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Acting on the column $(x,1{)}^T$ gives the faithful Matrix Lie group representationMultiplication and inversion areThus $G$ is the real affine group. Differentiating the matrices at the identity gives its Lie algebraWiththe matrix commutator is $[D,T]=T$, and $[D,D]=[T,T]=0$.

The left Maurer-Cartan form and its right analogue areFor a constant $h$, left translation leaves ${\theta }_L$ unchanged, and right translation leaves ${\theta }_R$ unchanged. Taking their dual vector fields givesThe first pair consists of left-invariant vector fields; the second consists of right-invariant vector fields. These are the invariant frames of the real affine group. Computing their Lie brackets of vector fields yieldsAll brackets of a field with itself vanish. The sign difference is necessary: evaluation at the identity identifies left-invariant vector fields with the matrix Lie algebra as a homomorphism, whereas right-invariant vector fields realize the opposite Lie algebra. Equivalently, $\xi \mapsto -{\xi }_R$ is a homomorphism for the same matrix commutator. There is no convention in which these particular coordinate fields both have the positive structure constant while retaining the usual Lie bracket of vector fields.

The Maurer-Cartan equation gives the same check. If ${\theta }_L=D{\ell }^D+T{\ell }^T$, thenFor ${\theta }_R=D{r}^D+T{r}^T$, insteadThus the left equation is $d{\theta }_L+{\theta }_L\wedge {\theta }_L=0$ and the right equation is $d{\theta }_R-{\theta }_R\wedge {\theta }_R=0$, explaining the opposite bracket signs directly.

Finally, the two given matrix curves are $\exp (ϵT)$ and $\exp (ϵD)$. Their left action on a general group element isDifferentiating at $ϵ=0$ gives $T_R$ and $D_R$ respectively. Hence Left translations on a Lie group are generated by right-invariant vector fields. More generally, the velocity of $\exp (t\xi )g$ is $(d{R}_g{)}_e\xi$, which is right-invariant because left and right multiplication commute. In the other direction, the flows $g\exp (t\xi )$ generate left-invariant vector fields and are right translations on a Lie group.

A Levi-Civita connection on a Riemannian manifold is a covariant derivative $D$ that is a torsion-free connection, $D_{X}Y-D_{Y}X=[X,Y]$, and a metric connection, $Xg(Y,Z)=g(D_{X}Y,Z)+g(Y,D_{X}Z)$. A covariant derivative is $C^{\infty }(M)$-linear in $X$, real linear in $Y$, and obeys $D_X(fY)=X(f)Y+f{D}_{X}Y$.

Combining metric compatibility for the cyclic triples and eliminating reversed derivatives with the torsion identity gives the Koszul formulaSince $g$ is nondegenerate, this determines $D_{X}Y$ uniquely. For existence, use the right-hand side to define $D_{X}Y$: expansion of the Lie bracket of vector fields shows it is $C^{\infty }(M)$-linear in $Z$, hence defines a smooth one-form, which the musical isomorphism converts to a smooth vector field. The same expansion shows $D_{fX}Y=f{D}_{X}Y$ and $D_X(fY)=X(f)Y+f{D}_{X}Y$. Subtracting the formula with $X,Y$ exchanged gives torsion zero; adding its versions for $Y,Z$ exchanged gives metric compatibility. Thus it is a Levi-Civita connection. Equivalently its coefficients in a manifold chart are the Christoffel symbolsThe intrinsically defined Koszul formula ensures these local expressions fit together. This proves the fundamental theorem of Riemannian geometry.

The product Riemannian metric on $M\times N$ isThe two factor tangent spaces are orthogonal, and the sum is positive definite. Lift $X$ from $M$ and $Y$ from $N$. We check $D_{X}Y$ against lifted local frame fields from both factors, which together span each product tangent space.

