Covariant and contravariant components: Exercises

Throughout, the "background" is ordinary Cartesian $x y$ -space, and a skew basis means basis vectors that are not all unit length and not all at $9 0^{\circ}$ . We reuse one running skew basis for the geometric problems so you build intuition on a single picture:

$e_{1} = (2, 0), e_{2} = (1, 2) .$

Figure — Covariant and contravariant components

The parallel (parallelogram) reading of a vector gives contravariant components $v^{i}$ (index up); the perpendicular (drop-a-shadow) reading gives covariant components $v_{i}$ (index down). Keep that picture in your head for every problem.

Level 1 — Recognition

Recall Solution L1.1

(a) Contravariant $v^{1}$ — it is literally an expansion coefficient (parallel projection). (b) Covariant $v_{2} = v \cdot e_{2}$ — the definition of a covariant component (perpendicular projection). (c) Covariant $v_{1}$ — "perpendicular shadow" is the orthogonal-projection rule. (d) Contravariant — "walk along the axes" is the parallelogram rule.

Recall Solution L1.2

$g_{11} = e_{1} \cdot e_{1} = 2 \cdot 2 + 0 \cdot 0 = 4$ . $g_{12} = e_{1} \cdot e_{2} = 2 \cdot 1 + 0 \cdot 2 = 2$ (and $g_{21} = g_{12} = 2$ by symmetry). $g_{22} = e_{2} \cdot e_{2} = 1 \cdot 1 + 2 \cdot 2 = 5$ . $g = (4225) .$ Why this first? The Metric tensor stores every length and angle of the basis — every later formula pulls from it.

Level 2 — Application

Recall Solution L2.1

(a) Contravariant — solve the parallelogram system $v^{1} e_{1} + v^{2} e_{2} = v$ : ${2 v^{1} + 1 v^{2} = 4 0 v^{1} + 2 v^{2} = 2 \Rightarrow v^{2} = 1, 2 v^{1} + 1 = 4 \Rightarrow v^{1} = \frac{3}{2} .$ So $v^{1} = \frac{3}{2}, v^{2} = 1$ . Why solve a system? Because "how far along each slanted axis" is exactly the coefficients that rebuild $v$ . (b) Covariant — dot with each basis vector (the definition): $v_{1} = v \cdot e_{1} = 4 \cdot 2 + 2 \cdot 0 = 8, v_{2} = v \cdot e_{2} = 4 \cdot 1 + 2 \cdot 2 = 8.$ So $v_{1} = 8, v_{2} = 8$ . Notice $v^{i} \neq = v_{i}$ — the basis is skew, so the two readings genuinely differ.

Recall Solution L2.2

$v_{1} = g_{11} v^{1} + g_{12} v^{2} = 4 \cdot \frac{3}{2} + 2 \cdot 1 = 6 + 2 = 8 ✓$ $v_{2} = g_{21} v^{1} + g_{22} v^{2} = 2 \cdot \frac{3}{2} + 5 \cdot 1 = 3 + 5 = 8 ✓$ Both match L2.1(b). This is the metric doing its job: it "lowers" an upper index by mixing in the geometry stored in $g_{ij}$ .

Level 3 — Analysis

Recall Solution L3.1

(a) $det g = 4 \cdot 5 - 2 \cdot 2 = 16$ . For a $2 \times 2$ matrix, $(a c b d)^{- 1} = \frac{1}{d e t} (d - c - b a)$ : $g^{- 1} = \frac{1}{16} (5 - 2 - 2 4) = (\frac{5}{16} - \frac{1}{8} - \frac{1}{8} \frac{1}{4}) .$ (b) Raising: $v^{1} = g^{11} v_{1} + g^{12} v_{2} = \frac{5}{16} \cdot 8 - \frac{1}{8} \cdot 8 = \frac{40}{16} - 1 = \frac{5}{2} - 1 = \frac{3}{2} ✓$ $v^{2} = g^{21} v_{1} + g^{22} v_{2} = - \frac{1}{8} \cdot 8 + \frac{1}{4} \cdot 8 = - 1 + 2 = 1 ✓$ Raising and lowering are perfect inverses — as they must be, since $g^{ik} g_{k j} = δ_{j}^{i}$ .

Recall Solution L3.2

$g^{- 1} g = (\frac{5}{16} - \frac{1}{8} - \frac{1}{8} \frac{1}{4}) (4225) = (\frac{20}{16} - \frac{2}{8} - \frac{4}{8} + \frac{2}{4} \frac{10}{16} - \frac{5}{8} - \frac{2}{8} + \frac{5}{4}) = (1001) .$ $δ_{j}^{i}$ is the Kronecker delta: $1$ when $i = j$ , $0$ otherwise — the machine that says "raising then lowering does nothing." See Tensors and index notation.

