Covariant and contravariant components: Question bank

This page targets the traps in Covariant and contravariant components. Prerequisites worth a click: Dual (reciprocal) basis, Tensors and index notation, Change of basis and transformation laws, Inner product spaces.

Ground rules (read this before the traps)

Everything below lives in 2 dimensions unless a prompt says otherwise: one skew basis ${e_{1}, e_{2}}$ drawn on a flat plane. Keeping it 2D lets us see every projection.

Figure — Covariant and contravariant components

Definition The dual (reciprocal) basis — and

why it exists Alongside ${e_{1}, e_{2}}$ define a second pair ${e^{1}, e^{2}}$ by the rule $e^{i} \cdot e_{j} = δ_{j}^{i} .$ Read literally: $e^{1}$ is perpendicular to $e_{2}$ (their dot is $0$ ) and scaled so $e^{1} \cdot e_{1} = 1$ ; likewise $e^{2} ⊥ e_{1}$ . Why can we always build it? Because the basis is non-parallel, "perpendicular to $e_{2}$ " picks a definite direction and the scaling condition fixes the length — a unique solution exists. The dual basis is exactly the tool that turns the shadow numbers $v_{i}$ into honest expansion coefficients, since $v = v_{i} e^{i} because v \cdot e_{j} = (v_{i} e^{i}) \cdot e_{j} = v_{i} δ_{j}^{i} = v_{j},$ which matches the definition $v_{j} = v \cdot e_{j}$ . So $v = v^{i} e_{i} = v_{i} e^{i}$ is not a coincidence — the dual basis was designed to make the second expansion work. The figure shows ${e^{i}}$ (amber) sitting rotated from ${e_{i}}$ (white).

True or false — justify

Every prompt below is a claim. The reveal says T/F and why — the "why" is the point.

A vector

v

has different covariant and contravariant components, so it is really two different vectors.

False.

v

is one geometric arrow that ignores coordinates;

v^{i}

and

v_{i}

are two readings of the same arrow (

v = v^{i} e_{i} = v_{i} e^{i}

), like giving one point both polar and Cartesian addresses.

If the basis is orthonormal, covariant and contravariant components are identical.

True. Then

g_{ij} = e_{i} \cdot e_{j} = δ_{ij}

, so lowering

v_{i} = g_{ij} v^{j}

does nothing and

v_{i} = v^{i}

. This is why school never mentions the distinction.

The metric

g_{ij}

is always the identity matrix.

False. It is the identity only for an orthonormal basis. For a skew or stretched basis its off-diagonal entries

e_{i} \cdot e_{j}

(

i \neq = j

) are nonzero and its diagonal entries encode axis lengths.

g_{ij}

is symmetric because the dot product is commutative.

True.

g_{ij} = e_{i} \cdot e_{j} = e_{j} \cdot e_{i} = g_{j i}

, so a real inner-product metric is always symmetric. In our 2D setting that means only 3 of its 4 entries are independent (

g_{12} = g_{21}

The covariant components

v_{i}

are the contravariant components of

v

in the dual basis.

True. Since

v = v_{i} e^{i}

, the numbers

v_{i}

are literally the parallel-expansion coefficients along

{e^{i}}

. "Covariant" and "contravariant" are the same idea viewed against two different bases.

Raising and then lowering an index gets you back your original components.

True.

g_{ij} (g^{j k} v_{k}) = (g_{ij} g^{j k}) v_{k} = δ_{i}^{k} v_{k} = v_{i}

because

g^{ij}

is defined as the inverse of

g_{ij}

and

δ_{i}^{k}

acts as the index filter. The two operations are exact inverses.

g^{ij}

is a genuinely new object you must compute separately from

g_{ij}

Half-false.

g^{ij}

is fully determined by

g_{ij}

— it is its matrix inverse (

g^{ik} g_{k j} = δ_{j}^{i}

). It is not independent data; it is the same geometry read the other way (and it only exists because the basis is non-degenerate).

Spot the error

Each line states a plausible but wrong move. The reveal names the flaw and the fix.

"To get the expansion coefficients of

v

, just compute

v \cdot e_{i}

Wrong in a skew basis.

v \cdot e_{i} = v_{i}

gives the covariant (perpendicular/shadow) numbers, not the expansion coefficients

v^{i}

. To get

v^{i}

solve

v^{i} e_{i} = v

or raise:

v^{i} = g^{ij} v_{j}

"The squared length is

v^{i} v^{i}

(sum both components squared)."

Wrong. Summing two up-indices secretly assumes

g_{ij} = δ_{ij}

. The invariant is

∣ v ∣^{2} = v^{i} v_{i} = g_{ij} v^{i} v^{j}

— one up paired with one down.

"Covariant components transform inversely to the basis, because 'co' means opposite."

Wrong. Covariant transforms with the basis (matrix

A

\tilde{v}_{i} = A^{j}_{i} v_{j}

); contravariant transforms with

A^{- 1}

, against the basis. The prefix describes behaviour under change of basis, and it is the reverse of the intuitive guess.

"Since indices are just labels, I can write

v^{i} w^{i}

and sum it."

