Question bank — Christoffel symbols — intro

False. Flat 2D space in polar coordinates has ${\Gamma }_{\theta \theta }^r=-r\ne 0$ yet zero curvature — the symbols report how the coordinate grid bends, not how space bends.

False. You can always make every $\Gamma$ vanish at a single chosen point by picking "normal coordinates"; flatness needs $\Gamma$ to stay zero (or its curvature combination to vanish) in a whole neighbourhood.

False. Under a coordinate change they pick up an inhomogeneous second-derivative term, so they violate the homogeneous Tensor Transformation Laws that define a tensor.

True in a coordinate (holonomic) basis, because ${\partial }_i{\mathbf{e}}_j={\partial }_j{\mathbf{e}}_i$ (mixed partials of position commute); in a non-coordinate basis it can fail.

True. The boxed formula ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{km}({\partial }_i{g}_{mj}+{\partial }_j{g}_{mi}-{\partial }_m{g}_{ij})$ builds them entirely from $g$ and its first derivatives — nothing else is needed.

True. There $g_{ij}={\delta }_{ij}$ is constant, so every ${\partial }_m{g}_{ij}=0$ and the formula gives $\Gamma =0$ everywhere.

True. Both carry the same inhomogeneous term, so subtracting them cancels the non-tensorial junk and the remainder transforms cleanly.

False. Raising/lowering with $g$ never fixes the transformation law; the extra second-derivative term is still there whatever the index placement.

The error ignores that in curvilinear coordinates the basis vectors also change; the true derivative is ${\nabla }_j{V}^k={\partial }_j{V}^k+{\Gamma }_{ij}^k{V}^i$, and dropping the $\Gamma$ term gives a quantity that isn't even a tensor.

The $\theta$ one is wrong: it needs the inverse metric $g^{\theta \theta }=1/r^2$, giving $\frac{1}{2}\cdot \frac{1}{r^2}\cdot 2r=1/r$, not $r$. Forgetting to contract with $g^{km}$ is the classic slip.

No — the minus sits on ${\partial }_m{g}_{ij}$, the derivative with respect to the dummy index $m$ (the one tied to $g^{km}$). The two cyclic terms with the free indices are the plusses.

False; the upper $k$ names the output basis vector while the lower $i,j$ name (which basis vector, moved in which direction). Only the two lowers may swap.

The blow-up is a coordinate artefact — the origin is a perfectly smooth flat point; polar coordinates just degenerate there because $\theta$ is undefined at $r=0$.

Constant speed is not zero acceleration; the velocity vector turns. The covariant derivative captures the real centripetal turning via ${\Gamma }_{\theta \theta }^r=-r$, giving $-r{\dot{\theta }}^2\ne 0$.

Because in practice the Metric Tensor is the one object we always have (from $d{s}^2$), while explicit basis vectors in an embedding may be unavailable — the formula frees us from needing an ambient space.

Because that symmetry comes from ${\partial }_i{\partial }_j(\text{position})={\partial }_j{\partial }_i(\text{position})$; in a non-coordinate frame the basis isn't obtained by differentiating a coordinate, so the mixed-partial argument fails.

Since $\Gamma$ can be made zero at any point by coordinates, no single-point value of $\Gamma$ can be geometric; only a combination that survives coordinate changes — the Riemann Curvature Tensor — measures true bending.

The plain derivative's transformation leaks a non-tensorial term; the $\Gamma$ term carries an equal and opposite non-tensorial term, and their sum cancels the junk, leaving a clean tensor.

The extra $-r{\dot{\theta }}^2$ is exactly ${\Gamma }_{\theta \theta }^r{\dot{\theta }}^2$; the geodesic equation turns "no real force" into these $\Gamma$-driven "fictitious force" terms demanded by the rotating basis.

You can kill $\Gamma$ at one point, but curvature forces $\partial \Gamma$ to stay nonzero nearby, so no coordinate system makes $\Gamma$ vanish everywhere at once — that impossibility is the invariant signature of curvature.

It diverges, but harmlessly: it flags that polar coordinates break down at the origin, not that geometry is singular — a Cartesian patch there has all $\Gamma =0$.

No. A diagonal metric still has derivatives like ${\partial }_r{g}_{\theta \theta }$, which feed off-diagonal-looking symbols such as ${\Gamma }_{r\theta }^{\theta }$; diagonality of $g$ does not make $\Gamma$ diagonal.

All zero, since every term in the formula is a derivative of $g$; a constant metric — even if non-orthonormal — gives a flat, $\Gamma$-free connection.

No. The sphere is genuinely curved, so its Riemann tensor is nonzero and no coordinates can globally flatten the connection — at best $\Gamma$ vanishes at one chosen point.

Not necessarily — they must agree up to a constant scaling that leaves all derivative ratios unchanged, since only derivatives of $g$ (weighted by $g^{-1}$) enter, so overall constant factors can differ.