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Here the Cartan subalgebra of a finite-dimensional complex semisimple Lie algebra is a maximal toral subalgebra: it is abelian, its adjoint operators are simultaneously diagonalizable, and its centralizer is itself. These standard properties give the root-space decompositionThe roots of a root system $\alpha$ are the nonzero linear functionals in ${\mathfrak{h}}^{\ast }$ for which ${\mathfrak{g}}_{\alpha }\ne 0$; $\Phi$ is their finite set. Each root space is one-dimensional. The zero weight space is the centralizer of $\mathfrak{h}$, hence exactly $\mathfrak{h}$. Thus the displayed sum is a direct sum of simultaneous eigenspaces of the Adjoint representation. It is the Cartan root-space decomposition meant here.

The bracket respects weight addition: $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subseteq {\mathfrak{g}}_{\alpha +\beta }$, where ${\mathfrak{g}}_0=\mathfrak{h}$ and a missing weight means the zero space. This follows by applying the Jacobi identity with $H\in \mathfrak{h}$ to the bracket of two root vectors. It explains how the vector-space decomposition encodes the Lie algebra structure.

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 base ring must be specified in the algebraic definition. For a commutative $k$-algebra $R$, work in ${\mathrm{End}}_k(R)$ and let $m_f$ mean multiplication by $f\in R$. Set $D_k^{-1}(R)=0$ and define recursivelyThese are the operators of order at most $i$. The order of a nonzero differential operator relative to a base ring is the least $i$ for which it belongs to $D_k^i(R)$; the zero operator can be assigned order $-\infty$. An operator of order zero commutes with all multiplications and is itself multiplication by $P(1)$, so $D_k^0(R)=R$. The union $D_k(R)={\bigcup }_{i\ge 0}{D}_k^i(R)$ is the ring of finite-order operators, with composition as multiplication. The identityand induction on the sum of orders give $D_k^p{D}_k^q\subseteq D_k^{p+q}$. Sums have order at most the maximum of their orders, proving that this union is a ring.

For an arbitrary commutative ring, the absolute version uses $k=\mathbb{Z}$, so the ambient endomorphisms are additive. For the polynomial ring in this question, the intended Weyl-algebra identification is the relative one with $k=\mathbb{C}$. Absolute operators would also include nonzero derivations of the coefficient field $\mathbb{C}$: a derivation on a rational function field in a transcendence basis over $\mathbb{Q}$ extends to its algebraic closure in characteristic zero, and hence to $\mathbb{C}$. Extending it coefficientwise to $\mathbb{C}[x]$ gives an order-one additive operator which is not complex-linear. Such an operator cannot belong to the complex Weyl algebra. This explains why the linearity convention matters.

Now take $R=\mathbb{C}[x_1,\dots ,x_n]$. Coordinate multiplication has order zero, and each partial derivative ${\partial }_i$ has order one because $[{\partial }_i,m_f]=m_{{\partial }_{i}f}$. They satisfy the defining Weyl algebra relations, giving an algebra homomorphism $A_n\to D_{\mathbb{C}}(R)$. Reordering spans $A_n$ by $x^{\alpha }{\partial }^{\beta }$. If an ordered expression acted as zero, commutation with the coordinates according to a maximal derivative multiindex would give multiplication by $\beta !f_{\beta }(x)=0$; removing the top coefficients successively proves that every coefficient is zero. Thus the homomorphism is injective, as in the ordered monomial basis of a Weyl algebra proof.

For surjectivity, let $P$ have order at most $m$, and put ${\delta }_{i}P=[P,m_{x_i}]$. The operations ${\delta }_i$ commute, since the coordinate multiplications commute. A product of more than $m$ of them kills $P$. Define polynomial coefficientsCommuting $P$ through $m_{x^{\alpha }}$ repeatedly gives the finite identityIt follows by induction on $|\alpha |$ from $P{m}_{x_i}=m_{x_i}P+{\delta }_{i}P$ and Pascal's identity. Terms with $|\beta |>m$ vanish. Sincefor $\beta \le \alpha$, we obtain $P(x^{\alpha })={\sum }_{|\beta |\le m}{a}_{\beta }(x){\partial }^{\beta }(x^{\alpha })$. The monomials are a complex basis of $R$, so complex linearity yields equality of the operators on all of $R$. HenceFor a nonzero ordered expression ${\sum }_{\beta }{a}_{\beta }(x){\partial }^{\beta }$, its order is exactly ${\max }_{a_{\beta }\ne 0}|\beta |$. The upper bound follows from composition; the lower bound follows by taking that many coordinate commutators to extract a nonzero multiplication operator. This describes the order filtration of a Weyl algebra:

