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Specify the displayed rule on left-invariant vector fields and extend by the Leibniz rule to an affine connection on the Lie group. In a left-invariant vector field frame $L_i$, the extension is ${\nabla }_{L_i}(z^j{L}_j)=L_i(z^j)L_j+\lambda z^j[L_i,L_j]$. Applying the bare bracket formula to arbitrary vector fields instead would violate the Leibniz rule unless $\lambda =1$. With $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity givesfor left-invariant arguments, where $B$ is the Killing form and $\mathrm{Ric}(X,Y)=\mathrm{tr}(Z\mapsto R(Z,X)Y)$. These identities determine the tensor fields everywhere. The endpoints $\lambda =0,1$ have zero curvature, with respectively left- and right-invariant parallel frames, and opposite torsion tensors. The midpoint $\lambda =1/2$ is torsion-free but need not have zero curvature.

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.

For a local basis $e_a$ of the tangent bundle, an affine connection has components defined by ${\nabla }_{e_c}{e}_b={\Gamma }^a{}_{bc}{e}_a$. If $V=V^b{e}_b$, then $({\nabla }_{e_c}V{)}^a=e_c(V^a)+{\Gamma }^a{}_{bc}{V}^b$. The connection components depend on the chosen basis and have an inhomogeneous transformation law; they are not themselves a tensor field. In a coordinate basis, a torsion-free connection has symmetry in the two lower indices.

For every contravariant slot of a tensor field, the tensor Lie derivative contributes a negative derivative of its generating vector field; every covariant slot contributes a positive one. The remaining term differentiates the tensor's components along the generator. This follows from the flow definition of the Lie derivative of a tensor field and its product and contraction rules, without choosing a affine connection.

The difference of two affine connections is a smooth tensor field of type $(1,2)$. Linearity over smooth functions holds in the first slot by the connection axioms. In the second slot, the two extra terms $X(f)Y$ cancel, so $\Delta (X,fY)=f\Delta (X,Y)$. This tensoriality makes the value depend only on $X_p,Y_p$. In coordinates its components are the differences of the Christoffel symbols, although either collection of connection coefficients individually is not a tensor.

A smooth endomorphism $A$ of the tangent bundle defines the tensor derivation $D_A$ by $D_{A}f=0$ and $D_{A}Y=A(Y)$. On a differential one-form, $D_A\omega =-\omega \circ A$. On a general tensor field, it acts by $A$ in each vector factor and by the negative dual action in each covector factor. The two actions cancel in each contracted pairing, proving compatibility with tensor contraction.

For the local flow ${\varphi }_t$ of a smooth vector field $X$, pull a tensor field $T$ back to the fixed base point before differentiating. Pullback uses $d{\varphi }_t^{-1}$ on vector factors and the dual of $d{\varphi }_t$ on covector factors. The chain rule gives ${\mathcal{L}}_{X}f=Xf$. In coordinates, $d{\varphi }_t^{-1}Y({\varphi }_t(x))=Y+t(dYX-dXY)+O(t^2)$, so ${\mathcal{L}}_{X}Y=[X,Y]$. Differentiation of pullback preserves tensor contractions and obeys the Leibniz rule for tensor products; hence this is the tensor derivation defining the Lie derivative of a tensor field.

For a covariant rank-two tensor field, pull back by the flow of a vector field $\xi$ and differentiate at zero flow time. Differentiating the evaluation point produces the first term above; differentiating the two coordinate Jacobians produces the remaining terms. In particular this formula gives the change in a metric tensor under an infinitesimal coordinate displacement. A perturbation defined relative to a fixed background consequently transforms as $q_{\mu \nu }\mapsto q_{\mu \nu }-{\mathcal{L}}_{\xi }{\bar{g}}_{\mu \nu }$ in the passive displacement convention. The formula applies to symmetric tensors as well as differential forms; tensor symmetry does not make the antisymmetric differential-form formula its definition.

For a smooth manifold $M$ of dimension $n$, the tangent space $T_{p}M$ can be defined as the space of derivations at $p$: linear maps $v$ on germs of smooth real functions satisfying $v(fg)=f(p)v(g)+g(p)v(f)$. Equivalently, $v$ is the velocity of a smooth curve through $p$, with curves identified when their coordinate velocities agree. A chart $(x^1,\dots ,x^n)$ supplies the basis ${\partial }_i{|}_p$, so $v=v^i{\partial }_i{|}_p$.

