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I've been asked to solve the following integral:

$$\int_{\mathbb{T}^2} xyz \, dw\wedge dy$$ where $\mathbb{T}^2\subset\mathbb{R}^4$ is the 2-torus defined by: $$w^2+x^2=y^2+z^2=1$$ I've tried "double" polar coordinates, but haven't reach anything. Any tips?

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I think "double" polar coordinates can solve this integral. I cannot understand where you are stuck. –  23rd Nov 14 '12 at 16:35
Mm the only problem is that the result I get is zero, I expected something cooler hehe, but I think I'll have to take it. –  Miguel Nov 14 '12 at 16:48
Is there an argument from symmetry showing this integral is $0$ without using polar coordinates? –  Michael Hardy Nov 14 '12 at 16:52

2 Answers 2

up vote 7 down vote accepted

Here is a simple symmetry argument: $(w,x,y,z) \mapsto (w,-x,y,z)$ is a symmetry of the torus which preserves $dw \wedge dy$ and maps $xyz$ to $-xyz$, so the integral $I$ has to satisfy $I=-I$, hence $I=0$.

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Choose local coordinates $(\theta, \phi) \in (0,2\pi) \times (0,2\pi)$ where $w = \cos\theta, x = \sin\theta, y = \cos\phi, z = \sin\phi$. Then you can just pullback the integral to $(0,2\pi) \times (0,2\pi)$ since the complement of this chart in $T^2$ is of measure zero.

We have $dw = -\sin\theta \, d\theta,\ dy = -\sin\phi \, d\phi$. So you end up with $$ \int_0^{2\pi} \int_0^{2\pi} \sin^2\theta\cos\phi\sin^2\phi \, d\theta \, d\phi, $$ which is a fairly straightforward integral ($u$-sub for the $\phi$ part and half angle for the $\theta$ part).

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yep, you get the same result with a similar method, thanks a lot! –  Miguel Nov 14 '12 at 16:49

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