Covariant and contravariant components: Worked examples

This page is the drill hall for the parent topic. We take the two rules — parallel projection (contravariant, index up) and perpendicular projection (covariant, index down) — and run them through every kind of basis geometry a problem can throw at you. If you never met these symbols before, read the parent first; here we only compute.

The scenario matrix

Every problem about these components is really one of the cells below. We will hit each cell with a labelled example.

Cell	Basis geometry	What is stressed	Example
A	Orthonormal (square)	The distinction collapses, $v_{i} = v^{i}$	Ex 1
B	Orthogonal but unequal lengths (rectangular)	$g_{ij}$ diagonal but $\neq = δ_{ij}$ ; scaling only	Ex 2
C	Skew, acute angle between axes	$g_{12} > 0$ ; full split	Ex 3
D	Skew, obtuse angle between axes	$g_{12} < 0$ ; sign flips in components	Ex 4
E	Dual (reciprocal) basis built explicitly	See where $v_{i}$ "lives"	Ex 5
F	Degenerate / limiting (axes → parallel)	$g$ becomes singular, $g^{- 1}$ blows up	Ex 6
G	Curvilinear (polar) — position-dependent basis	gradient is covariant	Ex 7
H	Word problem + exam twist (invariant length, wrong-pairing trap)	why up-with-down is forced	Ex 8

Prerequisites you may want open: Dual (reciprocal) basis, Inner product spaces, Change of basis and transformation laws, Curvilinear coordinates (polar, spherical).

Cell A — orthonormal basis (the sanity anchor)

Cell B — orthogonal but unequal lengths

Worked example Example 2 — rectangular paper (pure scaling)

Basis $e_{1} = (2, 0)$ , $e_{2} = (0, 3)$ — perpendicular, but long. Vector $v = (4, 3)$ .

Forecast: the axes are still perpendicular, so will $v_{i} = v^{i}$ ? (Careful — "perpendicular" is not "orthonormal".)

Step 1 — metric. $g_{11} = 4$ , $g_{12} = 0$ , $g_{22} = 9$ . So $g = (4009)$ , diagonal but not identity. Why this step? Off-diagonals are zero (perpendicular) yet diagonals $\neq = 1$ (long axes) — this is exactly the "scaling only" cell.

Step 2 — contravariant. $v^{1} (2) = 4 \Rightarrow v^{1} = 2$ ; $v^{2} (3) = 3 \Rightarrow v^{2} = 1$ . Why this step? Walking $2$ steps of $e_{1}$ and $1$ step of $e_{2}$ reaches $(4, 3)$ .

Step 3 — covariant. $v_{1} = v \cdot e_{1} = (4) (2) + (3) (0) = 8$ ; $v_{2} = (4) (0) + (3) (3) = 9$ . Why this step? Dotting picks up the axis length — that is where the factor $g_{ii}$ hides.

Verify: $v_{1} = g_{11} v^{1} = 4 \cdot 2 = 8$ ✓, $v_{2} = g_{22} v^{2} = 9 \cdot 1 = 9$ ✓. So even with perpendicular axes the two component sets differ ( $v^{i} \neq = v_{i}$ ) — the myth "perpendicular is enough" is busted.

Cell C — skew basis, acute angle

Figure — Covariant and contravariant components

Worked example Example 3 — the workhorse skew case

Basis $e_{1} = (1, 0)$ , $e_{2} = (1, 1)$ (angle between them is $45°$ , acute). Vector $v = (2, 1)$ . (This mirrors the parent's Example 1 — do it fully so you own it.)

Forecast: since $g_{12} = e_{1} \cdot e_{2} = 1 > 0$ , both component sets will differ and the covariant $v_{2}$ will exceed $v^{2}$ . Guess by how much.

Step 1 — metric. $g_{11} = 1$ , $g_{12} = 1$ , $g_{22} = 2$ : $g = (1112)$ , $det g = 1$ , so $g^{- 1} = (2 - 1 - 1 1)$ . Why this step? The $g_{12} = 1$ off-diagonal is the fingerprint of an acute skew — geometry first.

