LECTURE 1: DIFFERENTIAL FORMS

1. 1-forms on ℝ^{n}

In calculus, you may have seen the differential or exterior derivative df of a function f(x, y, z) defined to be

df = \frac{∂f}{∂x}dx + \frac{∂f}{∂y} dy + \frac{∂f}{∂z} dz.

The expression df is called a 1-form. But what does this really mean?

Definition: A smooth 1-form ϕ on ℝ^{n} is a real-valued function on the set of all tangent vectors to ℝ^{n}, i.e.,

ϕ : T ℝ^{n} → ℝ

with the properties that

1. ϕ is linear on the tangent space T_{x}ℝ^{n} for each x ∈ ℝ^{n}.

2. For any smooth vector field v = v(x), the function ϕ(v) : ℝ^{n} → ℝ is smooth.

Given a 1-form ϕ, for each x ∈ ℝ^{n} the map

ϕ_{x} : T_{x}ℝ^{n} → ℝ

is an element of the dual space (T_{x}ℝ^{n})^{∗}. When we extend this notion to all of ℝ^{n}, we see that the space of 1-forms on ℝ^{n} is dual to the space of vector fields on ℝ^{n}.

In particular, the 1-forms dx^{1}, …, dx^{n} are defined by the property that for any vector v = (v^{1}, …, v^{n}) ∈ T_{x}ℝ^{n},

dx^{i}(v) = v^{i}.

The dx^{i}’s form a basis for the 1-forms on ℝ^{n}, so any other 1-form ϕ may be expressed in the form

ϕ = ∑_{i=1}^{n} f_{i}(x) dx^{i}.

If a vector field v on ℝ^{n} has the form

v(x) = (v^{1}(x), …, v^{n}(x)),

then at any point x ∈ ℝ^{n},

ϕ_{x}(v) = ∑_{i=1}^{n} f_{i}(x) v^{i}(x).

2. p-forms on ℝ^{n}

The 1-forms on ℝ^{n} are part of an algebra, called the algebra of differential forms on ℝ^{n}. The multiplication in this algebra is called wedge product, and it is skew-symmetric:

dx^{i} ∧ dx^{j} = −dx^{j} ∧ dx^{i}.

One consequence of this is that dx^{i} ∧ dx^{i} = 0.

If each summand of a differential form ϕ contains p dx^{i}’s, the form is called a p-form. Functions are considered to be 0-forms, and any form on ℝ^{n} of degree p > n must be zero due to the skew-symmetry.

A basis for the p-forms on ℝ^{n} is given by the set

{dx^{i_{1}} ∧ ⋯ ∧ dx^{i_{p}} : 1 ≤ i_{1} < i_{2} < ⋯ < i_{p} ≤ n}.

Any p-form ϕ may be expressed in the form

ϕ = ∑_{|I|=p} f_{I} dx^{i_{1}} ∧ ⋯ ∧ dx^{i_{p}}

where I ranges over all multi-indices I = (i_{1}, …, i_{p}) of length p.

Just as 1-forms act on vector fields to give real-valued functions, so p-forms act on p-tuples of vector fields to give real-valued functions. For instance, if ϕ, ψ are 1-forms and v, w are vector fields, then

(ϕ ∧ ψ)(v, w) = ϕ(v)ψ(w) − ϕ(w)ψ(v).

In general, if ϕ_{1}, …, ϕ_{p} are 1-forms and v_{1} …, v_{p} are vector fields, then

(ϕ_{1} ∧ ⋯ ∧ ϕ_{p})(v_{1}, …, v_{p}) = ∑_{σ∈S_{p}} sgn(σ) ϕ_{1}(v_{σ(1)}) ϕ_{2}(v_{σ(2)}) ⋯ ϕ_{n}(v_{σ(n)}).

3. The exterior derivative

The exterior derivative is an operation that takes p-forms to (p + 1)-forms. We will first define it for functions and then extend this definition to higher degree forms.