The tangent bundle is the disjoint union $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (v)=p$. Its smooth structure is defined by the local local trivializationsOn overlapping charts, the change of trivialization isThese are smooth changes of coordinates on a $2n$-dimensional smooth manifold, linear and invertible in each fibre. Thus $TM$ is a rank-$n$ vector bundle, and its projection is a submersion. Fibrewise addition, scalar multiplication and the zero section are smooth. A smooth section of a vector bundle $X:M\to TM$, satisfying $\pi \circ X={\mathrm{id}}_M$, is exactly a vector field; locally $X=X^i{\partial }_i$, with components transforming by the displayed Jacobian matrix. Its action on functions is a derivation, and the commutator of two such derivations is their Lie bracket.

The dual bundle of $TM$ is the cotangent bundle $T^{\ast }M$. Its smooth sections of a vector bundle are differential 1-forms, locally $\alpha ={\alpha }_{i}d{x}^i$. The coefficient transformation is the inverse transpose of that for tangent vectors, ensuring that $\alpha (X)$ is a well-defined smooth function. Tensoring these two bundles gives the tensor bundlesTheir sections of a vector bundle are tensor fields; their transition functions of a vector bundle are the corresponding tensor products of the tangent and cotangent transformations. For example, the bundle $\mathrm{End}(TM)=TM\otimes T^{\ast }M$ contains fields of linear endomorphisms, while $S^2{T}^{\ast }M$ contains symmetric bilinear fields. Taking alternating covariant tensors gives ${\Lambda }^k{T}^{\ast }M$, whose sections are differential forms. The wedge product, exterior derivative and pullback of a differential form make these bundles central to integration, Generalized Stokes theorem and de Rham cohomology.

A nowhere-vanishing section of ${\Lambda }^n{T}^{\ast }M$ chooses an orientation of a smooth manifold; such a section exists exactly when $M$ is orientable. Without an orientation of a smooth manifold, one can still integrate sections of the density bundle $|{\Lambda }^n{T}^{\ast }M|$, whose coordinate changes use the absolute Jacobian determinant. A Riemannian metric is a positive-definite section $g$ of $S^2{T}^{\ast }M$. It identifies $TM$ with $T^{\ast }M$ by the musical isomorphisms $X\mapsto g(X,\cdot )$ and gives the Riemannian volume densityOn an oriented smooth manifold, this density corresponds to a volume form.

Another organizing construction is the frame bundle $FM$. A frame over $p$ is a linear isomorphism $u:{\mathbb{R}}^n\to T_{p}M$, and the right action $u\cdot A=u\circ A$ makes $FM$ a principal bundle with general linear group $GL(n,\mathbb{R})$ as its structure group. Given a group representation $\rho :GL(n,\mathbb{R})\to GL(V)$, the associated bundle $FM{\times }_{\rho }V$ is obtained from pairs $(u,v)$ by the relation $(uA,v)\sim (u,\rho (A)v)$. The standard representation reconstructs $TM$; dual, tensor and exterior representations reconstruct the associated bundles described above. A Riemannian metric reduces the frame bundle to the orthonormal frame bundle, with orthogonal group $O(n)$ as structure group; an orientation of a smooth manifold reduces it further to the special orthogonal group $SO(n)$. A global section of $FM$ is a global frame, so it exists exactly when $M$ is a parallelizable manifold. Local triviality does not imply a global trivialization: the Hairy ball theorem prevents even one nowhere-zero vector field on $S^2$.

A connection on a vector bundle differentiates its sections in tangent directions and defines parallel transport. For $TM$, the Levi-Civita connection is distinguished by compatibility with a Riemannian metric and vanishing torsion tensor. Its coefficients are not tensor components, because a coordinate change introduces second derivatives; the difference of two connections is a section of $T^{\ast }M\otimes \mathrm{End}(TM)$. The Riemann curvature tensor is a genuine tensor measuring the failure of covariant derivatives to commute.

Finally, a smooth map $f:M\to N$ gives the pullback tangent bundle $f^{\ast }TN$. Its sections are vector fields along a map, including the velocity of a curve and Jacobi fields along a geodesic. For an embedded submanifold $i:M\hookrightarrow N$, the derivative identifies $TM$ with a subbundle of $i^{\ast }TN$, and the quotient is the normal bundle. With an ambient Riemannian metric, the normal bundle is identified with the orthogonal complement of $TM$ in $i^{\ast }TN$; the second fundamental form records the normal component of the ambient derivative of tangent fields. These constructions connect the bundle description to both intrinsic and extrinsic differential geometry.

