Given the form $\Omega = -\sin\theta d \theta \wedge d\phi$.
Here $\theta$ and $\phi$ are the usual polar coordinates.
Since $\int_{S^2} \Omega=-4\pi$, the form is not exact on $S^2$.
Although $\Omega = d(\cos\theta d \phi)$, this only implies that $\Omega$ is exact on
$ S^2\setminus ( \text{North pole, South pole}) $.
Because the polar coordinates are not well defined at the North pole and the South pole, the form $\cos\theta d \phi$ is not defined at these poles.
My question:
Why the form $\Omega$ containing $d\phi$ is well defined on $S^2$?
I have a feeling that this form is a little bit similar to the function $\sin x/x$.
Although this function is not defined at $x=0$, it has a limit there.
Best Answer
The post you linked in the comment provides a beautiful definition of $\Omega$ that is valid on the whole of $S^2$, including the poles. However, this definition is a bit abstract, and it would be helpful to write it out more concretely using a coordinate system, as per peek-a-boo's second comment. I'll flesh the details for you in this answer.
So consider the natural embedding $i : S^2 \to \mathbb R^3$. On $\mathbb R^3$, we have the differential form $$ \Omega_{\mathbb R^3} = x dy \wedge dz + y dz \wedge dx + z dx \wedge dy,$$ which is manifestly smooth, and globally defined on $\mathbb R^3$.
If $p \in \mathbb R^3$ is the point $(x, y, z)$ and $v, w \in T_p \mathbb R^3$ are the tangent vectors $v_x \frac \partial {\partial x} + v_y \frac \partial {\partial y} + v_z \frac \partial {\partial z}$ and $w_x \frac \partial {\partial x} + w_y \frac \partial {\partial y} + w_z \frac \partial {\partial z}$ respectively, then it is easy to see that $$ (\Omega_{\mathbb R^3})_p(v, w) = x(v_y w_z - v_z - w_y) + y(v_z w_x - v_x w_z) + z(v_x w_y - v_y w_x) = (v \times w) . x, $$ which is the same as the abstract formula appearing in the linked post - but applied to points and tangents vectors on $\mathbb R^3$, rather than on $S^2$.
To construct a differential form $\Omega_{S^2}$ on $S^2$, we simply take the pullback with respect to the embedding $i : S^2 \to \mathbb R^3$: $$ \Omega_{S^2} = i^\star(\Omega_{\mathbb R^3}).$$ Having defined $\Omega_{S^2}$ in this way, we can be sure that $\Omega_{S^2}$ is well-defined (and smooth) on the whole of $S^2$. After all, this $\Omega_{S^2}$ is the pullback of a globally-defined smooth differential form on $\mathbb R^3$, so $\Omega_{S^2}$ must be well-defined and smooth on $S^2$.
If $p \in S^2$ and $v, w \in T_p(S^2)$, then $$ (\Omega_{S^2})_p(v, w) = (i^\star \Omega_{\mathbb R^3})_p(v, w) = (\Omega_{\mathbb R^3})_{i(p)}(i_\star(v), i_\star(w)) = (i_\star(v) \times i_\star(w)) . i(p),$$ so the abstract formula applies to points and tangents vectors on $S^2$ too, provided that you make the necessary identifications.
Finally, we should express $\Omega_{S^2}$ in spherical polar coordinates. The embedding $i: S^2 \to \mathbb R^3$ has the coordinate representation $$ i \ : \ (\theta, \phi) \mapsto (x(\theta, \phi), y(\theta, \phi), z(\theta, \phi)) = (\sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta). $$ So \begin{align} i^\star(dx) & = \cos\theta\cos\phi d\theta - \sin\theta\sin\phi d\phi \\ i^\star(dy) & = \cos\theta\sin\phi d\theta + \sin\theta\cos\phi d\phi \\ i^\star(dz) &= -\sin\theta d\theta.\end{align} Therefore, \begin{align} \Omega_{S^2} = & \ (\sin\theta \cos\phi)(\cos\theta\sin\phi d\theta + \sin\theta\cos\phi d\phi) \wedge (-\sin\theta d\theta) \\ & + (\sin\theta \sin\phi) (-\sin\theta d\theta) \wedge (\cos\theta\cos\phi d\theta - \sin\theta\sin\phi d\phi) \\ & + (\cos\theta) (\cos\theta\cos\phi d\theta - \sin\theta\sin\phi d\phi ) \wedge (\cos\theta\sin\phi d\theta + \sin\theta\cos\phi d\phi) \\ = & \ \sin \theta d\theta \wedge d\phi,\end{align} which agrees with your original expression, give or take a minus sign.
Of course, the spherical polar coordinates $(\theta, \phi)$ are only valid away from the two poles of the sphere, so the expression $\Omega_{S^2} = \sin \theta d\theta \wedge d\phi$ is only valid away from the two poles of the sphere. However, the preceding discussion shows that $\Omega_{S^2}$ is well-defined (and smooth) on the whole of $S^2$.