Worked examples — Metric tensor — raising - lowering indices

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The scenario matrix

Every metric you will ever lower/raise with falls into one of these case classes. The whole point of raising/lowering only changes with the shape of $g_{\mu \nu }$, so we organise by shape.

Cell	Case class	What is special about $g$	Covered by
A	Identity / Cartesian orthonormal	$g_{\mu \nu }={\delta }_{\mu \nu }$	Ex 1
B	Diagonal, all $+$, unequal	pure rescaling, no sign flip	Ex 2
C	Diagonal with a $-$ (indefinite)	one component flips sign	Ex 3
D	Coordinate-dependent diagonal (polar)	entries depend on position $r$	Ex 4
E	Off-diagonal (couples slots)	cross terms $g_{12}\ne 0$	Ex 5
F	Degenerate / zero row (breakdown)	$\det g=0$, no inverse exists	Ex 6
G	Rank-2 tensor, one slot at a time	mixed index $T^{\mu }{}_{\nu }$	Ex 7
H	Real-world word problem	reading $g$ off a physical setup	Ex 8
I	Exam twist: raise back / consistency	round-trip $g^{-1}g=1$	Ex 9

We will hit every cell. Each example says which cell it fills.

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Ex 1 — Cell A: the case where nothing happens

1. Write the rule. $V_{\mu }=g_{\mu \nu }{V}^{\nu }$. Why this step? It is the definition; we always start by naming which machine we run.
2. Do $\mu =1$: $V_1=g_{11}{V}^1+g_{12}{V}^2=(1)(3)+(0)(4)=3.$ Why this step? Sum over the repeated index $\nu$; only $g_{11}$ is non-zero in this row.
3. Do $\mu =2$: $V_2=g_{21}{V}^1+g_{22}{V}^2=(0)(3)+(1)(4)=4.$ Why this step? Same, second row.
4. Result: $V_{\mu }=(3,4)=V^{\mu }$.

Verify: the norm both ways: $g_{\mu \nu }{V}^{\mu }{V}^{\nu }=3^2+4^2=25$, and $V_{\mu }{V}^{\mu }=3\cdot 3+4\cdot 4=25$. Consistent, and equals ordinary $\Vert V{\Vert }^2$. ✓ This is exactly the Mistake the parent warns about: upper and lower look identical only here.

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Ex 2 — Cell B: pure rescaling (diagonal, all positive, unequal)

1. Lower. $V_1=g_{11}{V}^1=4\cdot 5=20$; $V_2=g_{22}{V}^2=9\cdot 2=18.$ Why this step? Diagonal metric ⇒ each lower component is just its own upper component scaled by the diagonal entry.
2. Build the inverse. For a diagonal matrix, $g^{\mu \nu }=\mathrm{diag}(1/4,1/9)$. Why this step? Raising needs $g^{\mu \nu }$, and for diagonal $g$ the inverse is entry-by-entry reciprocals.
3. Raise back. $V^1=g^{11}{V}_1=\frac{1}{4}\cdot 20=5$; $V^2=g^{22}{V}_2=\frac{1}{9}\cdot 18=2.$ Why this step? Round-trip must return the original — a built-in sanity check.

Verify: raising undid lowering: $(5,2)\to (20,18)\to (5,2)$. ✓ Also $g^{11}{g}_{11}=\frac{1}{4}\cdot 4=1={\delta }^1{}_1$. ✓ No sign flip because all diagonal entries are positive (that only happens in Cell C).

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Ex 3 — Cell C: the sign flip (indefinite metric)

1. Lower each slot. $p_0=g_{00}{p}^0=(-1)(5)=-5$; $p_1=(1)(3)=3$; $p_2=p_3=0.$ Why this step? The Minkowski spacetime metric is diagonal, so each slot uses one entry — but the time entry is $-1$, so the time component flips sign. That flip is the entire physical content of an indefinite metric.
2. Contract. $p_{\mu }{p}^{\mu }=p_0{p}^0+p_1{p}^1=(-5)(5)+(3)(3)=-25+9=-16.$ Why this step? The invariant length must be a valid contraction: one up, one down. This number is coordinate-independent.

Verify: using the metric directly, $g_{\mu \nu }{p}^{\mu }{p}^{\nu }=-E^2+p_x^2=-25+9=-16$. ✓ In particle physics this equals $-m^2$ (with $c=1$), so $m^2=16$, $m=4$. Sign is negative because the vector is timelike — exactly the right regime.

