4.10.10 · D5Advanced Topics (Elite Level)

Question bank — Metric tensor — raising - lowering indices

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Before we start, one shared vocabulary reminder so no symbol is unearned:


True or false — justify

Decide TRUE or FALSE and give the reason, then reveal.

and are always two names for the same list of numbers.
False. They agree only when (orthonormal Cartesian). In polar, Minkowski, or any curved metric the numbers genuinely differ.
The metric tensor is always symmetric, .
True. It is built from dot products, , and the dot product is symmetric, so swapping the basis vectors changes nothing.
The metric is a set of positive numbers because it measures lengths.
False. In Minkowski spacetime with signature the entry is negative; the metric is indefinite, and squared "lengths" can be negative (timelike) or zero (null).
Raising an index and then lowering it again returns the original components.
True. , because and are matrix inverses.
For a diagonal metric, lowering an index just multiplies each component by the matching diagonal entry.
True. With no off-diagonal terms, collapses to (no sum), one entry at a time.
If the metric has off-diagonal terms, you can still lower an index component by component.
False. mixes components; the cross term couples them and cannot be dropped.
The Kronecker delta is itself the metric of flat space.
False. is the mixed identity (1 up, 1 down) that comes from ; the flat metric is with both indices down. Position of indices matters.
is a fixed table of numbers no matter which coordinates you use.
False. Its components depend on the basis. In polar coordinates changes with position; only the underlying geometric object is invariant, not the number list.
In Minkowski space, lowering the time index of a 4-vector flips its sign.
True. With , ; the spatial components (with ) are unchanged.

Spot the error

Each line contains a mistake. Name it, then reveal the fix.

" in every coordinate system."
Wrong. The invariant norm is ; the naive sum-of-squares only works when .
" is the dot product of and ."
Illegal contraction — both indices are up. The invariant is , summing one up with one down.
"To raise an index, contract with ."
You must contract with the inverse . The down-indexed lowers; using it to raise is dimensionally and algebraically wrong unless is its own inverse.
", therefore ."
The raise step is inverted. Since , we get , not .
" lowers both indices at once."
An index can only appear twice (once up, once down) if it is summed; here appear as both free and dummy — a notation clash. Lower one index at a time with a fresh dummy: .
"Since , the metric equals its own inverse."
That relation just says they are matrix inverses; only when , e.g. Minkowski , but not for polar coordinates.
"In , the norm of is ."
The time part carries a minus sign: . Forgetting the signature is the classic slip.
"The metric is what defines vectors."
No — vectors exist without a metric. The metric adds geometry (lengths, angles, and the up/down translation). Vectors and their dual covectors are defined first; the metric merely connects them.

Why questions

Answer in one or two sentences, then reveal.

Why do we even need two kinds of components?
Because a general basis is not orthonormal: measuring a vector by counting basis steps () and by projecting onto basis directions () give different numbers, and both are useful.
Why is the metric defined as dot products of basis vectors specifically?
Because the squared length falls out of expanding ; the dot products are exactly what is needed to get lengths from components.
Why must the inverse metric raise, rather than the metric itself?
Lowering is multiplication by ; raising must undo it, and undoing a matrix multiply requires the matrix inverse .
Why does the summed (dummy) index always appear once up and once down?
A valid contraction pairs a vector slot with a dual slot (Dual space and covectors); this is the only combination that produces a coordinate-independent scalar, guaranteed by the Einstein summation convention.
Why does the -component get a factor when raised in polar coordinates?
The basis vector has length , so ; its inverse compensates for that stretched basis vector when converting .
Why do high-school physics classes never mention covariant vs contravariant?
They work exclusively in orthonormal Cartesian coordinates where , so and the distinction is invisible.
Why is the invariant dot product coordinate-independent even though and each change with coordinates?
The changes cancel: transforms oppositely to the pair , leaving the scalar fixed — that is the defining property of a genuine scalar.
Why do the Christoffel symbols involve derivatives of and not itself?
Raising/lowering uses the metric's values, but describing how the basis (and hence components) changes from point to point requires how the metric changes — its derivatives.

Edge cases

Handle the degenerate / limiting situations. Reveal after reasoning.

What happens to raising/lowering when (flat orthonormal)?
The operations become the identity: . This is the only regime where the distinction disappears, which is why it's the beginner's default.
What if the metric is singular (determinant zero) — can you still raise indices?
No. A zero determinant means does not exist, so there is no way to undo lowering; the geometry is degenerate and lengths/angles are not well defined there.
For a null (lightlike) vector in Minkowski space, what is its norm?
Exactly zero: even though . An indefinite metric allows non-zero vectors with zero length — impossible in Euclidean space.
Take the zero vector . What are its lowered components?
All zero: regardless of the metric. The zero vector is the same in every dialect.
In 2D polar coordinates, what happens to as ?
It blows up (), signalling the coordinate singularity at the origin — the direction is ill-defined at a point, not a failure of the geometry itself.
If a vector points purely along one orthonormal axis, does lowering change it?
Only by the diagonal entry for that axis: e.g. in Minkowski a purely-time vector lowers to — one flipped sign, others untouched.
What does with tell you about the space?
The metric is indefinite (like Minkowski spacetime); in a positive-definite inner product space the norm vanishes only for the zero vector.

Recall Quick self-audit before you leave

If you can't state why a contraction needs one up and one down index, and why raising needs the inverse metric, re-read the parent note's derivations — every trap above is one of those two rules in disguise.

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