Download Introduction to Differential Forms and more Study notes Vector Analysis in PDF only on Docsity!
Introduction to differential forms
Donu Arapura
May 6, 2016
The calculus of differential forms give an alternative to vector calculus which is ultimately simpler and more flexible. Unfortunately it is rarely encountered at the undergraduate level. However, the last few times I taught undergraduate advanced calculus I decided I would do it this way. So I wrote up this brief supplement which explains how to work with them, and what they are good for, but the approach is kept informal. In particular, multlinear algebra is kept to a minimum, and I don’t define manifolds or anything like that. By the time I got to this topic, I had covered a certain amount of standard material, which is briefly summarized at the end of these notes. My thanks to Jo˜ao Carvalho, John Crow, Mat´uˇs Goljer and Josh Hill for catching some typos.
Contents
1 1 -forms 2 1.1 1-forms................................ 2 1.2 Exactness in R^2............................ 3 1.3 Parametric curves.......................... 4 1.4 Line integrals............................. 4 1.5 Work.................................. 7 1.6 Green’s theorem for a rectangle................... 8 1.7 Exercise Set 1............................. 9
2 2 -forms 10 2.1 Wedge product............................ 10 2.2 2-forms................................ 11 2.3 Exactness in R^3 and conservation of energy............ 13 2.4 Derivative of a 2-form and divergence............... 14 2.5 Poincar´e’s lemma for 2-forms.................... 15 2.6 Exercise Set 2............................ 16
3 Surface integrals 18 3.1 Parameterized Surfaces........................ 18 3.2 Surface Integrals........................... 20
3.3 Surface Integrals (continued).................... 22 3.4 Length and Area........................... 24 3.5 Exercise Set 3............................. 27
4 Stokes’ Theorem 29 4.1 Green and Stokes........................... 29 4.2 Proof of Stokes’ theorem....................... 30 4.3 Cauchy’s theorem*.......................... 32 4.4 Exercise Set 4............................. 35
5 Gauss’ theorem 37 5.1 Triple integrals............................ 37 5.2 Gauss’ theorem............................ 37 5.3 Proof for the cube.......................... 38 5.4 Gravitational Flux.......................... 39 5.5 Laplace’s equation*.......................... 41 5.6 Exercise Set 5............................. 42
6 Beyond 3 dimensions* 44 6.1 Beyond 3D.............................. 44 6.2 Maxwell’s equations in R^4...................... 47
7 Further reading 49
A Review of multivariable calculus 50 A.1 Differential Calculus......................... 50 A.2 Integral Calculus........................... 52
1 1 -forms
1.1 1 -forms
A differential 1-form (or simply a differential or a 1-form) on an open subset of R^2 is an expression F (x, y)dx + G(x, y)dy where F, G are R-valued functions on the open set. A very important example of a differential is given as follows: If f (x, y) is C^1 R-valued function on an open set U , then its total differential (or exterior derivative) is
df =
∂f ∂x
dx +
∂f ∂y
dy
It is a differential on U. In a similar fashion, a differential 1-form on an open subset of R^3 is an expression F (x, y, z)dx + G(x, y, z)dy + H(x, y, z)dz where F, G, H are R-valued functions on the open set. If f (x, y, z) is a C^1 function on this set, then its total differential is
df =
∂f ∂x
dx +
∂f ∂y
dy +
∂f ∂z
dz
1.3 Parametric curves
Before discussing line integrals, we have to say a few words about parametric curves. A parametric curve in the plane is vector valued function C : [a, b] → R^2. In other words, we let x and y depend on some parameter t running from a to b. It is not just a set of points, but the trajectory of particle travelling along the curve. To begin with, we will assume that C is C^1. Then we can define the the velocity or tangent vector v = ( dxdt , dydt ). We want to assume that the particle travels without stopping, v 6 = 0. Then v gives a direction to C, which we also refer to as its orientation. If C is given by
x = f (t), y = g(t), a ≤ t ≤ b
then x = f (−u), y = g(−u), −b ≤ u ≤ −a
will be called −C. This represents the same set of points, but traveled in the opposite direction. Suppose that C is given depending on some parameter t,
x = f (t), y = g(t)
and that t depends in turn on a new parameter t = h(u) such that (^) dudt 6 = 0. Then we can get a new parametric curve C′
x = f (h(u)), y = g(h(u))
It the derivative (^) dudt is everywhere positive, we want to view the oriented curves C and C′^ as the equivalent. If this derivative is everywhere negative, then −C and C′^ are equivalent. For example, the curves
C : x = cos θ, y = sin θ, 0 ≤ θ ≤ 2 π
C′^ : x = sin t, y = cos t, 0 ≤ t ≤ 2 π
represent going once around the unit circle counterclockwise and clockwise re- spectively. So C′^ should be equivalent to −C. We can see this rigorously by making a change of variable θ = π/ 2 − t. It will be convient to allow piecewise C^1 curves. We can treat these as unions of C^1 curves, where one starts where the previous one ends. We can talk about parametrized curves in R^3 in pretty much the same way.