Use the Levi-Civita connection $\nabla$ of the Riemannian metric $g$ and fix the conventionThe Leibniz rule shows that this expression is linear over smooth functions in all three arguments: the differentiated coefficient terms in the first two derivatives cancel those in the bracket term. Thus $R$ is a section of ${\Lambda }^2{T}^{\ast }M\otimes \mathrm{End}(TM)$, the Riemann curvature tensor, and $R_4$ is a covariant tensor field. This convention gives positive sectional curvature on the round sphere.

Its algebraic symmetries areThe first identity follows immediately from the definition. For the second, apply $XY-YX-[X,Y]$ to $g(Z,W)$ and use compatibility of the Levi-Civita connection with the Riemannian metric. The left side is zero and the differentiated terms combine toFor the first Bianchi identity, use normal coordinates at a point $p$, so the Christoffel symbols vanish there. Torsion-freeness gives ${\Gamma }_{ij}^{\ell }={\Gamma }_{ji}^{\ell }$, andThe cyclic sum over $i,j,k$ cancels term by term. Since this is a tensor identity, it holds for arbitrary tangent vectors everywhere.

To prove pair interchange rather than assume it, abbreviate $R_4(a,b,c,d)$ to $R_{abcd}$ and put $A=R_{abcd}$, $B=R_{cdab}$. The first Bianchi identity and the two antisymmetries giveThe first Bianchi identity applied to $(d,b,a)$ with final argument $c$ gives $R_{dbca}+R_{adcb}=R_{badc}=A$. Hence $A=2B-A$, proving $A=B$.

There is also the differential second Bianchi identity,For its proof take commuting coordinate fields, normal at $p$. The Jacobi identity for operator commutators says $[{\nabla }_X,[{\nabla }_Y,{\nabla }_Z]]+\text{cyclic}=0$. On these commuting fields, $[{\nabla }_Y,{\nabla }_Z]=R(Y,Z)$; at $p$, the covariant derivatives of the coordinate arguments vanish. The commutator identity is therefore the displayed covariant derivative identity at $p$, and tensoriality establishes it everywhere.

For a two-dimensional plane $\sigma =\mathrm{span}\{X,Y\}\subset T_{p}M$, define the sectional curvature byThe denominator is the squared area of $X,Y$. Under an invertible change of plane basis, the alternating pairs in the numerator and the area in the denominator both acquire the square of the change-of-basis determinant. Thus the quotient depends only on $\sigma$. For an orthonormal basis of $\sigma$, it is simply $g(R(X,Y)Y,X)$.

The Ricci curvature is the tracewhere $(e_i)$ is any orthonormal basis of $T_{p}M$. This is independent of the basis because it is a trace. Applying the first Bianchi identity to $(e_i,X,Y)$ and pairing with $e_i$ proves that the summand difference for $\mathrm{Ric}(X,Y)-\mathrm{Ric}(Y,X)$ vanishes; the middle term $g(R(X,Y)e_i,e_i)$ is zero by the last-pair antisymmetry. Hence Ricci curvature is a symmetric bilinear form. For a unit vector $v=e_1$ extended to an orthonormal basis,For nonzero $X$, take $v=X/|X|$ and multiply the right side by $|X{|}^2$. This determines the entire Ricci curvature by the polarization identity:Finally, the scalar curvature is the metric trace of Ricci curvature. Taking its diagonal entries in an orthonormal basis and pairing the equal contributions from $(i,j)$ and $(j,i)$ yieldsIn particular, constant sectional curvature $K_0$ gives $\mathrm{Ric}=(n-1)K_{0}g$ and $\mathrm{Scal}=n(n-1)K_0$. In dimension one the traces are zero; there are no two-dimensional tangent planes and the sums are empty.

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 $V=0$.

Use sequential semicolon indices: $X^a{}_{;kl}={\nabla }_l({\nabla }_k{X}^a)$, where the second covariant derivative also differentiates the lower index of ${\nabla }_k{X}^a$. For a torsion-free connection, direct expansion givesSubtract the expression with $k,l$ exchanged. Commuting partial derivatives cancel the first term; the first-derivative terms cancel in pairs; symmetry of the lower Christoffel symbol indices cancels both terms containing ${\Gamma }^p{}_{kl}$. What remains isThis is the Ricci identity with the curvature convention printed on the PDF's first page. In terms of derivative operators it is $[{\nabla }_l,{\nabla }_k]X^a=R^a{}_{bkl}{X}^b$; writing the operator order explicitly avoids reversing the sign.

