(\alpha(x,y))}\end{pmatrix},\quad\nabla_{e_{2}}e_{1}=0$$ and $\displaystyle\omega_{12}=\bigg\langle\begin{pmatrix}\bigg(\frac{\cos(\alpha(x,y))}{\sin(\alpha(x,y))}+\frac{\cos(\alpha(x,y))}{\sin^{2}(\alpha(x,y))}\bigg)\alpha_{x}\\\frac{-\alpha_{x}}{\sin^{2}(\alpha(x,y))}\end{pmatrix},\begin{pmatrix}\frac{-\cos(\alpha(x,y))}{\sin(\alpha(x,y))}\\\frac{1}{\sin(\alpha(x,y))}\end{pmatrix}\bigg\rangle(dx+\cos(\alpha(x,y))dy)=\frac{-\alpha_{x}}{\sin(\alpha)}dx-\frac{\alpha_{x}\cos(\alpha)}{\sin(\alpha)}dy.$

By taking exterior derivative, I got $$d\omega_{12}=\frac{1}{\sin^{2}(\alpha)}\bigg(\alpha_{xy}\sin(\alpha)-\alpha_{x}\alpha_{y}\cos(\alpha)-\alpha_{xx}\cos(\alpha)\sin(\alpha)+\alpha_{x}^{2}\sin^{2}(\alpha)-\alpha_{x}^{2}\cos^{2}(\alpha)\bigg)dx\wedge dy$$ but $Ke^{1}\wedge e^{2}=K\sin(\alpha)dx\wedge dy$, so I don't get $\displaystyle K=\frac{-\alpha_{xy}}{\sin\alpha}$. Where did I make a mistake?

Answer 1

You could also keep everything in forms and use the Cartan structural equations (so you don't have to compute the Christoffel symbols). The Cartan Structural equations are:

$de_1^*=\omega_1^2\wedge e_2^*$

$de_2^*=-\omega_1^2\wedge e_1^*$

which imply $\omega_1^2=-\alpha_x dx$.

Then from $d\omega_1^2=-K e_1^*\wedge e_2^*$ you have:

$\alpha_{xy}=-K \sin(\alpha)$ which is what you want.

EDIT: some details:

Write $\omega_1^2=fdx+gdy$. The structural equations are

$-\sin(\alpha)\alpha_x dx\wedge dy=de_1^*=\omega_1^2\wedge e_2^*=f \sin\alpha dx\wedge dy$. Hence $f=-\alpha_x$.

The second equation is:

$\cos(\alpha)\alpha_x dx\wedge dy=de_2^*=-\omega_1^2\wedge \omega_1=(-f\cos(\alpha)-g)dx\wedge dy$, hence $g=0$ and $$\omega_1^2=-\alpha_xdx.$$

Finally $\alpha_{xy} dx\wedge dy=d\omega_1^2=-Ke_1^*\wedge e_2^*=-K\sin\alpha dx\wedge dy$ gives $K$.

Answer 2

You use the same symbol $d$ twice but they mean different things! The expression $d \omega_{12}$ is the exterior derivative of $\omega_{12}$ but $d e_1$ should be $\nabla e_1$ where $\nabla$ is the covariant derivative induced by the Riemannian metric. In any case, even if you thought that $d e_1$ is the regular derivative, I don't understand why $d e_1$ should be zero. Also, I think that you should get $+K$ and not $-K$ with standard conventions.

Allow me to use both upper/lower indices and the Einstein summation convention. The connection $1$-forms $\omega^{i}_j$ are defined by the equation

$$ \nabla_{X} e_i = \omega_i^j(X) e_j $$

where $\nabla$ is the covariant derivative. Then $d\omega^2_1 = K \, dA = K \, e^1 \wedge e^2$ where $(e^1,e^2)$ is the dual basis to $(e_1,e_2)$. Thus, you need to compute

$$ \left< \nabla_{e_1} e_1, e_2 \right>, \left< \nabla_{e_2} e_1, e_2 \right>, $$

set

$$ \omega^2_1 = \left< \nabla_{e_1} e_1, e_2 \right> e^1 + \left< \nabla_{e_2} e_1, e_2 \right> e^2 $$

and calculate $d\omega_1^2$.