1.4 Line integrals
Now comes the real question. Given a differential F dx + Gdy, when is it exact? Or equivalently, how can we tell whether a force is conservative or not? Checking that it’s closed is easy, and as we’ve seen, if a differential is not closed, then it can’t be exact. The amazing thing is that the converse statement is often (although not always) true:
THEOREM 1.4.1 If F (x, y)dx + G(x, y)dy is a closed form on all of R^2 with C^1 coefficients, then it is exact.
To prove this, we would need solve the equation df = F dx + Gdy. In other words, we need to undo the effect of d and this should clearly involve some kind of integration process. To define this, we first have to choose a parametric C^1 curve C. Then we define:
DEFINITION 1.4. ∫
C
F dx + Gdy =
∫ (^) b
a
[
F (x(t), y(t))
dx dt
dy dt
]
dt
If C is piecewise C^1 , then we simply add up the integrals over the C^1 pieces. Although we’ve done everything at once, it is often easier, in practice, to do this in steps. First change the variables from x and y to expresions in t, then replace dx by dxdt dt etc. Then integrate with respect to t. For example, if we parameterize the unit circle c by x = cos θ, y = sin θ, 0 ≤ θ ≤ 2 π, we see
− y x^2 + y^2
dx + x x^2 + y^2
dy = − sin θ(cos θ)′dθ + cos θ(sin θ)′dθ = dθ
and therefore ∫
C
y x^2 + y^2
dx +
x x^2 + y^2
dy =
∫ (^2) π
0
dθ = 2π
From the chain rule, we get
LEMMA 1.4. ∫
−C
F dx + Gdy = −
C
F dx + Gdy
If C and C′^ are equivalent, then ∫
C
F dx + Gdy =
C′
F dx + Gdy
While we’re at it, we can also define a line integral in R^3. Suppose that F dx + Gdy + Hdz is a differential form with C^1 coeffients. Let C : [a, b] → R^3 be a piecewise C^1 parametric curve, then
DEFINITION 1.4. ∫
C
F dx + Gdy + Hdz =
∫ (^) b
a
[
F (x(t), y(t), z(t)) dx dt
G(x(t), y(t), z(t)) dy dt
H(x(t), y(t), z(t)) dz dt
]
dt
Then we claim that df = F dx + Gdy. To see this, we differentiate using the fundamental theorem of calculus. The easy calculation is
∂f ∂y
∂y
∫ (^) y
0
G(x, t)dt
= G(x, y)
Slightly trickier is
∂f ∂x
∂x
∫ (^) x
0
F (x, 0)dt +
∂x
∫ (^) y
0
G(x, t)dt
= F (x, 0) +
∫ (^) y
0
∂G(x, t) ∂x
dt
= F (x, 0) +
∫ (^) y
0
∂F (x, t) ∂t
dt
= F (x, 0) + F (x, y) − F (x, 0) = F (x, y)
The same proof works if if we replace R^2 by an open rectangle. However, it will fail for more general open sets. For example,
−
y x^2 + y^2
dx +
x x^2 + y^2
dy
is C^1 1-form on the open set {(x, y) | (x, y) 6 = (0, 0)} which is closed. But it is not exact (see exercise 6). In more advanced treatments, this failure of closed forms to be exact can be measured by something called the de Rham cohomology of the set.
1.5 Work
Line integrals have many important uses. One very direct application in physics comes from the idea of work. If you pick up a rock off the ground, or perhaps roll it up a ramp, it takes energy. The energy expended is called work. If you’re moving the rock in straight line for a short distance, then the displacement can be represented by a vector d = (∆x, ∆y, ∆z) and the force of gravity by a vector F = (F 1 , F 2 , F 3 ). Then the work done is simply
−F · d = −(F 1 ∆x + F 2 ∆y + F 3 ∆z).
On the other hand, if you decide to shoot a rocket up into space, then you would have to take into account that the trajectory c may not be straight nor can the force F be assumed to be constant (it’s a vector field). However as the notation suggests, for the work we would now need to calculate the integral
c
F 1 dx + F 2 dy + F 3 dz
One often writes this as
−
c
F · ds
(think of ds as the “vector” (dx, dy, dz).)
