# Computing integral of $2$ - form on a torus

I am looking at problem 16-2 of Lee's Smooth Manifolds, second edition.

Problem 16 - 2: Let $\Bbb{T}^2 \subseteq \Bbb{R}^4$ be the two torus defined as the set of points $(w,x,y,z)$ such that $w^2 + x^2 = y^2 + z^2 = 1$, with the product orientation determined by the standard orientation on $\Bbb{S}^1$. Compute $\int_{\Bbb{T}^2} \omega$ where $\omega$ is the following two form on $\Bbb{R}^4$: $$\omega = xyz \hspace{1mm} dw \wedge dy.$$

Now if I want to evaluate such a two form do I need to care about the orientation? I am tempted to set up a map $F : \Bbb{R}^2 \to \Bbb{R}^4$ that sends $\varphi, \theta$ to $(\cos \varphi, \sin \varphi, \cos \theta, \sin \theta)$ and then taking the integral of $\omega$ on the torus to be $$\int_{[0,2\pi]^2} F^\ast \omega .$$

Is my reasoning correct, or do I need to care about the product orientation?

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## 2 Answers

To answer your question: yes you have to pay attention to the orientation. First of all you have to observe that $F:[0,2\pi)^2\to\mathbb{T}^2\subset\mathbb{R}^4$ is an isometry (the way you defined it), because of that then you can conclude as you wanted:

$$\int_{[0,2\pi)}F^\star\omega$$

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Isometry is irrelevant. All you need is an orientation-preserving parametrization. –  Ted Shifrin Jun 4 '13 at 3:25
You ate totally right, isometric is irrelevant, orientation-preserving is just needed –  Heberto del Rio Jun 10 '13 at 3:42

When dealing with integrals, one should always be concerned with orientation. You must check that your parametrization preserves the given orientation or at least switch the "sign" of your integral if it does not.

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