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For the displayed metric tensor, its inverse has components $g^{rr}=e^{-2\chi }$ and $g^{\theta \theta }=e^{-2\chi }/r^2$. The nonzero second-kind Christoffel symbols are ${\Gamma }^r{}_{rr}={\chi }^{\prime }$, ${\Gamma }^r{}_{\theta \theta }=-r-r^2{\chi }^{\prime }$ and ${\Gamma }^{\theta }{}_{r\theta }={\Gamma }^{\theta }{}_{\theta r}=1/r+{\chi }^{\prime }$. They follow by differentiating the two diagonal metric entries and raising the first index of the first-kind connection coefficients. Polar coordinates require treating $r=0$ separately.

For coordinates $(t,x,y,z)$ and line element $d{s}^2=c^{2}d{t}^2-a^2(t)d{\mathbf{x}}^2$, these are the nonzero Christoffel symbols. The prime means differentiation with respect to $t$, and coordinate zero is $t$, not $ct$. They follow either from the metric-derivative formula or the Euler-Lagrange equations for the affine geodesic Lagrangian. Both mixed lower-index orders must be counted to obtain the factor of two in the spatial geodesic equation.

Writing $f=1-2M(v)/r$, the radial Christoffel symbols are ${\Gamma }^v{}_{vv}=M/r^2$, ${\Gamma }^r{}_{vv}=fM/r^2+M^{\prime }/r$ and ${\Gamma }^r{}_{vr}=-M/r^2$. The angular couplings are ${\Gamma }^v{}_{\theta \theta }=-r$, ${\Gamma }^r{}_{\theta \theta }=-fr$, with ${\sin }^2\theta$ factors for the corresponding $\varphi \varphi$ terms, ${\Gamma }^{\theta }{}_{r\theta }={\Gamma }^{\varphi }{}_{r\varphi }=1/r$, ${\Gamma }^{\theta }{}_{\varphi \varphi }=-\sin \theta \cos \theta$ and ${\Gamma }^{\varphi }{}_{\theta \varphi }=\cot \theta$. The lower indices are symmetric. The plus sign of $M^{\prime }/r$ follows from the ingoing $+2dvdr$ cross term.

For a nondegenerate metric tensor and its Levi-Civita connection, contracting the Christoffel symbol formula gives ${\Gamma }^{\rho }{}_{\rho \mu }=\frac{1}{2}{g}^{ab}{\partial }_{\mu }{g}_{ab}$. The Jacobi determinant derivative formula identifies this with ${\partial }_{\mu }\log \sqrt{|\det g|}$. The volume density, rather than the determinant without an absolute value, works in every fixed metric signature.

For metric tensors related by $\tilde{g}=e^{2\omega }g$, differentiating the rescaled metric tensor in the Christoffel symbol formula gives the displayed difference, where ${\omega }_a={\partial }_a\omega$ and the index on ${\omega }^a$ is raised with $g$. Consequently ${\tilde{\nabla }}_{V}V={\nabla }_{V}V+2V(\omega )V-g(V,V){\mathrm{grad}}_g\omega$. The last term vanishes on a null vector, proving conformal preservation of null geodesic paths; for a non-null geodesic its orthogonal component generally changes the path.

For positive twice differentiable $f$, the only nonzero Christoffel symbols are ${\Gamma }_{yy}^x=-f{f}^{\prime }$ and ${\Gamma }_{xy}^y={\Gamma }_{yx}^y=f^{\prime }/f$. In the convention $R^d{}_{abc}={\partial }_b{\Gamma }_{ac}^d-{\partial }_c{\Gamma }_{ab}^d+{\Gamma }_{ac}^e{\Gamma }_{eb}^d-{\Gamma }_{ab}^e{\Gamma }_{ec}^d$, substitution gives $R^x{}_{yxy}=-f{f}^{\prime \prime }$. Dividing $R_{xyxy}$ by $g_{xx}{g}_{yy}-g_{xy}^2=f^2$ gives the Gaussian curvature $K=-f^{\prime \prime }/f$. Thus $f=x$ gives the flat metric in polar coordinates away from its axis, $f=\sin x$ gives unit positive curvature, and $f=\sinh x$ gives unit negative curvature.

The difference of two affine connections is a smooth tensor field of type $(1,2)$. Linearity over smooth functions holds in the first slot by the connection axioms. In the second slot, the two extra terms $X(f)Y$ cancel, so $\Delta (X,fY)=f\Delta (X,Y)$. This tensoriality makes the value depend only on $X_p,Y_p$. In coordinates its components are the differences of the Christoffel symbols, although either collection of connection coefficients individually is not a tensor.

