Trying to compute explicitly the curvature form of the unit sphere and got the following result:
The parametrization of the unit 2-sphere $ S^2$ is well known:
$ n=(\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$, let $$ e_1=\frac{\partial}{\partial\theta}=(\cos\theta\cos\phi, \cos\theta\sin\phi, -\sin\theta)$$ $$ e_2=\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}=(-\sin\phi, \cos\phi,0)$$ be an orthonormal frame defined on an open set U of $ S^2$, let $ \theta^1=d\theta, \theta^2=\sin\theta d\phi$ be the dual basis.
It's easy to compute that the Levi-Civita connection D on $ S^2$ gives the following (where P is the orthogonal projection to the tangent space of $ S^2$):
$$ D_{e_1}{e_1}=D_{e_1}{e_2}=0$$
$$ D_{e_2}{e_1}=P(\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}(\cos\theta\cos\phi, \cos\theta\sin\phi, -\sin\theta))=\cot\theta e_2$$
$$ D_{e_2}{e_2}=P(\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}(-\sin\phi, \cos\phi,0))=-\cot\theta e_1$$
So the connection form with respect to the frame $ \{e_1, e_2\}$ is:
$$ \omega=\begin{bmatrix}0 & \cot\theta \theta^2 \\-\cot\theta\theta^2 & 0 \end{bmatrix}=\begin{bmatrix}0 & \cos\theta d\phi \\-\cos\theta d\phi & 0 \end{bmatrix}. $$
Hence the curvature form is:
$$ \Omega=d\omega-\omega\wedge\omega=\begin{bmatrix}0 & -\sin\theta d\theta\wedge d\phi \\\sin\theta d\theta\wedge d\phi & 0 \end{bmatrix}$$
To verify let's compute the Gaussian curvature:
$$ k=R(e_1, e_2, e_1, e_2)=\Omega_1^2(e_1,e_2)=-\sin\theta d\theta\wedge d\phi(\frac{\partial}{\partial\theta}, \frac{1}{\sin\theta}\frac{\partial}{\partial\phi})=-1$$
But where is this negative sign coming from?
Best Answer
Oh just realized that I missed a negative sign at the last step: $$K=-R(e_1, e_2, e_1, e_2)=1$$