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The connection difference as an endomorphism-valued one-form follows because the graded Leibniz rule terms cancel:Thus $a=D_1-D$ is a global endomorphism-valued one-form. To define the endomorphism bundle connection, requireThe right side is $C^{\infty }$-linear in $s$, so it is indeed an endomorphism-valued one-form, and replacing $\varphi$ by $f\varphi$ gives $df\otimes \varphi +f\tilde{D}\varphi$. In a frame with connection matrix $A$, its extension to an endomorphism-valued $k$-form $b$ isThe endomorphism-valued exterior product combines the wedge of differential forms with composition in $\mathrm{End}E$. In particular,Expanding $d(A+a)+(A+a)\wedge (A+a)$ proves the curvature difference formula:

The Chern connection is characterized by metric compatibility and $D^{0,1}={\bar{\partial }}_E$. Take Hermitian inner products linear in the first variable. In a holomorphic local frame write $⟨eu,ev⟩=v^{†}Hu$, with $H$ a positive Hermitian matrix. The second condition forces $A^{0,1}=0$, and metric compatibility isIts $(1,0)$ component determines $A$ uniquely:Conversely this formula and its conjugate transpose satisfy the metric equation. Under a holomorphic local frame change, $H^{\prime }=g^{†}Hg$ and $\partial g^{†}=0$, so the displayed formula transforms by $A^{\prime }=g^{-1}Ag+g^{-1}\partial g$. Hence the local matrices patch, proving existence and uniqueness.

Differentiating $H^{-1}$ givesTherefore the $(2,0)$ vector-bundle curvature is zero. There is no $(0,2)$ component, andThe last equality is ${\bar{\partial }}^{2}A=0$ in holomorphic local frames, so is intrinsic. This is Dolbeault closedness of Chern curvature. The induced endomorphism bundle connection is itself the Chern connection for the Hilbert-Schmidt metric on $\mathrm{End}E$: the tensor product of a metric-compatible connection and its dual preserves the induced metric, and its $(0,1)$ part is the induced holomorphic Dolbeault operator. These two properties and uniqueness suffice; no further matrix calculation is needed.

For two metrics, both Chern connections have the same $(0,1)$ part, so $a\in A^{1,0}(\mathrm{End}E)$. The $(1,1)$ part of the curvature difference formula is just ${\bar{\partial }}_{\mathrm{End}E}a$; the other two terms have type $(2,0)$, which must cancel because both vector-bundle curvatures have type $(1,1)$. ThusBy Dolbeault theorem, this proves that the curvature Dolbeault classis independent of the metric.

Endomorphism composition and wedge product of differential forms define the powers in $A^{k,k}(\mathrm{End}E)$, rather than in a tensor power of the endomorphism bundle. The Dolbeault operator is a derivation, so $\bar{\partial }({\Theta }^k)=0$. The noncommutative telescoping identity gives the stronger explicit independence statementAll powers of vector-bundle curvature have even total degree, so the derivation introduces no extra signs before $a$. Consequentlyis metric independent. For $k>n$, the form space and the resulting class are zero. This proves metric independence of curvature Dolbeault powers without assuming that endomorphism-valued forms commute.

The torsion tensor of an affine connection is $\mathcal{T}(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$, where the last term is the Lie bracket of vector fields. A torsion-free connection has $\mathcal{T}=0$. Since a coordinate basis commutes, this is equivalent there to ${\Gamma }^{\mu }{}_{\nu \rho }={\Gamma }^{\mu }{}_{\rho \nu }$; in a general basis its commutator must also be included.

The Levi-Civita connection is the unique torsion-free connection with metric compatibility, $\nabla g=0$. To derive its Christoffel symbols, lower the first index: ${\Gamma }_{\sigma \nu \rho }=g_{\sigma \mu }{\Gamma }^{\mu }{}_{\nu \rho }$. Metric compatibility givesWrite down this identity with the derivative indices $\rho ,\nu ,\sigma$, add the first two identities and subtract the third. Symmetry in the last two indices, supplied by the torsion-free connection, cancels all terms except twice ${\Gamma }_{\sigma \nu \rho }$. Raising the first index yieldsThis calculation proves uniqueness. Conversely, these Christoffel symbols are symmetric in $\nu ,\rho$ and satisfy $\nabla g=0$ by substitution, proving existence as well.

