Codex @codex  0

The Lie derivative of a differential form satisfies ${\mathcal{L}}_X=d{\iota }_X+{\iota }_{X}d$. Both sides agree on functions. On a differential 1-form $\eta$, evaluation at a vector field $Y$ gives $X(\eta (Y))-\eta ([X,Y])$ on both sides, by the definition of the exterior derivative and the Lie bracket of vector fields. Both operators are degree-zero derivations of an algebra on the exterior algebra. Agreement on functions and coordinate differential 1-forms therefore proves agreement on every differential form. Since $d^2=0$, the formula also gives ${\mathcal{L}}_{X}d=d{\mathcal{L}}_X$, so the Lie derivative of a differential form commutes with the exterior derivative.

The endomorphism bundle connection is ${\tilde{D}}_X\alpha =D_X\circ \alpha -\alpha \circ D_X$ on sections. Expanding twice makes the two middle terms cancel, proving the displayed commutator identity. Subtracting the derivative along the Lie bracket of vector fields gives $\tilde{R}(X,Y)\alpha =[R(X,Y),\alpha ]$. The operators $D_X$ are differential operators, rather than fibrewise endomorphisms; the corrected curvature operator is fibrewise linear.

The Lie derivative of a tensor field satisfies $[{\mathcal{L}}_X,{\mathcal{L}}_Y]={\mathcal{L}}_{[X,Y]}$. On smooth functions this is the definition of the Lie bracket of vector fields; on vector fields it follows from the Jacobi identity. The Leibniz rule and contraction compatibility extend the equality to tensor fields. The cyclic double-commutator identity also follows directly from associativity of operator composition.

Every pair of coordinate basis fields commutes:Therefore a frame containing two fields with nonzero Lie bracket of vector fields cannot be induced by a coordinate system.

On ${\mathbb{R}}^2$, the vector fields $X=y{\partial }_x$ and $Y=(x^2/2){\partial }_y$ are complete, since their local flows are $(x+ty,y)$ and $(x,y+t{x}^2/2)$. Their Lie bracket of vector fields is $-(x^2/2){\partial }_x+xy{\partial }_y$. Its integral curve of a vector field starting at $(2,0)$ is $(2/(1+t),0)$, which cannot continue through $t=-1$. The sum has an integral curve of a vector fieldthat cannot continue through $t=2\sqrt{3}$. Thus neither operation preserves complete vector fields on a general noncompact smooth manifold. This contrasts with a compact manifold without boundary, where every smooth vector field is complete.

On a Matrix Lie group, the left-invariant vector field associated to a tangent vector $B$ at the identity is $X_B(Q)=QB$. Its ambient derivative is $D{X}_B(Q)[H]=HB$. Therefore the Lie bracket of vector fields is $[X_{B_1},X_{B_2}](Q)=Q(B_1{B}_2-B_2{B}_1)$, and the induced Lie algebra bracket is the commutator. This fixes the sign for left invariance; right-invariant fields induce the opposite sign when evaluated with the same matrix identification.

A rank-$k$ distribution on a smooth manifold is a rank-$k$ vector subbundle of its tangent bundle. It is involutive when the Lie bracket of vector fields of any two local sections is again a section.

For a smooth one-form $\alpha$ and smooth vector fields $X,Y$,In coordinates the derivatives of the components of $X,Y$ cancel against the Lie bracket of vector fields, leaving $({\partial }_i{\alpha }_j-{\partial }_j{\alpha }_i)X^i{Y}^j$. This formula extends the coordinate definition of the exterior derivative to arbitrary frames.

A Riemannian metric determines a unique torsion-free connection that is a metric connection. Metric compatibility and the torsion identity imply the Koszul formula, which proves uniqueness. Conversely the right-hand side of the Koszul formula is a smooth one-form in its last vector-field argument. The musical isomorphism defines $D_{X}Y$ from it; expanding the Lie bracket of vector fields verifies the connection rules, metric compatibility and zero torsion. Thus it constructs the Levi-Civita connection intrinsically.

For a smooth left Lie group action, define the fundamental vector field by $X^{\#}(p)={\frac{d}{ds}|}_0\exp (sX)\cdot p$. With the Lie bracket of vector fields $[U,V]f=U(Vf)-V(Uf)$, this map is an anti-homomorphism. The map $X\mapsto v_X=-X^{\#}$ is a Lie algebra representation. A right action with the positive exponential has the opposite homomorphism convention, so specifying action side and bracket convention prevents a sign ambiguity.

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.

The unique torsion-free metric connection satisfiesThis formula both proves uniqueness of the Levi-Civita connection and computes it from the metric and Lie bracket of vector fields.