The associated graded ring is thereforeThe symbol of ${\partial }_i$ is ${\xi }_i$. In a product of ordered operators, moving derivatives past coefficients introduces terms of lower differential order. Thus top-order symbols multiply as ordinary polynomials, and the ordered basis proves there are no further relations. In particular $[D^p,D^q]\subseteq D^{p+q-1}$, with $D^{-1}=0$: the top-order symbols of the two products agree and cancel. Unlike the Bernstein filtration, the order filtration's degree-zero part is the entire polynomial ring and its pieces are infinite-dimensional over $\mathbb{C}$.

Form the Rees ring of a filtered algebraThe central element $h$ is a non-zero-divisor. This ring is generated by $h$, the $x_i$, and $y_i=h{\partial }_i$, withIts quotient modulo $h$ is the associated graded ring just computed. Explicitly the degree-$m$ component of $h\mathcal{A}$ is $D^{m-1}{h}^m$, so the degree-$m$ quotient is $D^m/D^{m-1}$. The quotient modulo $h-1$ recovers $A_n$.

Since $\mathcal{A}/h\mathcal{A}$ is commutative, any two lifts $F,G\in \mathcal{A}$ of symbols have $[F,G]\in h\mathcal{A}$. DefineDivision by $h$ is unique because $h$ is a non-zero-divisor. If $F$ is replaced by $F+hU$, the quotient changes by $[U,G]$, which is zero modulo $h$; changing $G$ works the same way. Thus the bracket is well defined. Antisymmetry and bilinearity follow from the commutator. The identity $[F,GH]=[F,G]H+G[F,H]$ proves the Leibniz rule. For the Jacobi identity, lift each inner bracket by $[G,H]/h$; the cyclic sum of the outer expressions is the commutator Jacobi identity divided by $h^2$, hence zero. These verifications prove that the graded polynomial ring is a Poisson algebra. This is the Poisson bracket from a central deformation parameter construction.

The generators give $\{{\xi }_i,x_j\}={\delta }_{ij}$ and all coordinate-coordinate and symbol-symbol brackets zero. The Leibniz rule therefore determines the answer on every polynomial:The sign here follows the chosen convention $[P,Q]=PQ-QP$: in particular $\{{\xi }_i,x_j\}={\delta }_{ij}$, rather than $\{x_i,{\xi }_j\}={\delta }_{ij}$. For homogeneous symbols the bracket has order degree $p+q-1$, and can equivalently be written $\{{\sigma }_p(P),{\sigma }_q(Q)\}={\sigma }_{p+q-1}([P,Q])$.

Let ${\hat{T}}_a={\mathrm{ad}}_{T_a}$ on the Lie algebra itself. The Jacobi identity says, for every $X$,Therefore $[{\hat{T}}_a,{\hat{T}}_b]=c_{ab}^c{\hat{T}}_c$, proving that the displayed structure-constant matrices form the Adjoint representation of a Lie algebra.

The bilinear form ${\kappa }_{ab}=\mathrm{tr}({\hat{T}}_a{\hat{T}}_b)$ is the Killing form. It is symmetric by cyclicity of trace. Its invariance follows by expanding the two commutators:In components this is preciselyEquivalently, the adjoint action preserves the bilinear form, so it is an invariant tensor.

The invariant abelian subalgebra is an abelian ideal of a Lie algebra $\mathfrak{a}$. Assume it is nonzero, as required for the conclusion. For $X\in \mathfrak{a}$ and $Y\in \mathfrak{g}$, the map ${\mathrm{ad}}_X{\mathrm{ad}}_Y$ sends $\mathfrak{g}$ into $\mathfrak{a}$, because it is an ideal. It vanishes on $\mathfrak{a}$, because ${\mathrm{ad}}_Y$ preserves the ideal and ${\mathrm{ad}}_X$ kills it. Relative to $\mathfrak{a}$ and a vector-space complement, this map has zero diagonal blocks, so its trace is zero. ThusThis proves that abelian ideals lie in the radical of the Killing form. Choose a nonzero $X=x^a{T}_a$ in the ideal. Then ${\kappa }_{ba}{x}^a=0$ for every $b$, so the coordinate vector $x$ is a nonzero null eigenvector of the Killing matrix. The nonzero-ideal qualification cannot be dropped: the zero ideal is abelian even in a semisimple algebra with nondegenerate Killing form.

