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For a surface of revolution whose induced metric is $g=E(\rho )d{\rho }^2+{\rho }^{2}d{\varphi }^2$, with $\rho >0$ and $E>0$ smooth on a coordinate interval, define $\sigma (\rho )={\int }_{{\rho }_{\ast }}^{\rho }\sqrt{E(r)}dr/r$. Then ${\sigma }^{\prime }>0$ and the conformal factor ${\rho }^{-1}$ gives ${\rho }^{-2}g=d{\sigma }^2+d{\varphi }^2$. This proves the rescaled metric is locally the Euclidean metric without solving a curvature equation. Poles and places where $\rho$ is not a coordinate must be treated in other charts; the angular coordinate is also understood locally.

For a hypersurface with unit normal vector $n$, put $\epsilon =g(n,n)=\pm 1$, $P=1-\epsilon n\otimes n^{♭}$, and $K(X,Y)=g({\nabla }_{X}n,Y)$ for tangent vectors. The induced metric is nondegenerate precisely when the normal is non-null. With $R(X,Y)Z=[{\nabla }_X,{\nabla }_Y]Z-{\nabla }_{[X,Y]}Z$ and $R_{abcd}=g(e_a,R(e_c,e_d)e_b)$, the Gauss formula is ${\nabla }_{X}Y=D_{X}Y-\epsilon K(X,Y)n$. Substituting twice into the ambient Riemann curvature tensor and projecting tangentially gives the displayed Gauss equation. Its normal projection gives $g(n,R(X,Y)Z)=-(D_{X}K)(Y,Z)+(D_{Y}K)(X,Z)$, the curved-ambient Codazzi equation. A null hypersurface needs additional geometric data because its induced metric is degenerate.

With signature $(-,+,+,+)$, future unit normal $n$ and $K_{ab}=-P^c{}_a{P}^d{}_b{\nabla }_c{n}_d$, the Gauss–Codazzi equations become $(\perp R{)}_{abcd}={\mathcal{R}}_{abcd}+K_{ac}{K}_{db}-K_{ad}{K}_{cb}$ and $(\perp R{)}_{abcd}{n}^d=-D_a{K}_{bc}+D_b{K}_{ac}$, where in the second expression only the first three indices are projected. The intrinsic curvature and spatial covariant derivative use the spatial induced metric. The normal is timelike, so the normal-sign convention differs from a spacelike normal in a positive-definite ambient metric.

A geodesic is a curve whose tangent vector is parallel along itself under the Levi-Civita connection. With an affine parameter it has zero covariant acceleration:This is the Riemannian manifold version of moving straight ahead with constant speed. Its coordinate form is the geodesic equation. In Euclidean space it becomes $\ddot{\gamma }=0$, giving straight lines. On an embedded surface with its induced metric, the ordinary acceleration of such a curve is normal to the surface: all acceleration is used to follow the surface, rather than to turn within it.

On a unit round sphere, the great circle $\gamma (t)=(\cos t,\sin t,0)$ has $\ddot{\gamma }(t)=-\gamma (t)$, which is normal to the sphere. Two points at $t=0$ and $t=\pi /3$ can be joined along that circle by the short arc of length $\pi /3$ or the long arc of length $5\pi /3$. Each admits a constant-speed geodesic parametrization, yet only the short arc realizes their great-circle distance.

Every geodesic minimizes arc length on sufficiently short subarcs. A long segment can cease to be globally minimizing, as the sphere example shows; the first loss from a starting point is described by the cut locus. Thus “locally straight” and “shortest route between these endpoints” are distinct assertions. On a circular cylinder, different winding gives the same distinction, made explicit by Riemannian distance on a circular cylinder.

This provides a useful division of work. The geodesic equation predicts a path from an initial point and tangent vector; comparing arc lengths decides which candidate solves a global shortest-route problem. Variations of energy of a curve give the equation as an Euler-Lagrange equation, offering a proof outline and a connection with classical mechanics. ^[ref].

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 a smooth surface obtained by translating the plane graph $z=F(y)$ along the $x$ direction, the induced metric is $d{s}^2=d{x}^2+[1+F^{\prime }(y{)}^2]d{y}^2$. The transverse arc length coordinate $u(y)={\int }^y\sqrt{1+F^{\prime }(s{)}^2}ds$ makes it $d{s}^2=d{x}^2+d{u}^2$. This is a local isometry to the Euclidean plane, extending the intrinsic flatness of a circular cylinder to arbitrary smooth generating graphs.

