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A smooth curve tangent to a regular integrable distribution lies in one maximal leaf of a regular foliation. In a Frobenius theorem chart its transverse coordinates have zero derivative, so each short segment lies in one plaque. A finite chain of such segments covers any compact parameter interval and joins plaques in the same leaf. A single embedded local plaque only gives confinement while the curve remains in its chart. A dense immersed cylinder in a three-dimensional torus shows why global embedded confinement cannot generally replace the immersed statement.

A nowhere-zero differential one-form $n$ is hypersurface orthogonal if locally $n=Fdt$ with $F\ne 0$. Its kernel consists of tangent vectors to the level hypersurfaces of $t$. Frobenius theorem makes this equivalent to $n\wedge dn=0$, where $d$ is the exterior derivative. With a torsion-free Levi-Civita connection, this is $n_{[\alpha }{\nabla }_{\beta }{n}_{\gamma ]}=0$. For a timelike unit normal, projecting the derivative on both indices gives zero antisymmetric part, which explains why hypersurface orthogonality implies symmetric extrinsic curvature.

Unit normalization gives $n^{\nu }{\nabla }_{\mu }{n}_{\nu }=0$. Hence $K_{\mu \nu }=-P^{\alpha }{}_{\mu }{P}^{\beta }{}_{\nu }{\nabla }_{\alpha }{n}_{\beta }=-P^{\alpha }{}_{\mu }{\nabla }_{\alpha }{n}_{\nu }$. Set $F_{\mu \nu }={\nabla }_{\mu }{n}_{\nu }-{\nabla }_{\nu }{n}_{\mu }$. Frobenius theorem for a hypersurface normal gives $n_{\alpha }{F}_{\beta \gamma }+n_{\beta }{F}_{\gamma \alpha }+n_{\gamma }{F}_{\alpha \beta }=0$. Contracting with $n^{\alpha }{P}^{\beta }{}_{\mu }{P}^{\gamma }{}_{\nu }$ leaves $-P^{\beta }{}_{\mu }{P}^{\gamma }{}_{\nu }{F}_{\beta \gamma }=0$, precisely the vanishing antisymmetric part of the extrinsic curvature of a spatial hypersurface.

The Frobenius theorem gives the local criterion

A smooth distribution is involutive if its local smooth sections are closed under the Lie bracket of vector fields. The Frobenius theorem identifies this algebraic condition with being an integrable distribution.

A leaf is a maximal connected integral manifold of the smooth constant-rank integrable distribution defining a regular foliation. It carries its intrinsic manifold topology and is immersed in the ambient manifold. It need not be embedded or closed. Local Frobenius theorem coordinates make each leaf segment a plaque with constant transverse coordinates; global confinement to an immersed leaf follows by joining such plaques along tangent curves.

Every finite-dimensional real Lie algebra is the Lie algebra of a connected simply connected Lie group. One construction first uses the Ado theorem to embed it into a matrix Lie algebra, integrates the corresponding left-invariant distribution by the Frobenius theorem to a connected immersed Lie subgroup, and then takes a universal covering Lie group. The immersed subgroup need not be closed; taking its ambient closure may change its Lie algebra.

The null twist is ${\hat{\omega }}_{ab}={\hat{B}}_{[ab]}$. It vanishes for null generators orthogonal to a hypersurface by the Frobenius theorem.

For an affine null geodesic congruence, $B_{ab}={\nabla }_b{U}_a$ annihilates $U$ in both slots. Relative to a null partner, its antisymmetric part is its screen projection plus a term of the form $U_{[a}{q}_{b]}$. Hence zero null twist implies $U\wedge dU=0$, and Frobenius theorem makes $U_a$ locally proportional to a hypersurface gradient. Conversely, if $U_a=h{\nabla }_{a}S$, its antisymmetric derivative has a factor of $U$, so the screen projection vanishes. Geodesicity is essential to the converse from vanishing screen twist to the full Frobenius condition.

The null expansion, null shear and null twist of a congruence tangent to the generators of a Killing horizon all vanish there. If $k$ is its Killing generator and $\ell =fk$ is an affine generator, the screen-space projector kills every term proportional to $k$. The symmetric part of the projected derivative of $\ell$ therefore vanishes by the Killing equation. The antisymmetric part vanishes by Frobenius theorem, since $\ell$ is hypersurface-orthogonal. These conclusions extend through a regular bifurcation surface by continuity.