Now suppose $\kappa$ is invertible and let $C_{\kappa }={\kappa }^{bc}{t}_b{t}_c$ in any Lie algebra representation. In matrix notation invariance is $A_a^T\kappa +\kappa A_a=0$, where $(A_a{)}^d{}_b=c_{ab}^d$. Multiplying by the inverse form gives $A_a{\kappa }^{-1}+{\kappa }^{-1}{A}_a^T=0$. HenceNo commutativity of the representation matrices has been assumed. This proves the centrality of the quadratic Casimir operator formed with the inverse Killing form. Schur lemma makes it a scalar on every finite-dimensional complex irreducible representation.

For SU(2), make the normalization explicit. In a Hermitian angular-momentum basis with $[J_a,J_b]=i{ϵ}_{abc}{J}_c$, the structure-constant matrices obeyThus $C_{\kappa }=\frac{1}{2}{\sum }_a{J}_a^2$. In the compact anti-Hermitian basis $T_a=-i{J}_a$, the brackets have real structure constants ${ϵ}_{abc}$ and the Killing form of the SU(2) Lie algebra is instead $-2{\delta }_{ab}$; the inverse contraction gives the same $C_{\kappa }$. This sign change is a basis convention, not a different spectrum.

In the complex highest-weight basis $(H,E^{+},E^{-})$ used earlier, the same Killing form and its inverse areConsequentlyOn a highest vector of weight $m=2j$, the ladder calculation gives $E^{+}{E}^{-}v=mv$ and $E^{-}{E}^{+}v=0$. Its Casimir eigenvalue is thereforeCentrality gives the same value throughout the irreducible multiplet. This is the Killing-normalized SU(2) quadratic Casimir; the familiar angular-momentum operator ${\sum }_a{J}_a^2$ has eigenvalue $j(j+1)$ and is twice this inverse-Killing-form contraction.

Write the affine modes as $J_n^a=T_n^a$, and put $Q=Q^{\mathfrak{g}}$. The Sugawara construction requires a restricted module, meaning that sufficiently large positive current modes annihilate each fixed vector, and a noncritical level $2k+Q\ne 0$. A positive-energy highest-weight representation has the required restriction. DefineNormal ordering puts positive modes to the right, making each sum finite on a fixed finite-energy vector. The invariant sum over the current index $a$ is the quadratic Casimir operator construction for the affine current algebra.

Here is the normal ordering calculation responsible for its denominator. In commuting $S_m$ with $J_n^b$, the two affine central terms supply $-2kn{J}_{m+n}^b$. The remaining terms contain $i{f}^{abc}$ times the two ways to replace a current by $J_{p+n}^c$. Relabeling the summation index and using antisymmetry of $f^{abc}$ cancels the un-reordered terms. Reordering the modes that cross the creation/annihilation boundary leaves $n$ signed terms, whose two structure constants contract by $f^{abc}{f}^{abd}=Q{\delta }^{cd}$. Their sum is $-Qn{J}_{m+n}^b$. ThusThe adjoint-Casimir contribution is genuinely quantum. For an explicit check of its coefficient, let $|0⟩$ be the affine vacuum, with all $J_n^a|0⟩=0$ for $n\ge 0$. ThenThe first term acts by the adjoint Casimir operator $Q$, and only $p=1$ contributes $2k$ to the second. This also verifies that a current has conformal weight one under the displayed normalization.

Now commute the quadratic generators with one another. Acting on either current factor produces $(m-n)L_{m+n}^{\mathfrak{g}}$; the residual full contractions are scalar and vanish unless $m+n=0$. The Jacobi identity fixes their cubic dependence on $m$, and the plane vacuum convention $L_{-1}|0⟩=L_0|0⟩=L_1|0⟩=0$ fixes the combination $m^3-m$. To determine its coefficient explicitly, noteCommuting a single $J_1^a$ through the numerator gives $(2k+Q)J_{-1}^a|0⟩$: the two central contractions give $2k{J}_{-1}^a$, and the zero-mode commutator in the structure-constant term gives $Q{J}_{-1}^a$. The remaining $J_1^a$ contraction gives $k$ for each adjoint index. ThereforeFor an algebraic module without a positive inner product, the same calculation is its vacuum matrix element and fixes the universal central term. ConsequentlyThis is a representation of the Virasoro algebra, not merely its centerless part.