Step 2 — contravariant (parallel). Solve $v^{1} + v^{2} = 2, v^{2} = 1 \Rightarrow v^{1} = 1, v^{2} = 1$ . Look at the figure: the red dashed lines slide parallel to the axes to reach the tip. Why this step? Parallelogram rule = walk-along coefficients.

Step 3 — covariant (perpendicular). $v_{1} = v \cdot e_{1} = 2$ ; $v_{2} = v \cdot e_{2} = 2 + 1 = 3$ . In the figure these are the green perpendicular drops. Why this step? Definition $v_{i} = v \cdot e_{i}$ ; the drop onto the long, slanted $e_{2}$ is why $v_{2} = 3 > v^{2} = 1$ .

Verify: $v_{i} = g_{ij} v^{j}$ : $v_{1} = 1 \cdot 1 + 1 \cdot 1 = 2$ ✓, $v_{2} = 1 \cdot 1 + 2 \cdot 1 = 3$ ✓. Reverse: $v^{i} = g^{ij} v_{j}$ : $v^{1} = 2 \cdot 2 - 1 \cdot 3 = 1$ ✓, $v^{2} = - 2 + 3 = 1$ ✓.

Cell D — skew basis, obtuse angle (sign flip)

Worked example Example 4 — obtuse axes make

$g_{12} < 0$ Basis $e_{1} = (1, 0)$ , $e_{2} = (- 1, 1)$ (angle $135°$ , obtuse). Vector $v = (1, 2)$ .

Forecast: the axes now lean away from each other, so $g_{12} < 0$ . Predict the sign of the difference between $v_{1}$ and $v^{1}$ .

Step 1 — metric. $g_{11} = 1$ , $g_{12} = e_{1} \cdot e_{2} = (1) (- 1) + (0) (1) = - 1$ , $g_{22} = (- 1)^{2} + 1^{2} = 2$ . So $g = (1 - 1 - 1 2)$ , $det g = 1$ , $g^{- 1} = (2111)$ . Why this step? A negative $g_{12}$ is the fingerprint of an obtuse skew — this is a genuinely new cell.

Step 2 — contravariant. Solve $v^{1} - v^{2} = 1, v^{2} = 2 \Rightarrow v^{1} = 3, v^{2} = 2$ . Check: $3 (1, 0) + 2 (- 1, 1) = (1, 2)$ ✓. Why this step? Walk $3$ along $e_{1}$ , then $2$ along the backward-leaning $e_{2}$ — the figure shows the tip landing exactly on $v$ .

Step 3 — covariant. $v_{1} = v \cdot e_{1} = (1) (1) + (2) (0) = 1$ ; $v_{2} = v \cdot e_{2} = (1) (- 1) + (2) (1) = 1$ . Why this step? Perpendicular drop onto a backward-pointing axis can shrink the number — here $v_{1} = 1 < v^{1} = 3$ , the opposite sign of difference from Cell C.

Verify: $v_{i} = g_{ij} v^{j}$ : $v_{1} = 1 \cdot 3 + (- 1) \cdot 2 = 1$ ✓, $v_{2} = - 1 \cdot 3 + 2 \cdot 2 = 1$ ✓. So obtuse geometry ( $g_{12} < 0$ ) flips the relationship: covariant now undershoots contravariant.

Cell E — build the dual (reciprocal) basis explicitly

Worked example Example 5 — where the covariant numbers "live"

Reuse the acute basis of Ex 3: $e_{1} = (1, 0)$ , $e_{2} = (1, 1)$ . Build the dual basis $e^{1}, e^{2}$ and confirm $v = v_{i} e^{i}$ with $v_{1} = 2, v_{2} = 3$ .

Forecast: $e^{1}$ must be perpendicular to $e_{2}$ . Since $e_{2} = (1, 1)$ points up-right, guess the direction of $e^{1}$ .

Step 1 — dual defining rule. $e^{i} \cdot e_{j} = δ_{j}^{i}$ . In 2D the dual is the rows of $g^{- 1}$ times the old basis, or directly: $e^{1} ⊥ e_{2}$ , $e^{1} \cdot e_{1} = 1$ . Why this step? That perpendicularity is the whole reason covariant components exist — they are contravariant components in this basis.