For a finite-dimensional real Lie algebra $\mathfrak{h}$, the Lie third theorem asserts the existence of a connected simply connected Lie group whose Lie algebra is $\mathfrak{h}$. There are two complementary views of the construction.

Locally, the Baker--Campbell--Hausdorff formula defines multiplication on a sufficiently small neighbourhood of zero in $\mathfrak{h}$. Its leading terms are $X+Y+\frac{1}{2}[X,Y]$, its identity is zero, and inversion is $X\mapsto -X$. The convergent BCH series in finite dimensions and the Jacobi identity provide the local associative law. Continuing and globalizing this local group gives its simply connected integration; the globalization is a substantive step, rather than a claim that one exponential chart gives all of the group.

A matrix construction makes that global step more concrete. The Ado theorem gives a faithful finite-dimensional representationConsider the left-invariant distribution $D_A=A\iota (\mathfrak{h})$ on ${\mathrm{GL}}_N(\mathbb{R})$. It is involutive because $\iota (\mathfrak{h})$ is closed under commutators. The Frobenius theorem integrates it to its connected leaf through the identity. This leaf carries the intrinsic immersed-submanifold structure of the connected Lie subgroup generated by the matrix exponentials $e^{\iota (X)}$. Its tangent Lie algebra is exactly $\iota (\mathfrak{h})$; multiplication and inversion preserve the leaf and give its Lie group operations. Call this immersed group $H_0$.

Taking its universal covering Lie group givesAdo's faithfulness matters: the adjoint representation alone loses central elements. Also, the subgroup generated by the exponentials need not be closed in the ambient matrix group. For an irrational $a$, the map $t\mapsto (R_t,R_{at})$ gives a dense one-dimensional immersed subgroup of the two-dimensional torus. Its intrinsic group is $\mathbb{R}$, whereas its closure has a two-dimensional Lie algebra. Taking an ambient closure in the construction would therefore be incorrect.

Finally, any other connected Lie group with Lie algebra $\mathfrak{h}$ is obtained from this simply connected integration by a discrete central quotient, as in part (i). The simply connected integration is unique up to isomorphism, but an arbitrary corresponding group is not.

Work on a regular level surface, so $d\beta \ne 0$. A nonzero normal covector has the local form $n_a=f{\nabla }_a\beta$. Extend it as a normal to neighbouring level surfaces. Since the Levi-Civita connection is torsion-free,Antisymmetrizing the product with $n_c=f{\beta }_{;c}$ over three indices kills the symmetric second-derivative term and the term containing two copies of $d\beta$. HenceEquivalently $n\wedge dn=0$, the Frobenius theorem condition for hypersurface orthogonality.

For the second covariant derivative of a Killing vector, the curvature sign needs explicit care. Denote the curvature defined on the cover by $C^a{}_{bcd}$, and obtain its other index positions by ordinary raising and lowering with the metric tensor. With $D_{abc}=k_{a;bc}$, the Killing equation gives $D_{abc}=-D_{bac}$, while the supplied Ricci identity givesSuccessive use of these two facts givesSubstitute $D_{acb}=D_{abc}-C_a{}^d{}_{bc}{k}_d$. The curvature symmetries and first Bianchi identity then implyThus the identity consistent with the cover convention isThe plus-sign version is valid for the opposite curvature tensor $\tilde{R}=-C$. With ordinary metric index raising, the printed plus sign and the cover's curvature convention cannot both be used. This is the curvature-sign convention in Killing derivative identities, not a failure of the Killing equation.

An explicit sign check uses $d{s}^2=e^{2x}d{t}^2-d{x}^2-d{y}^2-d{z}^2$ and the timelike Killing vector $k={\partial }_t$. Here $k_t=e^{2x}$, ${\Gamma }^t{}_{tx}=1$ and ${\Gamma }^x{}_{tt}=e^{2x}$. Direct differentiation gives $k_{t;xx}=e^{2x}$, whereas the cover definition gives $C^t{}_{xxt}=-1$ and $C_{txx}{}^d{k}_d=-e^{2x}$. This also verifies the corrected sign without relying on a remembered curvature convention.