1.6 Green’s theorem for a rectangle
Let R be the rectangle in the xy-plane with vertices (0, 0), (a, 0), (a, b), (0, b). Let C be the boundary curve of the rectangle oriented counter clockwise. Given C^1 functions P (x, y), Q(x, y) on R, the fundamental theorem of calculus yields
∫ ∫
D
∂Q
∂x
dxdy =
∫ (^) b
0
[Q(a, y)) − Q(0, y)]dy =
C
Q(x, y)dy
∫ ∫
D
∂P
∂y dydx =
∫ (^) a
0
[P (x, b) − P (x, 0)]dx = −
C
P (x, y)dx
Subtracting yields Green’s theorem for R
THEOREM 1.6. ∫
C
P dx + Qdy =
R
∂Q
∂x
∂P
∂y
dxdy
Our goal is to understand, and generalize to 3 dimensions, the operation which takes the one form P dx+Qdy to the integrand on the right. In traditional vector calculus this is handled using the curl (∇×) which a vector field defined so that
∇ × (P i + Qj + Rk) · k =
∂Q
∂x
∂P
∂y
is the integrand of the right in Green’s theorem. In general, one can discover the formula for the other components of ∇ × (P i + Qj + Rk) by expressing the integrals of P i + Qj + Rk around the boundaries of rectangles in the xz and yz planes and rewriting them as double integrals. To make a long story short,
∇ × (P i + Qj + Rk) = (Ry − Qz )i + (Qx − Py )k + (Pz − Rx)j
(In practice, this is often written as a determinant
∇ × (P i + Qj + Rk) =
i j k ∂ ∂x
∂ ∂y
∂ ∂z P Q R
But this should really be treated as a memory aid and nothing more.)
2 2 -forms
2.1 Wedge product
The cross product of vectors u × v is a very useful operation in 3 dimensional geometry. It determines the area of the parallelogram containing these vectors and the plane containing it. While there is no direct analogue of the cross product in higher dimensions, there is an operation which determines the last two things. Given (row) vectors u, v ∈ Rn, define the matrix
u ∧ v =
(uT^ v − vT^ u) (1)
(∧ is pronounced “wedge”). In R^3
(a, b, c) ∧ (d, e, f ) =
0 ae − bd af − cd −ae + bd 0 bf − ce −af + cd −bf + ce 0
We can see that the nonzero entries are basically the same as for the cross product,
(a, b, c) × (d, e, f ) = (bf − ce)i + (−af + cd)j + (ae − bd)k
So these two operations are in some sense equivalent. The big difference is, of course, that the wedge product produces a matrix, but not just any matrix. Recall that a matrix A = (aij ) is a skew symmetric if AT^ = −A, i.e. aji = −aij. Let ∧^2 Rn^ denote the space of all skew symmetric n × n real matrices.
THEOREM 2.1.1 The wedge product of two vectors lies in ∧^2 Rn.
This should be clear when n = 3 from the above formula. In general, we can use standard facts from linear algebra to see that
1 2
(uT^ v − vT^ u)T^ =
(vT^ uT T^ − uT^ vT T^ )
(vT^ u − uT^ v) = −
(uT^ v − vT^ u)
The following properties are easy to check from the definition.
v ∧ u = −u ∧ v (2)
u ∧ u = 0 (3)
and c(u ∧ v) = (cu) ∧ v = u ∧ (cv) (4) u ∧ v + u ∧ w = u ∧ (v + w) (5)
In practice, we will use these rules rather than the definition (1) for calculations. To really be convinced that the wedge captures the essential geometric features of the cross product, we note the following non-obvious fact.
THEOREM 2.1.2 The product u ∧ v determines the area of the parallelogram spanned by u and v and the plane containing these vectors, when there is a unique such plane.
It will be useful to work in a bit more generality. Recall that a vector space V is an abstraction of Rn, where elements of V , to be thought of as vectors, can be added and multiplied by numbers and these operations satisfy standard rules of algebra. Given a vector space V , we can construct a new vector space of 2-vectors ∧^2 V = {
i
ui ∧ vi | ui, vi ∈ V }
where the rules (2), (3), (4) and (5) are imposed, and only those rules. To see how this compares to the earlier description. We can expand a vector in Rn^ as
(a 1 ,... , an) = a 1 e 1 +... + anen
uniquely, where e 1 = (1, 0 ,... , 0), e 2 = (0, 1 ,... , 0),...
