Download Notes on Differential Forms. Part 1: Forms on R N and more Lecture notes Differential Equations in PDF only on Docsity!
- What is a form? Since we’re not following the development in Guillemin and Pollack, I’d better write up an alternate approach. In this approach, we’re first going to define forms on Rn^ via unmotivated formulas, then prove that forms behave nicely, and only then go back and interpret forms as a kind of tensor with certain (anti)symmetry properties. On Rn, we start with the symbols dx^1 ,... , dxn, which at this point are pretty much meaningless. We define a multiplication operation on these symbols, denoted by a ∧, with the condition
dxi^ ∧ dxj^ = −dxj^ ∧ dxi.
Of course, we also want the usual properties of multiplication to also hold. If α, β, γ are arbitrary products of dxi’s, and if c is any constant, then
(α + β) ∧ γ = α ∧ γ + β ∧ γ α ∧ (β + γ) = α ∧ β + α ∧ γ (α ∧ β) ∧ γ = α ∧ (β ∧ γ) (1) (cα) ∧ β = α ∧ (cβ) = c(α ∧ β)
Note that the anti-symmetry implies that dxi^ ∧ dxi^ = 0. Likewise, if I = {i 1 ,... , ik} is a list of indices where some index gets repeated, then dxi^1 ∧ · · · ∧ dxik^ = 0, since we can swap the order of terms (while keeping track of signs) until the same index comes up twice in a row. For instance,
dx^1 ∧ dx^2 ∧ dx^1 = −dx^1 ∧ dx^1 ∧ dx^2 = −(dx^1 ∧ dx^1 ) ∧ dx^2 = 0.
- A 0-form on Rn^ is just a function.
- A 1-form is an expression of the form
i
fi(x)dxi, where fi(x) is a function and dxi is one of our meaningless symbols.
- A 2-form is an expression of the form
i,j
fij (x)dxi^ ∧ dxj^.
- A k-form is an expression of the form
I
fI (x)dxI^ , where I is a subset {i 1 ,... , ik}
of { 1 , 2 ,... , n} and dxI^ is shorthand for dxi^1 ∧ · · · ∧ dxik^.
- If α is a k-form, we say that α has degree k. For instance, on R^3 Date: April 14, 2016. 1
- 0-forms are functions
- 1-forms look like P dx + Qdy + Rdz, where P , Q and R are functions and we are writing dx, dy, dz for dx^1 , dx^2 , dx^3.
- 2-forms look like P dx ∧ dy + Qdx ∧ dz + Rdy ∧ dz. Or we could just as well write P dx ∧ dy − Qdz ∧ dx + Rdy ∧ dz.
- 3-forms look like f dx ∧ dy ∧ dz.
- There are no (nonzero) forms of degree greater than 3. When working on Rn, there are exactly
(n k
linearly independent dxI^ ’s of degree k, and 2 n^ linearly independent dxI^ ’s in all (where we include 1 = dxI^ when I is the empty list). If I′^ is a permutation of I, then dxI′ = ±dxI^ , and it’s silly to include both fI dxI^ and fI′ dxI′ in our expansion of a k-form. Instead, one usually picks a preferred ordering of {i 1 ,... , ik} (typically i 1 < i 2 < · · · < ik) and restrict our sum to I’s of that sort. When working with 2-forms on R^3 , we can use dx ∧ dz or dz ∧ dz, but we don’t need both. If α =
αI (x)dxI^ is a k-form and β =
βJ (x)dxJ^ is an `-form, then we define
α ∧ β =
I,J
αI (x)βJ (x)dxI^ ∧ dxJ^.
Of course, if I and J intersect, then dxI^ ∧ dxJ^ = 0. Since going from (I, J) to (J, I) involves kswaps, we have dxJ^ ∧ dxI^ = (−1)kdxI^ ∧ dxJ^ ,
and likewise β ∧ α = (−1)k`α ∧ β. Note that the wedge product of a 0-form (aka function) with a k-form is just ordinary multiplication.
- Derivatives of forms If α =
I
αI dxI^ is a k-form, then we define the exterior derivative
dα =
I,j
∂αI (x) ∂xj^ dx
j (^) ∧ dxI (^).
Note that j is a single index, not a multi-index. For instance, on R^2 , if α = xydx + exdy, then
dα = ydx ∧ dx + xdy ∧ dx + exdx ∧ dy + 0dy ∧ dy (2) = (ex^ − x)dx ∧ dy.
If f is a 0-form, then we have something even simpler:
df (x) =
∑ (^) ∂f (x) ∂xj^ dx
j (^) ,
which should look familiar, if only as an imprecise calculus formula. One of our goals is to make such statements precise and rigorous. Also, remember that xi^ is actually a function on
Exercise 1: On R^3 , there are interesting 1-forms and 2-forms associated with each vector field v(x) = (v 1 (x), v 2 (x), v 3 (x)). (Here vi is a component of the vector v, not a vector in its own right.) Let ω^1 v = v 1 dx+v 2 dy+v 3 dz, and let ω v^2 = v 1 dy∧dz+v 2 dz∧dx+v 3 dx∧dy. Let f be a function. Show that (a) df = ω^1 grad f, (b) dω^1 v = ω curl v^2 , and (c) dω v^2 = (div v)dx∧dy ∧dz, where grad, curl, and div are the usual gradient, curl, and divergence operations.
