20. Integration of differential forms on singular -cubes
This notebook is part of the Introduction to manifolds in SageMath by Andrzej Chrzeszczyk (Jan Kochanowski University of Kielce, Poland).
Remark. This notebook is an attempt to present a more friendly version of Chapter 4 of:
Michael Spivak. Calculus on manifolds, Addison Wesley, NY 1965,
with elementary examples in SageMath.
Remark. Readers interested in application of free software to illustrate the classical Stokes theorem should consult for example https://docs.sympy.org/latest/modules/vector/vector_integration.html
Warning. In the present notebook we extensively use the information on differential forms from notebook 14 and notebook 16.
Explicit description of some pullbacks
Let us start from some more explicit description of pullback of -forms on -dimensional open sets.
Consider the smooth map between open subsets in . Let be the coordinates on and - the coordinates on . For -form we have
where and .
In fact
(we have used the relations and the definition of determinant).
Singular -cubes
Let and be an open subset of such that
Let be a map of into an open set in ( and may be different). We will write this map simply as
(this map is a restriction of a map ).
The map will be called a singular -cube in . The word singular indicates that the map need not be one-to-one, though we will require that be in . For the singular -cubes are just parametrized curves and surfaces, respectively.
Standard -cubes
By the standard -cube we will mean the map defined by
so is an example of singular -cube with
For standard -cube we can define singular -cubes and (the i-th pieces of the boundary of ) as the maps
and
The singular -cubes and are called and faces of respectively.
Boundary of the singular -cube
In the definition of the boundary of a singular -cube we will use some auxiliary notions.
For arbitrary family of objects we can construct a vector space of formal linear combinations of elements as the family of functions , such that for all but finitely many . Vector space operations on such functions are defined as usual: if are formal linear combinations of elements of , then
The formal linear combination are usually written as
If we take as the family of all singular -cubes in we can define the space of -chains in as the space of formal linear combinations of singular -cubes in . Thus the space of -chains consists of (finite) linear combinations
$$ \sum a_i\phi_i,\quad \mbox{where } \phi_i\ \ \mbox{are singular}\ \ \text{$kParseError: KaTeX parse error: Expected 'EOF', got '}' at position 11: -cubes and}̲\ \ a_i\in R. $
Using the introduced notion we can define the boundary of the standard -cube as the -chain
In the case of singular -cube we define the and faces as
respectively and the boundary of as the -chain
Integrals of -forms on singular -cubes and chains
Let be a smooth -form on an open set containing . Assume that is of the form
where is a smooth function on .
We define first the integral of over
where the right hand side denotes the Riemann integral over the rectangle .
Using this relation we can define the integral of the smooth -form on over singular -cube :
With the help of (20.1) we can rewrite this formula:
where .
In the special case of standard -cube given by , we obtain
Finally we define the integral of -form of over -chain :
The last formula applied to -chain and -form gives
and in the case of -chain :
The definition (20.15) of the integral of on looks artificially, but we can give it a quite natural interpretation if we define the orientation of and .
We recall that the transition from one frame in to a second frame can be defined by the square matrix obtained from . The determinant of this matrix is always nonzero, and the set of all frames divides into two equivalence classes, each class containing all possible frames such that for any two of them the determinant of transition matrix is positive. Such equivalence classes are called orientation classes of frames in . To define an orientation means to fix one of these orientation classes. Thus, the oriented space is the space together with a fixed orientation class of frames.
Let be the unit outward (with respect to ) normal vector to or defined in (20.4),(20,5). Assume that the space has been oriented by . We can define the orientation of and prescribed by the unit outward normal vector to or in the following way.
Compare the frame ( omitted) with the frame , that orients the space .
If these are in the same orientation class, then (or ) has the orientation prescribed by the unit outward normal vector to (or ).
If these two frames are in different orientation classes, then we take the orientation of ( or ) opposite to that defined by as the one prescribed by the outward unit normal vector.
Defining we take the sum of all and but if or have the orientation opposite to that prescribed by the outward unit normal vector we take the corresponding integrals with minus sign.
Example 20.1
Consider the case .
If , then the integral over the boundary of defined in (20.15) takes the form
The orientation of the boundary of prescribed by the unit normal outward vector implies that:
on the frame is in the same orientation class as in ,
on the frame is in the same orientation class as in ,
on the frame is in the same orientation class as in ,
on the frame is in the same orientation class as in .
This is in accordance with the fact that in (*) the integrals over and are taken with the minus sign.
As we can see, in the case the formula (20.15) agrees with the orientation of boundary of prescribed by the unit outward normal vector and corresponds to going around the square counterclockwise.
Example 20.2
Consider the case .
If , then the integral over the boundary of defined in (20.15) takes the form
The orientation of the boundary of prescribed by the unit outward normal vector implies that:
on the frame is in the same orientation class as in ,
on the frame is in the same orientation class as in ,
on the frame is in the same orientation class as in ,
on the frame is in the same orientation class as in ,
on the frame is in the same orientation class as in ,
on the frame is in the same orientation class as in .
This is in accordance with the fact that in (**) the integrals over are taken with the minus sign.
As we can see, in the case the formula (20.15) agrees with the orientation of boundary of prescribed by the unit outward normal vector.
Integrals of -forms over and
General smooth -form on an open subset containing is of the form
( is omitted).
Consider the single summand .
Since the map is given by (20.4) and is constant, for we have
and analogously for , so if
If we have
The last equality follows from the fact, that if a function is constant with respect to , then .