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Ex 4 — Cell D: coordinate-dependent metric (polar)

1. Inverse metric. $g^{\mu \nu }=\mathrm{diag}(1,1/r^2)$. At $r=2$: $g^{\theta \theta }=1/4$. Why this step? Raising needs $g^{\mu \nu }$; here it depends on where you are — that is new versus Cells A–C.
2. Raise radial. $V^r=g^{rr}{V}_r=1\cdot 6=6.$ Why this step? $g^{rr}=1$, so the radial direction behaves like plain Cartesian.
3. Raise angular. $V^{\theta }=g^{\theta \theta }{V}_{\theta }=\frac{1}{4}\cdot 8=2.$ Why the $1/r^2$? Look at the red arrow in the figure: the basis vector ${\mathbf{e}}_{\theta }$ has physical length $r=2$, not $1$. A covariant component already carries a factor of that length, so raising must divide it back out — hence $1/r^2$.

Verify: the norm should be metric-independent of representation. $V_{\mu }{V}^{\mu }=V_r{V}^r+V_{\theta }{V}^{\theta }=6\cdot 6+8\cdot 2=36+16=52$. Cross-check with $g^{\mu \nu }{V}_{\mu }{V}_{\nu }=1\cdot 6^2+\frac{1}{4}\cdot 8^2=36+16=52$. ✓

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Ex 5 — Cell E: off-diagonal metric (coupled slots)

1. Lower, keeping BOTH terms. $V_1=g_{11}{V}^1+g_{12}{V}^2=2\cdot 3+1\cdot 1=7.$ $V_2=g_{21}{V}^1+g_{22}{V}^2=1\cdot 3+2\cdot 1=5.$ Why this step? Off-diagonal $g_{12}\ne 0$ couples the components — dropping the cross term is the classic error. Each lower component mixes all upper components.
2. Invert the metric. For $\left(\begin{array}{cc}a& b\\ b& d\end{array}\right)$, inverse $=\frac{1}{ad-b^2}\left(\begin{array}{cc}d& -b\\ -b& a\end{array}\right)$. Here $\det =2\cdot 2-1=3$, so $g^{\mu \nu }=\frac{1}{3}\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$. Why this step? Raising needs the true matrix inverse, not reciprocals of entries (that shortcut only works when off-diagonals vanish).
3. Raise back. $V^1=\frac{1}{3}(2\cdot 7-1\cdot 5)=\frac{1}{3}(9)=3$; $V^2=\frac{1}{3}(-1\cdot 7+2\cdot 5)=\frac{1}{3}(3)=1.$ Why this step? Round-trip must recover $(3,1)$.

Verify: $(3,1)\to (7,5)\to (3,1)$. ✓ And $g^{\mu \alpha }{g}_{\alpha \nu }$: row-1 = $\frac{1}{3}(2\cdot 2-1\cdot 1,2\cdot 1-1\cdot 2)=\frac{1}{3}(3,0)=(1,0)$. ✓ Norm: $V_{\mu }{V}^{\mu }=7\cdot 3+5\cdot 1=26$; and $g_{\mu \nu }{V}^{\mu }{V}^{\nu }=2\cdot 9+2\cdot 3\cdot 1+2\cdot 1=18+6+2=26$. ✓

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Ex 6 — Cell F: degenerate metric (the breakdown case)

1. Lowering still works. $V_1=1\cdot 3+2\cdot 1=5$; $V_2=2\cdot 3+4\cdot 1=10.$ Why this step? Lowering only multiplies by $g_{\mu \nu }$; no inverse needed, so it is always defined.
2. Try to raise. Compute $\det g=1\cdot 4-2\cdot 2=0$. Why this step? Raising requires $g^{\mu \nu }=g^{-1}$, and a matrix with zero determinant has no inverse. The raising operation is undefined.
3. Interpret. Row 2 is $2\times$ row 1: the second basis "direction" carries no independent length information. A genuine metric on an inner product space must be non-degenerate ($\det g\ne 0$) precisely so both directions of the translation exist.

Verify: $\det \left(\begin{array}{cc}1& 2\\ 2& 4\end{array}\right)=0$, so no inverse ⇒ raising impossible. This is the boundary/degenerate cell the reader must recognise: check $\det g\ne 0$ before assuming you can raise. ✓ (Notice $V_{\mu }=(5,10)$ is itself proportional $1:2$, echoing the degeneracy.)