In other words, the vectors ei form a basis of Rn. Any 2-vector can be expanded uniquely as an expression
∑^ n
i=
∑^ i−^1
j=
aij ei ∧ ej =
∑^ n
i=
∑^ n
j=
aij ei ∧ ej
where the coefficients satisfy aij = −aji. The upshot is that we may identify ∧^2 Rn^ with the space of skew symmetric n × n matrices as we did above.
2.2 2 -forms
A 2-form is an expression built using wedge products of pairs of 1-forms. On R^3 , this would be an expression:
F (x, y, z)dx ∧ dy + G(x, y, z)dy ∧ dz + H(x, y, z)dz ∧ dx
where F, G and H are functions defined on an open subset of R^3. Any wedge product of two 1-forms can be put in this format. For example, using the above rules, we can see that
(3dx + dy) ∧ (exdx + 2dy) = 3exdx ∧ dx + 6dx ∧ dy + exdy ∧ dx + 2dy ∧ dy = (6 − ex)dx ∧ dy
To be absolutely clear, we now allow c to be a function in (4). The real significance of 2-forms will come later when we do surface integrals. A 2-form will be an expression that can be integrated over a surface in the same way that a 1-form can be integrated over a curve.
2.3 Exactness in R^3 and conservation of energy
A C^1 1-form ω = F dx + Gdy + Hdz is called exact if there is a C^2 function (called a potential) such that ω = df. A 1-form ω is called closed if dω = 0, or equivalently if Fy = Gx, Fz = Hx, Gz = Hy
These equations must hold when
F = fx, G = fy , H = fz
Therefore:
THEOREM 2.3.1 Exact 1 -forms are closed.
We have a converse statement which is sometimes called “Poincar´e’s lemma”.
THEOREM 2.3.2 If ω = F dx + Gdy + Hdz is a closed form on R^3 with C^1 coefficients, then ω is exact. In fact if f (x 0 , y 0 , z 0 ) =
C ω, where^ C^ is any piecewise C^1 curve connecting (0, 0 , 0) to (x 0 , y 0 , z 0 ), then df = ω.
This can be rephrased in the language of vector fields. If F = F i + Gj + Hk is C^1 vector field representing a force, then it is called conservative if there is a C^2 real valued function P , called potential energy, such that F = −∇P. The theorem implies that a force F, which is C^1 on all of R^3 , is conservative if and only if ∇ × F = 0. P (x, y, z) is just the work done by moving a particle of unit mass along a path connecting (0, 0 , 0) to (x, y, z). To appreciate the importance of this concept, recall from physics that the kinetic energy of a particle of constant mass m and velocity
v =
dx dt
dy dt
dz dt
is
K =
m||v||^2 =
mv · v.
Also one of Newton’s laws says
m
dv dt
= F
If F is conservative, then we can replace it by −∇P above, move it to the other side, and then dot both sides by v to obtain
mv · dv dt
which can be simplified (exercise 6) to
d dt
(K + P ) = 0. (7)
This implies that the total energy K + P is constant.
2.4 Derivative of a 2 -form and divergence
Earlier we defined wedge products of pairs of vectors. Now we extend it to triples. Given three vectors u, v, w ∈ R^3 , we may think of the 3-vector u∧v ∧w as the oriented volume of parallelopiped with u, v, w as the first, second and third sides. Oriented volume is the usual volume if u, v, w is right-handed, otherwise it is minus the usual volume. With these rules, we see that
u ∧ v ∧ w = −v ∧ u ∧ w = v ∧ w ∧ u =...
and that u ∧ v ∧ w = 0
if any two of the vectors are equal. In addition, we have the distributive law
(u 1 + u 2 ) ∧ v ∧ w = u 1 ∧ v ∧ w + u 2 ∧ v ∧ w = 0
A 3-form is simply an expression
f (x, y, z)dx ∧ dy ∧ dz = −f (x, y, z)dy ∧ dx ∧ dz = f (x, y, z)dy ∧ dz ∧ dx =...
These are things that will eventually get integrated over solid regions. The important thing for the present is an operation which takes 2-forms to 3-forms once again denoted by “d”.
d(F dy ∧ dz + Gdz ∧ dx + Hdx ∧ dy) = (Fxdx + Fy dy + Fz dz) ∧ dy ∧ dz
- (Gxdx + Gy dy + Gz dz) ∧ dz ∧ dx
- (Hxdx + Hy dy + Hz dz) ∧ dx ∧ dy
This simplifies to
d(F dy ∧ dz + Gdz ∧ dx + Hdx ∧ dy) = (Fx + Gy + Hz )dx ∧ dy ∧ dz
It’s probably easier to understand the pattern after converting the above 2- form to the vector field V = F i + Gj + Hk. Then the coefficient of dx ∧ dy ∧ dz is the divergence ∇ · V = Fx + Gy + Hz So far we’ve applied d to functions (also called 0-forms) to obtain 1-forms, and then to 1-forms to get 2-forms, and finally to 2-forms.