For a complex semisimple Lie algebra, a Cartan subalgebra $\mathfrak{h}$ is a maximal abelian subalgebra of semisimple elements; equivalently it is the complexification of a maximal torus algebra in the compact real form. Its dimension is the rank of a semisimple Lie algebra, $r$. Choose an orthonormal Cartan basis with currents $H_n^i$. They have no structure-constant term:For $k\ne 0$, their abelian Sugawara construction isEach Cartan current has weight one under both $L^{\mathfrak{g}}$ and $L^{\mathfrak{h}}$, so $K_m=L_m^{\mathfrak{g}}-L_m^{\mathfrak{h}}$ commutes with all $H_n^i$, hence also with $L_n^{\mathfrak{h}}$. Subtracting the two commuting stress tensors yields the Coset constructionIn the intended nontrivial compact unitary theory at positive admissible level, this coset acts on a positive inner-product space, with $K_m^{†}=K_{-m}$. Its vacuum satisfies $K_n|0⟩=0$ for $n\ge -1$, soThus $c^{\mathfrak{g}}\ge r$. Also $Q>0$ for a compact simple Lie algebra and $k>0$, so $2k/(2k+Q)<1$, which proves the upper bound directly. Hence, under these physical representation hypotheses,The upper inequality is strict at finite positive level; $c^{\mathfrak{g}}$ tends to $\dim \mathfrak{g}$ as the level grows.

The positivity hypothesis is necessary and is not implied by the bare affine commutation relation. For example, at $k=0$ its trivial representation has all currents zero and gives $c^{\mathfrak{g}}=0<\mathrm{rank}\mathfrak{g}$. Even positive arbitrary levels need not satisfy the bound: with $\mathfrak{g}=\mathfrak{su}(2)$, $f^{abc}={ϵ}^{abc}$, $Q=2$ and $k=1/4$, the formula gives $c^{\mathfrak{g}}=3/5<1$. Such a level does not give the intended compact unitary positive-energy theory. Thus the last bound is a statement about nontrivial unitary admissible-level representations; it cannot hold for every representation in the unrestricted algebraic formulation. At $2k+Q=0$ the displayed Sugawara construction itself is undefined.

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.

Let $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. A real semisimple Lie algebra has nondegenerate $B$, and its invariance gives $B([X,Y],Z)=B(X,[Y,Z])$. Translating $B$ defines a bi-invariant pseudo-Riemannian metric $g=B$ on the Lie group.

For left-invariant vector fields, all derivatives of their pairwise inner products vanish. The Koszul formula reduces toNondegeneracy gives ${\nabla }_{X}Y=[X,Y]/2$. With $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity now givesTaking the trace of $X\mapsto R(X,Y)Z=-\frac{1}{4}\mathrm{ad}Z\mathrm{ad}Y(X)$ yields the Ricci tensorThus the Killing-form Einstein metric makes the group an Einstein manifold, allowing an indefinite metric. Its scalar curvature in this convention is $-\dim G/4$.

For a connected compact semisimple Lie group, $B$ is negative definite: the adjoint operators are skew-adjoint for an invariant positive inner product, so $B(X,X)=\mathrm{tr}((\mathrm{ad}X{)}^2)<0$ for $X\ne 0$. The conventional positive Riemannian metric is instead $g_{+}=-B$. Its Ricci tensor is $g_{+}/4$, and its sectional curvature isIt can vanish on commuting two-planes; positivity is not automatic in every direction. With the literal negative metric $B$, the corresponding sectional-curvature signs are reversed.

For a noncompact real semisimple group, a Cartan decomposition $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{p}$ has $B$ negative on $\mathfrak{k}$ and positive on $\mathfrak{p}$. The Killing metric is indefinite and cannot be made positive by an overall sign. Its Einstein constant remains $-1/4$ for $g=B$, but that does not assert a uniform sign for sectional curvature of all nondegenerate planes. For connected semisimple groups, negative-definite $B$ characterizes the compact case. Disconnected groups require a separate condition on the number of components before compactness follows.