If $Z$ is lifted from $M$, then $g(Y,Z)=g(X,Y)=0$, while $g(Z,X)$ depends only on $m$, so $Yg(Z,X)=0$. Also $[Y,Z]=[X,Y]=0$ and $[Z,X]$ is horizontal, hence orthogonal to $Y$. Every term in the Koszul formula is zero. If $W$ is lifted from $N$, $g(W,X)=g(X,Y)=0$ and $g(Y,W)$ depends only on $n$, so $Xg(Y,W)=0$. Now $[W,X]=[X,Y]=0$ and $[Y,W]$ is vertical, hence orthogonal to $X$. Again every term is zero. Thus $D_{X}Y$ is orthogonal to both spanning frame families, and positive definiteness yieldsThis calculation uses the lifts' independence of the other factor coordinates; a vector field with varying coefficients in those coordinates can have a nonzero mixed derivative.

A vector field is a smooth section $X:M\to TM$ of the tangent bundle. It differentiates smooth functions by $Xf(p)=X_p(f)$. Define the Lie bracket of vector fields intrinsically byExpanding this expression on a product $fh$ shows that the two mixed terms cancel, giving $[X,Y](fh)=f[X,Y]h+h[X,Y]f$. It is therefore a derivation, hence a vector field. In local coordinates,Its coefficients are smooth. The intrinsic definition depends on no chart, which proves coordinate independence of this formula and defines the bracket on the whole manifold.

For a Lie group $G$, let $L_g(h)=gh$ be Left translation on a Lie group. A left-invariant vector field satisfies $(d{L}_g{)}_h{X}_h=X_{gh}$ for every $g,h$. Evaluation $X\mapsto X_e$ is linear and injective, since $X_g=(d{L}_g{)}_e{X}_e$. Conversely any $B\in T_{e}G$ gives a smooth left-invariant vector field $X_B(g)=(d{L}_g{)}_{e}B$. Smoothness follows from smooth multiplication and its differential. ThusFor any diffeomorphism $F$, the intrinsic bracket identity on functions gives $F_{\ast }[X,Y]=[F_{\ast }X,F_{\ast }Y]$. Applying this naturality of the Lie bracket to $F=L_g$ proves that the bracket of two left-invariant vector fields remains left invariant. Evaluation at $e$ therefore defines the Lie algebra bracket on $\mathfrak{g}=T_{e}G$.

For the special orthogonal group, differentiation of $Q(t{)}^{T}Q(t)=I$ at $Q(0)=I$ gives $B^T+B=0$, so its tangent space is contained in the skew-symmetric matrices. Conversely, for any such $B$, the matrix exponential $Q(t)=e^{tB}$ satisfies $Q(t{)}^{T}Q(t)=I$ and lies in the determinant-one component, because $Q(0)=I$ and its determinant varies continuously in $\{\pm 1\}$. Its initial derivative is $B$, provingFor matrices, $X_B(Q)=QB$. Extend these fields to the open matrix group $\mathrm{GL}(n,\mathbb{R})$, where the differential of $Q\mapsto QB$ in direction $H$ is $HB$. HenceRestriction to the embedded special orthogonal group preserves this bracket because the fields are tangent. Evaluating at $I$ gives $[B_1,B_2]=B_1{B}_2-B_2{B}_1$, the required bracket on the Special orthogonal Lie algebra. The commutator is again skew symmetric. This calculation is differentiating left-invariant matrix fields; reversing the order of the two differentials would give the wrong sign.

A smooth vector field is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle, so $X_p\in T_{p}M$. In a manifold chart, $X=X^i{\partial }_i$ with smooth coefficients. It acts on a smooth function by $Xf=X^i{\partial }_{i}f$. For a diffeomorphism $\alpha :M\to N$, the pushforward of a vector field isThe Lie bracket of vector fields is the commutator of their actions on smooth functions:The second derivatives of $f$ cancel, so this is again a vector field. The corresponding Jacobi identity is $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$.

A local flow of $X$ is a smooth map $(t,p)\mapsto {\varphi }_t(p)$ on an open neighbourhood of $\{0\}\times M$ satisfying ${\varphi }_0(p)=p$ and ${\partial }_t{\varphi }_t(p)=X_{{\varphi }_t(p)}$. On a sufficiently small neighbourhood and time interval, each ${\varphi }_t$ is a diffeomorphism onto its image, with inverse ${\varphi }_{-t}$. Uniqueness of integral curves of a vector field gives ${\varphi }_{s+t}={\varphi }_s\circ {\varphi }_t$ wherever both sides are defined. These are local statements; no assumption of a complete vector field is required.