Level 4 — Synthesis

Recall Solution L4.1

(a) Cartesian truth: $∣ v ∣^{2} = 4^{2} + 2^{2} = 16 + 4 = 20$ . (b) Contraction (up with down): using $v^{1} = \frac{3}{2}, v^{2} = 1, v_{1} = 8, v_{2} = 8$ , $v^{i} v_{i} = v^{1} v_{1} + v^{2} v_{2} = \frac{3}{2} \cdot 8 + 1 \cdot 8 = 12 + 8 = 20 ✓$ The scalar is basis-independent because we paired one upper with one lower index. (c) Both up (illegal): $v^{1} v^{1} + v^{2} v^{2} = (\frac{3}{2})^{2} + 1^{2} = \frac{9}{4} + 1 = \frac{13}{4} = 3.25 \neq = 20$ . This is meaningless — it secretly assumes $g = I$ . See Inner product spaces for why the metric must sit between the components.

Recall Solution L4.2

The dual basis is $e^{i} = g^{ij} e_{j}$ . Compute each as a Cartesian vector: $e^{1} = g^{11} e_{1} + g^{12} e_{2} = \frac{5}{16} (2, 0) - \frac{1}{8} (1, 2) = (\frac{10}{16} - \frac{1}{8}, - \frac{2}{8}) = (\frac{1}{2}, - \frac{1}{4}) .$ $e^{2} = g^{21} e_{1} + g^{22} e_{2} = - \frac{1}{8} (2, 0) + \frac{1}{4} (1, 2) = (- \frac{1}{4} + \frac{1}{4}, \frac{1}{2}) = (0, \frac{1}{2}) .$ Quick sanity: $e^{1} \cdot e_{1} = \frac{1}{2} \cdot 2 + (- \frac{1}{4}) \cdot 0 = 1 ✓$ ; $e^{1} \cdot e_{2} = \frac{1}{2} \cdot 1 - \frac{1}{4} \cdot 2 = 0 ✓$ . Now rebuild with covariant components $v_{1} = 8, v_{2} = 8$ : $v_{1} e^{1} + v_{2} e^{2} = 8 (\frac{1}{2}, - \frac{1}{4}) + 8 (0, \frac{1}{2}) = (4, - 2) + (0, 4) = (4, 2) = v ✓$ So covariant components ARE contravariant components with respect to the dual basis — the promised symmetry.

Level 5 — Mastery

Recall Solution L5.1

(a) Contravariant — against the basis. If basis vectors double in length, you need half as much of each to reach the same tip: $\tilde{v}^{1} = \frac{1}{2} v^{1} = \frac{3}{4}, \tilde{v}^{2} = \frac{1}{2} v^{2} = \frac{1}{2} .$ Check: $\tilde{v}^{1} \tilde{e}_{1} + \tilde{v}^{2} \tilde{e}_{2} = \frac{3}{4} (4, 0) + \frac{1}{2} (2, 4) = (3, 0) + (1, 2) = (4, 2) = v ✓$ . These scaled by $A^{- 1} = \frac{1}{2} I$ — contravariant components fight the basis change. (b) Covariant — with the basis. $\tilde{v}_{i} = v \cdot \tilde{e}_{i} = v \cdot (2 e_{i}) = 2 v_{i}$ : $\tilde{v}_{1} = 2 \cdot 8 = 16, \tilde{v}_{2} = 2 \cdot 8 = 16.$ These scaled by $A = 2 I$ — covariant components cooperate with the basis change. See Change of basis and transformation laws. (c) Length invariance: $\tilde{v}^{i} \tilde{v}_{i} = \frac{3}{4} \cdot 16 + \frac{1}{2} \cdot 16 = 12 + 8 = 20$ — identical to L4.1. The $A$ and $A^{- 1}$ factors cancel exactly. That cancellation is the entire reason the up–down staircase exists.

Recall Solution L5.2

(a) Covariant (natural) gradient — differentiate, index stays down: $\partial_{1} f = \frac{\partial ( r ^{2} )}{\partial r} = 2 r, \partial_{2} f = \frac{\partial ( r ^{2} )}{\partial θ} = 0.$ So $\partial_{i} f = (2 r, 0)$ . Why down? A gradient measures "change per unit coordinate step" — it eats a displacement, so it is a covector. See Curvilinear coordinates (polar, spherical). (b) Raise with the inverse metric $g^{ij} = (10 0 1/ r^{2})$ : $(\nabla f)^{1} = g^{11} \partial_{1} f = 1 \cdot 2 r = 2 r, (\nabla f)^{2} = g^{22} \partial_{2} f = \frac{1}{r ^{2}} \cdot 0 = 0.$ So $(\nabla f)^{i} = (2 r, 0)$ . (c) The $θ$ -component is $0$ because $f = r^{2}$ does not vary with angle. Had it varied, the factor $1/ r^{2}$ would have been essential — it converts "per radian" into "per unit arc length." This is exactly the raising/lowering machinery made concrete; in Cartesian coordinates $g^{ij} = δ^{ij}$ hides all of it.

Recall One-line self-tests before you leave

Skew basis, $v \cdot e_{i}$ gives which components? ::: Covariant $v_{i}$ (down).
To rebuild $v$ from $v_{i}$ , which basis? ::: The dual basis $e^{i}$ : $v = v_{i} e^{i}$ .
Which contraction is invariant, $v^{i} v_{i}$ or $v^{i} v^{i}$ ? ::: $v^{i} v_{i}$ (one up, one down).
Double the basis lengths — what happens to $v^{i}$ ? ::: They halve ( $A^{- 1}$ ); covariant $v_{i}$ double ( $A$ ).