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Differential forms and the exterior derivative.

We can now apply the constructions of the previous section to the tangent space to a manifold. We define a $ k$-form on the tangent space at $ x \in M$ to be an element of

$\displaystyle \Lambda^k T^*_x M. $

We want to define $ k$-form fields' in the same way we define vector fields except that we do not call them $ k$-form fields we call the differentiable $ k$-forms or sometimes just $ k$-forms. Choose co-ordinates $ (U, \psi)$ on $ M$. Then $ \omega(x) $ in $ \Lambda^k(T^*_xM)$ can be be written as

$\displaystyle \omega(x) = \sum_{i_1, \dots, i_k} \frac{1}{k!} \omega_{i_1,\dots,i_k} d\psi^{i_1} \wedge \dots \wedge d\psi^{i_k} $

at each $ x \in U$. Hence we have defined a function

$\displaystyle \omega_{i_1,\dots,i_k} \colon U \to \mathbb{R}$

for each set of $ k$ indices. We call these functions the components of $ \omega$ with respect to the co-ordinate chart. The components satisfy the anti-symmetry conditions in the previous section. We can also define the $ \omega_{i_1, \dots, i_k}$ as

$\displaystyle \omega_{i_1, \dots, i_k} = \omega(\frac{\partial\phantom{\psi^{i_1}}}{\psi^{i_1}}, \dots, \frac{\partial\phantom{\psi^{i_k}}}{\psi^{i_k}}). $

We define a smooth differential form by

Definition 5.1 (Differential form.)   A differential form $ \omega$ is smooth if its components with respect to a collection of co-ordinate charts whose domains cover $ M$ are smooth.

We have the usual Lemma

Lemma 5.1   If a differential form is smooth then its components with respect to any co-ordinate chart are smooth.

We denote by $ \Omega^k(M)$ the set of all smooth differentiable $ k$ forms on $ M$. Notice that $ \Omega^0(M)$ is just $ C^\infty(M)$ the space of all smooth functions on $ M$.

Using the equation

$\displaystyle \omega_{i_1, \dots, i_k} = \omega(\frac{\partial\phantom{\psi^{i_1}}}{\psi^{i_1}}, \dots, \frac{\partial\phantom{\psi^{i_k}}}{\psi^{i_k}}). $

for the components of the differential form we can calculate the way the components change if we use another co-ordinate chart $ (V, \chi)$. We have

$\displaystyle \frac{\partial\phantom{\psi^i}}{\psi^i} = \sum_{a=1}^n \frac{\partial \chi^a}{\partial \psi^i} \frac{\partial\phantom{\chi^a}}{\chi^a} $

and substituting this into the formula gives

$\displaystyle \omega_{i_1, \dots, i_k} =\sum_{a_1, \dots, a_k=1}^n (\frac{\par... ...ots \frac{\partial \chi^{a_n}}{\partial \psi^{i_n}}) \omega_{a_1, \dots, a_k}. $

The usual derivative on functions defines a linear differential operator

$\displaystyle d \colon \Omega^0(M) \to \Omega^1(M). $

As well as being linear $ d$ satisfies the Leibniz rule:

$\displaystyle d(fg) = fdg + (df)g. $

We want to prove

Proposition 5.3   If the dimension of $ M$ is $ n$ then there are unique linear maps

$\displaystyle d \colon \Omega^p(M) \to \Omega^{p+1}(M) $

for all $ p = 0, \dots, n-1$ satisfying:
1.
If $ p=0$ $ d$ is the usual derivative,
2.
$ d^2 = 0$, and
3.
$ d(\omega\wedge\rho) = (d\omega)\wedge\rho + (-1)^p \omega \wedge (d\rho)$ where $ \omega \in \Omega^p(M)$ and $ \rho \in \Omega^q(M)$.

Proof. We define $ d$ recursively. We have the ordinary definition of $ d$ if $ p=0$. We assume that we have it defined for all $ p < k$ and that the conditions (i), (ii) and (iii) hold when ever they make sense. Consider a $ k$ form $ \omega$. Let $ (U, \psi)$ be a co-ordinate chart and let

$\displaystyle \omega = \sum_{i_1 \dots i_k} \frac{1}{k!} \omega_{i_1 \dots i_k} d\psi^{i_1} \wedge \dots \wedge d\psi^{i_k}. $

Then define

$\displaystyle \omega_i = \sum_{i_2 \dots i_k} \frac{1}{k!} \omega_{i i_{2} \dots i_k} d\psi^{i_2} \wedge \dots \wedge d\psi^{i_k} $

so that

$\displaystyle \omega = \sum_{i=1}^k \omega_i \wedge d\psi^i. $

Notice that $ \omega_i$ is uniquely defined by this equation because

$\displaystyle \omega_i = \iota_{\frac{\partial\phantom{\psi^i}}{\psi^i}}\omega. $

Consider now another choice of co-ordinates $ (V, \chi)$. We have

$\displaystyle \omega = \sum_{a=1}^k \omega_a \wedge d\chi^a $

where

$\displaystyle \omega_a = \iota_{\frac{\partial\phantom{\chi^a}}{\chi^a}}\omega. $