{0-forms} d (^) //
{1-forms} d (^) //
{2-forms} d (^) //
{3-forms}
{functions}
{vector fields}
∇× //
{vector fields}
{functions}
The real power of this notation is contained in the following simple-looking formula
This doesn’t solve the problem but it simplifies it, because ω(x, y, 0) doesn’t depend on z. Differentiating both sides of the last equation, shows that ω(x, y, 0) is also closed. So if we can find ξ 1 so that
ω(x, y, 0) = dξ 1
then ξ = ξ 1 + Hz ω
solves the original problem. To find ξ 1 , we reduce the problem further by inte- grating out y and then x as above, to get
dHy ω(x, y, 0) = ω(x, y, 0) − ω(x, 0 , 0)
dHz ω(x, 0 , 0) = ω(x, 0 , 0) − ω(0, 0 , 0)
So now we have reduced the problem to finding ξ 2 such that ω(0, 0 , 0) = dξ 2 ; this is easy (exercise 7). Then using the above equations, we see that
ξ = ξ 2 + Hz ω + Hy ω(x, y, 0) + Hxω(x, 0 , 0)
gives a solution.
2.6 Exercise Set 2
Let α = xdx + ydy + zdz, β = zdx + xdy + ydz and γ = xydz in the following problems.
- Calculate
(a) α ∧ β (b) α ∧ γ (c) β ∧ γ (d) (α + γ) ∧ (α + γ)
- Calculate
(a) dα (b) dβ (c) d(α + γ) (d) d(xα)
- Given ω = f dx + gdy + hdz such that ω ∧ dz = 0, what can we conclude about f, g and h?
- (a) Let ω = F dx + Gdy + Hdz and let C be the straight line connecting (0, 0 , 0) to (x 0 , y 0 , z 0 ) show that ∫
C
ω =
0
(F (x 0 t, y 0 , z 0 ) + G(x 0 , y 0 t, z 0 ) + H(x 0 , y 0 , z 0 t))dt
(b) Use this to prove theorem 2.3.
- Check that the following 1-forms are exact, and express them as df.
(a) dx + 2ydy + 3z^2 dz (b) zy cos(xy)dx + zx cos(xy)dy + sin(xy)dz.
- Check that equations (6) and (7) in the text are equivalent.
- Prove the last step of the proof of theorem 2.5.1 that a 2-form with con- stant coefficients is exact. Hint: observe that d(xdy) = dx ∧ dy etc.
- Find a solution of dξ = ω, when ω = zdx ∧ dz + dy ∧ dz.
are linearly independent. In this case, once we pick an ordering of the variables (say u first, v second) an orientation is determined by the normal
n =
Tu × Tv ||Tu × Tv ||
T u
T v
n
FIGURE 1
S
v=constant
u=constant
If we look at the examples given earlier. (1) is smooth. However there is a slight problem with our examples (2) and (3). Here Tθ = 0, when φ = 0 in example (2) and when r = 0 in example (3). To deal with scenario, we will consider a surface smooth if there is at least one smooth parameterization for it. Let C be a closed C^1 curve in R^2 and D be the interior of C. D is an example of a surface with a boundary C. In this case the surface lies flat in the plane, but more general examples can be constructed by letting S be a parameterized surface (^)
x = f (u, v) y = g(u, v) z = h(u, v) (u, v) ∈ D ⊂ R^2
then the image of C in R^3 will be the boundary of S. For example, the boundary
of the upper half sphere S
x = sin(φ) cos(θ) y = sin(φ) sin(θ) z = cos(φ) 0 ≤ φ ≤ π/ 2 , 0 ≤ θ < 2 π
is the circle C given by
x = cos(θ), y = sin(θ), z = 0, 0 ≤ θ ≤ 2 π
In what follows, we will need to match up the orientation of S and its boundary curve. This will be done by the right hand rule: if the fingers of the right hand point in the direction of C, then the direction of the thumb should be “up”.
n
S
C
FIGURE 2
3.2 Surface Integrals
Let S be a smooth parameterized surface
x = f (u, v) y = g(u, v) z = h(u, v) (u, v) ∈ D
with orientation corresponding to the ordering u, v. The symbols dx etc. can be converted to the new coordinates as follows
dx =
∂x ∂u
du +
∂x ∂v
dv