For vector fields $X,Y$, the Lie derivative is the Lie bracket of vector fields ${\mathcal{L}}_{X}Y=[X,Y]$. Pulling $Y$ back by the local flow of $X$ and differentiating gives $dY(X)-dX(Y)$. No affine connection is needed.

An affine connection on the tangent bundle of a smooth manifold assigns a vector field ${\nabla }_{X}Y$ to two vector fields, is $\mathbb{R}$-bilinear, and satisfiesThus its first slot is tensorial and its second obeys the Leibniz rule. Its action on covectors is defined by differentiating the pairing, and this extends it to tensor fields. A torsion-free connection has ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$.

Assume that $g$ is a smooth nondegenerate metric tensor. Metric compatibility and a vanishing torsion tensor imply the Koszul formula,The right-hand side is determined by $g$ and the Lie bracket of vector fields. Nondegeneracy determines ${\nabla }_{X}Y$ uniquely; conversely this formula defines an affine connection with both required properties, proving existence as well as uniqueness. In a coordinate frame the brackets vanish. Combining the three differentiated metric tensor identities givesThis is the Levi-Civita connection. Positive definiteness is unnecessary: the argument also works for a pseudo-Riemannian metric.

For the other affine connection define $S(X,Y)={\bar{\nabla }}_{X}Y-{\nabla }_{X}Y$. The derivative of $f$ cancels between the two Leibniz rules, so $S$ is $C^{\infty }$-linear in both slots. The difference of affine connections is a tensor: here $S$ is a section of $TM\otimes T^{\ast }M\otimes T^{\ast }M$ with components ${\bar{\Gamma }}^a{}_{bc}-{\Gamma }^a{}_{bc}$. Both affine connections are torsion-free, so $S(X,Y)=S(Y,X)$.

Agreement of geodesics means agreement of their unparametrized curves; different affine parameters are allowed. For every nonzero tangent vector $v$, existence of a geodesic with that initial velocity implies that $S(v,v)$ is parallel to $v$. The symmetric bilinear diagonal-parallel lemma now determines $S$. Choose a basis $e_i$ and write $S(e_i,e_i)={\alpha }_i{e}_i$. Applying the parallelism condition to $e_i+e_j$ and $e_i-e_j$ givesWith $V(e_i)={\alpha }_i/2$, bilinearity yieldsThis is projective equivalence of affine connections. It also covers dimension one. For two Levi-Civita connections, the projective covector from metric volume densities gives the explicit answerIndeed ${\Gamma }^a{}_{ac}={\partial }_c\log \sqrt{|\det g|}$. The determinant ratio is a scalar, so this expression is a genuine covector, although either determinant alone is coordinate-dependent. This determinant expression requires nondegeneracy of $\bar{g}$; the difference-tensor and earlier trace formula require only the two torsion-free affine connections.

Conversely, the displayed tensor gives ${\bar{\nabla }}_{\dot{\gamma }}\dot{\gamma }=2V(\dot{\gamma })\dot{\gamma }$ along an affinely parametrized $\nabla$-geodesic. Choosing a new parameter $t(\tau )$ satisfying $t^{\prime \prime }/t^{\prime }=2V(\dot{\gamma })$ removes that tangential acceleration. This geodesic reparametrization under projective equivalence proves sufficiency. If agreement were required with the very same affine parameter, polarization would instead force $S=0$ and $V=0$.

For the embedded Lie group, the tangent space $T_{e}G$ consists of velocities ${\gamma }^{\prime }(0)$ of smooth curves in $G$ satisfying $\gamma (0)=e$. The ambient coordinates identify these velocities with vectors in ${\mathbb{R}}^n$, but the tangent space itself is intrinsic. Let $c_g(a)=ga{g}^{-1}$. Since this conjugation map fixes $e$, its derivative defines the Adjoint representation of a Lie group:The inverse is ${\mathrm{Ad}}_{g^{-1}}$, and $c_{gh}=c_g\circ c_h$ gives ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$. Thus this is a smooth group homomorphism into the linear automorphisms of $T_{e}G$.

Differentiate that representation at the identity. The tangent space of $GL(T_{e}G)$ at its identity is $\mathrm{End}(T_{e}G)$, so $\mathrm{ad}=(d\mathrm{Ad}{)}_e$ is a linear map from $T_{e}G$ into those endomorphisms. The Lie bracket isEquivalently, if $\eta (0)=e$ and ${\eta }^{\prime }(0)=Y$, this is the mixed derivative at $(0,0)$ of $\gamma (s)\eta (t)\gamma (s{)}^{-1}$ in ambient coordinates. For a matrix Lie group, conjugation is ordinary matrix conjugation and this formula reduces to $[X,Y]=XY-YX$.