The lower-index rule follows by applying the commutator to the scalar contraction $Y_a{X}^a$. Its second derivatives commute, so the product rule implies $Y_{a;kl}-Y_{a;lk}=-R^p{}_{akl}{Y}_p$. The derivative commutator is a derivation on tensor products, since the cross terms cancel. It therefore acts once on every index, with a positive sign on each upper index and a negative sign on each lower index. For an arbitrary tensor of valence $(r,s)$,Local sums of decomposable tensors establish the formula for every tensor field, rather than only for a single tensor product.

In a coordinate chart the Lie derivative of a function and Lie derivative of a vector field areThe Lie derivative of a tensor field is the extension that respects tensor products and contractions. For a covector field $\alpha$, differentiate the scalar pairing $\alpha (Y)$ to obtain $({\mathcal{L}}_X\alpha )(Y)=X(\alpha (Y))-\alpha ([X,Y])$. In components this givesThus for a mixed rank-two tensor field, the coordinate tensor Lie derivative isEach upper index contributes a negative derivative of the generator and each lower index a positive one. Explicitly, for type $(m,n)$,These rules follow by expanding in coordinate vector and covector bases and using the product rule; they define the extension independently of that basis. For a symmetric connection, replacing the partial derivatives by covariant derivatives leaves these expressions unchanged because the connection terms cancel.

For the commutator identity, $[{\mathcal{L}}_X,{\mathcal{L}}_Y]f=[X,Y](f)$ on functions. On a vector field $Z$, the Jacobi identity gives $[X,[Y,Z]]-[Y,[X,Z]]=[[X,Y],Z]$. The difference $D=[{\mathcal{L}}_X,{\mathcal{L}}_Y]-{\mathcal{L}}_{[X,Y]}$ is a tensor derivation respecting contractions. Since it vanishes on functions and vectors, pairing a covector with every vector shows it vanishes on covectors. Product expansions then show it vanishes on every tensor. This proves the commutator identity for Lie derivatives:A Killing vector generates infinitesimal isometries, meaning ${\mathcal{L}}_{K}g=0$, equivalently the Killing equation ${\nabla }_a{K}_b+{\nabla }_b{K}_a=0$ for the metric Levi-Civita connection. Linearity in the generator gives ${\mathcal{L}}_{aK+bL}g=0$ for real constants $a,b$ whenever $K,L$ are Killing. The identity above likewise gives ${\mathcal{L}}_{[K,L]}g=[{\mathcal{L}}_K,{\mathcal{L}}_L]g=0$. Thus the Killing fields form a real Lie algebra; arbitrary nonconstant coefficients would introduce additional derivative terms.

For the rotating star, the hypothesis that there are exactly two independent Killing fields means $[T,\Phi ]=aT+b\Phi$ with constant coefficients. Axial symmetry supplies a periodic rotation flow; with the stated asymptotic normalization its generator is $\Phi$ and its period is $2\pi$. Pullback by this flow acts on the Killing algebra, so $e^{2\pi \mathrm{ad}\Phi }=1$. In the basis $(T,\Phi )$, $\mathrm{ad}\Phi$ has matrixIts real eigenvalues are $-a$ and zero, so the exponential identity forces $a=0$. The remaining matrix squares to zero; its exponential is then $1+2\pi \mathrm{ad}\Phi$, forcing $b=0$. The two-dimensional Killing algebra with a periodic generator is abelian result therefore gives $\boxed{[T,\Phi ]=0}$.

This uses genuine periodic axial symmetry and the two-dimensional Killing algebra. Mere limits of vector components do not by themselves justify differentiating those limits to evaluate their bracket; the periodic-flow argument avoids that gap. With the usual differentiable asymptotically flat falloff one can also take the bracket at infinity and use independence of the limiting time and angular generators to set both constants to zero.