Step 2 — solve for $e^{1} = (a, b)$ . Perp to $e_{2} = (1, 1)$ : $a + b = 0$ . And $e^{1} \cdot e_{1} = a = 1$ . So $a = 1, b = - 1$ : $e^{1} = (1, - 1)$ . Why this step? Two scalar conditions fix the 2 unknowns exactly.

Step 3 — solve for $e^{2} = (c, d)$ . Perp to $e_{1} = (1, 0)$ : $c = 0$ . And $e^{2} \cdot e_{2} = c + d = 1 \Rightarrow d = 1$ . So $e^{2} = (0, 1)$ . Why this step? Same recipe, second axis.

Step 4 — reassemble $v$ . $v_{1} e^{1} + v_{2} e^{2} = 2 (1, - 1) + 3 (0, 1) = (2, - 2 + 3) = (2, 1) = v$ ✓. Why this step? This is the punchline: the covariant numbers $(2, 3)$ are the walk-along coefficients in the dual grid, reproducing the same arrow.

Verify: $e^{1} \cdot e_{1} = 1$ , $e^{1} \cdot e_{2} = (1) (1) + (- 1) (1) = 0$ , $e^{2} \cdot e_{1} = 0$ , $e^{2} \cdot e_{2} = 1$ — all $δ_{j}^{i}$ ✓. And $2 (1, - 1) + 3 (0, 1) = (2, 1)$ ✓.

Cell F — degenerate / limiting (axes collapse together)

Worked example Example 6 — what breaks when axes go parallel

Family $e_{1} = (1, 0)$ , $e_{2} = (1, ε)$ with $ε \to 0^{+}$ (the second axis tips down toward the first). Watch the metric.

Forecast: as $ε \to 0$ the two axes merge. Do you expect $g$ to stay invertible? What happens to $g^{- 1}$ ?

Step 1 — metric as a function of $ε$ . $g_{11} = 1$ , $g_{12} = 1$ , $g_{22} = 1 + ε^{2}$ . So $det g = (1) (1 + ε^{2}) - 1^{2} = ε^{2}$ . Why this step? The determinant is the "volume" of the parallelogram spanned by the axes; it should vanish exactly when they align.

Step 2 — inverse blows up. $g^{- 1} = \frac{1}{ε ^{2}} (1 + ε^{2} - 1 - 1 1)$ . As $ε \to 0$ every entry $\to \infty$ . Why this step? Raising indices needs $g^{- 1}$ ; if the basis degenerates, raising becomes impossible — mathematically encoded by $det g \to 0$ .

Step 3 — geometric meaning. With $v = (0, 1)$ : contravariant needs $v^{1} (1, 0) + v^{2} (1, ε) = (0, 1)$ , giving $v^{2} = 1/ ε \to \infty$ , $v^{1} = - 1/ ε \to - \infty$ . Why this step? To reach a tip off the near-parallel axes you must walk enormous, opposite distances — the components explode even though $v$ is unit length.

Verify: $det g = ε^{2}$ ; at $ε = 0.1$ , $det g = 0.01$ , and $v^{2} = 1/0.1 = 10$ , $v^{1} = - 10$ . Sanity: $- 10 (1, 0) + 10 (1, 0.1) = (0, 1)$ ✓. Rule learned: a basis is usable iff $det g \neq = 0$ ; degeneracy = the metric goes singular.

Cell G — curvilinear (polar): the gradient is covariant

Worked example Example 7 — polar coordinates, position-dependent metric

Polar coordinates $x^{1} = r$ , $x^{2} = θ$ with line element $d s^{2} = d r^{2} + r^{2} d θ^{2}$ , so the metric is $g_{ij} = (10 0 r^{2})$ . Take $f (r, θ) = r^{2}$ . Find the covariant gradient $\partial_{i} f$ and the contravariant (true vector) gradient $(\nabla f)^{i}$ .

Forecast: guess whether the $θ$ -component of the raised gradient carries a factor of $r^{2}$ or $1/ r^{2}$ .

