This post is a continuation on Connection matrix for the Poincare disk.
The ingredients are the Poincaré disk $D = \{x = (x,y) \in \mathbb{R}^2 : |(x,y)| < 1\}$. An orthonormal frame for $\mathbf{D}$ is $$e_1 = \frac{1}{2} (1 - |x|^2) \partial_x, \qquad e_2 = \frac{1}{2} (1 - |x|^2) \partial_y. $$ Find the connection matrix $ \omega = [\omega_j^{i}]$ relative to the orthonormal frame $e_1, e_2$ of the Riemannian connection $\nabla$ on the Poincare disk.
The dual frame satisfies $\theta^{i} (e_j ) = \delta_{j}^{i}$. So that means $$ \theta^{1} = \frac{2}{ 1 - |x|^2} dx, \qquad \theta^{2} = \frac{2}{ 1 - |x|^2} dy. $$We compute their differentials, $$d\theta^1 = -\frac{2y}{(1-|x|^2)^2}dx\wedge dy,$$and $$d\theta^2 = \frac{2x}{(1-|x|^2)^2}dx\wedge dy.$$
1st question: In the cited post, the OP said that we have no torsion. How does he knows that without a connection defined? I know that for a Riemannian manifold there is always a unique Riemannian connection. So, putting a metric gives a Riemannian connection right? I will just share that the reverse question, given a connection is there a Riemannian metric for which it is the Levi-Civita connection? is When can a connection Induce a Riemannian metric for which it is the Levi-Civita connection?. So I will proceed assuming that we are dealing with this Riemannian connection and that we are torsion free.
Now, since the connection $\nabla$ is compatible with the metric, its connection matrix $[\omega_j^i]$ is skew-symmetric, i.e. $$\omega_i^j = -\omega^i_j.$$From this we take that $\omega_1^1 = \omega_2^2 = 0$. To compute the remaining terms we use the defining equation that relates the orthonormal frame with what we are looking for: $$d\theta^i + \sum_j\omega_j^i \wedge \theta^j = 0, \ \ i = 1,2.$$For $i=1$ we write $\omega_2^1 = a_1\theta^1 + a_2\theta^2$ so that we have $$d\theta^1 + \omega_1^1\wedge \theta^1 + \omega_2^1 \wedge \theta^2 = 0 \Leftrightarrow a^1\theta^1 \wedge \theta^2 = -d\theta^1,$$which reads $$a_1\frac{4}{(1-|x|^2)^2}dx \wedge dy=\frac{2y}{(1-|x|^2)^2}dx\wedge dy,$$which means $a_1 = y/2$. Writing $\omega_1^2 = b_1\theta^1 + b_2\theta^2$ we already have $b_1 = -y/2$. To compute $a_2$ and $b_2$ we go for $$d\theta^2 + \omega^2_1 \wedge \theta^1 + \omega_2^2\wedge \theta^2 = 0 \Leftrightarrow \omega^2_1 \wedge \theta^1 = -d\theta^2.$$Which means $$b_2\frac{4}{(1-|x|^2)^2}dy\wedge dx = -\frac{2x}{(1-|x|^2)^2}dx \wedge dy,$$so $b_2 = x/2$. From the skew-symmetry we see that $a_2 = -x/2$. Then $$\omega_2^1 = \frac{y}{2} \theta^1 - \frac{x}{2}\theta^2, \ \ \omega_1^2 = -\frac{y}{2} \theta^1 + \frac{x}{2}\theta^2.$$To compute the Gaussian curvature $K$ we use Theorema Egregium, $$d\omega_2^1 = K\theta^1 \wedge \theta^2.$$However, computing $$d\omega_2^1 = \frac{1}{2}(dy\theta^1) - \frac{1}{2}(dx\theta^2) = -\frac{2(1+x^2-y^2)}{(1-|x|^2)^2}dx \wedge dy$$whereas $$\theta^1 \wedge \theta^2 = \frac{4}{(1-|x|^2)^2}dx \wedge dy.$$However this shows that $K = -\frac{(1+x^2-y^2)}{2}$ which is incorrect since the result is supposed to be $K = -1$.
I wonder if anyone can point out the mistake.
ps.: from the first cited post, there was someone suggesting to avoid using the $x,y$ coordinates but I could not do the computations without using them. I also question how would one do in that way?
Thanks in advance!
By the Riemannian connection, the original OP intended the Levi-Civita connection, yes, which is the unique torsion-free metric-compatible connection.
First, your computations of $d\theta^i$ are off by a factor of $2$. I suspect you forgot the original coefficients of $2$. So, as the original post and its comments concluded, we have $\omega_2^1 = y\theta^1 - x\theta^2$, with no factors of $1/2$.
Second, you are not differentiating correctly: You are forgetting to differentiate the coframe forms. $d(y\,\theta^1) = dy\wedge\theta^1 + y\,d\theta^1 = dy\wedge\theta^1 - y\,\omega^1_2\wedge\theta^2$. The intent of my suggestion was to rewrite this as $$d(y\,\theta^1) = \frac{1-|z|^2}2\theta^2\wedge\theta^1 - y^2\theta^1\wedge\theta^2 = -\left(\frac{1-|z|^2}2+y^2\right)\theta^1\wedge\theta^2.$$ Similarly, as you can check, $$d(-x\,\theta^2) = -\left(\frac{1-|z|^2}2+x^2\right)\theta^1\wedge\theta^2,$$ from which you indeed conclude that $d\omega_2^1=(-1)\theta^1\wedge\theta^2$, and so $K=-1$.