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One Forms

Basis Oneforms

A oneform is like a machine that when fed with
a vector spits out a number which depends linearly
on the input. We can define dx^{μ} to be the basis
oneform which gives 1 when you input the basis
tangent vector, ∂_{μ}, and 0 when you input ∂_{ν} for
μ ≠ ν. Thus,
[dx^{μ}](∂_{ν}) = (∂/∂x^{ν})(x^{μ}) = ∂x^{μ}/∂x^{ν} = δ^{μ}_{ν}
This should not be interpreted as ∂[∂_{ν}]/∂x^{μ} but
rather as (dx^{μ}) 'acting' on (∂_{ν}).
The basis oneforms are the gradients of the
scalar coordinate functions (i.e. x^{μ} = x^{μ}(p)
where (U,p) is a local chart. Equivalently
we can say φ: U > R^{n} where φ consists of n
real valued functions, x^{μ}). We can show this
as follows:
∇ := (∂/∂x^{ν})e^{ν}
∇x^{μ} = (∂x^{μ}/∂x^{ν})e^{ν}
Where ∂x^{μ}/∂x^{ν} are the components of the gradient
(a dual vector) and e^{ν} are the dual basis vectors.
Therefore,
∇x^{μ} = δ^{μ}_{ν}e^{ν}
= e^{μ}
≡ dx^{μ}
Transformation Laws

vectors oneforms
 
∂_{μ} > ∂'_{μ} = (∂x^{μ}/∂x'^{μ})∂_{μ} dx^{μ} > dx'^{μ} = (∂x'^{μ}/∂x^{μ})dx^{μ}
V^{μ} > V'^{μ} = (∂x'^{μ}/∂x^{μ})V^{μ} ω_{μ} > ω'_{μ} = (∂x^{μ}/∂x'^{μ})ω_{μ}
V'^{x} = (∂x'/∂x)V^{x} + (∂x'/∂y)V^{y}
V'^{y} = (∂y'/∂x)V^{x} + (∂y'/∂y)V^{y}
     
 V'^{x}  =  ∂x'/∂x ∂x'/∂y  V^{x} 
 V'^{y}   ∂y'/∂x ∂y'/∂y  V^{y} 
     
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' 
 
Oneform 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 oneforms have a contravariant transformation
law.
Oneforms

Oneforms are also known as covariant/dual vectors.
They are often written as ω:
df ≡ ω = ω_{μ}dx^{μ}
Where,
ω_{μ} = {∂f/∂x^{1},∂f/∂x^{2}, ...}
and,
dx^{μ} = {dx^{1},dx^{2}, ...}
Therefore,
df = ω = (∂f/∂x^{1})dx^{1} + (∂f/∂x^{2})dx^{2} ...
= (∂f/∂x^{μ})dx^{μ}
Example: Conversion oneform in polar coorinates to
oneform 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/√(x^{2} + y^{2}) = x/r = cosθ
∂r/∂y = y/√(x^{2} + y^{2}) = y/r = sinθ
∂θ/∂x = y/(x^{2} + y^{2}) = y/r^{2} = sinθ/r
∂θ/∂y = x/(x^{2} + y^{2}) = x/r^{2} = cosθ/r
dr = (x/r)dx + (y/r)dy
dθ = (y/r^{2})dx + (x/r^{2})dy
The product of a oneform with a vector is given by:
v = v^{1}(∂/∂x'^{1}) + v^{2}(∂/∂x'^{2}) + ...
= v^{1}∂_{1} + v^{2}∂_{1} + ...
= Σv^{μ}∂_{μ}
^{μ}
ω = ω_{1}dx^{1} + ω_{2}dx^{2} + ...
= Σω_{μ}dx^{μ}
^{ν}
vω = Σv^{μ}∂_{μ}(Σω_{ν}dx^{ν})
^{μ} ^{ν}
= Σv^{μ}ω_{ν}δ^{ν}_{μ}
^{μν}
= Σv^{μ}ω_{μ} ∈ R
^{μ}
Example: Product of vector and oneform 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))
= cos^{2}θdr(∂u/∂r)  (1/r)cosθsinθdr(∂u/∂θ)
 rsinθcosθdθ(∂u/∂r) + sin^{2}θdθ(∂u/∂θ)
= 1
(sinθdr + rcosθdθ)(sinθ(∂u/∂r) + (1/r)cosθ(∂u/∂θ))
= sin^{2}θdr(∂u/∂r) + (1/r)sinθcosθdr(∂u/∂θ)
 rcosθsinθdθ(∂u/∂r) + cos^{2}θdθ(∂u/∂θ)
= 1
Exterior Derivative

df(V) := Vf
= V^{μ}∂_{μ}f
= (∂f/∂x^{μ})V^{μ}
The reasoning for this can be found by considering
the directional derivative d/dλ = (dx^{μ}/dλ)∂_{μ}.
df[(dx^{μ}/dλ)(∂_{μ})] = df[(dx^{μ}/dλ)(∂/∂x^{μ})]

^ = 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
(0form). df is therefore a 1form (dual vector).
Therefore, df acting on the tangent vector dx^{μ}/dλ)(∂_{μ})
produces a scalar, df/dλ.
Example:
f(x,y) = x^{2}y at point (1,1) in the direction (1,1).
= 2xy + x^{2}
= (2,1) at (1,1)
(1,1) => (√2/2,√2/2) after normalization
(2,1).(√2/2,√2/2) = √2/2
^ ^ ^
  
∂f/∂x^{μ} dx^{μ}/dλ df/dλ
= ∂_{μ}f = V^{μ}
In terms of the oneform we get:
df(V) = V^{μ}(∂f/∂x^{μ})
= (∂f/∂x^{μ})V^{μ}
= 2xy(√2/2) + x^{2}(√2/2)
= √2/2 at (1,1) as before.
Infinitesimal Version of the Chain Rule

From before we had:
df(V) = (∂f/∂x^{μ})V^{μ}
Now,
dx^{μ}(V) = dx^{μ}(V^{ν}∂_{ν})
= V^{ν}dx^{μ}(∂_{ν})
= V^{ν}δ^{μ}_{ν}
= V^{μ}
Therefore,
df(V) = (∂f/∂x^{μ})dx^{μ}(V)
Or,
df = (∂f/∂x^{μ})dx^{μ}
In the context of oneforms, dx = (∂x/∂x^{μ})dx^{μ}
is the infinitesimal version of the chain rule.
We can see the relationship as follows:
df/dx^{μ} = (∂f/∂g(x^{μ}))(dg(x^{μ})/dx^{μ}) where f = f(g(x^{μ}))
Therefore, rearranging:
df/dg(x^{μ}) = (∂f/∂x^{μ})(dx^{μ}/dg(x^{μ}))
multiply by dg(x^{μ}) to get:
df = (∂f/∂x^{μ})dx^{μ}
Strictly speaking when we talk about infinitesimal
displacements we should be specifying differential
forms in the equations and not dx^{μ}. This is because
displacement is a scalar that can only be obtained
by contracting a oneform with a tangent vector.
For example, the line segment, ds^{2} = η_{μν}dx^{μ}dx^{ν},
should really be ds^{2} = η_{μν}dx^{μ}dx^{ν}. However, most
references do not distinguish between 'dx', the
informal notion of an infinitesimal displacement,
and 'dx^{μ}', the rigorous notion of a basis oneform
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}
= cos^{2}θ(dr)^{2} + r^{2}sin^{2}θ(dθ)^{2} + sin^{2}θ(dr)^{2} + r^{2}cos^{2}θ(dθ)^{2}
= (dr)^{2} + r^{2}(dθ)^{2}