Its geodesics therefore lift to straight lines in $(x,u)$ coordinates. Those that are graphs over $x$ satisfy $u=ax+b$, while the additional family $x=\text{constant}$ cannot be written as $y(x)$. The same graph formula follows from the Beltrami identity for the arc-length integrand $\sqrt{1+[1+F^{\prime }(y{)}^2]y^{\prime 2}}$. These straight lines are locally minimizing; on cylinders with a closed generating curve, global identifications can create several different winding paths between two points.

A circular cylinder of radius $\rho$ has induced metric locally equal to $d{u}^2+d{z}^2$ with $u=\rho \phi$. Its Riemann curvature tensor therefore vanishes, despite its nonzero extrinsic curvature. With outward unit normal and convention $K_{ij}=e_i^a{e}_j^b{\nabla }_a{n}_b$, its trace is $1/\rho$ and its principal curvatures are $1/\rho$ and zero. Its averaged mean curvature is $1/(2\rho )$. The angular identification changes global topology, not local flatness.

The line element of a Riemannian metric is its local expression for squared infinitesimal length. Along a parametrized curve $x(t)$, the arc length is $\int \sqrt{g_{ij}(x(t)){\dot{x}}^i{\dot{x}}^j}dt$. For a circular cylinder of radius $R$ parametrized by angle $\theta$ and axial position $z$, the induced metric gives $d{s}^2=R^{2}d{\theta }^2+d{z}^2$. Unrolling with $u=R\theta$ yields the Euclidean form $d{u}^2+d{z}^2$, explaining the straight unrolled descriptions of geodesics on a circular cylinder.

For a unit timelike normal $n$, the induced metric is $h_{\mu \nu }=g_{\mu \nu }+n_{\mu }{n}_{\nu }$. The spatial covariant derivative projects the derivative and all free indices with the spatial projection tensor $P^{\mu }{}_{\nu }={\delta }^{\mu }{}_{\nu }+n^{\mu }{n}_{\nu }$. Metric compatibility gives ${\nabla }_{\lambda }{h}_{\mu \nu }=({\nabla }_{\lambda }{n}_{\mu })n_{\nu }+n_{\mu }{\nabla }_{\lambda }{n}_{\nu }$; projecting kills the normal factor in each term. Thus the spatial covariant derivative is compatible with the induced metric.

For a positive lapse function $\alpha$, shift vector ${\beta }^i$ and spatial induced metric ${\gamma }_{ij}$, the block metric has $g_{00}=-{\alpha }^2+{\gamma }_{ij}{\beta }^i{\beta }^j$ and $g_{0i}={\gamma }_{ij}{\beta }^j$. Its block determinant is $g=-{\alpha }^2\det \gamma$. Therefore the metric volume tensor contracted with the future unit normal restricts to the spatial volume tensor: $n^{\mu }{ϵ}_{\mu ijk}=\sqrt{\det \gamma }[ijk]$.

For a minimally coupled scalar field with spatial induced metric $h_{ij}$, the field's derivative along the future unit normal is $\Pi$. If the shift vector enters as $d{x}^i-N^{i}dt$, then $\Pi =(\dot{\varphi }+N^i{\partial }_i\varphi )/N$. The canonical momentum of the density is ${\pi }_{\varphi }=\sqrt{h}\Pi$. Confusing this density with the normal derivative loses a volume factor.

Keep the unit-normal sign explicit: $n^a{n}_a=\epsilon \in \{+1,-1\}$. Since the hypersurface is spacelike its normal is timelike. ConstructChoose the sign of n for the desired time orientation. The pullback of h is the nondegenerate induced metric. The second fundamental form isIts sign reverses when n reverses. It is tangential in both indices and symmetric: the antisymmetric part of the derivative of a normalized gradient vanishes after both tangential projections. Equivalently it is the projected Hessian of f divided by the normalizing factor. This is the normal-change definition of extrinsic curvature.

For tangent vector fields X,Y, let D be the induced Levi-Civita connection. Orthogonality to n gives the derivative decompositionUse the curvature convention of the preceding solution, $R(X,Y)Z=[{\nabla }_X,{\nabla }_Y]Z-{\nabla }_{[X,Y]}Z$. Substitute the decomposition twice and take its tangential component. Terms differentiated normally cancel, while the two normal components contribute the quadratic shape terms:Lowering the first curvature index gives the Gauss equation for a nonnull hypersurfaceAll four displayed indices on the left are tangential.

