# Closed non-exact $2$-form on $T = S^1 \times S^1$

I would like to find a closed non-exact $2$-form on $T = S^1 \times S^1$.

Here are my thoughts:

Since $d\theta$ and $d\varphi$ are closed non-exact $1$-forms an obvious candidate is $d\theta \wedge d \varphi$. This is closed since

$$d(d\theta \wedge d \varphi) = d\theta \wedge d \varphi \wedge d \theta + d\theta \wedge d \varphi \wedge d \varphi = 0$$

But how can I show that it is not exact?

(I'm sure it is indeed not exact)

I had the idea that by contradiction assume that it is exact. Then by Stokes' theorem

$$\int_T d\theta \wedge d \varphi = \int_T d \psi = \int_{\partial T} \psi = \int_\varnothing \psi = 0$$

But I don't see why this should be a contradiction...

## 1 Answer

To achieve a contradiction, show by some other means that the integral is nonzero. Since $d\theta \wedge d\phi$ is in fact the standard volume form on $T$, this shouldn't be too difficult.

• I know that $d\theta \wedge d \varphi$ is "the volume form of the torus" which probably means that integrating over it yields the surface are but I don't know how to actually prove it. – a student Jan 24 '15 at 2:27
• Since $T=S^1\times S^1$, Fubini's Theorem tells you that $$\int_T d\theta\wedge d\phi = \left(\int_{S^1} d\theta\right)\left(\int_{S^1} d\phi\right) = (2\pi)^2.$$ – Ted Shifrin Jan 24 '15 at 14:25
• @TedShifrin But the comment shortly before theorem 2.6.1. in this book here seems to suggest that $d(x,y)\neq dx \wedge dy$ so how can you apply Fubini here? Is there another version of the Fubini theorem for wedge products? – a student Jan 25 '15 at 1:14
• Check the definitions carefully. – Ted Shifrin Jan 25 '15 at 1:17
• @TedShifrin Which? Of $dx \wedge dy$ or of $dx dy$? – a student Jan 25 '15 at 2:03