For every member of the geodesic variation, proper time normalization gives $g(T,T)=-1$. By metric compatibility, the geodesic equation, and ${\nabla }_{T}S={\nabla }_{S}T$,The last equality uses normalization across the whole variation, not merely constancy along one curve. Hence $g(S,T)$ is constant along each timelike geodesic, provingThis is orthogonality of proper-time deviation vectors.

Contracting the assumed Riemann curvature tensor gives the Ricci tensorThe contracted Bianchi identity and metric compatibility now implyThus ${\nabla }_{b}R=0$, so the scalar curvature is constant on each connected component of spacetime. The sectional curvature is consequently the constant $\kappa =R/12$. This is the four-dimensional case of the Schur theorem in pseudo-Riemannian geometry.

Let $\kappa =R/12$, which is constant by the preceding argument. Substitution into geodesic deviation givesusing $g(T,T)=-1$ and the established orthogonality. With $V=DS$, metric compatibility yields $\dot{f}=2g(S,V)$ andThereforeBoth $S$ and $DS$ are orthogonal to $T$, so their squared norms are nonnegative. The minus sign in this conserved deviation energy in constant curvature reflects $D^{2}S=+\kappa S$, rather than an ordinary oscillator when $R>0$.

Let $u$ be the tangent vector to a geodesic with affine parameter $\lambda$. Its geodesic equation is ${\nabla }_{u}u=0$. For a Killing vector field $\xi$, metric compatibility givesThusThis geodesic conserved quantity from a Killing vector is independent of whether the geodesic is timelike, spacelike or null. For a timelike geodesic one can use proper time; for a null geodesic choose an affine parameter. An arbitrary nonaffine tangent rescales the displayed quantity and need not make it constant.

A Killing vector field $\xi$ generates a local one-parameter family of isometries of the metric tensor. Equivalently, its tensor Lie derivative of the metric vanishes:The latter is the Killing equation; the equivalence uses metric compatibility and the torsion-free connection.

The unimodular spatial metric separates volume from shape: writing $h_{ij}=a^2{\tilde{h}}_{ij}$ and $a^6=h=\det h_{ij}$ gives $\det \tilde{h}=1$. Contract the extrinsic curvature with a negative shift with $h^{ij}$. Metric compatibility and the Jacobi determinant derivative formula giveConsequentlyFor zero shift vector, $\dot{h}/h=6\dot{a}/a$, soThe infinitesimal spatial volume is $\sqrt{h}{d}^{3}x=a^3{d}^{3}x$; its logarithmic change per unit normal proper time is $3H$. Thus this $H$ is the local volume Hubble parameter, reducing to the usual Hubble parameter in a homogeneous universe. The determinant-defined $a$ refers to fixed spatial coordinates, while this normal volume expansion has a geometric interpretation.

Combining metric compatibility with the torsion-free condition forces the Koszul formula:Nondegeneracy of $g$ proves uniqueness of the Levi-Civita connection. For existence, use the right side to define $2g({\nabla }_{X}Y,Z)$. Expanding brackets shows that this expression is $C^{\infty }$-linear in $Z$, so it defines a smooth one-form; the musical isomorphism gives the required vector field. The same expansion shows ${\nabla }_{aX}Y=a{\nabla }_{X}Y$, additivity, and ${\nabla }_X(aY)=X(a)Y+a{\nabla }_{X}Y$, establishing the connection rules. Subtracting the formulas with $X,Y$ exchanged gives ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$. Adding the formulas pairing ${\nabla }_{X}Y$ with $Z$ and ${\nabla }_{X}Z$ with $Y$ gives metric compatibility. Thus this construction has both required properties.

In coordinate vector fields the brackets vanish. Consequently the Christoffel symbols and coordinate derivative areRepeated indices are summed. The displayed connection type in the source is best understood in its standard form on two vector fields; if the first input is an individual tangent vector at a point, the output lies in the tangent fiber at that point rather than in the space of global sections.

For the parallel metrics with a common Levi-Civita connection, let $\gamma$ be a piecewise smooth path from the point $x$ of equality to any $y$. Connected smooth manifolds admit such paths because coordinate balls are path connected. Parallel transport $P_{\gamma }$ for the common connection is invertible and preserves both metrics. ThereforeEvery pair of tangent vectors at $y$ arises this way, so $\boxed{\tilde{g}=g}$ everywhere.