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Ex 7 — Cell G: a rank-2 tensor, one slot at a time

1. Rule for one slot. $T^{\mu }{}_{\nu }=g_{\nu \alpha }{T}^{\mu \alpha }$ — the metric pairs with the second index only. Why this step? Raise/lower one index at a time; each $g$ converts exactly one slot (parent §4).
2. Column $\nu =0$ (uses $g_{00}=-1$): $T^0{}_0=g_{00}{T}^{00}=-2$; $T^1{}_0=g_{00}{T}^{10}=-4.$ Why this step? Lowering the $0$-slot multiplies that whole column by $-1$.
3. Column $\nu =1$ (uses $g_{11}=1$): $T^0{}_1=T^{01}=3$; $T^1{}_1=T^{11}=5.$ Why this step? $g_{11}=1$ leaves the $1$-column unchanged.
4. Result: $T^{\mu }{}_{\nu }=\left(\begin{array}{cc}-2& 3\\ -4& 5\end{array}\right).$

Verify: the trace $T^{\mu }{}_{\mu }=T^0{}_0+T^1{}_1=-2+5=3$ — this contracted scalar is invariant. Direct check: $g_{\mu \nu }{T}^{\mu \nu }=g_{00}{T}^{00}+g_{11}{T}^{11}=-2+5=3$. ✓

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Ex 8 — Cell H: real-world word problem

1. Read off the metric. Perpendicular axes with scales $2$ and $3$: $g_{\mu \nu }=\mathrm{diag}(2^2,3^2)=\mathrm{diag}(4,9)$. Why this step? $g_{\mu \nu }={\mathbf{e}}_{\mu }\cdot {\mathbf{e}}_{\nu }$; each basis vector's squared physical length sits on the diagonal.
2. True squared length. $\Vert V{\Vert }^2=g_{\mu \nu }{V}^{\mu }{V}^{\nu }=4\cdot 3^2+9\cdot 2^2=36+36=72.$ Why this step? The invariant length uses the metric, not the naive $\sum (V^{\mu }{)}^2$ — that would give the wrong $9+4=13$.
3. Length. $\Vert V\Vert =\sqrt{72}=6\sqrt{2}\approx 8.485$ metres.
4. Covariant components. $V_1=g_{11}{V}^1=4\cdot 3=12$; $V_2=g_{22}{V}^2=9\cdot 2=18.$

Verify: cross-check length via covariant form: $V_{\mu }{V}^{\mu }=12\cdot 3+18\cdot 2=36+36=72$. ✓ Naive (wrong) answer would be $13$ — the metric matters, exactly the Mistake point.

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Ex 9 — Cell I: exam twist, the consistency round-trip

1. State the key identity. $g^{\mu \beta }{g}_{\beta \nu }={\delta }^{\mu }{}_{\nu }$ (the inverse-metric defining relation). Why this step? This is the ONE fact that makes raising undo lowering — the metric and its inverse are true matrix inverses.
2. Apply it. $g^{\mu \beta }(g_{\beta \nu }{V}^{\nu })=(g^{\mu \beta }{g}_{\beta \nu })V^{\nu }={\delta }^{\mu }{}_{\nu }{V}^{\nu }=V^{\mu }.$ Why this step? Associativity of matrix multiplication lets us collapse the pair to $\delta$, and the Kronecker delta simply relabels.
3. Numeric spot-check with $(a,b)=(3,1)$: from Ex 5 we already saw $(3,1)\to (7,5)\to (3,1)$. ✓

Verify: For this specific $g$, $g^{\mu \beta }{g}_{\beta \nu }=\frac{1}{3}\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\delta$. ✓

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Recall Answers

1. Cell A ($g=\delta$, Cartesian orthonormal) — the metric is the identity, so the machine changes nothing.
2. $p_{\mu }=(-5,3,0,0)$, $p_{\mu }{p}^{\mu }=-16$.
3. Off-diagonal $g_{12}\ne 0$ couples slots: $V_1=g_{11}{V}^1+g_{12}{V}^2$ mixes both.
4. $\det g\ne 0$ (non-degenerate) — otherwise no inverse $g^{\mu \nu }$ exists (Cell F).

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Connections

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

Case Map