A Poisson manifold is a smooth manifold $P$ with a bilinear bracket on $C^{\infty }(P)$ that is antisymmetric, satisfies the Jacobi identity and obeys $\{f,gh\}=\{f,g\}h+g\{f,h\}$. Equivalently it has a bivector $\Pi$ defining $\{f,g\}=\Pi (df,dg)$ with vanishing Schouten bracket $[\Pi ,\Pi ]$; nondegeneracy is not required.

For a symplectic manifold, define the Hamiltonian vector field by ${\iota }_{X_f}\omega =df$ and put $\{f,g\}=\omega (X_f,X_g)$. Nondegeneracy gives a unique $X_f$, antisymmetry is inherited from $\omega$, and $d(gh)=gdh+hdg$ proves the Leibniz rule. Moreover ${\mathcal{L}}_{X_f}\omega =d{\iota }_{X_f}\omega +{\iota }_{X_f}d\omega =0$ by Cartan's magic formula. Since $X_{f}g=-\{f,g\}$,For completeness, evaluate $d\omega =0$ on $X_f,X_g,X_h$. Its three derivative terms sum to minus the cyclic sum of $\{f,\{g,h\}\}$, and its three bracket terms sum to minus the same cyclic sum, using the displayed identity. Hence that sum is zero. This proves the Jacobi identity for the Poisson bracket, not merely the Jacobi identity modulo constants. The inverse of a closed nondegenerate two-form therefore defines a Poisson manifold.

The Symplectic Darboux theorem says that near every point of a $2n$-dimensional symplectic manifold there are coordinates $(q^1,\dots ,q^n,p_1,\dots ,p_n)$ withThus every symplectic structure has the same local normal form, despite possible global topological differences.

Write ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$ for the fundamental vector field of a left Lie group action, and use ${\iota }_{X_f}\omega =df$ throughout. Such fundamental fields satisfy $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$. A moment map has linear components ${\mu }_{\xi }=⟨\mu ,\xi ⟩$ satisfyingFor the action to be Hamiltonian, these components can be chosen equivariantly, or, infinitesimally,This identifies the span of the component Hamiltonians with the Lie algebra image, rather than identifying the entire infinite-dimensional Poisson algebra $C^{\infty }(P)$ with $\mathfrak{g}$. For an actual isomorphism with $\mathfrak{g}$, injectivity is also needed; a faithful infinitesimal action supplies it.

Preservation of $\omega$ alone gives $d{\iota }_{{\xi }_P}\omega =0$, not necessarily exactness. For example, translation generated by ${\partial }_q$ on the symplectic two-torus has contraction $dp$, whose period is nonzero. That action has no globally defined component Hamiltonian. Once the contractions are exact, choose Hamiltonians linearly in $\xi$. On each connected component of $P$, the discrepancyis constant: apply $[X_f,X_g]=-X_{\{f,g\}}$ and the fundamental-field bracket. The Jacobi identity makes $\kappa$ a Lie algebra two-cocycle. Replacing ${\mu }_{\xi }$ by ${\mu }_{\xi }+b(\xi )$ changes it to $\kappa (\xi ,\eta )-b([\xi ,\eta ])$. Thus the exactness condition and vanishing of this moment-map equivariance obstruction are precisely the required infinitesimal conditions.

For a semisimple Lie algebra, both hold. First Cartan's magic formula givesSemisimplicity implies $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so every generator is a sum of commutators and has a globally exact contraction. This proves Hamiltonian existence for a semisimple symplectic action.

Next use the nondegenerate Killing form to define $D$ by $B(D\xi ,\eta )=\kappa (\xi ,\eta )$. Antisymmetry makes $D$ skew for $B$. The cocycle identity and invariance of $B$ imply $D[\xi ,\eta ]=[D\xi ,\eta ]+[\xi ,D\eta ]$, so $D$ is a derivation. Choose $Z$ byThen, using cyclicity of trace and $[D,\mathrm{ad}\xi ]=\mathrm{ad}(D\xi )$,Hence $D=\mathrm{ad}Z$ and $\kappa (\xi ,\eta )=B(Z,[\xi ,\eta ])$. Taking $b(\xi )=B(Z,\xi )$ removes the cocycle and proves semisimple moment-map equivariance without merely invoking a cohomology theorem. For a connected group, integrate the infinitesimal equivariance along its one-parameter subgroups to obtain group equivariance. A disconnected group can require an additional discrete-equivariance check.