Step 1 — read off the metric. $g_{11} = 1$ , $g_{22} = r^{2}$ (angles at radius $r$ span arc length $r d θ$ ), $g_{12} = 0$ . Inverse: $g^{11} = 1$ , $g^{22} = 1/ r^{2}$ . Why this step? Here $g$ depends on position — this is the essential new feature versus Cells A–F.

Step 2 — covariant gradient. $\partial_{r} f = 2 r$ , $\partial_{θ} f = 0$ . So $\partial_{i} f = (2 r, 0)$ . Why this step? $\partial_{i} f = \partial f / \partial x^{i}$ naturally carries a lower index — differentiation w.r.t. a contravariant coordinate produces a covector (see Tensors and index notation).

Step 3 — raise to get the true gradient vector. $(\nabla f)^{i} = g^{ij} \partial_{j} f$ : $(\nabla f)^{1} = g^{11} \partial_{r} f = 2 r$ ; $(\nabla f)^{2} = g^{22} \partial_{θ} f = (1/ r^{2}) (0) = 0$ . Why this step? Only after raising with $g^{ij}$ does the gradient point the way steepest ascent physically points (General relativity — raising and lowering indices).

Verify: For $f = r^{2}$ , physically $∣\nabla f ∣ = ∣ df / d r ∣ = 2 r$ (radial). Compute $∣\nabla f ∣^{2} = g_{ij} (\nabla f)^{i} (\nabla f)^{j} = 1 \cdot (2 r)^{2} + r^{2} \cdot 0 = 4 r^{2}$ , so magnitude $= 2 r$ ✓. In Cartesian $f = x^{2} + y^{2}$ , $\nabla f = (2 x, 2 y)$ , $∣\nabla f ∣ = 2 x^{2} + y^{2} = 2 r$ ✓ — same invariant.

Cell H — word problem + exam twist (why up pairs with down)

Worked example Example 8 — the "wrong pairing" trap

Word problem. A surveyor lays out a skew grid $e_{1} = (1, 0)$ , $e_{2} = (1, 1)$ . A pipeline runs as $v = (2, 1)$ . She needs the pipeline's true length-squared (an inner-product invariant). A rushing student computes $\sum_{i} (v^{i})^{2}$ . Show why that is wrong and give the right number.

Forecast: naïve $\sum (v^{i})^{2} = 1^{2} + 1^{2} = 2$ . But the honest length of $(2, 1)$ is $5$ . Which is right?

Step 1 — the invariant rule. A coordinate-free scalar must contract one up with one down: $∣ v ∣^{2} = v^{i} v_{i} = g_{ij} v^{i} v^{j}$ . Why this step? Only up-with-down survives a change of basis ( $A$ and $A^{- 1}$ cancel). Two-ups-summed depends on the (wrong) assumption $g = δ$ .

Step 2 — correct contraction. From Ex 3: $v^{i} = (1, 1)$ , $v_{i} = (2, 3)$ . Then $v^{i} v_{i} = (1) (2) + (1) (3) = 5$ . Why this step? This is the pairing the rule demands.

Step 3 — the trap's number. $\sum_{i} (v^{i})^{2} = 1 + 1 = 2$ ; $\sum_{i} (v_{i})^{2} = 4 + 9 = 13$ . Neither equals $5$ — both are basis-dependent junk. Why this step? To show concretely that both same-level sums fail.

Verify: Background-Cartesian truth: $v = (2, 1)$ , $∣ v ∣^{2} = 2^{2} + 1^{2} = 5$ ✓. Equals $v^{i} v_{i} = 5$ ✓. The pipeline is $5 \approx 2.236$ units. Exam moral: always sum a staircase, one index up + one down.

Recall Which cell am I in? (fast triage)

Diagonal $g$ with all $1$ s? ::: Cell A — components coincide. Diagonal $g$ , entries $\neq = 1$ ? ::: Cell B — perpendicular but scaled; $v_{i} \neq = v^{i}$ . $g_{12} > 0$ ? ::: Cell C — acute skew; covariant overshoots. $g_{12} < 0$ ? ::: Cell D — obtuse skew; covariant undershoots. $det g \to 0$ ? ::: Cell F — degenerate; $g^{- 1}$ blows up, no raising possible. $g$ depends on position? ::: Cell G — curvilinear; gradient must be raised.