A linear connection on a manifold is a rule $(X,Y)\mapsto {\nabla }_{X}Y$ for differentiating vector fields which is additive, is $C^{\infty }$-linear in $X$, is real-linear in $Y$, and obeys ${\nabla }_X(fY)=X(f)Y+f{\nabla }_{X}Y$. It extends to covector fields and arbitrary tensor fields by the Leibniz rule and compatibility with contraction. In coordinates,This is the covariant derivative; the affine connection terms correct the changes of the coordinate basis. A symmetric affine connection has zero torsion tensor, $T(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]=0$, or ${\Gamma }^c{}_{ab}={\Gamma }^c{}_{ba}$ in a coordinate basis. Symmetry does not imply metric compatibility; no metric is required for this question.

Additivity of each covariant derivative immediately gives additivity of ${\Delta }_{ab}$. Expanding the second derivative of a tensor product givesSubtract the expression with $a,b$ exchanged. The two mixed terms cancel, proving that the curvature commutator is a tensor derivation:For a scalar, ${\nabla }_a{\nabla }_{b}f={\partial }_a{\partial }_{b}f-{\Gamma }^c{}_{ab}{\partial }_{c}f$. Commuting partial derivatives and using the symmetric affine connection gives ${\Delta }_{ab}f=0$. Consequently ${\Delta }_{ab}(fU)=f{\Delta }_{ab}U$: the commutator is linear over smooth functions, not merely over constant scalars. Write $U=U^c{e}_c$ in a local basis. The tensor-product rule gives ${\Delta }_{ab}U=U^c{\Delta }_{ab}{e}_c$, showing that its value at a point depends only on $U$ at that point. Thus it defines the Riemann curvature tensor byThe antisymmetrized second covariant derivative is itself tensorial in the derivative indices, so this pointwise map transforms as a tensor. The formula or the defining commutator gives $R_{abc}{}^d=-R_{bac}{}^d$, hence $R_{abc}{}^d=R_{[ab]c}{}^d$ with normalized antisymmetrization.

Apply the commutator to the scalar contraction ${\omega }_c{U}^c$. Its action vanishes, soSince $U$ is arbitrary, ${\Delta }_{ab}{\omega }_c=-R_{abc}{}^d{\omega }_d$. Repeated use of the tensor-product rule gives one plus curvature action for an upper index and one minus action for each lower index. In particular,The type notation in the PDF means one contravariant and two covariant indices, not a fractional tensor order.

For the first Bianchi identity, put ${\omega }_c={\nabla }_{c}f$ and $H_{bc}={\nabla }_b{\nabla }_{c}f$. The symmetric Hessian of a scalar field satisfies $H_{bc}=H_{cb}$. Therefore the six third-derivative terms cancel inSubstituting the covector curvature action gives $(R_{abc}{}^d+R_{bca}{}^d+R_{cab}{}^d){\nabla }_{d}f=0$. At a chosen point the gradient of a smooth scalar can be any covector, soThe second form follows from the already proved first-pair antisymmetry.

For the second Bianchi identity, the same third-derivative cancellation can be organized as an operator identity. On vector-valued components set $D_a={\partial }_a+{\Gamma }_a$, with $({\Gamma }_a{)}^e{}_d={\Gamma }^e{}_{ad}$. Although a full second covariant derivative includes a term $-{\Gamma }^f{}_{ab}{D}_f$ for its lower derivative index, that term cancels on antisymmetrization because the affine connection is symmetric. Thus $[D_b,D_c]={\mathcal{R}}_{bc}$, where $({\mathcal{R}}_{bc}{)}^e{}_d=R_{bcd}{}^e$ acts by multiplication. Expanding the three nested operator commutators makes all six triple products cancel:Equivalently, the cyclic sum of ${\partial }_a{\mathcal{R}}_{bc}+[{\Gamma }_a,{\mathcal{R}}_{bc}]$ vanishes. These are the covariant derivative terms on the upper and last lower curvature indices. The two remaining lower indices add $-{\Gamma }^f{}_{ab}{R}_{fcd}{}^e-{\Gamma }^f{}_{ac}{R}_{bfd}{}^e$. In the cyclic sum these additions cancel in pairs, using ${\Gamma }^f{}_{ab}={\Gamma }^f{}_{ba}$ and $R_{fcd}{}^e=-R_{cfd}{}^e$. Hence the Bianchi identities from covariant derivative commutators giveBoth identities here apply to any torsion-free linear affine connection; the second derivation does not require choosing a metric affine connection.