Exercise 2: A form ω is called closed if dω = 0, and exact if ω = dν for some other form ν. Since d^2 = 0, all exact forms are closed. On Rn^ it happens that all closed forms of nonzero degree are exact. (This is called the Poincare Lemma). However, on subsets of Rn^ the Poincare Lemma does not necessarily hold. On R^2 minus the origin, show that ω = (xdy − ydx)/(x^2 + y^2 ) is closed. We will soon see that ω is not exact.
- Pullbacks Suppose that g : X → Y is a smooth map, where X is an open subset of Rn^ and Y is an open subset of Rm, and that α is a k-form on Y. We want to define a pullback form g∗α on X. Note that, as the name implies, the pullback operation reverses the arrows! While g maps X to Y , and dg maps tangent vectors on X to tangent vectors on Y , g∗^ maps forms on Y to forms on X.
Theorem 3.1. There is a unique linear map g∗^ taking forms on Y to forms on X such that the following properties hold:
(1) If f : Y → R is a function on Y , then g∗f = f ◦ g. (2) If α and β are forms on Y , then g∗(α ∧ β) = (g∗α) ∧ (g∗β). (3) If α is a form on Y , then g∗(dα) = d(g∗(α)). (Note that there are really two different d’s in this equation. On the left hand side d maps k-forms on Y to (k + 1)-forms on Y. On the right hand side, d maps k forms on X to (k + 1)-forms on X. )
Proof. The pullback of 0-forms is defined by the first property. However, note that on Y , the form dyi^ is d of the function yi^ (where we’re using coordinates {yi} on Y and reserving x’s for X). This means that g∗(dyi)(x) = d(yi^ ◦ g)(x) = dgi(x), where gi(x) is the i-th component of g(x). But that gives us our formula in general! If α =
I
αI (y)dyI^ , then
(6) g∗α(x) =
I
αI (g(x))dgi^1 ∧ dgi^2 ∧ · · · ∧ dgik^.
Using the formula (6), it’s easy to see that g∗(α ∧ β) = g∗(α) ∧ g∗(β). Checking that g∗(dα) = d(g∗α) in general is left as an exercise in definition-chasing.
Exercise 3: Do that exercise!
An extremely important special case is where m = n = k. The n-form dy^1 ∧ · · · ∧ dyn^ is called the volume form on Rn.
Exercise 4: Let g is a smooth map from Rn^ to Rn, and let ω be the volume form on Rn. Show that g∗ω, evaluated at a point x, is det(dgx) times the volume form evaluated at x.
Exercise 5: An important property of pullbacks is that they are natural. If g : U → V and h : V → W , where U , V , and W are open subsets of Euclidean spaces of various dimensions, then h ◦ g maps U → W. Show that (h ◦ g)∗^ = g∗^ ◦ h∗.
Exercise 6: Let U = (0, ∞) × (0, 2 π), and let V be R^2 minus the non-negative x axis. We’ll use coordinates (r, θ) for U and (x, y) for V. Let g(r, θ) = (r cos(θ), r sin(θ)), and let h = g−^1. On V , let α = e−(x^2 +y^2 )dx ∧ dy. (a) Compute g∗(x), g∗(y), g∗(dx), g∗(dy), g∗(dx ∧ dy) and g∗α (preferably in that order). (b) Now compute h∗(r), h∗(θ), h∗(dr) and h∗(dθ). The upshot of this exercise is that pullbacks are something that you have been doing for a long time! Every time you do a change of coordinates in calculus, you’re actually doing a pullback.
- Integration Let α be an n-form on Rn, and suppose that α is compactly supported. (Being com- pactly supported is overkill, but we’re assuming it to guarantee integrability and to allow manipulations like Fubini’s Theorem. Later on we’ll soften the assumption using partitions of unity.) Then there is only one multi-index that contributes, namely I = { 1 , 2 ,... , n}, and α(x) = αI (x)dx^1 ∧ · · · ∧ dxn. We define
(7)
Rn
α :=
Rn
αI (x)|dx^1 · · · dxn|.
The left hand side is the integral of a form that involves wedges of dxi’s. The right hand side is an ordinary Riemann integral, in which |dx^1 · · · dxn| is the usual volume measure (sometimes written dV or dnx). Note that the order of the variables in the wedge product, x^1 through xn, is implicitly using the standard orientation of Rn. Likewise, we can define the integral of α over any open subset U of Rn, as long as α restricted to U is compactly supported. We have to be a little careful with the left-hand-side of (7) when n = 0. In this case, Rn
is a single point (with positive orientation), and α is just a number. We take
α to be that
number.
Exercise 7: Suppose g is an orientation-preserving diffeomorphism from an open subset U of Rn^ to another open subset V (either or both of which may be all of Rn). Let α be a compactly supported n-form on V. Show that ∫
U
g∗α =
V
α.