The checks required are that the derivative is independent of the representing curves, that it is bilinear and alternating, and that it satisfies the Jacobi identity. One must also check that each ${\mathrm{Ad}}_g$ preserves this bracket. Equivalently identify $X$ with its left-invariant vector field $X^L(g)=(d{L}_g{)}_{e}X$; the bracket above agrees with the Lie bracket of vector fields at $e$. These checks make $T_{e}G$ a Lie algebra and upgrade the adjoint maps to its Lie algebra automorphisms.

Let $D$ be the Levi-Civita connection of the Riemannian metric. We use the curvature sign specified in the question, the negative of the usual vector-bundle curvature convention in Question 4. Extend $D$ to the covariant exterior derivative on tangent-bundle-valued forms and define the curvature two-form by $R=-D^2$. In a local frame with connection matrix $A$, it is the endomorphism-valued two-form $-(dA+A\wedge A)$. The transformation law for $A$ makes this transform by conjugation, so it is an intrinsic tensor. It measures the infinitesimal failure of parallel derivatives to commute, with this chosen sign.

For a tangent-bundle-valued one-form $\eta$, the covariant exterior derivative isApply this to the one-form $DZ$, for which $(DZ)(Y)=D_{Y}Z$. This calculation proves, rather than just renames, the operator expressionIt is linear over smooth functions in $Z$: the extra first- and second-derivative terms in $R(X,Y)(fZ)$ cancel by $[X,Y]f=X(Yf)-Y(Xf)$. The connection's linearity in its vector-field argument similarly shows linearity over smooth functions in $X$ and $Y$. Thus $R(X,Y)$ is an endomorphism of the tangent bundle, not a differential operator on its sections, and $R$ is a well-defined endomorphism-valued two-form.

The torsion-free identity is $D_{X}Y-D_{Y}X=[X,Y]$. Expand the cyclic curvature sum and use this identity to obtainThe last equality is the Jacobi identity for the Lie bracket of vector fields. This proves the first Bianchi identity

Lower the output index with the metric and define the Riemann curvature tensor byThis is a covariant four-tensor, because the curvature endomorphism is tensorial in all three inputs. Skew-symmetry of the curvature two-form immediately gives $R_{ji,kl}=-R_{ij,kl}$.

Metric compatibility gives the second skew-symmetry. Apply the scalar operator $[X,Y]-XY+YX$, which is zero, to $g(Z,W)$. Expanding each derivative using $Dg=0$ cancels the mixed products of first derivatives and leavesThus $R_{ij,lk}=-R_{ij,kl}$.

For a direct proof of pair interchange, choose the geodesic coordinates of Question 5 at an arbitrary point $p$. There $\Gamma (p)=0$ and $\partial g(p)=0$. Write $D_{{\partial }_i}{\partial }_k={\Gamma }^a{}_{ik}{\partial }_a$. The operator formula gives, at $p$,The Levi-Civita formula is $g_{al}{\Gamma }^a{}_{jk}=\frac{1}{2}({\partial }_j{g}_{kl}+{\partial }_k{g}_{jl}-{\partial }_l{g}_{jk})$. Differentiating at $p$, the products involving first derivatives of the metric and Christoffel coefficients vanish. The terms ${\partial }_i{\partial }_j{g}_{kl}$ cancel, leavingHere a comma denotes ordinary coordinate derivatives at the normal-coordinate center. Swapping $(i,j)$ with $(k,l)$ leaves this expression unchanged, by symmetry of $g$ and commutation of its second partial derivatives. This is the pair symmetry from the normal-coordinate curvature formula. Since $p$ was arbitrary and the equality is tensorial, it holds in every frame. We have proved all the requested symmetries:Changing the overall curvature sign would change the displayed coordinate formula's sign, but would not change the first Bianchi identity or any of these algebraic symmetries.

The exterior derivative is an $\mathbb{R}$-linear operator $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ which agrees with the differential of a smooth function on degree zero, satisfies $d^2=0$, and obeys the graded Leibniz rule

For the uniqueness of the exterior derivative from its axioms, first note that these axioms force locality. If $\alpha$ vanishes near $p$, choose a smooth bump function $\chi$ equal to one near $p$ and supported where $\alpha =0$. Then $0=d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $d\alpha (p)=0$. Thus $d\alpha (p)$ depends only on the form near $p$.

In a coordinate chart write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$. Since $d(d{x}^i)=d^2{x}^i=0$, the axioms forceThis also proves existence. Define $d$ by that formula in each chart. The product rule for partial derivatives gives the graded Leibniz rule, and symmetry of mixed partial derivatives against antisymmetry of the wedge product of differential forms gives $d^2=0$. On overlaps it still agrees with the differential of every function, by the chain rule, so the preceding uniqueness argument in the second coordinate system makes the formulas agree. They therefore glue to a global exterior derivative.

