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Integration of differential forms

Let $ U \subset
\mathbb{R}^n$ and $ \psi \colon
\mathbb{R}^n \to
\mathbb{R}^n$ be a diffeomorphism. Then it is well known that if $ f \colon \psi(U)\to
\mathbb{R}$ is a function then

$\displaystyle \int_U f\circ\psi \vert\det\left({\frac{\partial \psi^i}{\partial x^j}}\right)\vert dx^1\dots dx^n =
\int_{\psi(U)} f
dx^1\dots dx^n.
$

In this formula we regard $ dx^1 \dots dx^n$ as the symbol for Lebesgue measure. However it is very suggestive of the notation for differential forms developed in the previous section.

If $ \omega$ is a differential $ n$ form on $ V = \psi(U)$ then we can write it as

$\displaystyle \omega(x) = f(x) dx^1 \wedge \dots \wedge dx^n.
$

If we pull it back with the diffeomorphism $ \phi$ then, as we seen before,

$\displaystyle \phi^*(\omega) =
f(x) \det\left({\frac{\partial \psi^i}{\partial x^j}}\right) dx^1 \wedge \dots
\wedge dx^n.
$

So differential $ n$ forms transform by the determinant of the jacobian of the diffeomorphism and Lebesgue measure transforms by the absolute value of the determinant of the jacobian of the diffeomorphism. We define the integral of a differential $ n$ form by

$\displaystyle \int_V \omega = \int_V f(x) dx^1\dots dx^n
$

when $ \omega = f(x) dx^1 \wedge \dots \wedge dx^n$. Alternatively we can write this as

$\displaystyle \int_V \omega = \int_V \omega(\frac{\partial\phantom{x^1}}{\partial x^1},\dots,\frac{\partial\phantom{x^n}}{\partial x^n})
dx^1\dots dx^n.
$

Call a diffeomorphism $ \psi \colon U \to V$ orientation preserving if

$\displaystyle \det\left({\frac{\partial \psi^i}{\partial x^j}}\right)(x) >0
$

for all $ x \in U$. Then we have

Proposition 5.5   If $ \psi\colon U \to \psi(U)$ is an orientation preserving diffeomorphism and $ \omega$ is a differential $ n$ form on $ \psi(U)$ then

$\displaystyle \int_{\psi(U)} \omega = \int_U \psi^*(\omega).
$

We can use this proposition to define the integral of differential forms on a manifold. Let $ \{(U_\alpha , \psi_\alpha )\}_{\alpha \in I}$ be a covering of $ M$ by co-ordinate charts. Choose a partition of unity $ \phi_\alpha $ subordinate to $ U_\alpha $. Then if $ \omega$ is a differential $ n$ form we can write

$\displaystyle \omega = \sum_{\alpha } \phi_\alpha \omega
$

where the support of $ \phi_a\omega $ is in $ U_\alpha $. First we define the integral of each of the forms $ \phi_\alpha \omega$

$\displaystyle \int_M \phi_a\omega =\int_{\psi_\alpha (U_\alpha )}
(\psi^{-1})^*(\phi_a\omega).
$

Then we define the integral of $ \omega$ to be

$\displaystyle \int_M\omega = \sum_{\alpha \in I} \int_{\psi_\alpha (U_\alpha )}
(\psi^{-1})^*(\omega).
$

We have to show that this is independent of all the choices we have made. So let us take another open cover $ \{(V_\b , \chi_\b )\}_{\b\in J}$ with partition of unity $ \rho_\b $. Then we have

$\displaystyle \sum_{\alpha \in I} \int_{\psi_\alpha (U_\alpha )} (\psi^{-1})^*(\phi_a\omega)=$ $\displaystyle \sum_{\alpha \in I} \int_{\psi_\alpha (U_\alpha )} (\psi^{-1})^*(\sum_{\b\in J} \rho_b\phi_a\omega)$    
$\displaystyle =$ $\displaystyle \sum_{\alpha \in I} \int_{\psi_\alpha (U_\alpha )} (\psi^{-1})^*((\sum_{\b\in J} \rho_b)\phi_a\omega)$    
$\displaystyle =$ $\displaystyle \sum_{\alpha \in I}\sum_{\b\in J} \int_{\psi_\alpha (U_\alpha )} (\psi^{-1})^* (\rho_b\phi_a\omega).$    

The differential forms $ \rho_b\phi_a\omega$ have support in $ U_\alpha \cap V_\b $ so we have

$\displaystyle \int_{\psi_\alpha (U_\alpha )} (\psi^{-1})^* (\rho_b\phi_a\omega)$ $\displaystyle =\int_{\psi_\alpha (U_\alpha \cap V_\b )} (\psi^{-1})^* (\rho_b\phi_a\omega)$    
  $\displaystyle =\int_{\psi_\alpha (U_\alpha \cap V_\b )} (\psi_\alpha ^{-1})^* (\rho_b\phi_a\omega)$    

If the diffeomorphism

$\displaystyle \chi_\b\circ {\psi_\alpha ^{-1}}_{I\psi(U_\alpha \cap V_b)}
$

is orientation preserving then we have

$\displaystyle \int_{\psi_\alpha (U_\alpha \cap V_\b )} (\psi_\alpha ^{-1})^* (\rho_b\phi_a\omega)$ $\displaystyle =\int_{\chi_\b (U_\alpha \cap V_\b )} (\chi_\b ^{-1})^* (\rho_b\phi_a\omega)$ $\displaystyle =\int_{\chi_\b (U_\alpha )} (\chi_\b ^{-1})^* (\rho_b\omega).$    

So we can complete the calculation above and have

$\displaystyle \sum_{\alpha \in I} \int_{\psi_\alpha (U_\alpha )} (\psi^{-1})^*(...
...ha \omega)= \sum_{\b\in J} \int_{\chi_\b (U_\b )} (\chi^{-1})^*(\chi_\b\omega).$    

All this calculation rests on the fact that

$\displaystyle \chi_\b\circ {\psi_\alpha ^{-1}}_{\vert\psi(U_\alpha \cap V_b)}
$

is an orientation preserving diffeomorphism. In general this will not be the case. We have to introduce the notion of an oriented manifold and and oriented co-ordinate chart. Before we can do that we need to discuss orientations on a vector space.


next up previous contents
Next: Orientation. Up: Differential forms. Previous: Pulling back differential forms   Contents
Michael Murray
1998-09-16