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My text book asks me to show some examples of differential forms on a $n$-dimensional Torus.

What I have done: Let's take for example $S^1$. I'm confused since I saw differential forms given by a restriction of differential forms of $\mathbb{R}^2$. But I also saw them expressed like $\partial \theta$. For example on $T_2$ I saw one expressed as $\partial \theta \wedge \partial \eta$ that integrated on the whole torus gives the volume.

How can them be expressed in a proper way? Via immersion in $\mathbb{R}^m$? Via $\partial$angles multiplied for some infinitely derivable functions over the torus as in the euclidean case? Thanks!

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  • $\begingroup$ Hi, good luck with differential manifolds. I think everything is there : en.wikipedia.org/wiki/Differential_form#Integration " A general 1-form is a linear combination of these differentials ... " $\endgroup$ – reuns Jan 9 '16 at 6:40
  • $\begingroup$ Is $sin(2\pi \theta)\partial \theta + sin^2(2\pi \theta)cos(2\pi \theta)\partial\eta$ a differential form on $T_2$? $\endgroup$ – Maffred Jan 9 '16 at 17:31
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    $\begingroup$ yes math.colorado.edu/~jnc/lecture1.pdf I'm not an expert, but think to "what is it useful for ?" why do they annoy me with all these abstract definitions ? think first to the differential 1-forms $df$ of $\mathbb{R}^n$ and then of the sphere and to the corresponding result of $\int_a^b df$ and $\int_a^b g \ df$. realize that from a differential form we can define a minimal length between $a$ and $b$ : $\min_{\gamma:a\to b} \left| \int_\gamma df \right|$, stretching (by a smooth real valued function) a differential form is in some sort equivalent to stretch the manifold itself. $\endgroup$ – reuns Jan 9 '16 at 18:20
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$\newcommand{\Reals}{\mathbf{R}}$Let $\Lambda \subset \Reals^{n}$ be a lattice (discrete additive subgroup) of full rank. If $p:\Reals^{n} \to T = \Reals^{n}/\Lambda$ is the quotient map, then every differential $k$-form $\omega$ on $T$ pulls back to a $\Lambda$-invariant $k$-form $p^{*}\omega$ on $\Reals^{n}$. Conversely, every $\Lambda$-invariant $k$-form on $\Reals^{n}$ arises in this way, i.e., "pushes forward" to a $k$-form on $T$.

In the case of $S^{1} = \Reals/2\pi\mathbf{Z}$, for example, $0$-forms are in bijective correspondence with $2\pi$-periodic functions on $\Reals$, and $1$-forms correspond to $1$-forms $f(x)\, dx$ with $f$ a real-valued function on $\Reals$ of period $2\pi$. Every smooth $1$-form is closed, but it's a pleasant exercise to show the $1$-form corresponding to $f(x)\, dx$ is exact if and only if $$ \int_{0}^{2\pi} f(x)\, dx = 0. $$

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  • $\begingroup$ Thank you very much! I'm trying to prove the last statement: of course $f(x)dx$ is closed. If it is exact then exists $F$ such that $F'(x)=f(x)$ and $\partial F = f(x)dx$. The integral is thus equal to $F(2\pi)-F(0)$ and since we are on $S^1$ it is $0$. Is that right? How to reverse? $\endgroup$ – Maffred Jan 9 '16 at 19:44
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    $\begingroup$ Yes, that's right. For the converse, put $F(x) = \int_{0}^{x} f(t)\, dt$ and use the integral condition to deduce $F$ is $2\pi$-periodic (hence descends to a smooth function $F$ on the circle). $\endgroup$ – Andrew D. Hwang Jan 9 '16 at 21:35

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