Contract with $h^{ac}{h}^{bd}$ and use $h^{ab}=g^{ab}-\epsilon n^a{n}^b$ for tangential contractions. The double-normal term vanishes by curvature antisymmetry and each mixed contraction is ambient Ricci curvature. With $K=h^{ab}{K}_{ab}$ this givesThis also records which metric and curvature conventions enter the sign.

In vacuum without a cosmological constant, the ambient Ricci tensor and scalar vanish. A totally umbilic hypersurface has $K_{ab}=\lambda h_{ab}$, so $K=3\lambda$ and $K_{ab}{K}^{ab}=3{\lambda }^2$. Therefore the scalar curvature of an umbilic vacuum hypersurface is $6\epsilon {\lambda }^2$.

To express the requested nonnegative version, take Lorentz signature $(+---)$, so the timelike unit normal has $\epsilon =+1$ and h is the induced negative-definite spatial metric. ThenIf instead one uses signature $(-+++)$ and its positive-definite spatial metric with the same curvature-slot convention, the scalar is $-6{\lambda }^2$. Replacing h by $-h$ reverses its scalar curvature. Thus the printed inequality has a signature/induced-metric convention built into it; the general formula above makes that choice explicit rather than losing the sign. The metric in the separate final question explicitly uses the mostly-plus signature.

Let $D$ be the ambient Levi-Civita connection and $\nabla$ the one for the induced metric on $M$. The Gauss formula iswhere the second fundamental form $II(X,Y)=(D_{X}Y{)}^{\perp }$ is normal. The tangential part of $D$ is metric-compatible and torsion-free, so the existence and uniqueness of the Levi-Civita connection identifies it with $\nabla$.

For a normal field $\xi$, the Weingarten formula defines the shape operator in a normal direction $A_{\xi }$ and the normal connection byDifferentiate $⟨\xi ,Y⟩=0$ using metric compatibility to getThe symmetry of the second fundamental form follows from the ambient torsion-free condition:since the Lie bracket of vector fields tangent to $M$ is tangent. Thereforeso every normal-direction shape operator is self-adjoint.

The paper uses $R(X,Y)=D_{[X,Y]}-[D_X,D_Y]$, so use the Gauss equation with reversed curvature convention. Applying the Gauss formula and Weingarten formula twice givesSubtracting in the stated curvature order yieldsPair with a tangent field $W$ and use the defining relation for the shape operator in a normal direction:This is the Gauss equation in the supplied sign convention. If one defines curvature by $[D_X,D_Y]-D_{[X,Y]}$ instead, the two quadratic terms have the opposite signs.

For an orthonormal frame $X,Y$ on a surface, positive Gaussian curvature with the convention here is $K=⟨R^M(X,Y)X,Y⟩$. If the ambient space is Euclidean and $\xi$ is a unit normal, the equation becomesThe Koszul formula determines $\nabla$ from the induced metric, and hence determines its Riemann curvature tensor and $K$. Therefore the Gaussian curvature depends only on that metric, even though its expression as a product of principal curvatures appears extrinsic. This is Theorema Egregium.

Write $q_{\alpha \beta }=g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ for the induced metric. By definition of the spatial covariant derivative,The spacetime Levi-Civita connection is metric compatible, so $\nabla g=0$. The remaining terms contain either $P_{\alpha }{}^{\sigma }{n}_{\sigma }$ or $P_{\beta }{}^{\lambda }{n}_{\lambda }$, each zero because $n^{\mu }{n}_{\mu }=-1$. ThereforeThis verifies metric compatibility of the spatial covariant derivative directly, without assuming that the unprojected derivative of $q$ vanishes.

On the sphere, both cosmic time and the comoving coordinate $r$ are constant. The positive spatial induced metric obtained from the FLRW metric is therefore $d{\ell }^2=a^2{r}^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)$. Its area element is the square root of its determinant:Integrating over the spherical coordinates givesThus $ar$ is the areal radius. In a curved spatial slice it need not equal the proper radial distance $a{\int }_0^r(1-k{r}^{\prime 2}{)}^{-1/2}d{r}^{\prime }$; the angular part of the FLRW metric, rather than a Euclidean radial-distance assumption, establishes the surface area.

Use the Levi-Civita connection and the Riemann curvature tensor conventionfor orthonormal $T,E$. Thus the round sphere has positive sectional curvature. Work on a fixed compact parameter interval $[a,b]$ inside the given open interval; this keeps all differentiations and endpoint evaluations well-defined. The base geodesic is regular and has constant speed, so a constant rescaling of $t$ makes it unit-speed. Put $T={\partial }_t\varphi$, $J={\partial }_s\varphi$, $D_t={\nabla }_T$, $D_s={\nabla }_J$ and evaluate at $s=0$. The torsion-free connection gives $D_{s}T=D_{t}J$ because the two parameter derivatives commute. The geodesic equation is $D_{t}T=0$.