It is easy to check that on $ U \cap V$ we have

$\displaystyle \omega_a = \sum_{a=1}^n \frac{\partial \psi^i}{\partial \chi^a} \omega_i. $

Then if the proposition is to be true we must have

$\displaystyle d\omega =$ $\displaystyle d ( \sum_{i=1}^k \omega_i \wedge d\psi^i)$    
$\displaystyle =$ $\displaystyle \sum_{i=1}^k d\omega_i \wedge d\psi^i$    

This defines a differential $ k+1$ form on the open set $ U$. On the open set $ V $ it is defined by

$\displaystyle \sum_{a=1}^k d\omega_a \wedge d\chi^a $

and we need to check that these two agree. We have

$\displaystyle d\omega_a =$ $\displaystyle d(\sum_{i=1}^n \frac{\partial \psi^i}{\partial \chi^a} \omega_i)$    
$\displaystyle =$ $\displaystyle \sum_{b, i=1}^n \frac{\partial ^2 \psi^i}{ \partial \chi^b \parti... ...wedge\omega_i + \sum_{i=1}^n \frac{\partial \psi^i}{\partial \chi^a} d\omega_i.$    

Hence

$\displaystyle \sum_{a=1}^n d\omega_a \wedge d\chi^a = \sum_{i, a, b =1}^n \frac... ...m_{i, a=1}^n \frac{\partial \psi^i}{\partial \chi^a} d\omega_i \wedge d\chi^a. $

The first term vanishes because the partial derivative is symmetric in $ a$ and $ b$ and the wedge product is anti-symmetric. Hence we have

$\displaystyle \sum_{a=1}^n d\omega_a \wedge d\chi^a = \sum_{i =1}^n d\omega_i \wedge d\chi^i $

as required. Clearly condition (i) is still true. For (ii) note that $ dd\omega =d(\sum_{i=1}^n d\beta_i \wedge d\psi^i) = \sum_{i=1}^n dd\beta_i \wedge d\psi^i = 0$. For the final condition let $ \rho =\sum_{i=1}^n\rho_i\wedge d\psi^i$. Assume that $ \rho $ has degree $ q$. Then

$\displaystyle \omega\wedge \rho =\sum_i^n \left( \sum_{j=1}^n (-1)^q \omega_i\wedge\rho_j\wedge \psi^j\right)\wedge \psi^i $

so that

$\displaystyle d(\omega\wedge \rho) =\sum_i^n \left( \sum_{j=1}^n (-1)^q d(\omega_i\wedge\rho_j\wedge d\psi^j)\right)\wedge d\psi^i. $

Then applying the result for degrees lower than $ k$ we have

$\displaystyle d(\omega_i\wedge\rho_j\wedge d\psi^j) = d(\omega_i)\wedge \rho_j \wedge d\psi^j + (-1)^{p-1} \omega_i \wedge d\rho_j \wedge d\psi^j. $

Putting this altogether we have

$\displaystyle d(\omega\wedge \rho) =$ $\displaystyle \sum_{j=1}^n \left[(-1)^q d(\omega_i)\wedge \rho_j \wedge d\psi^j + (-1)^{p+ q-1} \omega_i \wedge d\rho_j \wedge d\psi^j\right] \wedge d\psi^i$    
$\displaystyle =$ $\displaystyle \sum_{j=1}^n d(\omega_i)\wedge d\psi^i \wedge \rho_j \wedge d\psi^j + (-1)^{q} \omega_i \wedge d\psi^i\wedge d\rho_j \wedge d\psi^j$    
$\displaystyle =$ $\displaystyle d(\omega) \wedge\rho + (-1)^q \omega \wedge d\rho.$    

as required. $ \qedsymbol$

Note that this proposition implies that if

$\displaystyle \omega = \sum_{i_1, \dots, i_k} \frac{1}{k!} \omega_{i_1,\dots,i_k} d\psi^{i_1} \wedge \dots \wedge d\psi^{i_k}. $

then we have

$\displaystyle d\omega = \sum_{i_1, \dots, i_k} \frac{1}{k!} d\omega_{i_1,\dots,i_k}\wedge d\psi^{i_1} \wedge \dots \wedge d\psi^{i_k} $

or

$\displaystyle d\omega = \sum_{i_0,i_1, \dots, i_k} \frac{1}{k!} \frac{\omega_{i... ...l \psi^{i_0}} {d\psi^{i_0}}\wedge d\psi^{i_1} \wedge \dots \wedge d\psi^{i_k}. $

It is a useful exercise to reprove this lemma using this definition of the exterior derivative.

Example 5.2   Recall from 5.1 the way in which we identified one-forms and two-forms on $ \mathbb{R}^3$ with vectors. It follows that differentiable one and two forms on $ \mathbb{R}^3$ can be identified with vector-fields. Similarly zero and three forms are functions. With these identifications it is straightforward to check that the exterior derivative of zero, one and two forms corresponds to the classical differential operators grad, curl and div.

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