Visual walkthrough — Metric tensor — raising - lowering indices

Step 0 — What is a "component"? (the thing we're translating)

The subtle bit: if the basis arrows are not at right angles, the honest recipe "how many ${\mathbf{e}}_1$, how many ${\mathbf{e}}_2$" is not the same as "how far does $\mathbf{V}$ reach along each arrow". Those are two different questions, and each gets its own list of numbers. That is the whole story of this page.

• The red arrows are the basis ${\mathbf{e}}_1,{\mathbf{e}}_2$ — deliberately not perpendicular.
• The yellow arrow is $\mathbf{V}$.
• The dashed blue lines are the parallelogram rule: slide along ${\mathbf{e}}_1$, then along ${\mathbf{e}}_2$, to reach the tip. Their lengths give the contravariant numbers $V^1,V^2$.

Step 1 — The second list: projection instead of recipe

WHY a dot product answers "how far along". For any two arrows, $\mathbf{V}\cdot {\mathbf{e}}_{\mu }=|\mathbf{V}||{\mathbf{e}}_{\mu }|\cos \theta$, and $|\mathbf{V}|\cos \theta$ is exactly the length of $\mathbf{V}$'s shadow on ${\mathbf{e}}_{\mu }$ (times $|{\mathbf{e}}_{\mu }|$). So the dot product is the "reach along" measurement — that is why we chose it and not, say, an area.

WHAT IT LOOKS LIKE. Drop a perpendicular from the tip of $\mathbf{V}$ onto each basis arrow.

• Yellow: the vector $\mathbf{V}$ (unchanged from Step 0).
• Green dashed lines: the perpendicular drops. The foot of each drop marks the shadow.
• Compare with Step 0's blue parallelogram: the two constructions land at different places on the arrows. That mismatch is why $V^{\mu }\ne V_{\mu }$ in general.

Step 2 — When the two lists agree (build intuition on the easy case)

WHY. With right angles, sliding along ${\mathbf{e}}_1$ is the perpendicular drop onto ${\mathbf{e}}_1$ — the parallelogram becomes a rectangle, and its side is the shadow. With unit length there is no rescaling either.

This is the ordinary Cartesian world of high school — which is precisely why nobody ever told you there were two kinds of components. The moment the axes tilt or stretch, the two lists split apart, and we need a translator. That translator is the metric.

Step 3 — Introduce the metric: a table of arrow dot-products

WHY this table is exactly what we need. It records the two things that broke Step 2's easy world: lengths of the arrows (diagonal entries) and their tilt (off-diagonal entries). Nothing more, nothing less. It is symmetric, $g_{\mu \nu }=g_{\nu \mu }$, because $\mathbf{a}\cdot \mathbf{b}=\mathbf{b}\cdot \mathbf{a}$.

• Diagonal cells (blue) = squared lengths of each arrow.
• Off-diagonal cells (yellow) = the shared tilt ${\mathbf{e}}_1\cdot {\mathbf{e}}_2=|{\mathbf{e}}_1||{\mathbf{e}}_2|\cos \varphi$, where $\varphi$ is the angle between the arrows.
• When $\varphi =90^{\circ }$, $\cos \varphi =0$, the yellow cells vanish, and we're back in Step 2. See how the table contains the easy case.

Step 4 — Derive lowering: turn the recipe into shadows

WHAT / WHY, line by line. Start from the definition of the shadow (Step 1), then substitute the recipe (Step 0):

Term by term:

• $\mathbf{V}\cdot {\mathbf{e}}_{\mu }$ — the shadow question we want the answer to.
• $V^{\nu }{\mathbf{e}}_{\nu }$ — replace $\mathbf{V}$ by its recipe (the numbers we have).
• pull the scalar $V^{\nu }$ out front — dot product is bilinear, numbers slide out.
• ${\mathbf{e}}_{\nu }\cdot {\mathbf{e}}_{\mu }$ is by definition an entry of the metric table.

So the metric is the machine that converts the list you have (recipe, upper) into the list you want (shadow, lower).

The picture shows the arithmetic for $V_1$: each contravariant number $V^{\nu }$ gets multiplied by the tilt/length stored in row $1$ of the table, and the results are summed. When the metric is off-diagonal the $V^2$ term leaks into $V_1$ — that leak is the tilt.

Step 5 — Derive raising: you must use the inverse table

WHY the inverse and not $g_{\mu \nu }$ again. Lowering multiplied by $g_{\mu \nu }$. To go back we must cancel that multiplication — and the thing that cancels a matrix is its inverse. Multiply the lowering formula by $g^{\alpha \mu }$ and sum:

The loop $V^{\mu }\stackrel{g_{\mu \nu }}{\to }{V}_{\mu }\stackrel{g^{\mu \nu }}{\to }{V}^{\mu }$ returns you home — because $g^{\mu \alpha }{g}_{\alpha \nu }={\delta }^{\mu }{}_{\nu }$ is exactly "table then undo-table = do nothing".

Step 6 — Every case, on one plate (degenerate & signed metrics)

We must never leave a scenario undrawn. Three regimes cover them all:

• Left panel (green): identity — the two lists sit on top of each other.
• Middle (blue): diagonal stretch — perpendicular but rescaled, arrows still line up in direction, numbers rescale.
• Right (red): a negative sign flips the covariant version onto the other side — the hallmark of an indefinite metric.

The degenerate warning: if some $g_{\mu \nu }$ had determinant $0$ (a squashed basis where one direction collapses), the inverse $g^{\mu \nu }$ would not exist and raising would be impossible. A usable metric must be non-degenerate — that's the price of two-way translation.

The one-picture summary

One diagram, the entire walkthrough: contravariant $V^{\mu }$ (recipe / parallelogram) on the left, covariant $V_{\mu }$ (shadows / perpendicular drops) on the right, and the metric table $g_{\mu \nu }$ (down = lower) and its inverse $g^{\mu \nu }$ (up = raise) as the two arrows crossing between them.

Connections

• Metric tensor — raising - lowering indices (parent)
• Einstein summation convention
• Inner product spaces
• Dual space and covectors
• Line element and ds^2
• Minkowski spacetime
• Tensors — definition and transformation laws
• Christoffel symbols