I think this is a very simple question but I'm not really confident in mathematics (even if I like it very much)
Let's fix a cube $[0,1]^3$ in $R^3$ and identify opposite sides, so as to construct a torus T.
I'd like to construct a base for $\Omega^1(T)$ (the one forms) and to understand which 1-forms are closed and which are exact.
1) is it possible to do that?
I have in mind that the torus has inheritaed a 1-form $dx$ from the embedding in $R^3$ (if we interpret the one forms as vector fields, it is the 1-form associated to the vector field that has coordinates (1,0,0) in $R^3$), but I'm not sure of that and I don't know how to construct this one form in a mathematical coherent way. The fact that this torus hasn't got a global patch confuses me...so...
2) what is this form dx on the torus?
Each point ${\bf x}\in T:={\mathbb R}^3/{\mathbb Z}^3$ has a tangent space $T_{\bf x}$ which is a copy of ${\mathbb R}^3$, and the standard basis $({\bf e}_1,{\bf e}_2,{\bf e}_3)$ of ${\mathbb R}^3$ can be taken over by translation. The dual basis $({\bf e}^*_1,{\bf e}_2^*,{\bf e}_3^*)$ consists of the three coordinate differentials $dx_1$, $dx_2$, $dx_3$, and these make sense as a basis of $T_{\bf x}^*$ as well.
It follows that locally a $1$-form on $T$ assumes the form $$\omega=f_1({\bf x}) dx_1+f_2({\bf x}) dx_2+f_3({\bf x}) dx_3\tag{1}$$ with smooth coefficient functions $f_i$. To make this form well defined on $T$ the $f_i$ have to be well defined on $T$, which means that they have to be $1$-periodic in each variable $x_k$ $\>(1\leq k\leq3)$ when you consider the letter ${\bf x}$ in $(1)$ as a point in ${\mathbb R}^3$.
The form $\omega$ is closed when it has locally a potential. The condition for this is that $$d\omega=0\ ,$$ or that the curl of the force field ${\bf F}=(f_1,f_2,f_3)$ vanishes identically.
If this condition is fulfilled one may ask whether $\omega$ is exact. Exact means that the field ${\bf F}$ has a global potential $u:\ T\to {\mathbb R}$ which is well defined on $T$. In this case ${\bf F}=\nabla u$, or $$\omega=du\ .$$ The condition for this to happen is that the the integral of $\omega$, resp. ${\bf F}$, along three generating cycles of $H^1(T)$ is zero. Therefore it suffices to compute the three integrals $$\int_0^1 f_i(t\>{\bf e}_i)\ dt\qquad(1\leq i\leq3)$$ and to check whether they all vanish.