A smooth tensor field of type $(r,s)$ is a smooth section of $(TM{)}^{\otimes r}\otimes (T^{\ast }M{)}^{\otimes s}$, where the factors are the tangent bundle and cotangent bundle. The flow definition of the Lie derivative of a tensor field isHere pullback applies $d{\varphi }_t^{-1}$ to each vector factor and the dual of $d{\varphi }_t$ to each covector factor, evaluating $T$ at ${\varphi }_t(p)$. Thus all tensors being differentiated lie in the same fibre over $p$.

For a smooth function, ${\varphi }_t^{\ast }f=f\circ {\varphi }_t$, and the chain rule givesFor a vector field $Y$, in local coordinates the expansions ${\varphi }_t(x)=x+tX(x)+O(t^2)$ and $d{\varphi }_t(x)=I+td{X}_x+O(t^2)$ giveConsequentlyPullback preserves tensor products and tensor contractions, so differentiation makes this definition a tensor derivation. It therefore agrees on every tensor field with the Lie derivative of a tensor field determined by these two formulas.

Now put ${\psi }_t=\alpha \circ {\varphi }_t\circ {\alpha }^{-1}$. The chain rule givesThus ${\psi }_t$ is the local flow of ${\alpha }_{\ast }X$. In the case $M=N$, if ${\alpha }_{\ast }X=X$, uniqueness of integral curves of a vector field makes ${\psi }_t={\varphi }_t$ on their common domains, or $\alpha \circ {\varphi }_t={\varphi }_t\circ \alpha$. Conversely, differentiating this commuting identity at zero gives $d{\alpha }_p{X}_p=X_{\alpha (p)}$, hence ${\alpha }_{\ast }X=X$. This proves that diffeomorphism invariance of a vector field is equivalent to commuting with its local flow, with every identity understood on the domain where its compositions exist.

Let $i:M\hookrightarrow N$ be the inclusion. The restriction of a connection to an embedded submanifold is the pullback connection on $i^{\ast }TN=TN{|}_M$. Concretely, for a tangent vector field $X$ on $M$ and a local section $S$ of $TN{|}_M$, extend $S$ smoothly off $M$ and define ${\nabla }_{X}S$ by differentiating the extension in direction $di(X)$. Two extensions differ by a field whose coefficients vanish on $M$. Their derivatives along every tangent curve in $M$ vanish too, and the connection's coefficient terms are multiplied by the zero field. The answer is therefore independent of the extension. Dependence only on the value of the first argument follows from the connection's linearity over smooth functions.

The induced Riemannian metric is positive definite, so the fibrewise orthogonal projection $\pi :TN{|}_M\to TM$ is smooth. For tangent fields define the projected ambient connection $D_{X}Y=\pi ({\nabla }_{X}Y)$. It is real-linear and satisfiesbecause $\pi Y=Y$. Thus $D$ is a Koszul connection on $M$.

The Levi-Civita connection is characterized by the torsion-free connection and metric connection conditionsThese determine it uniquely. For the ambient Levi-Civita connection, the Gauss formula is ${\nabla }_{X}Y=D_{X}Y+II(X,Y)$, withLinearity over smooth functions in $X$ is immediate. In $Y$, the additional $X(f)Y$ term is tangent and is killed by $I-\pi$. Thus tensoriality makes $II$ a bilinear map $TM{\times }_{M}TM\to \nu M$ into the normal bundle. Its antisymmetric part isbecause the Lie bracket of vector fields tangent to $M$ is tangent. This proves the symmetry of the second fundamental form in an arbitrary Riemannian ambient manifold.

Write the curvature operators as $R^N(v,w)y={\nabla }_v{\nabla }_{w}y-{\nabla }_w{\nabla }_{v}y-{\nabla }_{[v,w]}y$ and $R^M$ with $D$ in place of $\nabla$. To match the four-slot notation in the requested identity, useThis convention is stated because permuting the slots can change the displayed signs.

