Wolfram Alpha:

```One Forms
---------

Basis One-forms
---------------

A one-form is like a machine that when fed with
a vector spits out a number which depends linearly
on the input.  We can define dxμ to be the basis
one-form which gives 1 when you input the basis
tangent vector, ∂μ, and 0 when you input ∂ν for
μ ≠ ν.  Thus,

[dxμ](∂ν) = (∂/∂xν)(xμ) = ∂xμ/∂xν = δμν

This should not be interpreted as ∂[∂ν]/∂xμ but
rather as (dxμ) 'acting' on (∂ν).

The basis one-forms are the gradients of the
scalar coordinate functions (i.e. xμ = xμ(p)
where (U,p) is a local chart.  Equivalently
we can say φ: U -> Rn where φ consists of n
real valued functions, xμ).  We can show this
as follows:

∇ := (∂/∂xν)eν

∇xμ = (∂xμ/∂xν)eν

Where ∂xμ/∂xν are the components of the gradient
(a dual vector) and eν are the dual basis vectors.
Therefore,

∇xμ = δμνeν

= eμ

≡ dxμ

Transformation Laws
-------------------

vectors                     one-forms
-------                     ---------
∂μ -> ∂'μ = (∂xμ/∂x'μ)∂μ   dxμ -> dx'μ = (∂x'μ/∂xμ)dxμ

Vμ -> V'μ = (∂x'μ/∂xμ)Vμ   ωμ -> ω'μ = (∂xμ/∂x'μ)ωμ

V'x = (∂x'/∂x)Vx + (∂x'/∂y)Vy

V'y = (∂y'/∂x)Vx + (∂y'/∂y)Vy

-   -     -             -  -  -
| V'x | = | ∂x'/∂x ∂x'/∂y || Vx |
| V'y |   | ∂y'/∂x ∂y'/∂y || Vy |
-   -     -             -  -  -

Vector components have a contravariant transformation
law.

∂x' = (∂x/∂x')∂x + (∂y/∂x')∂y

∂y' = (∂x/∂y')∂x + (∂y/∂y')∂y

-       -     -    -  -             -
| ∂x' ∂y' | = | ∂x ∂y || ∂x/∂x' ∂x/∂y' |
-       -      -    - | ∂y/∂x' ∂y/∂y' |
-             -

Basis vectors have a covariant transformation law.

∂φ/∂x' = (∂x/∂x')(∂φ/∂x) + (∂y/∂x')(∂φ/∂y)

∂φ/∂y' = (∂x/∂y')(∂φ/∂x) + (∂y/∂y')(∂φ/∂y)

-             -     -           -  -             -
| ∂φ/∂x' ∂φ/∂y' | = | ∂φ/∂x ∂φ/∂y || ∂x/∂x' ∂x/∂y' |
-             -     -           - | ∂y/∂x' ∂y/∂y' |
-             -

One-form components have a covariant transformation
law.

dx' =  (∂x'/∂x)dx + (∂x'/∂y)dy

dy' =  (∂y'/∂x)dx + (∂y'/∂y)dy

-   -     -             -  -  -
| dx' | = | ∂x'/∂x ∂x'/∂y || dx |
| dy' |   | ∂y'/∂x ∂y'/∂y || dy |
-   -     -             -  -  -

Basis one-forms have a contravariant transformation
law.

One-forms
---------

One-forms are also known as covariant/dual vectors.
They are often written as ω:

df ≡ ω = ωμdxμ

Where,

ωμ = {∂f/∂x1,∂f/∂x2, ...}

and,

dxμ = {dx1,dx2, ...}

Therefore,

df = ω = (∂f/∂x1)dx1 + (∂f/∂x2)dx2 ...

= (∂f/∂xμ)dxμ

Example: Conversion one-form in polar coorinates to
one-form in xy plane.

x = rcosθ

y = rsinθ

dx = (∂x/∂r)dr + (∂x/∂θ)dθ

dy = (∂y/∂r)dr + (∂y/∂θ)dθ

dx = cosθdr - rsinθdθ

dy = sinθdr + rcosθdθ

Likewise,

dr = (∂r/∂x)dx + (∂r/∂y)dy

dθ = (∂θ/∂x)dx + (∂θ/∂y)dy

∂r/∂x = x/√(x2 + y2) = x/r = cosθ

∂r/∂y = y/√(x2 + y2) = y/r = sinθ

∂θ/∂x = -y/(x2 + y2) = -y/r2 = -sinθ/r

∂θ/∂y = x/(x2 + y2) = x/r2 = cosθ/r

dr = (x/r)dx + (y/r)dy

dθ = (-y/r2)dx + (x/r2)dy

The product of a one-form with a vector is given by:

v = v1(∂/∂x'1) + v2(∂/∂x'2) + ...