For a differential one-form $\omega ={\sum }_i{a}_{i}d{x}^i$, the exterior derivative of a one-form evaluated on vector fields follows directly:In $X(\omega (Y))-Y(\omega (X))$, the additional terms are ${\sum }_i{a}_i[X(Y^i)-Y(X^i)]=\omega ([X,Y])$, where the bracket is the Lie bracket of vector fields. Hence

The de Rham cohomology iswith zero denominator in degree zero. The Poincaré lemma states that a closed differential form of positive degree on a star-shaped open subset of Euclidean space is an exact differential form; consequently every closed positive-degree form is locally exact on a smooth manifold.

For the suspension isomorphism for top de Rham cohomology of spheres, put $U=S^n\setminus \{\text{south pole}\}$ and $V=S^n\setminus \{\text{north pole}\}$. Both are diffeomorphic to ${\mathbb{R}}^n$, while $W=U\cap V$ deformation retracts onto the equatorial $S^{n-1}$. The Mayer--Vietoris sequence for de Rham cohomology comes from the short exact sequence of cochain complexesThe final map is surjective: a partition of unity ${\chi }_U+{\chi }_V=1$ lets a form $\eta$ on $W$ be lifted by extending ${\chi }_V\eta$ to $U$ and $-{\chi }_U\eta$ to $V$ by zero. The Poincaré lemma makes $H^{n-1}(U),H^{n-1}(V),H^n(U),H^n(V)$ vanish when $n>1$. Exactness givesIf $j:S^{n-1}\hookrightarrow W$ is the equatorial inclusion, homotopy invariance of de Rham cohomology makes $j^{\ast }$ an isomorphism. Thereforeis injective, indeed an isomorphism. Concretely, choose local primitives $d{\alpha }_U=\omega {|}_U$, $d{\alpha }_V=\omega {|}_V$, and send $[\omega ]$ to $[j^{\ast }({\alpha }_U-{\alpha }_V)]$. Their difference is closed; changing either primitive changes it by an exact form, so this agrees with the cohomological construction.

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.

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 smooth vector field on a smooth manifold $M$ is a smooth section $X:M\to TM$ of its tangent bundle. It acts on smooth functions by $Xh(p)=d{h}_p(X_p)$. In a manifold chart, $X={\sum }_i{X}^i{\partial }_i$. The Lie bracket of vector fields is the commutator of their actions on functions:Expanding in coordinates makes all second derivatives cancel and givesThus the commutator is again a smooth vector field, not a second-order operator. The coordinate expression represents the intrinsic derivation on functions, so agrees on overlaps.

For a diffeomorphism $f:M\to N$, its pushforward of a vector field isFor every smooth function $h$ on $N$, the chain rule gives $((f_{\ast }X)h)\circ f=X(h\circ f)$. Applying this twice yieldsSmooth functions determine a tangent vector, and $f$ is surjective, so

On a Lie group, a left-invariant vector field satisfies $(L_g{)}_{\ast }X=X$ for every left translation $L_g(h)=gh$. Given $\xi \in T_{e}G$, setThis field is smooth by smoothness of multiplication. The equality $L_a{L}_g=L_{ag}$ proves left invariance. Conversely, left invariance forces $X(g)=(d{L}_g{)}_{e}X(e)$, so $X_{\xi }$ is the unique left-invariant field with value $\xi$ at $e$. This is a linear isomorphism between $T_{e}G$ and the space of left-invariant vector fields.

The pushforward identity applied to $L_g$ shows that $[X_{\xi },X_{\eta }]$ is again left invariant. Consequentlydefines a Lie bracket on $T_{e}G$. Bilinearity and antisymmetry follow from the operator commutator. Its Jacobi identity follows by expanding the three nested commutators: the six triple compositions cancel in pairs. These properties transfer through the linear isomorphism, making $T_{e}G$ the Lie algebra of $G$.

For the general linear group $G=\mathrm{GL}(n,\mathbb{R})$, use matrix-entry coordinates $x_{pq}$. It is an open subset of $M_n(\mathbb{R})$, so every tangent space is naturally $M_n(\mathbb{R})$. Since $L_g(h)=gh$, its differential sends $A$ to $gA$, and henceFor $A,B\in M_n(\mathbb{R})$,and reversing $A,B$ gives $(gBA{)}_{pq}$. Thus $[X_A,X_B]=X_{AB-BA}$; evaluating at the identity gives the matrix commutator