To prove the static Killing field is a Ricci eigenvector, use the permitted alternative of adapted coordinates. Hypersurface orthogonality supplies hypersurfaces orthogonal to $k$. Choose spatial coordinates on one such hypersurface and carry them along the flow of the Killing vector, using its flow parameter as $t$. The flow preserves both the metric tensor and $k$, so it preserves orthogonality to $k$. Consequently the mixed metric components vanish, and the Killing equation makes all coefficients independent of $t$:The only nonzero Christoffel symbols involving time can bein particular ${\Gamma }^0{}_{ij}={\Gamma }^i{}_{0j}={\Gamma }^0{}_{00}=0$. In $C_{0i}$, the derivative of the connection trace vanishes because ${\Gamma }^a{}_{0a}=0$; the other derivative is either a time derivative of a static coefficient or a derivative of ${\Gamma }^j{}_{0i}=0$. Every connection-product term contains a vanishing coefficient of these same types. Thus $C_{0i}=0$. The one-form $C_{ab}{k}^b=C_{a0}$ has only a time component, and is proportional to $k_a$. ThereforeChanging the overall curvature sign leaves this zero identity unchanged.

Finally a perfect fluid in signature $(+---)$ has $T_{ab}=(\rho +p)u_a{u}_b-p{g}_{ab}$ and $u^a{u}_a=1$. Contract the Einstein field equations with $k^b$, and antisymmetrize with $k_a$. Terms proportional to $g_{ab}$ disappear, as does the Ricci term just proved. This leavesTwo nonzero timelike vectors cannot be orthogonal: in the fluid's rest local inertial frame, $k\cdot u=k^0\ne 0$. The hypothesis $\rho +p>0$ therefore forces $u_{[a}{k}_{b]}=0$. Choosing $k$ future-directed gives the normalized perfect-fluid alignment in a static spacetime resultThe enthalpy hypothesis matters: when $\rho +p=0$, the stress-energy tensor is proportional to the metric tensor and does not select a velocity.

The Frobenius theorem says that hypersurface orthogonality of the nonzero covector $U_a$ is equivalent to $U_{[a}{\nabla }_b{U}_{c]}=0$, or locally $U_a=h{\nabla }_{a}S$ for some nonzero $h$. This immediately makes the screen projection of ${\nabla }_{[b}{U}_{a]}$ zero, since its antisymmetric part is a wedge product with $\nabla S$, which is proportional to $U$. Thus hypersurface orthogonality gives zero null twist.

For the converse, put $B_{ab}={\nabla }_b{U}_a$. Nullness gives $U^a{B}_{ab}=\frac{1}{2}{\nabla }_b(U^2)=0$, and the affine geodesic equation gives $B_{ab}{U}^b=0$. Resolve both tensor slots in the basis consisting of $U,N$ and the screen. These two identities exclude covector factors of $N$. Consequentlyfor some covector $q$. If $\hat{\omega }=0$, wedging this identity with $U$ makes $U_{[a}{\nabla }_b{U}_{c]}=0$. Applying Frobenius theorem completes the proof:This is the null twist vanishes exactly for hypersurface-orthogonal geodesics criterion; the geodesic assumption ensures that the screen criterion implies the full Frobenius condition. The resulting hypersurfaces are null, and the congruence follows their null generators.

Away from a bifurcation set, the normal to a Killing horizon is its nonzero Killing vector field $K$. Any affine tangent to the same generators has the form $k=hK$. In the optical tensor, derivatives of $h$ contribute a factor of $K_a$ and disappear under the screen-space projector. Consequentlyby the Killing equation. The trace and trace-free symmetric part are therefore zero: $\theta =0$ and ${\sigma }_{ab}=0$.

For the antisymmetric part, locally the horizon normal is proportional to a gradient. The restriction of its dual one-form to every tangent direction on the horizon is zero. Taking its exterior derivative and evaluating on two screen tangents therefore gives zero as well; equivalently this is the hypersurface-orthogonality consequence of the Frobenius theorem. Since both screen indices are tangential, $B_{[ab]}=0$. Thus the optical scalars on a Killing horizon satisfyNo affine normalization of the Killing vector field itself was assumed. At a regular bifurcation surface, $K$ vanishes but the smooth null generators do not; the conclusions for their optical quantities follow by continuity along each horizon branch.