If $\eta$ is a normal field and $x$ is tangent, metric compatibility applied to $⟨\eta ,x⟩=0$ gives the tangential derivative of a normal field identityUsing the Gauss formula and this identity,Here the normal part of ${\nabla }_v(D_{w}y)$ pairs to zero with $x$. Interchange $v,w$ and subtract; the bracket term obeys $⟨{\nabla }_{[v,w]}y,x⟩=⟨D_{[v,w]}y,x⟩$. It follows thatRearranging proves the Gauss equation in a curved ambient manifold:All expressions are tensorial, so choosing local extensions of the four tangent vectors proves the pointwise identity at $P$. With flat Euclidean ambient space, $R^N=0$ and this reduces to the usual Gauss equation.

On smooth functions, the Lie derivative of a function satisfiesOn an arbitrary vector field $W$, the Lie derivative of a vector field givesThe middle equality follows from the Jacobi identity for the Lie bracket of vector fields. One can verify that identity without assuming the result being proved: view vector fields as derivations acting on smooth functions, expand their commutators, and cancel the six compositions. Equality as derivations implies equality as vector fields.

For the second identity, let $A,B,C$ be any operators on either smooth functions or vector fields. Associativity of composition givesEach term cancels in the sum. Applying this to $A={\mathcal{L}}_X$, $B={\mathcal{L}}_Y$, $C={\mathcal{L}}_Z$ proves both requested identities on both kinds of arguments:The commutator identity for Lie derivatives expresses that the Lie derivative of a tensor field represents the Lie bracket of vector fields.

The Lie derivative of a function is differentiation along the flow, while the Lie derivative of a vector field compares a vector with its pullback by that flow. Their index-free expressions areHere $X(f)$ is the directional derivative of the smooth function, and $[X,Y]$ is the Lie bracket of vector fields, characterized by $[X,Y](f)=X(Y(f))-Y(X(f))$. To see the second expression directly, if ${\varphi }_t$ is the local flow of $X$, differentiating $(d{\varphi }_t{)}^{-1}Y\circ {\varphi }_t$ at $t=0$ gives $dY(X)-dX(Y)=[X,Y]$. This uses the flow definition of the Lie derivative of a tensor field.

The real vector space of symmetric matrices has dimension $n(n+1)/2$. On its invertible part the derivative of the determinant isAt a determinant-one symmetric matrix, this derivative is nonzero: the symmetric direction $C=B$ gives value $n$. The regular level set theorem therefore yieldsThis applies to every signature component, not only positive-definite matrices.

The special linear congruence action on symmetric matrices preserves symmetry because $(AB{A}^T{)}^T=AB{A}^T$, and preserves determinant because $\det (AB{A}^T)=(\det A{)}^2\det B=1$. The identity acts trivially andThese verify a smooth left Lie group action of the special linear group. By Sylvester's law of inertia, it also preserves the numbers of positive and negative eigenvalues.

For $n=2$, parametrize the symmetric matrices byThus $M$ is exactly the two-sheeted hyperboloidThe two sheets consist respectively of positive-definite and negative-definite matrices. Each is preserved by the special linear congruence action on symmetric matrices. They are individually transitive: on the positive sheet $B=A{A}^T$ with $A=B^{1/2}$ and $\det A=1$; on the negative sheet use $B=-A{A}^T$. The stabilizer of either $I$ or $-I$ is $SO(2)$.

Differentiating the left Lie group action along $\exp (sX)$ gives the fundamental vector fieldFor the three matrices in the question, their coordinate components areA sign convention matters here. With the usual Lie bracket of vector fields $[U,V]f=U(Vf)-V(Uf)$, the fundamental fields of this left action obey $[u_X,u_Y]=-u_{[X,Y]}$. Indeed, for linear coordinate fields $u_X(z)=R_{X}z$, the bracket has coefficient $(R_Y{R}_X-R_X{R}_Y)z$. To obtain a Lie algebra representation rather than an anti-representation, use the infinitesimal left-action sign conventionAn explicit answer is thereforeThese are tangent to the two-sheeted hyperboloid: applying each to $Q=t^2-x^2-y^2$ gives zero. For example,They can be written on either sheet using $t=\pm \sqrt{1+x^2+y^2}$:Here the omitted ${\partial }_t$ component is determined by tangency, and derivatives of $t(x,y)$ must be included when computing brackets in these coordinates.

