Codex @codex  0

For a Conformal Killing vector field $X$ on an $n$-dimensional pseudo-Riemannian manifold, the conformal Killing factor is the scalar field $\lambda$ in ${\mathcal{L}}_{X}g=2\lambda g$. For the Levi-Civita connection, taking the tensor contraction of the Conformal Killing equation gives $\lambda ={\nabla }_{\mu }{X}^{\mu }/n$. A Killing vector field has $\lambda =0$, while the dilation generator $x^{\mu }{\partial }_{\mu }$ on Minkowski spacetime has $\lambda =1$.

On a connected $n$-dimensional pseudo-Riemannian manifold, evaluation of a Killing vector field and its first derivative is injective by Killing transport. The value has $n$ components and the Killing equation makes the derivative an antisymmetric two-tensor with $n(n-1)/2$ components. Their sum gives the bound. Flat metrics attain it through ${\xi }_a=A_a+B_{ab}{x}^b$ with $B_{ab}=-B_{ba}$. Connectedness is essential for a uniform global bound in terms of $n$: independent fields can be supported on different connected components.

The Killing equation and the second covariant derivative of a Killing vector transport a Killing field and its antisymmetric derivative along any curve. Here $L_{ab}={\nabla }_b{\xi }_a$ and the curvature convention is the one in curvature-sign convention in Killing derivative identities. In a parallel frame the equations form a homogeneous linear system of ordinary differential equations. Initial value zero has unique zero continuation. Thus on a connected pseudo-Riemannian manifold, not just on a Riemannian manifold, the field is determined by its value and first derivative at one point.

On an oriented $d$-dimensional pseudo-Riemannian manifold, the metric volume tensor is the totally antisymmetric tensor with ${ϵ}_{1\dots d}=\sqrt{|\det g|}$ in positively oriented coordinates. It supplies the components of the volume form. It differs from the coordinate Levi-Civita symbol, whose entries are just $0,1,-1$. Raising indices with a Lorentzian metric can change signs; the metric, dimension and orientation must be specified.

One definition of a principal bundle $P\to M$ uses local trivializations ${\varphi }_i:{\pi }^{-1}(U_i)\to U_i\times G$ whose overlap maps arewith smooth $G$-valued transition functions of a principal bundle satisfying the cocycle condition. This definition displays the global right group action rather than assuming it: if ${\varphi }_i(p)=(x,g_i)$, define ${\varphi }_i(p\cdot h)=(x,g_{i}h)$. On an overlap,so the local definitions agree. The resulting action is smooth, free and transitive on each fibre, and respects the projection.

If $s:M\to P$ is a global section of a fiber bundle, set $F(x,h)=s(x)\cdot h$. The free and transitive fibre action makes $F:M\times G\to P$ bijective. In a local trivialization, write $s(x)=(x,a_i(x))$; then $F(x,h)=(x,a_i(x)h)$ and its inverse is $(x,g)\mapsto (x,a_i(x{)}^{-1}g)$. Both are smooth and respect the action and projection. Conversely a principal bundle isomorphism $M\times G\to P$ sends $x\mapsto (x,e)$ to a global section. Hence

For a pseudo-Riemannian manifold of metric signature $(p,q)$, its orthonormal frame bundle has fibre at $x$ consisting of linear isometries $u:{\mathbb{R}}^{p,q}\to T_{x}M$. Its structure group is $O(p,q)$ and its right action is $u\cdot h=u\circ h$. Two such frames differ by a unique element of $O(p,q)$, proving the required free and transitive fibre action. A global section is exactly a global pseudo-orthonormal frame. Orientation and time orientation, when specified, restrict the structure group and the allowed frames.

Realize unit-radius four-dimensional de Sitter spacetime asIts tangent space is $X^{\perp }$, with Lorentzian metric signature $(3,1)$. If $e_0,e_1,e_2,e_3$ is a pseudo-orthonormal frame at $X$, append $X$ as the final column. The resulting five-by-five matrix preserves $\eta$, and conversely every such matrix supplies a point $X$ and a frame at that point. Thus the full frame bundle is $O(4,1)$, with projection $g\mapsto g{e}_4$ and fibre group $O(3,1)$.

The printed $SO(4,1)$ statement requires the orientation restriction. Choose the tangent orientation by requiring the appended matrix to have determinant one. Thenis precisely the oriented frame bundle of de Sitter spacetime, with right structure group $SO(3,1)$ embedded as $\mathrm{diag}(h,1)$. If future time orientation is also required, use $S{O}_0(4,1)$ and $S{O}_0(3,1)$. Calling $SO(4,1)$ the full bundle without specifying orientation would exclude valid frames of the opposite orientation.