The local flow of $X$ is the smooth map $\Phi :D\to M$, where $D\subset \mathbb{R}\times M$ is the maximal open domain of initial-point/time pairs, whose curve $t\mapsto \Phi (t,q)={\varphi }_t(q)$ is the maximal integral curve of a vector field through $q$, with ${\varphi }_0(q)=q$. Smooth dependence in the ordinary differential equation theorem gives joint smoothness and openness of $D$. Each time slice is a diffeomorphism between its domain and image, with inverse ${\varphi }_{-t}$.

Here is the group law, including its domain qualification. Fix $s,q$ where the relevant trajectories exist. Both curvessolve the same autonomous ordinary differential equation and have value ${\varphi }_s(q)$ at $t=0$. The uniqueness from the first part therefore giveson their common interval. These are local identities; neither the vector field nor its local flow has been assumed complete.

For the commuting local flows ${\varphi }_t$ and ${\psi }_s$, differentiate ${\varphi }_t({\psi }_s(q))={\psi }_s({\varphi }_t(q))$ with respect to $s$ at zero. The chain rule givesThus the pushforward of a vector field $Y$ by the $X$ flow equals $Y$. Interchanging the two fields gives the analogous identity for $X$ under ${\psi }_s$.

Now let ${\mathcal{D}}_q=\mathrm{span}(X_q,Y_q)$. Pointwise linear independence makes this a rank-two smooth distribution. The identity just proved implies $[X,Y]=0$: in coordinates, differentiating $D_q{\varphi }_t{Y}_q=Y_{{\varphi }_t(q)}$ at $t=0$ gives $DX(q)Y_q=DY(q)X_q$, precisely the vanishing Lie bracket of vector fields. The bracket of two local sections is also in $\mathcal{D}$, sinceand the analogous terms for $[fX,gX]$ and $[fY,gY]$ remain in the same span. Therefore $\mathcal{D}$ is an involutive distribution.

The Frobenius theorem says that a smooth constant-rank involutive distribution has local coordinates $(u^1,\dots ,u^n)$ in which it is spanned by ${\partial }_{u^1},{\partial }_{u^2}$, and has unique maximal connected integral manifolds through its points. These are the leaves of a regular foliation. Let $S$ be the maximal leaf through $p$. The global confinement to an immersed leaf gives the answer: $S$ is a two-dimensional immersed integral manifold and every smooth $\mathcal{D}$-tangent curve starting at $p$ stays in $S$.

For the curve assertion, in each Frobenius chart the transverse coordinate functions $u^3,\dots ,u^n$ have zero derivative along such a curve. Hence its segment in that chart stays in one plaque. On any compact subinterval of the curve's parameter interval, a finite chain of overlapping such segments connects the initial plaque to the last; all belong to the same maximal leaf. Exhausting the parameter interval proves the assertion for every time.

The two commuting local flows also give the leaf parametrization directly near $p$:Its partial derivatives are ${\partial }_{s}F=X_F$ and ${\partial }_{t}F=Y_F$, by the pushforward identity. Its differential has rank two. The constant rank theorem makes a sufficiently small image an embedded local plaque with tangent plane $\mathcal{D}$. This is a concrete local construction of the integral manifold.

There is a global distinction in the word “submanifold”. If it is required to mean an embedded submanifold, the assertion for all times is not valid in general. On the three-dimensional torus ${\mathbb{R}}^3/{\mathbb{Z}}^3$, takeThe fields are independent and commute, and their dense immersed cylinder in a three-dimensional torus through zero isIt is an injectively immersed cylinder $\mathbb{R}\times (\mathbb{R}/\mathbb{Z})$ and is dense in the torus. Indeed, fixing $x\mathrm{mod}1$ and varying $s$ by integers makes $\alpha s\mathrm{mod}1$ dense by an irrational rotation of the circle; $t$ is unrestricted. Every point of this leaf is reached by a tangent curve $r\mapsto (rs,rt,\alpha rs)\mathrm{mod}{\mathbb{Z}}^3$. An embedded surface containing it would, in a submanifold chart near one of its points, be a closed coordinate plane containing a dense subset of the ambient open chart, an impossibility. Thus the global conclusion uses an immersed leaf; the embedded conclusion is local, for curve segments remaining in a suitable chart.

