Objectives:
• Contravariant and covariant vectors, one-forms.
Reading: Schutz chapter 3; Hobson chapter 3
We have had
If
~ and
~
·
~
= η
αβ
V α V β .
~ are four-vectors then
~ with components
V α = A α + B α , is also a four-vector. Therefore
~
·
~
= η
αβ
( A
α
+ B
α
) ¡ A
β
+ B
β ¢ ,
= η
αβ
A
α
A
β
=
~
·
~
+
+
~
·
η
~
αβ
+
A
~
α
B
β
·
+
~
+
η
~
αβ
·
B
α
A
α
+ η
αβ
B
α
B
β
,
Since η
αβ is symmetric then
~
·
~
=
~
·
~
, so
~
·
~
=
~
·
~
+ 2
~
·
~
+
~
·
Since
~
·
~
,
~
·
~ and
~
·
~ are all scalars, then
~
·
~
= η
αβ
A α B β (4.1) is also a scalar, i.e. invariant between all inertial frames. This defines the scalar product of two vectors.
~
·
~
= 0 = ⇒
~ and
~ orthogonal. Null vectors are self-orthogonal.
14

LECTURE 4.
VECTORS
With the following basis vectors (4D versions of ~i , ~j ,
~k
):
~e
0
= (1 , 0 , 0 , 0) ,
~e
1
= (0 , 1 , 0 , 0) ,
~e
~e
2
3
= (0 , 0 , 1 , 0) ,
= (0 , 0 , 0 , 1) ,
15 we can write for frames S and S ′ :
~
= A α ~e
α
= A α ′
~e
α ′
.
These express the frame-independent nature of any fourvector, just as we write a to represent a three-vector.
Substituting
A
α
= Λ
α
β ′
A
β ′
, then
Λ α
β ′
A β ′
~e
α
= A α ′
~e
α ′
, and re-labelling dummy indices, β ′ → α ′ , α → β ,
¡ ~e
α ′
− Λ
β
α ′
~e
β
¢ A
α ′
= 0 .
Since
~ is arbitrary, the term in brackets must vanish, i.e.
~e
α ′
= Λ
β
α ′
~e
β
.
(4.2)
Comparing with
A α ′
= Λ α ′
β
A β , we see that the components transform “oppositely” to the basis vectors, hence these are often called “contravariant vectors” and superscripted indices are called “contravariant indices”.
Note that indices are lowered on basis vectors to fit raised indices on components.
Consider the gradient ∇ φ = ( ∂φ/∂x 0 , ∂φ/∂x 1 , ∂φ/∂x 2 , ∂φ/∂x 3 ), where φ is a scalar function of the coordinates. Is it a vector?
The chain rule gives dφ =
∂φ
∂x β dx
β
,

LECTURE 4.
VECTORS 16 and on differentiating wrt x α ′
∂φ
∂x α ′
=
∂φ
∂x β
∂x β
∂x α ′
.
But x β = Λ β
γ ′ x γ ′ so
∂x β
∂x α ′
= Λ
β
γ ′
δ
γ ′
α ′
= Λ
β
α ′
.
Therefore
∂φ
∂x α ′
= Λ
β
α ′
∂φ
∂x β
(4.3)
Thus the components of the gradient ∇ φ do not transform like the components of four-vectors, instead they transform like basis vectors.
Quantities like ∇ φ are called “covariant vectors” or “covectors” or “one-forms”, the latter emphasizing their difference from vectors.
I will write one-forms with tildes such as ˜ . Like vectors, one-forms can be defined by their transformation, i.e. if quantities p
α transform as p
α ′
= Λ β
α ′ p
β
.
then they are components of a one-form ˜ .
(4.4)
One-forms are written with subscripted indices, also known as “covariant” indices. Do not confuse with “Lorentz covariance”.
Given a one-form ˜ and a vector
~
, consider the quantity: p
α
A α .
Because of the “contra” and “co” transformations, this is a scalar. In a more frame-independent way we can write this as ˜ (
~
). Thus a one-form is a “machine” that produces a scalar from a vector. Equally, a vector is a machine that produces a scalar from a one-form, A (˜ ).
p
α
A α is one number. Why?
One-forms are best thought of as a series of parallel surfaces. The number of such surfaces crossed by a vector is the scalar. One-forms cannot be thought as “arrows” because they do not transform in the same way as vectors. Oneforms do not crop up in orthonormal bases (e.g. Cartesian coordinates or unit vectors in polar coordinates ˆ , θ
ˆ
) because in that one case they transform identically to vectors. They cannot be avoided in GR.

LECTURE 4.
VECTORS
17
A set of basis vectors ~e
β
α :
α
( ~e
β
) = δ
α
β
, (4.5) because then
˜ (
~
) = [ p
α
α ] ¡ A β ~e
β
¢ ,
= p
α
A β ω α ( ~e
β
) ,
= p
α
A β δ α
β
,
= p
α
A α , as required.
One can then show that basis one-forms transform like vector components, i.e.
α ′
= Λ α ′
β
ω β .
(4.6)
Given a vector are:
~
= A α ~e
α and one-form ˜ = p
α
ω α the four transformations
A
α ′
α ′ p
~e
α ′
α ′
= Λ
α ′
β
A
β
,
= Λ α ′
β
β ,
= Λ β
α ′ p
β
,
= Λ β
α
~e
β
.
As long as you remember that vector components have superscripted indices and one-form components have subscripted indices, and balance free and dummy indices properly, it should be straightforward to remember these relations.