Covariant and contravariant components: Visual walkthrough

Step 1 — A blank slanted grid, and one honest arrow

WHAT. Draw two basis arrows that are not the usual square ones: $e_{1}$ points right, $e_{2}$ points up-and-right at a slant. Then drop a single arrow $v$ somewhere in the plane.

WHY. On ordinary square graph paper the two readings we're about to meet are secretly identical, so the problem is invisible. To see the two readings split apart, we must slant the paper on purpose. A basis is just the pair of reference arrows we agree to measure everything against.

PICTURE. Look at the figure. The pale grid lines run parallel to the two basis arrows — that is your slanted graph paper. The arrow $v$ (magenta) is a real geometric object: it exists before we choose any numbers.

Figure — Covariant and contravariant components

Step 2 — Reading #1: slide along the axes (contravariant)

WHAT. Ask: how far do I walk along $e_{1}$ , then along $e_{2}$ , to land on the tip of $v$ ? Those two travel-distances are the numbers $v^{1}$ and $v^{2}$ .

WHY. This is the parallelogram rule — the most natural way to build any arrow out of two others. The little raised numbers $v^{1}, v^{2}$ (index up) are the contravariant components. They are defined by $v = steps along e_{1} v^{1} e_{1} + steps along e_{2} v^{2} e_{2} .$

PICTURE. The dashed lines form a parallelogram whose sides run parallel to the axes. Follow the orange path: walk $v^{1}$ along $e_{1}$ , turn, walk $v^{2}$ along a copy of $e_{2}$ — you arrive exactly at the tip. Here that path is $1$ step then $1$ step, so $v^{1} = 1, v^{2} = 1$ .

Step 3 — Reading #2: drop a perpendicular (covariant)

WHAT. Now ignore the parallelogram. Instead, for each axis, drop a right-angle shadow of $v$ straight onto that axis. The strength of that shadow is $v_{i} = v \cdot e_{i}$ .

WHY. The dot product $v \cdot e_{i}$ literally measures the perpendicular shadow of $v$ on the direction of $e_{i}$ (times $e_{i}$ 's length). It answers a different question than Step 2: not "how far do I slide?" but "how much does $v$ agree with this axis?" The lowered numbers $v_{1}, v_{2}$ (index down) are the covariant components: $v_{i} = our arrow v \cdot axis i e_{i} .$

PICTURE. The violet dashed drop-lines meet each axis at a right angle (little square marks). Notice the shadow feet land at different places than the parallelogram corners of Step 2. Here $v_{1} = v \cdot e_{1} = 2$ and $v_{2} = v \cdot e_{2} = 2 + 1 = 3$ .

Step 4 — The overlay: why they disagree

WHAT. Put both readings on the same picture: the parallelogram corners (Step 2) and the perpendicular feet (Step 3) at once.

WHY. Seeing them together makes the whole subject click. They differ only because the axes are not at $90°$ . If $e_{1} ⊥ e_{2}$ and both had length $1$ , the parallelogram would be a rectangle and its corners would sit exactly on the perpendicular feet — the two readings would merge.

PICTURE. The orange dots (parallel) and violet dots (perpendicular) do not coincide. Watch the gap: it is the visual signature of a non-orthonormal basis. The inner product is the only tool that can quantify that gap — which is our Step 5.

Step 5 — The bridge appears: define the metric

WHAT. We now derive the link between the two readings. Start from the definition of the covariant number and substitute the contravariant expansion: $v_{i} = v \cdot e_{i} = (v^{j} e_{j}) \cdot e_{i} = v^{j} just numbers! (e_{j} \cdot e_{i}) .$

WHY. Covariant numbers are defined by dotting with $e_{i}$ , and we already wrote $v = v^{j} e_{j}$ (Step 2) — so we simply plug one into the other. The dot products of basis arrows with each other pop out on their own. Give that table a name: $g_{ij} \equiv e_{i} \cdot e_{j} ⟹ v_{i} = g_{ij} v^{j}$

$g_{ij}$ — the metric tensor: a small symmetric table storing every length and angle of the basis.
$g_{ij} v^{j}$ — reads "for each axis $i$ , mix the contravariant numbers using the geometry table" → out comes the covariant number.