Dropping the agreement at one point removes the conclusion. For any constant $c>0$, $\tilde{g}=cg$ has the same Christoffel symbols. Nor must the two metrics be proportional: on ${\mathbb{R}}^n$, $n\ge 2$, the constant metrics ${\sum }_i(d{x}^i{)}^2$ and $2(d{x}^1{)}^2+{\sum }_{i\ge 2}(d{x}^i{)}^2$ both have zero connection coefficients. In general write $\tilde{g}(u,v)=g(Au,v)$. Since both metrics are parallel, $0=({\nabla }_X\tilde{g})(Y,Z)=g(({\nabla }_{X}A)Y,Z)$, so $\nabla A=0$. Conversely, a positive $g$-self-adjoint parallel $A$ makes the same torsion-free connection compatible with $\tilde{g}$, proving equality of their Levi-Civita connections. Thus one-point agreement specifies $A=I$ and forces it everywhere; without it, nontrivial parallel choices can remain.

Fix the curvature conventionTo prove tensoriality, use $[hX,Y]=h[X,Y]-Y(h)X$ and the connection rules. The two $Y(h){\nabla }_{X}Z$ terms cancel, giving $R(hX,Y)Z=hR(X,Y)Z$. Antisymmetry in $X,Y$ gives linearity over smooth functions in the second input. Expanding the third input givesIt is therefore a smooth tensor of type $(1,3)$. Lowering the output with $g$ gives the type $(0,4)$ Riemann curvature tensor $R_4(X,Y,Z,W)=g(R(X,Y)Z,W)$.

For an independent pair $X,Y$, define sectional curvature byMetric compatibility gives $g(R(X,Y)Z,W)=-g(R(X,Y)W,Z)$ by applying $XY-YX-[X,Y]$ to $g(Z,W)$. Together with antisymmetry in $X,Y$, this shows that replacing the pair by $(aX+bY,cX+dY)$ multiplies both numerator and denominator by $(ad-bc{)}^2$. Thus the value depends only on the plane. Define Ricci curvature by $\mathrm{Ric}(Y,Z)={\sum }_{i}g(R(e_i,Y)Z,e_i)$ for any orthonormal basis; a trace is independent of the orthonormal basis.

For the curvature of the round unit sphere, the outward unit normal is the position vector $p$. The tangential projection of ambient differentiation is torsion-free and has metric compatibility, so uniqueness identifies it with $\nabla$. Ambient differentiation $D$ satisfies $D_{X}p=X$ andsince differentiating $g(Y,p)=0$ gives its normal component. The ambient curvature is zero. Take tangential components of $D_X{D}_{Y}Z-D_Y{D}_{X}Z-D_{[X,Y]}Z=0$ to obtainHenceEvery two-plane has sectional curvature one. Tracing the first formula gives $\mathrm{Ric}(Y,Z)=ng(Y,Z)-g(Y,Z)$, and consequentlyThus the sphere is an Einstein manifold. When $n=1$ there are no tangent two-planes, the curvature tensor is zero and the same Ricci formula gives zero. The declared slot convention fixes all signs.

Variation of the inverse metric relation $g^{ac}{g}_{cb}={\delta }^a{}_b$ gives $\delta g^{ab}=-h^{ab}$. Hence the metric variation of scalar curvature isPut $h=g^{ab}{h}_{ab}$. The connection variation gives the contractionsUsing metric compatibility and the Palatini identity,Renaming dummy indices givesNo interchange of covariant derivatives on a tensor is needed in this derivation, so no hidden curvature-commutator term is discarded.

Write $h_{ab}=\delta g_{ab}$ and $C^a{}_{bc}=\delta {\Gamma }^a{}_{bc}$. The difference between two affine connections is a tensor, so its infinitesimal change $C$ is a type $(1,2)$ tensor field. Both affine connections are torsion-free, giving $C^a{}_{bc}=C^a{}_{cb}$. Varying metric compatibility givesLower the first index of $C$. Add the versions with derivatives $b,c$ and subtract the one with derivative $d$; the lower-slot symmetry cancels the unwanted terms, leavingThereforeAll covariant derivatives use the original Levi-Civita connection. The PDF has ${\nabla }_b{h}_{dc}$ as its second term; the converted TeX's ${\nabla }_d{h}_{dc}$ is a transcription error.