Analogously we can compute the integral over , so
and
Stokes theorem on the standard -cube
Consider a smooth -form (20.17) on an open subset containing and let be the standard -cube defined in (20.3). Let us compute
By linearity we can compute each summand
separately. Let us focus on first. We have
Since the integrand does not depend on it is equal to its integral , so
In the last equality we have used (20.19) and (20.20) for .
For we have
Since the integrand does not depend on it is equal to its integral , so
In the last equality we have used (20.19) and (20.20) for .
Now we are ready to consider the complete form (20.17). First compute the exterior differential
and next the integral
As a consequence of (20.18)-(20.20) we obtain
Thus we have checked that
Recall, that in (20.15) we defined the boundary integral of as so we get:
Formulation of Stokes theorem on the standard -cube.
If is a smooth -form (20.17) on an open set containing , is the standard -cube defined in (20.3) and if the boundary integral is defined by (20.15), then
Example 20.3
Using Stokes theorem compute the integral where denotes the boundary of the unit cube , oriented by the outward unit normal.
Using Stokes theorem we can replace the integral by where
so we start from definition of 2-form and computing its exterior derivative .
Computing we use (20.13) and the iterated integral
Formulation of Stokes theorem on singular -cubes
Assume that is an open subset of , let be a smooth -form on and be singular -cube. Recall that in (20.16) we defined the boundary integral of over as .
The Stokes theorem on singular -cubes reads as follows
Before the proof we check that for arbitrary singular -cube and -form on the following relation holds true
Let us compute both sides separately.
From the properties of pullback we know, that implies and analogously . Thus
On the other hand from (20.15)
so both sides of (20.24) are equal.
The formula (20.23) is a consequence of (20.22) and (20.24), since
Extending the integration domains
Assume that are smooth functions on and define the set
Consider the mapping
One can prove that
is one-to-one and .
If is a smooth map, then is a singular 2-cube, so for 1-forms on some neighborhood of Stokes formula takes the form
If we define and then
so we can use Stokes theorem for domains of type (20.25).
Thus Stokes theorem can be applied to domains whose parameterizations can be reparameterized into unit cubes.
Remark. In practice, computing integrals for of the form on we do not use reparametrizations but the formula
One can prove that Stokes theorem can be applied to analogous domains in higher dimensions, for example
for some smooth functions .
Example 20.4
Compute where denotes the 2-dimensional unit sphere, which can be considered as the (reparametrized version of) singular 2-cube
We define the 2-form first.
To use (reparametrized version of) (20.11) we need the pullback .
To obtain the final result we use the iterated integral .
Example 20.5
Compute where is the part of the elliptic cone bounded by the planes , which may be considered as (reparametrized version of) the singular 2-cube
Let us define 2-form and its pullback under .
We use (reparametrized version of) (20.11) and replace the double integral by the iterated integral
Example 20.6
Compute the integral , where is the sphere
First we define the 2-form ,
and its exterior differential:
Using spherical coordinates , the sphere can be considered as reparametrized singular -cube .
So if we apply (reparametrized version of) (20.12), and the relation , we obtain the following form of the iterated integral:
Example 20.7
Using Stokes theorem, compute the integral over the unit sphere . (In Example 20.4 we computed the integral using (20.11)).
As in the previous example we start from defining the 2-form and its exterior derivative.
We can use the value of Jacobian from previous example,
and obtain the following iterated integral:
Example 20.8
Compute the integral , where is the boundary of hemisphere .
Define the 1-form and 2-form .
Let us compute both sides of Stokes formula. Note that using spherical coordinates we can define the hemisphere as the (reparametrized) singular 2-cube
The corresponding parts of can be defined as follows:
corresponds to so ,
corresponds to so ,
corresponds to so ,
corresponds to so .
Since by (20.16) , the integral is zero and last two integrals cancel each other, the boundary integral reduces to .
Define the pullback of under ,
and use (reparametrized version of) (20.11), to obtain the boundary integral: .
Now we use the map ,
and the pullback .
We can apply (reparametrized version of) (20.11) to obtain the integral of over the hemisphere:
Example 20.9
Compute the integral over the tetrahedron bounded by the coordinate planes and the plane .
As usual we start from defining 2-form and 3-form .
Since the tetrahedron can be described by the inequalities
which is a three-dimensional version of (20.25), by Stokes theorem, the boundary integral is equal to
Example 20.10
Compute the integral where is the boundary of the triangle with vertices .
We start from defining 1-form and 2-form .
First we compute the integral of over the triangle using the pullback of under .
The iterated integral is equal to.
The boundary of the triangle can be considered as the 1-chain , where are (reparametrized) singular 1-cubes (parametric curves). Note that the counterpart of reduces to a point.
Let us define the space for the parameter ,
next the singular 1-cubes and the pullbacks :
According to (reparametrized version of) (20.16), the integral of 1-form over 1-chain is equal to
Stokes theorem for chains
Extending Stokes theorem to -chains is easy, since for -chain , where are singular -cubes in an open set and for -form in , we have by (20.14) and (20.23)
Remark. In case of problems, some iterated integrals can be also computed with the help of https://www.wolframalpha.com/
For example the command (without #):
in wolframalpha window returns
Remark. Computing iterated integrals, SageMath (or strictly speaking Maxima) may need some assumptions, for example:
What's next?
Take a look at the notebook Connection.