Integrals: Why Change from M to $\phi(M)$?

[davidge]

Yes, I know that I have already created another thread on this subject before. But, in this one, I would like to ask specifically why should we change from $M$ to $\phi (M)$ in the integral below?

$$ \int_M (\partial_\nu w_\mu - \partial_\mu w_\nu) \ dx^\nu \wedge dx^\mu = \int_{\phi (M)} (\partial_\nu w_\mu - \partial_\mu w_\nu) \ d^2x$$

Why is it needed to do so? And why after doing that change, we have to substitute $dx^\nu \wedge dx^\mu$ by $d^2x$ in the integrand?

I guess this is because $M$ is a manifold while $\phi (M)$ is a function. So we need to change the integrand to something that "lives" in the function space, namely $d^2x$. Did I guess right?

 

[haushofer]

This expression is wrong. On the RHS, no indices are contracted. And without any reference telling us what phi is, I don't think you'll get many reactions :)

 

[stevendaryl]

Yes, I know that I have already created another thread on this subject before. But, in this one, I would like to ask specifically why should we change from $M$ to $\phi (M)$ in the integral below?

$$ \int_M (\partial_\nu w_\mu - \partial_\mu w_\nu) \ dx^\nu \wedge dx^\mu = \int_{\phi (M)} (\partial_\nu w_\mu - \partial_\mu w_\nu) \ d^2x$$

Why is it needed to do so? And why after doing that change, we have to substitute $dx^\nu \wedge dx^\mu$ by $d^2x$ in the integrand?

I guess this is because $M$ is a manifold while $\phi (M)$ is a function. So we need to change the integrand to something that "lives" in the function space, namely $d^2x$. Did I guess right?

Can you either post a link to wherever you saw this, or reproduce some of the context? I can't understand what your equality means. What is \phi(M)?

 

[davidge]

Thanks haushofer and stevendaryl. I got the answer to this question a while ago