= v1∂1 + v2∂1 + ...

= Σvμ∂μ
μ

ω = ω1dx1 + ω2dx2 + ...

= Σωμdxμ
ν
vω = Σvμ∂μ(Σωνdxν)
μ     ν
= Σvμωνδνμ
μν
= Σvμωμ ∈ R
μ

Example:  Product of vector and one-form in polar
coordinates.

dx = cosθdr - rsinθdθ

dy = sinθdr + rcosθdθ

∂u/∂x = (∂u/∂r)(∂r/∂x) + (∂u/∂θ)(∂θ/∂x)

= cosθ(∂u/∂r) - (1/r)sinθ(∂u/∂θ)

∂u/∂y = (∂u/∂r)(∂r/∂y) + (∂u/∂θ)(∂θ/∂y)

= sinθ(∂u/∂r) + (1/r)cosθ(∂u/∂θ)

(cosθdr - rsinθdθ)(cosθ(∂u/∂r) - (1/r)sinθ(∂u/∂θ))

= cos2θdr(∂u/∂r) - (1/r)cosθsinθdr(∂u/∂θ)

- rsinθcosθdθ(∂u/∂r) + sin2θdθ(∂u/∂θ)

= 1

(sinθdr + rcosθdθ)(sinθ(∂u/∂r) + (1/r)cosθ(∂u/∂θ))

= sin2θdr(∂u/∂r) + (1/r)sinθcosθdr(∂u/∂θ)

- rcosθsinθdθ(∂u/∂r) + cos2θdθ(∂u/∂θ)

= 1

Exterior Derivative
-------------------

df(V) := Vf

= Vμ∂μf

= (∂f/∂xμ)Vμ

The reasoning for this can be found by considering
the directional derivative d/dλ = (dxμ/dλ)∂μ.

df[(dxμ/dλ)(∂μ)] = df[(dxμ/dλ)(∂/∂xμ)]
----------
^       = df[(d/dλ)]
|
V       = (d/dλ)[f]

= df/dλ

In other words, df when given a vector returns
the directional derivative in that direction.

It is worth noting that f is a scalar function
(0-form).  df is therefore a 1-form (dual vector).
Therefore, df acting on the tangent vector dxμ/dλ)(∂μ)
produces a scalar, df/dλ.

Example:

f(x,y) = x2y at point (1,1) in the direction (-1,1).

= 2xy + x2

= (2,1) at (1,1)

(-1,1) => (-√2/2,√2/2) after normalization

(2,1).(-√2/2,√2/2) = -√2/2
^         ^           ^
|         |           |
∂f/∂xμ   dxμ/dλ       df/dλ
= ∂μf     = Vμ

In terms of the one-form we get:

df(V) = Vμ(∂f/∂xμ)

= (∂f/∂xμ)Vμ

= 2xy(-√2/2) + x2(√2/2)

= -√2/2 at (1,1) as before.

Infinitesimal Version of the Chain Rule
---------------------------------------

df(V) = (∂f/∂xμ)Vμ

Now,

dxμ(V) = dxμ(Vν∂ν)

= Vνdxμ(∂ν)

= Vνδμν

= Vμ

Therefore,

df(V) = (∂f/∂xμ)dxμ(V)

Or,

df = (∂f/∂xμ)dxμ

In the context of one-forms, dx = (∂x/∂xμ)dxμ
is the infinitesimal version of the chain rule.
We can see the relationship as follows:

df/dxμ = (∂f/∂g(xμ))(dg(xμ)/dxμ) where f = f(g(xμ))

Therefore, rearranging:

df/dg(xμ) = (∂f/∂xμ)(dxμ/dg(xμ))

multiply by dg(xμ) to get:

df = (∂f/∂xμ)dxμ

Strictly speaking when we talk about infinitesimal
displacements we should be specifying differential
forms in the equations and not dxμ.  This is because
displacement is a scalar that can only be obtained
by contracting a one-form with a tangent vector.
For example, the line segment, ds2 = ημνdxμdxν,
should really be ds2 = ημνdxμdxν. However, most
references do not distinguish between 'dx', the
informal notion of an infinitesimal displacement,
and 'dxμ', the rigorous notion of a basis one-form
given by the gradient of a coordinate function.

Example:

(ds)2 = (dx)2 + (dy)2

= ((∂x/∂r)dr + (∂x/∂θ)dθ)2 + ((∂y/∂r)dr + (∂y/∂θ)dθ)2

= (cosθdr - rsinθdθ)2 + (sinθdr + rcosθdθ)2

= cos2θ(dr)2 + r2sin2θ(dθ)2 + sin2θ(dr)2 + r2cos2θ(dθ)2

= (dr)2 + r2(dθ)2
```