The Euclidean metric is the constant Riemannian metric on ${\mathbb{R}}^n$ given in Cartesian coordinates by $g={\sum }_i(d{x}^i{)}^2$. Its Levi-Civita connection has zero Christoffel symbols in those coordinates, so it is flat and its geodesics are straight lines. The Euclidean norm is the pointwise length determined by this metric.

Fermi coordinates transport a local frame along a timelike worldline. Along a timelike geodesic, a parallel-transported orthonormal frame defines Fermi normal coordinates in which the metric tensor is Minkowskian and the Christoffel symbols vanish on the worldline. Curvature first appears in terms quadratic in the spatial distance from it, so finite laboratories retain tidal forces.

The four-acceleration is the covariant derivative of four-velocity along a massive observer's worldline: $A^{\mu }=d{U}^{\mu }/d\tau +{\Gamma }_{\nu \rho }^{\mu }{U}^{\nu }{U}^{\rho }$. In Cartesian Minkowski spacetime coordinates the Christoffel symbols vanish, leaving the ordinary proper time derivative. Differentiating the constant velocity norm gives $U\cdot A=0$; in the instantaneous rest frame, $A=(0,\mathbf{a})$, so $A\cdot A=-a^2$ with $+---$ metric signature. The magnitude $a$ is the proper acceleration.

Rolling a sheet into a circular cylinder bends it without changing its internal lengths and angles. Fitting that same sheet smoothly onto a sphere requires distortion of its first fundamental form. Gaussian curvature distinguishes these situations: it measures the intrinsic curvature of a smooth surface, even though its simplest definition uses how the surface bends in Euclidean space.

The Gaussian curvature at a point is the determinant of the shape operator, or the product of the two principal curvatures:Here $I=Ed{u}^2+2Fdudv+Gd{v}^2$ is the first fundamental form and $II=ed{u}^2+2fdudv+gd{v}^2$ is the second fundamental form, with $S=-dN$. Regularity makes $EG-F^2>0$. Reversing the unit normal negates $e,f,g$ and both principal curvatures, leaving Gaussian curvature unchanged. ^[ref]

The sign describes bending at the point: positive Gaussian curvature means the principal curvatures have the same nonzero sign, as at an elliptic point; negative Gaussian curvature means opposite signs, as on a saddle; zero Gaussian curvature means at least one principal curvature vanishes. Zero Gaussian curvature at a single point does not imply that a neighborhood is a plane. ^[ref]

Four worked models make the signs concrete:

The patches use a common length unit, with $R=1$ for the sphere and circular cylinder. The marked red and blue curves have principal directions at the black point; the arrow is the chosen unit normal. For physical lengths, principal curvature and mean curvature have units of inverse length, while Gaussian curvature has units of inverse squared length. Rescaling a surface by $\lambda >0$ divides its principal curvatures by $\lambda$ and its Gaussian curvature by ${\lambda }^2$. Writing the saddle as $h=(x^2-y^2)/(2L)$ restores an explicit length scale: $K(0)=-1/L^2$.

The Theorema Egregium says that Gaussian curvature depends only on the first fundamental form and its first two coordinate derivatives. A local isometry therefore preserves Gaussian curvature, even if the shape operator and the principal curvatures change. ^[ref]

Here is the outline, rather than a full proof. The Gauss formula splits second derivatives of a parametric surface into tangential Christoffel symbols and the second fundamental form. The Christoffel symbols are determined by the first fundamental form and its first derivatives. Commuting the next partial derivatives gives the Gauss equation, which identifies $\det S$ with the intrinsic sectional curvature computed from the Riemann curvature tensor. Thus the apparently extrinsic product $k_1{k}_2$ is determined by the Riemannian metric. ^[ref]

One can see the distinction directly by rolling a plane strip with coordinates $(u,v)$ into the circular cylinder $X(u,v)=(R\cos (u/R),R\sin (u/R),v)$. Both have $I=d{u}^2+d{v}^2$, so the map is a local isometry, and is one-to-one on an open strip of width less than $2\pi R$. Extending $u$ over all real numbers wraps around periodically; it does not give a global isometry between the entire plane and the entire circular cylinder. A nonempty open patch of a sphere cannot be flattened by a local isometry into a plane, because its Gaussian curvature $1/R^2$ cannot become zero. The obstruction concerns preservation of the first fundamental form, rather than a particular drawing or choice of seam.