Differentiating the length of a curve once and twice, using metric compatibility, givesandLet $A=D_{s}J$. The curvature commutator gives $D_s{D}_{t}J=D_{t}A+\mathcal{R}(J,T)J$, and the skew-adjoint curvature symmetry gives $⟨\mathcal{R}(J,T)J,T⟩=-⟨\mathcal{R}(J,T)T,J⟩$. Integration by parts therefore proves the moving-endpoint second variation of Riemannian arc length:Equivalently, with $W=J-⟨J,T⟩T$, the integral is $I(W,W)$, the Riemannian index form. The endpoints of this geodesic variation generally move, so the displayed boundary term must be retained. For a base speed $c$ instead of one, the boundary and integral have an overall factor $1/c$, and the subtracted square inside the integral is $⟨D_{t}J,T{⟩}^2/c^2$. Returning to the original parameter, every ruling has constant speed, so $L_{[a,b]}(s)=(b-a)|{\partial }_t\varphi (0,s)|$ and the full open-interval length is $2\delta |{\partial }_t\varphi (0,s)|$. ConsequentlyThe preceding curvature formula, with the base-speed factor $1/c$ when necessary, therefore evaluates the original full length derivative using any compact subinterval. No extension to the missing endpoints is required.

Differentiating $D_{t}T=0$ in $s$ yields the Jacobi field equationThe same ruling curves are geodesics of the surface with its induced metric, because the tangential part of their ambient acceleration is zero. Write its second fundamental form as $II$ and its induced Levi-Civita connection as $D^N$. In particular $II(T,T)=0$. The scalar $f=⟨J,T⟩$ satisfies $f^{\prime \prime }=0$, by the Jacobi field equation, so $W=J-fT$ is a Jacobi field both intrinsically and in the ambient manifold. In a regular surface chart $h=|W|>0$, and $E=W/h$ is the perpendicular unit tangent. Along a two-dimensional geodesic, $D_t^{N}E=0$: its derivative is orthogonal to both $E$ and the parallel field $T$. The intrinsic Jacobi field equation consequently givesFor the ambient derivative, the Gauss formula gives $D_{t}W=h^{\prime }E+hII(T,E)$. Applying the same norm differentiation used above to $h=|W|$ givesComparing these identities proves the curvature comparison for a geodesically ruled surface:This is also the Gauss equation in a curved ambient manifold with $II(T,T)=0$. It applies locally to each regular patch of the surface.

For strict inequality, take the Euclidean embedded submanifold parametrized by $\varphi (t,s)=(t,s,ts)$. Each $t$-curve is a straight-line geodesic. Its first fundamental form has coefficients $E_0=1+s^2$, $F_0=st$, $G_0=1+t^2$; the normal $(-s,-t,1)/\sqrt{1+t^2+s^2}$ gives second fundamental form coefficients $e=0$, $f_0=1/\sqrt{1+t^2+s^2}$, $g_0=0$. HenceThus equality can fail everywhere, although every ruling is an ambient geodesic.

The length differentiation requires a nonconstant base ruling; a regular parametrized surface patch guarantees this. If “image of a smooth map” is read literally without a rank assumption, a degenerate member can fail even to have a first length derivative. For example, in the flat cylinder $\mathbb{R}\times S^1$, the map $\varphi (t,s)=(st,e^{is})$ for $|t|<\delta$, $|s|<4\pi$ has geodesic rulings and a two-dimensional open image: every nonzero-$s$ point has full-rank parametrization, and the sole zero-$s$ image point also has a full-rank preimage at $s=2\pi ,t=0$. Yet $L({\gamma }_s)=2\delta |s|$ is not differentiable at zero. The variation formula therefore applies to nonconstant members, as usual. The curvature inequality still holds on the whole image: regular values are dense by Sard theorem, and both curvatures extend continuously to the remaining points.

Let ${\gamma }_{ij}$ denote the spatial induced metric and $\gamma =\det {\gamma }_{ij}$. Its conformally rescaled version ${\tilde{\gamma }}_{ij}={\gamma }^{-1/3}{\gamma }_{ij}$ has determinant one. The Jacobi determinant derivative formula consequently givesThus the determinant measures the volume-changing part of the spatial metric, while the unit-determinant part contains shape changes. Metric compatibility for the spatial covariant derivative also gives ${\gamma }^{ij}{D}_i{N}_j=D_i{N}^i$.