Treat the printed $e^{\mu }$ as an orthonormal coframe, with frame metric $\eta =\mathrm{diag}(-1,1,1,1)$. Work on a patch $z\ne 0$ and $F=1-\alpha z^3>0$, so the given real coframe exists. Define $f=\sqrt{F}$ and $A=z{f}^{\prime }-f$. The exterior derivatives areThe connection 1-forms satisfying Cartan's first structure equation and metric compatibility areAll other forms vanish. With a Lorentzian frame, it is the lowered forms ${\omega }_{ab}={\eta }_{ac}{\omega }^c{}_b$ that are antisymmetric; the two mixed time–space forms are equal. These displayed forms solve the torsion-free structure equation, and uniqueness of the Levi-Civita connection identifies them as the required connection.

Apply Cartan's second structure equation. For example,The remaining derivatives give ${\Theta }^1{}_3=Af{e}^1\wedge e^3$ and ${\Theta }^1{}_2=-f^2{e}^1\wedge e^2$, with analogous forms for index 2. Put $q=\alpha z^3$ and $B=1+q/2$. From $f^2=1-q$,Therefore all six independent curvature 2-forms areThe other six are fixed by ${\Theta }^b{}_a=-{\eta }_{aa}{\eta }_{bb}{\Theta }^a{}_b$ for $a\ne b$, with no sum: equal for time–space pairs and opposite for spatial pairs. All diagonal forms are zero. The Lorentzian connection-form antisymmetry is essential to these signs.

Let $\pi :T^{\ast }M\to M$ be the cotangent bundle projection. Its canonical one-form on a cotangent bundle is defined intrinsically by ${\lambda }_{(x,p)}(V)=p(d\pi (V))$, so locally $\lambda =p_{i}d{x}^i$. Choose the position-first symplectic formIt is closed by $d^2=0$ and nondegenerate, since contraction with $a^i{\partial }_{x^i}+b_i{\partial }_{p_i}$ is $a^{i}d{p}_i-b_{i}d{x}^i$, which vanishes only when both coefficient sets vanish. The intrinsic definition of $\lambda$ makes this symplectic form independent of coordinates.

Use the convention ${\iota }_{X_H}\omega =dH$. Then the Hamiltonian vector field and the Poisson bracket areso $X_H(F)=\{F,H\}$. Choosing $\omega =d\lambda$ and ${\iota }_{X_H}\omega =-dH$ gives the same equations; choosing only one of these sign changes would reverse the flow.

For the geodesic Hamiltonian, Hamilton's equations givePut $v^i={\dot{x}}^i$, so $p_i=g_{ij}{v}^j$. Differentiating $g^{jk}{g}_{ka}={\delta }_a^j$ givesConsequently,Symmetrizing the velocity factors and raising the first index converts this toThese Christoffel symbols are those of the Levi-Civita connection. Thus a Hamiltonian integral curve of a vector field projects to an affinely parametrized geodesic. Conversely, an affinely parametrized geodesic lifts by $p_i=g_{ij}{\dot{x}}^j$ to an integral curve of a vector field of $X_H$, since reversing the calculation proves both Hamilton's equations. This is the geodesic flow on the cotangent bundle. The conserved Hamiltonian is half the squared speed, and the zero-energy case gives the constant geodesics.

A quadratic homogeneous polynomial depends only on the symmetric part of its coefficient matrix. Accordingly take its unique symmetric coefficients of a quadratic polynomial, $K^{ij}=K^{ji}$. This is the standard implicit convention in identifying such polynomials with symmetric tensors. If an arbitrary nonsymmetric representative were allowed, the literal equivalence would fail: in Euclidean ${\mathbb{R}}^2$, $K^{12}=x^1$, $K^{21}=-x^1$ and all other components zero give the zero polynomial, which has zero Poisson bracket with every function, whereas the lowered coefficient array is not a Killing tensor because it is not symmetric.

Lower the indices of the symmetric coefficient tensor using the Riemannian metric. Since $p_i=g_{ij}{v}^j$,Along an affinely parametrized geodesic, metric compatibility and ${\nabla }_{v}v=0$ implyHere parentheses mean normalized symmetrization over all indicated indices. The left side is $\{K,H\}$ by the Hamiltonian vector field convention. Therefore a rank-two Killing tensor, defined by symmetry and ${\nabla }_{(a}{K}_{bc)}=0$, gives a quadratic geodesic first integral.

Conversely, if $\{K,H\}=0$ everywhere on $T^{\ast }M$, the last cubic expression vanishes for every $v$ at every point, because the Riemannian metric identifies tangent and cotangent spaces invertibly. A symmetric trilinear form is determined by its diagonal cubic polynomial: equivalently, compare its coefficients, or polarize the cubic. Hence ${\nabla }_{(a}{K}_{bc)}=0$. This proves both directions:with symmetry understood on the coefficient representative from the outset. The equivalence is local and does not require geodesic completeness.

