# Finding a basis for the cohomology vector space of 1-forms in the 2-torus, $H^1 (T^2)$

I would like help in understanding where I am going wrong here:

If I consider the 2-torus $T^2 = S^1 \times S^1$ with an atlas $(\theta_1,\theta_2)$, I can define 2 closed 1-forms

$\omega_1 = d\theta_1, \omega_2 = d\theta_2$

which are closed, $d\omega_i = 0$ for $i=1,2$ since $d^2 = 0$ i.e. nilpotent. What I do not understand is why they are inequivalent (and hence used as the basis which span $H^1(T^2)$). From the definition of the cohomology equivalence classes, these two 1-forms are inequivalent if they do not differ by an exact 1-form i.e.

$\omega_1 - \omega_2 = d\alpha$

has no solution for $\alpha$. However if I set

$\alpha = \theta_1 - \theta_2$

then in the coordinate basis

$d\alpha = \frac{\partial \alpha}{\partial \theta_i} d\theta_i = d\theta_1 - d\theta_2$

as required. Where is the problem here?

Neither $\theta_{1}$ nor $\theta_{2}$ is a (global, continuous) function on the torus, only on the universal cover. Integrating "$d\theta_{i}$" over a closed path parallel to the $\theta_{i}$ axis gives $1$, while the integral of an exact form would be $0$.
• Sorry, but I don't understand what you mean by $\theta_1$ and $\theta_2$ not being global,continuous functions on the torus? Also, when you integrate over the closed path parallel to the $\theta_i$ axis for either $i=1,2$, how do you get 1? Commented Apr 5, 2016 at 13:50
• For definiteness, view $S^{1} = \mathbf{R}/\mathbf{Z}$. If $\theta$ is the standard Cartesian coordinate on $\mathbf{R}$, then $d\theta$ is invariant under the translation $\theta \mapsto \theta + 1$, but $\theta$ itself isn't. Consequently, the closed $1$-form $d\theta$ descends to the circle, but the real-valued function $\theta$ does not. On the torus, you have an analogous picture "in each factor". [...] Commented Apr 5, 2016 at 14:06
• If you prefer to think of $S^{1}$ as the set of unit vectors in the plane, you can think of "$\theta$" as the polar angle function. But $\theta$ isn't a continuous, real-valued function on the circle (why not?), even though its differential is well-defined. In fact, $d\theta = \dfrac{x\, dy - y\, dx}{x^{2} + y^{2}}$ is smooth away from the origin. Commented Apr 5, 2016 at 14:09
• Is the reason why $\theta$ is not continuous the fact that the charts that make up the atlas are open sets e.g. $(0,2\pi), (-\pi,\pi)$? In that case can I even define any global function on $S^1$ at all (that is not a constant function)? Also, this seems to imply that for an r-form to be considered exact, it must be exact everywhere in the manifold. Is this correct? Sorry that I didn't use any of your more recent answer as I found it more confusing. Commented Apr 5, 2016 at 14:48
• 1. There is no continuous choice of $\theta$ on the circle because the integral of $d\theta$ over the circle is not zero. 2. The space of continuous, real-valued functions on $S^{1}$ is infinite-dimensional. While $\theta$ is only defined mod $2\pi$, the functions $x = \cos \theta$ and $y = \sin\theta$ are both continuous, as is any continuous function of $x$ and $y$. 3. Yes, exactness is global, i.e., an $r$-form $\omega$ on a manifold $M$ is exact iff there exists an $(r - 1)$-form $\eta$ on $M$ such that $\omega = d\eta$. Commented Apr 5, 2016 at 16:09