Wrong. Only an up–down contraction

v^{i} w_{i}

is coordinate-free (a true scalar).

v^{i} w^{i}

changes value when you change basis, so it is not a meaningful invariant — always keep the staircase, one up + one down.

"The gradient of a scalar is a contravariant vector, since it 'points' somewhere."

Wrong.

\partial_{i} f = \partial f / \partial x^{i}

carries a lower index — it is naturally a covector. To make it a contravariant displacement-type vector you must raise it:

(\nabla f)^{i} = g^{ij} \partial_{j} f

(see General relativity — raising and lowering indices).

"In polar coordinates the covariant and contravariant gradient are the same, like in Cartesian."

Wrong. In curvilinear coordinates the metric is not the identity (

g_{θ θ} = r^{2}

), so raising an index rescales entries. The Cartesian coincidence is special, not general.

g_{ij} v^{i} v^{j}

and

g^{ij} v_{i} v_{j}

give different lengths."

Wrong. Both equal

∣ v ∣^{2}

. Since

v_{i} = g_{ij} v^{j}

, substitution shows

g^{ij} v_{i} v_{j} = g^{ij} g_{ik} v^{k} g_{j l} v^{l} = δ_{k}^{j} g_{j l} v^{k} v^{l} = g_{ij} v^{i} v^{j}

. Same scalar (the

δ

filters

j \to k

Why questions

Why do we even need two kinds of components for one vector?

Because in a non-orthonormal basis "how much

e_{1}

is in

v

" has two honest answers — slide-along-axes (parallel) vs cast-a-shadow (perpendicular) — and they disagree. Orthonormal bases hide this because both agree.

Why is the dot product

e_{i} \cdot e_{j}

the right thing to name "the metric"?

Because when you lower an index the substitution

v_{i} = (v^{j} e_{j}) \cdot e_{i} = v^{j} (e_{j} \cdot e_{i})

makes

e_{i} \cdot e_{j}

appear unavoidably. It is exactly the data (lengths and angles) needed to convert up-indices to down-indices.

Why must an invariant scalar pair one up with one down index?

Under a basis change the up-index transforms with

A^{- 1}

and the down-index with

A

; their product cancels (

A^{- 1} A = I

, i.e.

δ

), leaving a number that all observers agree on. Two same-height indices do not cancel.

Why are the names "contra" and "co" attached to the components and not the vector?

The vector is fixed; only its numbers respond to a basis change. "Contra" describes numbers that move opposite the basis to keep

v

fixed; "co" describes numbers that move along with it.

Why does the dual basis vector

e^{1}

point perpendicular to

e_{2}

Because

e^{1} \cdot e_{2} = δ_{2}^{1} = 0

forces perpendicularity, and

e^{1} \cdot e_{1} = 1

fixes its length. This is precisely what makes

v_{i} = v \cdot e_{i}

the coefficients along

{e^{i}}

Why does the whole covariant/contravariant machinery matter physically?

In curved spacetime and curvilinear frames the metric is genuinely non-trivial, so raising/lowering is the only correct way to build coordinate-independent quantities like lengths, energies, and physical laws.

Edge cases

What happens to the covariant/contravariant distinction as the basis becomes orthonormal?

It collapses smoothly: off-diagonal

g_{ij} \to 0

and diagonal

\to 1

, so

v_{i} \to v^{i}

and

g^{ij} \to g_{ij} \to δ_{ij}

. The two readings merge into the single "school" component.

What if the two basis vectors point in nearly the same direction (almost parallel)?

The basis becomes nearly degenerate:

det g \to 0

, so

g^{ij}

blows up and contravariant components become huge and numerically unstable. Geometrically, a small change in the arrow needs enormous coefficients along near-parallel axes.

What if two basis vectors are actually parallel (linearly dependent)?

Then it is not a basis at all:

g

is singular,

g^{- 1}

does not exist, and you cannot raise indices. Some vectors have no expansion and others have infinitely many — the whole framework requires a genuine basis.

For a zero vector, what are its covariant and contravariant components?

All zero in both, in any basis:

0 = 0 \cdot g^{ij} = g_{ij} \cdot 0

. The zero vector is the one case where "which projection?" never matters.

If a basis is orthogonal but not normalized (axes perpendicular but different lengths), does

v_{i} = v^{i}

No. Then

g_{ij}

is diagonal but not the identity (

g_{ii} = ∣ e_{i} ∣^{2} \neq = 1

), so

v_{i} = g_{ii} v^{i}

rescales each component. Orthogonality kills the off-diagonal mixing but not the length rescaling.

In one dimension, is there any difference between covariant and contravariant?

Yes if the single basis vector isn't unit length:

g_{11} = ∣ e_{1} ∣^{2}

, so

v_{1} = g_{11} v^{1}

. The distinction survives even in 1D whenever the axis is stretched.

Recall Final gut-check

Say in one breath: contravariant = parallel projection = index up = transforms against the basis ( $A^{- 1}$ ); covariant = perpendicular projection = index down = transforms with the basis ( $A$ ); the metric $g_{ij}$ walks you down the staircase and $g^{ij}$ walks you up.