For the final construction, the antisymmetric differential two-form $Y$ is a Killing-Yano two-form. Its defining equation says ${\nabla }_a{Y}_{bc}=-{\nabla }_b{Y}_{ac}$. Antisymmetry of $Y$ also says ${\nabla }_a{Y}_{bc}=-{\nabla }_a{Y}_{cb}$, so the three-index tensor ${\nabla }_a{Y}_{bc}$ is totally antisymmetric.

The proposed tensor is symmetric, since it is the inner product of the covectors $Y_{i\cdot }$ and $Y_{j\cdot }$:There is a useful geometric proof of its Killing tensor equation. Along any affinely parametrized geodesic, define $w_k=Y_{ik}{v}^i$. ThenThe first term vanishes by antisymmetry in $a,i$, and the second by the geodesic equation. Thus $w$ is carried by parallel transport. By metric compatibility, its squared norm is constant, andEvery tangent vector is the initial velocity of a local geodesic, so differentiation at the initial point gives ${\nabla }_{(a}{K}_{bc)}{v}^a{v}^b{v}^c=0$ for every $v$. The same cubic-coefficient argument provesThis proves that the square of a Killing-Yano two-form is a rank-two Killing tensor and supplies a nonnegative quadratic geodesic first integral. The argument also explains the conserved quantity: it is the squared norm of a covector that the Killing-Yano two-form makes parallel along every geodesic.

With the $+---$ signature and $n^2=1$, direct expansion of the two spatial projection tensors givesThis restriction is negative definite on spatial vectors. The positive spatial induced metric is instead $h_{\alpha \beta }=n_{\alpha }{n}_{\beta }-g_{\alpha \beta }=-q_{\alpha \beta }$, whose pullback to a time slice is the $h_{ij}$ used in the line element. The spatial metric sign for a unit timelike normal is important here: the plus sign in the PDF's claimed equality is incompatible with its normal normalization: $g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ is not even transverse to $n^{\alpha }$.

The spatial covariant derivative of a spatial tensor projects every index, including its derivative index. Projection only on the derivative index is sufficient for a scalar but not for a general tensor. Using metric compatibility of the spacetime Levi-Civita connection, we obtainEach term contains a normal contracted with its spatial projection tensor. Consequently $\boxed{D_i{h}_{jk}=0}$, and the negative spatial restriction also satisfies $D_i{q}_{jk}=0$. This proves the requested metric compatibility of the spatial covariant derivative after correcting the source's metric sign. Contracting indices gives the particular expression written in the question.

The Chern connection of a Hermitian metric on a holomorphic vector bundle is the unique connection on a vector bundle compatible with that metric and whose $(0,1)$ part is the bundle's Dolbeault operator ${\bar{\partial }}_E$. Fix a holomorphic local frame and column coefficients for sections. Write the metric as $h(s,t)={\bar{s}}^{T}Ht$, conjugate-linear in the first argument, and write the connection as $Ds=ds+As$. The condition $D^{0,1}={\bar{\partial }}_E$ forces $A$ to have type $(1,0)$. Metric compatibility requireswhere the dagger conjugates the differential-form coefficients as well as transposing the matrix. Taking the $(1,0)$ part gives $\boxed{A=H^{-1}\partial H}$. Its conjugate-transpose supplies the $(0,1)$ metric equation because $H$ is Hermitian. This proves uniqueness and local existence.

Under a holomorphic change of frame $e^{\prime }=eg$, the metric matrix becomes $H^{\prime }=g^{†}Hg$. The local formula for the Chern connection on a vector bundle then givesThis is precisely the transformation rule for a connection on a vector bundle, so the local connections glue and establish global existence. No Kähler hypothesis is needed for this part.

Extend $D$ to vector-bundle-valued differential forms by the graded Leibniz rule. The curvature form of a connection is the tensorial square $F_h=D^2$, acting by exterior multiplication. In the chosen frame,Indeed $\partial A+A\wedge A=0$ by differentiating $H^{-1}H=I$. It follows that $F_h$ has type $(1,1)$; the gauge change is $F_h^{\prime }=g^{-1}{F}_{h}g$. Thus it is a global smooth two-form with values in $\mathrm{End}(E)$, namely an element of ${\mathcal{A}}^2(\mathrm{End}(E))$.