Definition: If f : ℝ^{n} → ℝ is differentiable, then the exterior derivative of f is the 1-form df with the property that for any x ∈ ℝ^{n}, v ∈ T_{x}ℝ^{n},

df_{x}(v) = v(f),

i.e., df_{x}(v) is the directional derivative of f at x in the direction of v.

LIE GROUPS AND THE METHOD OF THE MOVING FRAME 3

It is not difficult to show that

df = ∑_{i=1}^{n} \frac{∂f}{∂x^{i}} dx^{i}.

The exterior derivative also obeys the Leibniz rule

d(fg) = g df + f dg

and the chain rule

d(h(f)) = h′(f) df.

We extend this definition to p-forms as follows: Definition: Given a p-form ϕ = ∑_{|I|=p}f_{I} dx^{i_{1}} ∧ ⋯ ∧ dx^{i_{p}}, the exterior derivative dϕ is the (p + 1)-form

dϕ = ∑_{|I|=p} df_{I} ∧ dx^{i_{1}} ∧ ⋯ ∧ dx^{i_{p}}.

If ϕ is a p-form and ψ is a q-form, then the Leibniz rule takes the form

d(ϕ ∧ ψ) = dϕ ∧ ψ + (−1)^{p}ϕ ∧ dψ.

Very Important Theorem: d^{2} = 0. i.e., for any differential form ϕ,

d(dϕ) = 0.

Proof: First suppose that f is a function, i.e., a 0-form. Then

d(df) = d(∑_{i=1}^{n} \frac{∂f}{∂x^{i}} dx^{i})

= ∑_{i,j} \frac{∂^{2}f}{∂x^{i} ∂x^{j}} dx^{j} ∧ dx^{i}

= ∑_{i<j}(\frac{∂^{2}f}{∂x^{j} ∂x^{i}} − \frac{∂^{2}f}{∂x^{i} ∂x^{j}})dx^{i} ∧ dx^{j}

= 0

because mixed partials commute. Next, note that dx^{i} really does mean d(x^{i}), where x^{i} is the ith coordinate function. So by the argument above, d(dx^{i}) = 0. Now suppose that

ϕ = ∑_{|I|=p} f_{I} dx^{i_{1}} ∧ ⋯ ∧ dx^{i_{p}}.

Then by the Leibniz rule,

d(dϕ) = d( ∑_{|I|=p} df_{I} ∧ dx^{i_{1}} ∧ ⋯ ∧ dx^{i_{p}})

= ∑_{|I|=p}[d(df_{I}) ∧ dx^{i_{1}} ∧ ⋯ ∧ dx^{i_{p}} − df_{I} ∧ d(dx^{i_{1}}) ∧ ⋯ ∧ dx^{i_{p}} + …]

= 0. □

Definition: A p-form ϕ is closed if dϕ = 0. ϕ is exact if there exists a (p − 1)-form η such that ϕ = dη.

By the Very Important Theorem, every exact form is closed. The converse is only partially true: every closed form is locally exact. This means that given a closed p-form ϕ on an open set U ⊂ ℝ^{n}, any point x ∈ U has a neighborhood on which there exists a (p − 1)-form η with dη = ϕ.

4. Differential forms on manifolds

Given a smooth manifold M, a smooth 1-form ϕ on M is a real-valued function on the set of all tangent vectors to M such that

1. ϕ is linear on the tangent space T_{x}M for each x ∈ M.

2. For any smooth vector field v on M, the function ϕ(v) : M → ℝ is smooth.

So for each x ∈ M, the map

ϕ_{x} : T_{x}M → ℝ

is an element of the dual space (T_{x}M)^{∗}.

Wedge products and exterior derivatives are defined similarly as for ℝ^{n}. If f : M → ℝ is a differentiable function, then we define the exterior derivative of f to be the 1-form df with the property that for any x ∈ M, v ∈ T_{x}M,

df_{x}(v) = v(f).