A Poisson manifold is a smooth manifold with an antisymmetric contravariant tensor field $\pi =\frac{1}{2}{\pi }^{ij}{\partial }_i\wedge {\partial }_j$ such that the bracketsatisfies the Jacobi identity. Bilinearity, antisymmetry and the Leibniz rule follow directly from this formula. The Poisson bivector is allowed to be degenerate.

Apply the Jacobi identity to coordinate functions $x^i,x^j,x^k$. Necessity gives the coordinate Jacobi condition for a Poisson bivector:For sufficiency, expand $\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}$. The terms differentiating $\pi$ are $J^{ijk}{f}_i{g}_j{h}_k$. All terms containing second derivatives cancel. For example, those containing $g_{j\ell }$ have coefficient $({\pi }^{i\ell }{\pi }^{jk}+{\pi }^{k\ell }{\pi }^{ij})f_i{h}_k$, which is antisymmetric under $j\leftrightarrow \ell$ after contraction with $f_i{h}_k$, whereas $g_{j\ell }$ is symmetric. The same argument handles the second derivatives of $f$ and $h$. Thus the displayed condition is both necessary and sufficient.

A symplectic manifold has a smooth nondegenerate closed differential form $\Omega$ of degree two. Write $\Omega =\frac{1}{2}{\Omega }_{ij}d{x}^i\wedge d{x}^j$ and choose the explicit inverse convention ${\pi }^{ik}{\Omega }_{kj}={\delta }_j^i$. This defines a smooth antisymmetric bivector. Differentiating the inverse matrix givesSubstitution into the Jacobi coefficient, followed by relabeling summed indices, givessince the parentheses are the components of $d\Omega$. Hence every symplectic manifold carries a Poisson structure. The opposite overall convention for the inverse bracket gives the same conclusion. Nondegeneracy and closure together are crucial; an arbitrary invertible antisymmetric tensor need not satisfy the Jacobi identity.

For a finite-dimensional Lie algebra $\mathfrak{g}$, its dual space ${\mathfrak{g}}^{\ast }$ carries the Lie-Poisson bracketwhere $d{f}_{\ell },d{g}_{\ell }$ are identified with elements of $\mathfrak{g}$. If $[e_i,e_j]=c_{ij}{}^k{e}_k$ and $x_i(\ell )=\ell (e_i)$, this reads $\{x_i,x_j\}=c_{ij}{}^k{x}_k$. The Jacobi coefficient iswhich vanishes by the Lie algebra Jacobi identity.

In particular, the dual of $\mathfrak{so}(3)$ is ${\mathbb{R}}^3$ with the rotational Lie-Poisson structure on R3,The Poisson bivector has rank two away from the origin and rank zero at the origin. The function $x^2+y^2+z^2$ is a Casimir function, since its bracket with every function vanishes. The centred spheres are its two-dimensional symplectic leaves, and the origin is a point leaf. This is a Poisson manifold but not a symplectic manifold: its bivector is not invertible, and an odd-dimensional manifold cannot have a nondegenerate alternating bilinear form.

First verify the difference of affine connections is a tensor. Linearity over smooth functions in $X$ follows from the affine connection axioms. In $Y$, the two derivative terms cancel:Thus $A$ is a smooth $(1,2)$-tensor field, and $A_p(u,v)$ depends only on the tangent vectors $u,v$ at $p$. For any smooth curve, its two covariant accelerations satisfyIf $A(X,Y)=-A(Y,X)$, its diagonal values vanish. The two geodesic equations are therefore identical, and the connections have the same parametrized geodesics.

For necessity, the required local geodesic existence theorem is this: for every smooth affine connection, $p\in M$ and $v\in T_{p}M$, a unique geodesic exists on some interval about zero with $\gamma (0)=p$ and $\dot{\gamma }(0)=v$. Indeed the geodesic equation is the smooth first-order ordinary differential equation on position and velocityso local existence and uniqueness apply. If the two connections have the same parametrized geodesics, apply the acceleration identity to this geodesic at zero to obtain $A_p(v,v)=0$ for every $p,v$. Expanding $A_p(u+v,u+v)=0$ gives $A_p(u,v)+A_p(v,u)=0$. ThereforeThis is the criterion that parametrized geodesics determine the symmetric part of an affine connection. It concerns equality with the same affine parameters. Agreement merely of unparametrized images is weaker: adding $\alpha (X)Y+\alpha (Y)X$ to a connection, for a smooth one-form $\alpha$, changes acceleration by a multiple of the velocity, which can be absorbed by reparametrization. Such a difference need not be antisymmetric.