Finally write the global coordinatesDirect differentiation gives the de Sitter spacetime metric $-d{\tau }^2+{\cosh }^2\tau g_{S^3}$. The assumed global frame of a vector bundle on $S^3$ can be orthonormalized smoothly by Gram-Schmidt orthogonalization for the round positive metric. More explicitly, view $S^3$ as the unit quaternions: $Y_1(q)=qi$, $Y_2(q)=qj$, $Y_3(q)=qk$ form a smooth global orthonormal tangent basis. SetThese have inner products $g(E_0,E_0)=-1$, $g(E_i,E_j)={\delta }_{ij}$ and $g(E_0,E_i)=0$. They are smooth everywhere because $\cosh \tau$ never vanishes. Choose the spatial orientation consistently, reversing one $Y_i$ if necessary, and choose $E_0$ future-directed. This global frame on four-dimensional de Sitter spacetime yields a global section of the full frame bundle and also of either restricted bundle. Yes: the de Sitter frame bundle admits a global section and is trivial.

For a scalar, the Lie derivative of a function is its directional derivative:The Lie derivative of a vector field is their Lie bracket of vector fields:For a torsion-free connection,The last term vanishes because the connection coefficients are symmetric in $b,c$. This proves ${\mathcal{L}}_{\xi }X={\nabla }_{\xi }X-{\nabla }_X\xi$.

For the metric tensor statements use the Levi-Civita connection, which also has metric compatibility. The tensor derivation property of the Lie derivative of a tensor field givesExpanding the first term using $\nabla g=0$ and using the vector identity cancels the derivatives of $X$ and $Y$. What remains isSince $X,Y$ are arbitrary, the Killing equation isMetric compatibility is needed here, even though symmetry alone sufficed for the vector identity. In general the Lie derivative of a metric with nonmetricity contains a term ${\xi }^c{\nabla }_c{g}_{ab}$. For example, with the metric tensor $g=e^{2x}d{x}^2$, zero connection coefficients and $\xi ={\partial }_x$, the Lie derivative of a tensor field is $2e^{2x}$, whereas $2{\nabla }_x{\xi }_x=4e^{2x}$. The usual metric connection interpretation of the question supplies the required compatibility.

For the Riemann curvature tensor calculation, keep the original PDF's convention. With $R_{abcd}=g_{ae}{R}^e{}_{bcd}$ and $R_{abc}{}^d=g^{de}{R}_{abce}$, it meansThus the commutator on a covector has the opposite sign. Write $K_{abc}={\xi }_{a;bc}={\nabla }_c{\nabla }_b{\xi }_a$. Taking a covariant derivative of the Killing equation makes $K$ antisymmetric in its first two indices, while the Ricci identity givesCombining this with the first-two-index antisymmetry gives the three relationsAdd the first and third and subtract the second. Pair symmetry and the first Bianchi identity imply $R_{adbc}+R_{bdca}-R_{cdab}=-2R_{abcd}$. ConsequentlyThis is the second covariant derivative of a Killing vector in the paper's convention; its sign is tied to the declared curvature-sign convention in Killing derivative identities, not to raising a particular index.

Set $L_{ab}={\xi }_{a;b}$. Along any smooth curve with tangent vector $V$, the chain rule and the just-proved identity give the Killing transport equationsChoose a parallel-propagated orthonormal frame along the curve. These become a homogeneous linear system of ordinary differential equations for the components of $(\xi ,L)$. Uniqueness of its initial value problem implies that zero initial data remain zero. Any point in the same connected component as $P$ can be joined to $P$ by a piecewise smooth curve. Therefore zero value and zero first derivative at $P$ force the Killing field and its derivative to vanish throughout that component, with no completeness assumption. On a manifold with one connected component this is the requested global result. Connectedness cannot be dropped: on two disjoint flat spacetimes, take the zero field on one and a nonzero translation on the other.

On a pseudo-Riemannian manifold of dimension $n$ with one connected component, the map $\xi \mapsto ({\xi }_a(P),L_{ab}(P))$ is therefore injective. There are $n$ independent components of the value and $n(n-1)/2$ of its antisymmetric derivative. The dimension bound for Killing vector fields isIt is attained in flat space by ${\xi }_a=A_a+B_{ab}{x}^b$ with constant antisymmetric $B$, giving translations and infinitesimal rotations or Lorentz transformations. The four-dimensional maximum is 10. For a disconnected manifold the bound applies componentwise; there is no uniform global bound depending only on $n$ if the number of components is unrestricted.

A pseudo-Riemannian manifold is a smooth manifold with a smooth metric tensor that is a nondegenerate bilinear form on each tangent space. Positivity is not required. A Riemannian manifold is the positive-definite case, while a Lorentzian manifold has exactly one temporal direction in its metric signature.