In coordinates $(\theta ,z)$ on a circular cylinder of radius $a$, the induced metric is $a^{2}d{\theta }^2+d{z}^2$. Its coefficients are constant, so the Christoffel symbols vanish and the affinely parametrized geodesic equation is ${\theta }^{\prime \prime }=z^{\prime \prime }=0$. The nonconstant solutions are generators when $c=0$, circular parallels when $d=0$, and helices when both are nonzero. Unrolling the cylinder with coordinate $u=a\theta$ makes all these curves straight lines in the Euclidean metric. Unit-speed parametrization additionally requires $a^2{c}^2+d^2=1$.

For a worked helix on a circular cylinder of radius $1$, choose $\theta (t)=2\pi t,z(t)=2t$ for $0\le t\le 1$. Unrolling gives the straight segment $(u,z)=(2\pi t,2t)$, so its arc length is $\sqrt{4{\pi }^2+4}$. Its acceleration in space points normally to the surface, consistent with the geodesic equation.

This geodesic is not globally shortest between its endpoints: they share the same angular position, and the vertical generator has length $2$. The helix minimizes length among paths with the same endpoints and one full winding, because their unrolled lifts join $(0,0)$ to $(2\pi ,2)$. In general, geodesics minimize sufficiently short subarcs; changing winding can produce a shorter route. Keeping this local/global distinction makes unrolling useful without mistaking every straight lift for an unrestricted shortest path.

For smooth stationary fields on ${\mathbb{R}}^3$, set ${\Gamma }^i{}_{tt}=-E^i$ and ${\Gamma }^i{}_{tj}={\Gamma }^i{}_{jt}={ϵ}^i{}_{jk}{B}^k$, with all other Christoffel symbols zero. This torsion-free connection makes the equation $\ddot{\mathbf{x}}=\mathbf{E}+2\mathbf{B}\times \dot{\mathbf{x}}$ a geodesic equation with time as affine parameter. Arbitrary smooth fields suffice; a metric realization imposes additional potential conditions.

For smooth spatial $\varphi$ and $g(c)=-c^2{e}^{2\varphi /c^2}d{t}^2+{\delta }_{ij}d{x}^{i}d{x}^j$, the only nonzero Christoffel symbols are ${\Gamma }^t{}_{ti}={\Gamma }^t{}_{it}=c^{-2}{\partial }_i\varphi$ and ${\Gamma }^i{}_{tt}=e^{2\varphi /c^2}{\partial }_i\varphi$. Consequently the Levi-Civita connection converges smoothly on compact sets as $c\to \infty$ although the metric itself diverges in these coordinates. The limiting torsion-free connection has only ${\Gamma }^i{}_{tt}={\partial }_i\varphi$ and gives $d^2{x}^i/d{t}^2=-{\partial }_i\varphi$ when $t$ is an affine parameter. With $R^a{}_{bcd}={\partial }_c{\Gamma }^a{}_{db}-{\partial }_d{\Gamma }^a{}_{cb}+\cdots$, its Ricci tensor is $(\Delta \varphi )dt\otimes dt$: quadratic connection terms vanish and $R^i{}_{tjt}={\partial }_j{\partial }_i\varphi$. Thus zero Ricci tensor is the Laplace equation, while full flatness would require the entire Hessian to vanish.

At any point of a Riemannian manifold, choose geodesic normal coordinates and orthonormalize their coordinate fields smoothly. At the point, the metric is the identity and its first derivatives vanish, so the orthonormalization coefficients have zero first derivatives. The Christoffel symbols also vanish there. This gives a local Riemannian orthonormal frame with the displayed property, useful for invariant identities evaluated at one point.

Use the source's $+---$ metric signature and $x^0=\eta$, with conformal time derivatives denoted by primes. For this vector cosmological perturbation, the metric tensor and inverse metric through first order areSpatial indices on perturbations are raised with the background Kronecker delta. The Christoffel symbol formula isBecause $a=a(\eta )$ only, it givesFor the other requested Christoffel symbols, only the time derivative of the spatial metric tensor contributes:ThusFinally contract the first upper and first lower spatial indices, summing over $i$:Since $J_{ii}=2{\partial }_i{F}_i=0$ by the transverse vector field condition on $\mathbf{F}$,The useful Christoffel symbol traces are consequently ${\Gamma }_{\mu 0}^{\mu }=4\mathcal{H}$ and ${\Gamma }_{\mu j}^{\mu }=0$ to first order: the first uses $J_{ii}^{\prime }=0$ and the second uses the contraction just proved.

The official notes, PDF pp. 11–13, handwritten pp. 4–6 derive all four geodesic equations and list the twelve nonzero Christoffel symbols up to symmetry of the lower indices, using time coordinate $t$.