PICTURE. Each cell of the metric grid shows an overlap $e_{i} \cdot e_{j}$ . Diagonal cells $g_{11}, g_{22}$ are squared lengths; off-diagonal $g_{12} = g_{21}$ measures the slant. For our basis: $g = (1112) .$

Step 6 — The return trip: raising with the inverse metric

WHAT. To go back from covariant to contravariant, undo the metric with its inverse $g^{ij}$ : $v^{i} = g^{ij} v_{j}, g^{ij} g_{j k} = δ_{k}^{i} .$

WHY. $g_{ij}$ is a matrix; the only honest way to reverse a matrix multiply is to multiply by its inverse. The symbol $δ_{k}^{i}$ (Kronecker delta, $1$ if $i = k$ else $0$ ) is the identity table — " $g$ then $g^{- 1}$ gives back exactly what you started with."

PICTURE. The figure shows the two arrows of the derivation: $g_{ij}$ points down the staircase (raise→lower), $g^{ij}$ points back up. For our basis $g^{- 1} = (2 - 1 - 1 1)$ , and $g^{ij} v_{j}$ returns $(2, 3) \to (1, 1)$ .

Step 7 — The degenerate case: square paper (why school never warned you)

WHAT. Redraw everything with $e_{1} ⊥ e_{2}$ , both length $1$ — an orthonormal basis.

WHY. This is the sanity check every claim must survive. Now $e_{i} \cdot e_{j} = δ_{ij}$ , so the metric collapses to the identity, and $v_{i} = g_{ij} v^{j} = δ_{ij} v^{j} = v^{i}$ . The two readings fuse into one.

PICTURE. The parallelogram becomes a rectangle; its corners land on top of the perpendicular feet. Orange and violet dots coincide — the gap of Step 4 has vanished. This is precisely why ordinary vectors in school need only one set of components.

Recall Edge cases at a glance

Square paper collapses metric to identity? ::: Yes — $g_{ij} = δ_{ij}$ , so $v_{i} = v^{i}$ . If the two axes were parallel (same direction)? ::: The basis fails: no parallelogram closes, $g$ is singular (no inverse), you cannot raise indices. Longer axis $e_{2}$ but still perpendicular? ::: $g$ stays diagonal but not identity; readings differ by a scale factor per axis, no cross-mixing.

The one-picture summary

Everything on one canvas: the arrow $v$ , its two readings, and the metric bridge that ties them. Trace the loop — parallel slide gives $v^{i}$ , perpendicular shadow gives $v_{i}$ , and $g_{ij}$ / $g^{ij}$ walk you between them. This exact machinery reappears in curvilinear coordinates and in general relativity, where the metric is no longer constant.

Recall Feynman retelling of the whole walkthrough

I drew slanted graph paper and put one arrow on it. I asked two fair questions. Question one: how far do I walk along each slanted line to reach the arrow's tip? — that gave me the "up-index" numbers, the contravariant ones $(1, 1)$ . Question two: how big is the arrow's straight-down shadow on each line? — that gave me the "down-index" numbers, the covariant ones $(2, 3)$ . On slanted paper these two answers disagree, and the disagreement is caused entirely by the axes not being at ninety degrees. So I built a tiny translation table by dotting the basis arrows with each other — the metric $g$ . Multiplying by $g$ turns the up-numbers into the down-numbers; multiplying by its inverse turns them back. Finally I straightened the paper into squares: the parallelogram became a rectangle, the two answers snapped together, and the whole mystery disappeared — which is exactly why nobody in school ever told me there were two kinds of components.