Work over a coordinate neighborhood $U\subset B$ with coordinates $x^i$, and choose a basis $T_a$ of the Lie algebra, with $[T_a,T_b]=c^c{}_{ab}{T}_c$. Put $A=A_i^a{T}_{a}d{x}^i$. For each $T_a$ let $R_a$ be its right-invariant vector field on the fiber group. In matrix notation $R_a(\gamma )=T_a\gamma$, andThis minus sign is the right-invariant vector fields realize the opposite Lie algebra rule. It is essential here. Define the coordinate horizontal lifts of a principal connection byTheir projections are ${\pi }_{\ast }{H}_i={\partial }_i$, so they are linearly independent and there are exactly $\dim B$ of them. Also $d\gamma (H_i)=-A_i\gamma$, whenceThey therefore span the horizontal distribution of a principal connection on this trivialization.

The coefficients $A_i^a$ depend only on the base variables, so taking Lie brackets giveswhere $F_{ij}={\partial }_i{A}_j-{\partial }_j{A}_i+[A_i,A_j]$ and $F=\frac{1}{2}{F}_{ij}d{x}^i\wedge d{x}^j=dA+A\wedge A$. Since the right-invariant vector fields form a basis on each fiber, $\boxed{[H_i,H_j]=0\text{ for all }i,j⟺F=0}$. This is the local coordinate version of vanishing curvature of a principal connection and, by the Frobenius theorem, integrability of the horizontal distribution.

The coordinate qualification is necessary: horizontal lifts of arbitrary noncommuting base fields need not commute even for a flat principal connection. Nor does flatness guarantee a globally defined coordinate frame or a global horizontal section; holonomy of a connection can obstruct the latter. If the requested frame were read globally, the trivial flat bundle $S^2\times G$ would already be a counterexample: restricting a global horizontal frame to the identity-fiber section would trivialize $T{S}^2$, contrary to the Hairy ball theorem. The construction on each coordinate trivialization supplies the intended result without that global claim.

Locally write the null hypersurface as $F=0$. Its generators are tangent to its raised normal, so on it $U_a=h{\nabla }_{a}F$ for a nonzero scalar $h$. The antisymmetric derivative issince the Hessian of $F$ is symmetric for the torsion-free Levi-Civita connection. Each term contains $\nabla F$, which is proportional to $U$ and is killed by the screen-space projector. HenceThis is the null version of hypersurface orthogonality in the Frobenius theorem. A null hypersurface has no independent normal direction outside its tangent space: its null vector normal also generates it. That is why the same argument applies to the generators' null twist.

For the spherical null expansions in double-null coordinates, the cross-sectional area of a symmetry sphere is $A=4\pi r^2$. For the two affine null geodesic congruences, the definition of null expansion therefore givesThere is no extra derivative of $f$ in these formulas: the null expansion differentiates the transverse area, not the normalization of the tangent. Equivalently, the projected metric is $q_{AB}=r^2{\gamma }_{AB}$ and its optical tensor is $(k(r)/r)q_{AB}$, or $(\ell (r)/r)q_{AB}$ for the other congruence. It is purely trace, so both null shears vanish.

The null twist vanishes because the covector tangents are gradients of $-U$ and $-V$, and hence their exterior derivatives vanish. Each congruence is orthogonal to a family of null hypersurfaces. This also identifies the Frobenius theorem reason for zero null twist.

Differentiate the normalization of the timelike unit normal with the metric-compatible covariant derivative:Thus the derivative of the normal is already orthogonal to the normal in its second index. Expanding the second spatial projection tensor in the definition of the extrinsic curvature of a spatial hypersurface givesTo prove symmetry using Frobenius theorem, set $F_{\alpha \beta }={\nabla }_{\alpha }{n}_{\beta }-{\nabla }_{\beta }{n}_{\alpha }$. Hypersurface orthogonality is equivalent toContract this with $n^{\alpha }{P}^{\beta }{}_{\mu }{P}^{\lambda }{}_{\nu }$. The last two terms vanish because a projected normal vanishes, while $n^{\alpha }{n}_{\alpha }=-1$ in the first term. HenceThe antisymmetric part of the doubly projected covariant derivative of $n$ is therefore zero. This proves that hypersurface orthogonality implies symmetric extrinsic curvature:No geodesic assumption for the normal congruence is needed; its normal acceleration may be nonzero.