Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

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 $I=0$.

share|improve this answer
add comment

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).

share|improve this answer
    
yep, you get the same result with a similar method, thanks a lot! –  Miguel Nov 14 '12 at 16:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.