Put $\mu =GM/c^2$, $A=1-2\mu /r$ and $B=1+2\mu /r$. For an affinely parametrized geodesic, use the unhalved quadratic Lagrangian $\mathcal{L}=-c^{2}A{\dot{t}}^2+B{\dot{\mathbf{r}}}^{,2}$. Its Euler-Lagrange equations are unchanged by this overall normalization. Since $t$ is a cyclic coordinate, $d(-2c^{2}A\dot{t})/d\tau =0$, equivalentlyThe spatial derivatives give $\partial \mathcal{L}/\partial {\dot{x}}^i=2B{\dot{x}}^i$ and $\partial \mathcal{L}/\partial x^i=-c^2{\dot{t}}^2{A}_i+{\dot{\mathbf{r}}}^{,2}{B}_i$, where $A_i={\partial }_{i}A$. Therefore

For the connection coefficients, take $x^0=ct$, so $g_{00}=-A$ and $g_{ij}=B{\delta }_{ij}$. Substituting the diagonal inverse metric into the Christoffel symbol formula gives all nonzero component types, including lower-index symmetry:All other component types vanish. If the time coordinate is $t$ rather than $ct$, replace ${\Gamma }^i{}_{00}$ by ${\Gamma }^i{}_{tt}=c^2{\Gamma }^i{}_{00}$; the displayed time–space mixed coefficients are unchanged. Spatial indices in these formulas are raised with the background Euclidean metric.

For the circular orbit let $u=\mu /R$ and $\Omega =d\varphi /dt=\omega /\dot{t}$. Both $A$ and $B$ are constant along it. The radial spatial equation gives $(1+u){\omega }^2=(GM/R^3){\dot{t}}^2$, so, treating the displayed metric coefficients algebraically, ${\Omega }^2=GM/[R^3(1+u)]$. The weak-field approximation securely determines the leading Keplerian orbit resultThe time normalization is $-c^2=-c^{2}A{\dot{t}}^2+B{R}^2{\omega }^2$, yielding $\dot{t}=1+3u/2+O(u^2)$ and $P_{\tau }/P_t=1-3u/2+O(u^2)$. One must distinguish the proper-time angular frequency in the supplied path from the coordinate-time angular frequency. The formal factor $1+u$ in the frequency equation comes from treating the truncated metric as exact; a complete relative post-Newtonian correction to the orbital period would also require the omitted higher-order temporal metric terms.

Use the curvature sign conventionwhich has the derivative ordering printed in the source for $R^i{}_{0j0}$. A Christoffel symbol is of order $\mu /r^2$, so its derivative is of order $\mu /r^3$, whereas a product of two connections is of order ${\mu }^2/r^4$. Their ratio is $\mu /r\ll 1$. The metric is stationary, hence the temporal derivative term vanishes. ConsequentlyHere $g_{00}=-1+2\mu /r$, and ${\partial }_i{\partial }_j(1/r)=(3x_i{x}_j-r^2{\delta }_{ij})/r^5$. The weak-field tidal curvature in a stationary metric is

Printed sign inconsistency. The source's positive half-Hessian and its final negative-prefactor expression are the opposites of this result. A concrete check at $(r,0,0)$ is ${\Gamma }^1{}_{00}\simeq \mu /r^2$, giving ${\partial }_1{\Gamma }^1{}_{00}=-2\mu /r^3$, while the printed final expression gives $+2\mu /r^3$. Thus both printed equalities cannot hold with its stated derivative ordering. Reversing the definition of the Riemann curvature tensor would give the printed final sign, but would also require reversing the curvature term in the geodesic deviation equation. The convention used here and in the next part gives the physical radial stretching and transverse compression.

Take the unhalved geodesic LagrangianThe cyclic coordinates $t,\varphi$ give $d(2c^2\alpha \dot{t})/d\lambda =0$ and $d(2r^2{\sin }^2\theta \dot{\varphi })/d\lambda =0$. The other two Euler-Lagrange equations areThese reproduce the supplied geodesic equations. Expanding the derivatives and comparing with ${\ddot{x}}^{\mu }+{\Gamma }^{\mu }{}_{\nu \rho }{\dot{x}}^{\nu }{\dot{x}}^{\rho }=0$, in coordinates $(t,r,\theta ,\varphi )$, gives all nonzero Christoffel symbols:Here ${\alpha }^{\prime }=-2r/r_S^2$. This is the static patch of de Sitter spacetime, with $0\le r<r_S$ for the stationary observers used below. Apparent divergences at the polar axis, origin or horizon belong to the chosen coordinate chart; they are not themselves evidence of a curvature singularity.