A local basis for the space of 1-forms on M can be described as before in terms of any local coordinate chart (x^{1}, …, x^{n}) on M, and it is possible to show that the coordinate-based notions of wedge product and exterior derivative are in fact independent of the choice of local coordinates and so are well-defined.

More generally, suppose that M_{1}, M_{2} are smooth manifolds and that F : M_{1} → M_{2} is a differentiable map. For any x ∈ M_{1}, the differential dF (also denoted F_{∗}) : T_{x}M_{1} → T_{F (x)}M_{2} may be thought of as a vector-valued 1-form, because it is a linear map from T_{x}M_{1} to the vector space T_{F (x)}M_{2}. There is an analogous map in the opposite direction for differential forms, called the pullback and denoted F^{∗}. It is defined as follows.

Definition: If F : M_{1} → M_{2} is a differentiable map, then

1. If f : M_{2} → ℝ is a differentiable function, then F^{∗}f : M_{1} → ℝ is the function

(F^{∗}f)(x) = (f ∘ F)(x).

2. If ϕ is a p-form on M_{2}, then F^{∗}ϕ is the p-form on M_{1} defined as follows: if v_{1}, …, v_{p} ∈ T_{x}M_{1}, then

(F^{∗}ϕ)(v_{1}, …, v_{p}) = ϕ(F_{∗}(v_{1}), …, F_{∗}(v_{p})).

In terms of local coordinates (x^{1}, …, x^{n}) on M_{1} and (y^{1}, …, y^{m}) on M_{2}, suppose that the map F is described by

y^{i} = y^{i}(x^{1}, …, x^{n}), 1 ≤ i ≤ m.

Then the differential dF at each point x ∈ M_{1} may be represented in this coordinate system by the matrix

[\frac{∂y^{i}}{∂x^{j}} ] .

The dx^{j}’s are forms on M_{1}, the dy^{i}’s are forms on M_{2}, and the pullback map F^{∗} acts on the dy^{i}’s by

F^{∗}(dy^{i}) = ∑_{j=1}^{n} \frac{∂y^{i}}{∂x^{j}} dx^{j}.

The pullback map behaves as nicely as one could hope with respect to the various operations on differential forms, as described in the following theorem.

Theorem: Let F : M_{1} → M_{2} be a differentiable map, and let ϕ, η be differential forms on M_{2}. Then

1. F^{∗}(ϕ + η) = F^{∗}ϕ + F^{∗}η.

2. F^{∗}(ϕ ∧ η) = F^{∗}ϕ ∧ F^{∗}η.

3. F^{∗}(dϕ) = d(F^{∗}ϕ).

5. The Lie derivative

The final operation that we will define on differential forms is the Lie derivative. This is a generalization of the notion of directional derivative of a function.

Suppose that v(x) is a vector field on a manifold M, and let φ : M×(−ε, ε) → M be the flow of v. This is the unique map that satisfies the conditions

\frac{∂φ}{∂t}(x, t) = v(φ(x, t))

φ(x, 0) = x.

In other words, φ_{t}(x) = φ(x, t) is the point reached at time t by flowing along the vector field v(x) starting from the point x at time 0.

Recall that if f : M → ℝ is a smooth function, then the directional derivative of f at x in the direction of v is

v(f) = lim_{t→0} \frac{f(φ_{t}(x)) − f(x)}{t}

= lim_{t→0} \frac{(φ_{t}^{∗}(f) − f)(x)}{t}.

Similarly, given a differential form ϕ we define the Lie derivative of ϕ along the vector field v(x) to be

L_{v}ϕ = lim_{t→0} \frac{φ_{t}^{∗}ϕ − ϕ}{t}.

Fortunately there is a practical way to compute the Lie derivative. First we need the notion of the left-hook of a differential form with a vector field. Given a p-form ϕ and a vector field v, the left-hook v ϕ of ϕ with v (also called the interior product of ϕ with v) is the (p − 1)-form defined by the property that for any w_{1}, …, w_{p−1} ∈ T_{x}ℝ^{n},

(v ϕ)(w_{1}, …, w_{p−1}) = ϕ(v, w_{1}, …, w_{p−1}).

