How do you proove that a torus is an orientable 2-dim manifold? Let the torus $M := \{(x,y,z)\in \mathbb{R}^3 | (\sqrt{x^2+y^2}-R)^2+z^2=r^2 \}$ with $r<R$
Afaik, you can prove that a manifold is orientable by calculating if the derivative of the determinant of the two compsited mappings are positive, i.e. $\det(\Psi_2^{-1}\circ\Psi_1)^\prime > 0$. But if I'm not given any mapping functions, how would I go about it?
The paremeterized equations should be:
$$x=(R + r\cos v)\cos u$$
$$y=(R + r\cos v)\sin u$$
$$z=r\sin v$$
Do I use the parametric equations to calculate these mappings, or is there a better way to do this?
And about the two dimensionality: what is required here?
 A: Your torus is already embedded in $\mathbb{R}^3$ which you can orient canonically (choose as positive the basis $e_1,e_2,e_3$).
$\mathbb{R}^3$ has also a metric given by the standard scalar product.
This metric induces a volume form namely $\nu = dx^1\wedge dx^2 \wedge dx^3$.
If you can find a normal unit vector along the torus $N$, then the contracted form $\iota_N \nu $ will be a volume form for the torus (so it will be orientable).
I recall that by definition $\iota_N \nu (X_1,X_2) = \nu(N, X_1,X_2)$ so $\forall X_1,X_2 \in T_p M $ linearly independent tangent vectors  you have that $ \iota_N \nu (X_1,X_2)>0$ (definition of volume form).
So the point is to find a vector field $N:M\to \mathbb{R}^3 $ such that $N$ is orthogonal to the tangent space of the torus and also of unit length.
Can you complete the answer by yourself? Here is the solution

$$N(u,v) = \frac {\psi_u(u,v)\times \psi_v(u,v)}{||\psi_u(u,v)\times \psi_v(u,v)||}  $$ where $\psi(u,v) = (x(u,v),y(u,v),z(u,v))$ is the local parametrization

By the way, another method to prove that the torus is orientable is that since it is diffeomorphic to $\mathbb{S}^1\times \mathbb{S}^1$ and the circle $\mathbb{S}^1$ is orientable then also the torus it is.
