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I'm really trying to understand differential forms so any critiques at all of the argument below would be greatly appreciated.

I want to show that $dx, dy\in \Omega^1(\mathbb{R}^2)$ induce well-defined forms $\omega_1,\omega_2 \in \Omega^1(T^2), T^2 = \mathbb{R}^2/\mathbb{Z}^2$. Then compute $d\omega_i$ and $\int_{C_j}\omega_i$, where $C_j$ is the image of the $j$-axis under the projection $\mathbb{R}^2\rightarrow\mathbb{R}^2/\mathbb{Z}^2$.

What I have:

Identify $T^2$ with $S^1\times S^1$ via some orientation preserving diffeomorphism. Let $\phi_{1}:(0,1)^2\rightarrow T^2$ be given by $(t,s)\mapsto(\cos2\pi t,\sin2\pi t,\cos2\pi s, \sin2\pi s)$ and $\phi_{2}:(-\frac{1}{2},\frac{1}{2})^2\rightarrow T^2$ defined the same. These are diffeomorphisms whose inverses are identical in overlap so we have an atlas. Then the pullback $(\phi^{-1}_i)^*(dx)$ is defined for $p\in T^2$ as $$(\phi^{-1}_i)^*(dx)(p) = dx(\phi_i^{-1}(p))\circ d(\phi_i^{-1})_p,$$ so since the $\phi_i^{-1}$ agree on the overlap, this form is well defined on all of $T^2$. Similarly we get a pulled back $(\phi_i^{-1})^*(dy)$. Then the exterior derivative of both is zero since $d(\phi_i^{-1})^*(dx) = (\phi_i^{-1})^*(d^2x) = 0$. Lastly, to take the integral, we can take restrictions of $\phi_2$ to parametrize the $C_j$ so that, for example, $$\int_{C_1}\omega_1 = \int_{(0,1)\times\{0\}}(\phi_2)^*(\phi_2^{-1})^*(dx) = 1. $$ and $$\int_{C_2}\omega_1 = \int_{\{0\}\times(0,1)}(\phi_2)^*(\phi_2^{-1})^*(dx) = 0. $$

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  • $\begingroup$ I don't think you've quite covered everything in the torus. But the first thing you should think about here is that both $dx$ and $dy$ are invariant under the action of $\Bbb Z^2$ on $\Bbb R^2$, hence descend to well-defined $1$-forms on $\Bbb R^2/\Bbb Z^2$. $\endgroup$ – Ted Shifrin May 9 '17 at 1:26
  • $\begingroup$ @TedShifrin Hmm, how so? I think the only parts left out of $\phi_1$ are two circles $\{a\}\times S^1$ and $S^1\times\{a\}$. Also, I see that, but I was just wondering if what I did made sense, so I could make sure I'm understanding how to manipulate forms. So, I guess, assuming I did cover everything in the torus with some $\phi_i$ satisfying what I used above, would the argument hold? $\endgroup$ – Juan Chi May 9 '17 at 2:28
  • $\begingroup$ @TedShifrin Also, for your method, I'm trying to see exactly how the descending would work out. So $dx(r)\in \Lambda^1 (T_r (\mathbb{R}^2)^*)$ for $r\in\mathbb{R}$, but we want $dx([r])\in \Lambda^1 (T_{[r]} (\mathbb{R}^2/\mathbb{Z}^2)^*)$ for $[r]\in\mathbb{R}^2/\mathbb{Z}^2$. So how would one go about finding an expression for the descended $dx$ to see that it lies in the space we want it to? $\endgroup$ – Juan Chi May 9 '17 at 2:46
  • $\begingroup$ One way to think about this is that $\pi\colon\Bbb R^2\to T^2$ is a local diffeomorphism. We want to define a $1$-form $\omega$ on $T^2$. Given a point $q\in T^2$ and $w\in T_q(T^2)$, choose $p\in\Bbb R^2$ and $v\in T_p\Bbb R^2$ so that $\pi(p)=q$ and $d\pi(p)(v)=w$. Then you compute $dx(p)(v)$ to evaluate $\omega(q)(w)$. Now why is this well-defined? Go through that carefully. $\endgroup$ – Ted Shifrin May 11 '17 at 20:38

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