Prove there is only one $2$-form $p^*\omega = dx\wedge dy$

Question

I am new with forms and pullbacks and admit differential geometry is not my best area. I'm trying to solve the next problem.

Let be $(x, y)$ coordenates on $\mathbb{R}^2$. Let $p:\mathbb{R}^2\rightarrow\mathbb{R}^2/\mathbb{Z}^2=\mathbb{T}^2$ the projection. Show that there is only one $2$-form $\omega$ on $\mathbb{T}^2$ such that \begin{equation} p^*\omega = dx\wedge dy \end{equation} Is this form closed? is this form exact?

I did a demostration by contradiction but I'm not sure if it's correct.

Suppose there is another $2$-form $\theta$ such that $p^*\theta= dx\wedge dy $ then $p^*(\omega - \theta) = dx\wedge dy - dx\wedge dy = 0$

because $p$ is no null then $\omega-\theta=0$. It's too simple and I doubt that it works, any help will be apreciate.

Answer 1

$p:\mathbb{R}^2\rightarrow\mathbb{T}^2$ is a local diffeomorphism and it is surjective. It implies that for every $x\in \mathbb{T}^2$ there exists $x'\in\mathbb{R}^2$ such that $p(x')=x$ and $dp_{x'}:T_{x'}\mathbb{R}^2\rightarrow T_{p(x)}\mathbb{T}^2$ is an isomorphism.

Suppose that $p^*\omega=p^*\theta$, let any $x\in \mathbb{T}^2, u,v\in \mathbb{T}^2$ and $x'\in\mathbb{R}$, $u',v'\in T_{x'}\mathbb{R}^2$ such that $p(x')=x$, $dp_{x'}(u')=u, dp_{x'}(v')=v$; $0=p^*(\omega-\theta)_{x'}(u',v')=(\omega-\theta)_{p(x')}(dp_{x'}(u'),dp_{x'}(v'))=(\omega-\theta)_x(u,v)=0$ implies that $u=v$.

Thus the form is unique, the form is closed since $\mathbb{T}^2$ is $2$-dimensional and a the differential of a $2$-form is $3$-form which is zero since an alternated $3$-form is zero on a $2$-dimensional vector space.

The $2$-form is not exact since it is a volume form.

In fact $\omega$ exists since $\mathbb{T}^2$ is the quotient of $\mathbb{R}^2$ by $f(x,y)=(x+1,y)$ and $g(x,y)=(x,y+1)$ and $f^*(dx\wedge dy)=g^*(dx\wedge dy)=dx\wedge dy$.

Answer 2

Because $p$ is no null

What does this even mean? Perhaps the proof is valid, but we won't be able to tell unless you define what it means for a map to be null.

Anyway, your idea is good: you are using the linearity of $p^*$ to reduce the problem to showing that $\ker p^* = 0$. This is indeed true (so $p^*$ is injective)! To prove it, you need to use the fact that $p$ is a surjective submersion:

Suppose $p^* \alpha = 0$ for some 2-form $\alpha$ on $\mathbb{T}^2$. Then, let $v, w$ be any two tangent vectors at any point $y \in \mathbb{T}^2$. Since $p$ is surjective, let $x \in \mathbb{R}^2$ such that $p(x) = y$. Since $p$ is a submersion, let $v', w' \in T_x(\mathbb{R}^2)$ such that $dp_x(v') = v$ and $dp_x(w') = w$. Then, by definition of pullback, $$0 = (p^* \alpha)_x (v', w') = \alpha_y(dp_x(v'), dp_x(w')) = \alpha_y(v', w').$$ Since $y,v,w$ were arbitrary, $\alpha = 0$, as desired.

Next, you should include a construction of the 2-form $\omega$ in question (so that you know such a 2-form actually exists, not just that there is at most one such); this will help you answer the next part of the question.