There is a sign inconsistency in the PDF. Contracting its defining expression for $K_{ij}$, with its threading convention $d{x}^i-N^{i}dt$ and its future normal $n^{\mu }=N^{-1}(1,N^i)$, givesnot the displayed trace with a minus sign before the divergence. This is the trace of extrinsic curvature with a negative shift. For a concrete check, take ${\gamma }_{ij}={\delta }_{ij}$, $N=1$ and $N^i=b{x}^i$. The defining formula gives $K_{ij}=-b{\delta }_{ij}$ and $K=-3b$, whereas the printed trace would give $+3b$. Reversing the shift convention can produce the other trace formula, but also changes the shift terms in $K_{ij}$ and the normal; the two conventions cannot be combined.

For the requested expansion interpretation, set the shift vector to zero. The sign discrepancy then disappears. A small coordinate volume has proper volume proportional to $\sqrt{\gamma }$, and proper time along its normal is $ds=Ndt$. Its expansion rate isThereforeThis is the local volume Hubble parameter, the mean of the three local directional expansion rates. In a homogeneous isotropic background $\gamma =a^6$, it reduces to $\dot{a}/(Na)$. The interpretation does not require the local expansion to be isotropic: the trace-free part of the extrinsic curvature can still describe shear.

Write $h_{ij}={}^{(3)}{g}_{ij}$ for the induced metric and lower the shift vector with it, $N_i=h_{ij}{N}^j$. In the stated negative-shift convention the four-dimensional components and inverse components areMultiplying the given future unit normal by this metric tensor gives $n_{\mu }=(-N,0,0,0)$. In particular the tangential normal components vanish, but their covariant derivatives do not:The relevant Christoffel symbol isFor the second equality, the last parenthesis is $2h_{k\ell }{}^{(3)}{\Gamma }^{\ell }{}_{ij}$, so its contraction subtracts the two spatial connection terms in the symmetrized derivative of $N_i$. ThereforeThis is extrinsic curvature with a negative shift. Both the future-normal orientation and the definition $K=-\nabla n$ matter: replacing the threading by $d{x}^i+N^{i}dt$ reverses the shift terms. The spatial induced metric is generally a function of time as well as position; restricting it to time-independent data would simply make the displayed time derivative vanish.

For local surface coordinates $u^1,u^2$ and metric coefficients $g_{ij}$, affinely parametrized geodesics satisfyThese are the geodesic equation and Christoffel symbols. They follow by applying the Euler-Lagrange equation to the energy $\frac{1}{2}{g}_{ij}{\dot{u}}^i{\dot{u}}^j$. Equivalently, in an embedded surface a constant-speed geodesic has acceleration normal to the surface.

Parameterize a circular cylinder by $(a\cos \theta ,a\sin \theta ,z)$, where $a>0$. Its induced metric is $a^{2}d{\theta }^2+d{z}^2$, with constant coefficients, so all Christoffel symbols vanish in this chart. ThusThese geodesics on a circular cylinder are straight generators when $c=0$, circles around the cylinder when $d=0$, and helices when $cd\ne 0$. Constant curves arise if both vanish. For unit-speed nonconstant curves, $a^2{c}^2+d^2=1$. This description is global after allowing the angle to run through arbitrary multiples of $2\pi$.

The unrolled coordinate $u=a\theta$ exhibits the intrinsic flatness of a circular cylinder: every displayed family is a straight line in $(u,z)$. A circle is therefore a geodesic despite bending in ambient three-dimensional space; its acceleration is normal to the cylinder.

Write $q_{\alpha \beta }=g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ for the induced metric. The spatial covariant derivative projects its derivative index and both tensor indices:Four-dimensional metric compatibility removes the derivative of $g$. Each remaining term contains an undifferentiated normal on a projected index and hence vanishes. ThereforeThis establishes metric compatibility of the spatial covariant derivative directly from its projected definition.

Differentiating the normal's normalization givesso $n^{\mu }{\nabla }_{\nu }{n}_{\mu }=0$. Consequently the second projection in the extrinsic curvature definition is redundant:Finally differentiate $P^{\lambda }{}_{\nu }={\delta }^{\lambda }{}_{\nu }+n^{\lambda }{n}_{\nu }$ and project:The first term vanishes by $P^{\nu }{}_{\beta }{n}_{\nu }=0$, and the second is precisely the defining curvature. The positive sign follows from the specified positive-expansion convention, rather than from an independent projector convention.