For instance,

\frac{∂}{∂x} (dx ∧ dy + dz ∧ dx) = dy − dz.

Now according to Cartan’s formula, the Lie derivative of ϕ along the vector field v is

L_{v}ϕ = v dϕ + d(v ϕ).

Exercises

1. Classical vector analysis avoids the use of differential forms on ℝ^{3} by converting 1-forms and 2-forms into vector fields by means of the following one-to-one correspondences. (ε_{1}, ε_{2}, ε_{3} will denote the standard basis ε_{1} = [1, 0, 0], ε_{2} = [0, 1, 0], ε_{3} = [0, 0, 1].)

f_{1} dx^{1} + f_{2} dx^{2} + f_{3} dx^{3} ⟷ f_{1} ε_{1} + f_{2} ε_{2} + f_{3} ε_{3}

f_{1} dx^{2} ∧ dx^{3} + f_{2} dx^{3} ∧ dx^{1} + f_{3} dx^{1} ∧ dx^{2} ⟷ f_{1} ε_{1} + f_{2} ε_{2} + f_{3} ε_{3}

Vector analysis uses three basic operations based on partial differentiation:

1. Gradient of a function f:

grad(f) = ∑_{i=1}^{3} \frac{∂f}{∂x^{i}} ε_{i}

2. Curl of a vector field v = ∑_{i=1}^{3}v^{i}(x) ε_{i}:

curl(v) = (\frac{∂v^{3}}{∂x^{2}} − \frac{∂v^{2}}{∂x^{3}} ) ε_{1} + (\frac{∂v^{1}}{∂x^{3}} − \frac{∂v^{3}}{∂x^{1}} ) ε_{2} + (\frac{∂v^{2}}{∂x^{1}} − \frac{∂v^{1}}{∂x^{2}} ) ε_{3}

3. Divergence of a vector field v = ∑_{i=1}^{3}v^{i}(x) ε_{i}:

div(v) = ∑_{i=1}^{3} \frac{∂v^{i}}{∂x^{i}}

Prove that all three operations may be expressed in terms of exterior derivatives as follows:

1. df ↔ grad(f)

2. If ϕ is a 1-form and ϕ ↔ v, then dϕ ↔ curl(v).

3. If η is a 2-form and η ↔ v, then dη ↔ div(v) dx^{1} ∧ dx^{2} ∧ dx^{3}.

Show that the identities

curl(grad(f)) = 0 div(curl(v)) = 0

follow from the fact that d^{2} = 0.

2. Let f and g be real-valued functions on ℝ^{2}. Prove that

df ∧ dg = |\begin{matrix}\frac{∂f}{∂x} & \frac{∂f}{∂y} \\ \frac{∂g}{∂x} & \frac{∂g}{∂y} \end{matrix} | dx ∧ dy.

(You may recognize this from the change-of-variables formula for double integrals.)

3. Suppose that ϕ, ψ are 1-forms on ℝ^{n}. Prove the Leibniz rule

d(ϕ ∧ ψ) = dϕ ∧ ψ − ϕ ∧ dψ.

4. Prove the statement above that if F : M_{1} → M_{2} is described in terms of local coordinates by

y^{i} = y^{i}(x^{1}, …, x^{n}), 1 ≤ i ≤ m

then

F^{∗}(dy^{i}) = ∑_{j=1}^{n} \frac{∂y^{i}}{∂x^{j}} dx^{j}.

5. Let (r, θ) be coordinates on ℝ^{2} and (x, y, z) coordinates on ℝ^{3}. Let F : ℝ^{2} → ℝ^{3} be defined by

F (r, θ) = (cos θ, sin θ, r).

Describe the differential dF in terms of these coordinates and compute the pullbacks F^{∗}(dx), F^{∗}(dy), F^{∗}(dz).