Let ν denote a form on V , as represented by a form α on Uj. We then write α = ψ∗ j (ν). As with the polar-cartesian exercise above, writing a form in a particular set of coordinates is technically pulling it back to the Euclidean space where those coordinates live. Note that ψi = ψj ◦ gij , and that ψ i∗ = g∗ ij ◦ ψ∗ j , since the realization of ν in Ui is (by equation (9)) the pullback, by gij , of the realization of ν in Uj. This also tells us how to do calculus with forms on manifolds. If μ and ν are forms on X, then
- The wedge product μ ∧ ν is the form whose realization on Ui is ψ∗ i (μ) ∧ ψ∗ i (ν). In other words, ψ∗ i (μ ∧ ν) = ψ i∗ μ ∧ ψ i∗ ν.
- The exterior derivative dμ is the form whose realization on Ui is d(ψ i∗ (μ)). In other words, ψ i∗ (dμ) = d(ψ∗ i μ).
Exercise 8: Show that μ ∧ ν and dμ are well-defined.
Now suppose that we have a map f : X → Y of manifolds and that α is a form on Y. The pullback f ∗(α) is defined via coordinate patches. If φ : U ⊂ Rn^ → X and ψ : V ⊂ Rm^ → Y are parametrizations of X and Y , then there is a map h : U → V such that ψ(h(x)) = f (φ(x)). We define f ∗(α) to be the form of X whose realization in U is h∗^ ◦ (ψ∗α). In other words,
(10) φ∗(f ∗α) = h∗(ψ∗α).
An important special case is where X is a submanifold of Y and f is the inclusion map. Then f ∗^ is the restriction of α to X. When working with manifolds in RN^ , we often write down formulas for k-forms on RN^ , and then say “consider this form on X”. E.g., one might say “consider the 1-form xdy − ydx on the unit circle in R^2 ”. Strictly speaking, this really should be “consider the pullback to S^1 ⊂ R^2 by inclusion of the 1-form xdy − ydx on R^2 ,” but (almost) nobody is that pedantic!
- Integration on oriented manifolds Let X be an oriented k-manifold, and let ν be a k-form on X whose support is a compact subset of a single coordinate chart V = ψi(Ui), where Ui is an open subset of Rk. Since X is oriented, we can require that ψi be orientation-preserving. We then define
(11)
X
ν =
Ui
ψ i∗ ν.
Exercise 9: Show that this definition does not depend on the choice of coordinates. That is, if∫ ψ 1 , 2 : U 1 , 2 → V are two sets of coordinates for V , both orientation-preserving, that
U 1
ψ∗ 1 ν =
U 2
ψ∗ 2 ν. If a form is not supported in a single coordinate chart, we pick an open cover of X consisting of coordinate neighborhoods, pick a partition-of-unity subordinate to that cover,
and define (^) ∫
X
ν =
X
ρiν.
We need a little bit of notation to specify when this makes sense. If α = αI (x)dx^1 ∧ · · · ∧ dxk is a k-form on Rk, let |α| = |αI (x)|dx^1 ∧ · · · ∧ dxk. We say that ν is absolutely integrable if each |ψ∗ i (ρiν)| is integrable over Ui, and if the sum of those integrals converges. It’s not hard to show that being absolutely integrable with respect to one set of coordinates and partition of unity implies absolute integrability with respect to arbitrary coordinates and partitions
of unity. Those are the conditions under which
X
ν unambiguously makes sense. When X is compact and ν is smooth, absolute integrability is automatic. In practice, we rarely have to worry about integrability when doing differential topology. The upshot is that k-forms are meant to be integrated on k-manifolds. Sometimes these are stand-alone abstract k-manifolds, sometimes they are k-dimensional submanifolds of larger manifolds, and sometimes they are concrete k-manifolds embedded in RN^. Finally, a technical point. If X is 0-dimensional, then we can’t construct orientation-
preserving maps from R^0 to the connected components of X. Instead, we just take
X
α = ∑
x∈X
±α(x), where the sign is the orientation of the point x. This follows the general principle
that reversing the orientation of a manifold should flip the sign of integrals over that manifold.
Exercise 10: Let X = S^1 ⊂ R^2 be the unit circle, oriented as the boundary of the unit disk.
Compute
X
(xdy − ydx) by explicitly pulling this back to R with an orientation-preserving
chart and integrating over R. (Which is how you learned to do line integrals way back in calculus.) [Note: don’t worry about using multiple charts and partitions of unity. Just use a single chart for the unit circle minus a point.]
Exercise 11: Now do the same thing one dimension up. Let Y = S^2 ⊂ R^3 be the unit
sphere, oriented as the boundary of the unit ball. Compute
X
(xdy ∧dz +ydz ∧dx+zdx∧dy)
by explicitly pulling this back to a subset of R^2 with an orientation-preserving chart and integrating over that subset of R^2. As with the previous exercise, you can use a single coordinate patch that leaves out a set of measure zero, which doesn’t contribute to the integral. Strictly speaking this does not follow the rules listed above, but I’ll show you how to clean it up in class.