Direct matrix multiplication givesDirect differentiation of the displayed vector fields gives exactlyFor instance, in ambient $(t,x,y)$ coordinates, $[v_1,v_3]=(-2x,-2t,0)=v_2$. These are the defining relations of the sl2 Lie algebra. The three fields are linearly independent over constant real coefficients: if $a{v}_1+b{v}_2+c{v}_3$ vanishes on a sheet, its ${\partial }_x$ coefficient $(-a+c)y-2bt$ forces $b=0,c=a$, and its ${\partial }_y$ coefficient then equals $-2at$, forcing $a=c=0$. Thus the representation is faithful. Their pointwise span need only have dimension two, consistent with the dimension of each sheet.

A Poisson bivector is a smooth antisymmetric contravariant two-tensorwhose bracket on smooth functions,satisfies the Jacobi identity. Bilinearity and antisymmetry are immediate, and the product rule makes the bracket a derivation in each argument. Together these properties define a Poisson manifold. Unlike a symplectic form, the Poisson bivector need not be nondegenerate.

Apply the Jacobi identity to the coordinate functions. Since $\{x^i,x^j\}={\omega }^{ij}$,Changing ${\omega }^{im}$ to $-{\omega }^{mi}$ and rearranging the three summands gives the printed coordinate Jacobi condition for a Poisson bivector:This is also sufficient: expanding the Jacobiator of three arbitrary smooth functions, all terms involving second derivatives cancel in pairs by antisymmetry. The remaining coefficient of ${\partial }_{i}F{\partial }_{j}G{\partial }_{k}H$ is the coordinate-function Jacobiator displayed above.

For a Lie algebra, the natural global space carrying the proposed linear bracket is its dual space ${\mathfrak{g}}^{\ast }$. If $v^i$ is the chosen basis, define its linear coordinate function by $x^i(\ell )=\ell (v^i)$. The Lie-Poisson bracket iswhere $d{F}_{\ell },d{G}_{\ell }$ are elements of $\mathfrak{g}=({\mathfrak{g}}^{\ast }{)}^{\ast }$. On a general manifold the same coordinate expression gives a local construction; a global one requires compatible transition rules. The use of ${\mathfrak{g}}^{\ast }$ supplies that compatibility intrinsically.

Here ${\omega }^{ij}=c_r^{ij}{x}^r$ and ${\partial }_m{\omega }^{jk}=c_m^{jk}$. The left side of the coordinate Jacobi condition for a Poisson bivector is thereforeThe final coefficient is the negative of the coefficient of $v^r$ in the Lie algebra identity $[v^i,[v^j,v^k]]+[v^j,[v^k,v^i]]+[v^k,[v^i,v^j]]=0$. Thus the Lie-Poisson bracket satisfies the Jacobi identity.

It remains to find the Lie algebra structure constants for the printed rotation fields. Distinguish their original spatial coordinates $(x,y,z)$ from the coordinates $(x^1,x^2,x^3)$ on the dual space. Use the conventional Lie bracket of vector fields. Their component vectors are $(0,-z,y)$, $(z,0,-x)$ and $(-y,x,0)$, respectively. For example,Similarly,The negative sign is essential: these are the fundamental fields of a left rotation action with the stated Lie bracket of vector fields, and consequently have the infinitesimal left-action sign convention discussed above. The first field here has component $y{\partial }_z$, as printed in the PDF.

The Lie-Poisson bracket on the dual space consequently hasFor the evolution convention $\dot{F}=\{F,H\}$, the Hamiltonian function has derivatives $(2A{x}^1,2B{x}^2,2C{x}^3)$. Substitution gives the quadratic rotational Lie-Poisson dynamicsThese are Euler-type Hamilton's equations on a noncanonical Poisson manifold. When $A=1/(2I_1)$, $B=1/(2I_2)$ and $C=1/(2I_3)$ with positive principal inertias, they are the Euler equations for a torque-free rigid body in body angular-momentum coordinates. As a check, $H$ is conserved by antisymmetry of the Poisson bracket, and $C_0=(x^1{)}^2+(x^2{)}^2+(x^3{)}^2$ is a Casimir function of a Poisson manifold. Direct differentiation of $C_0$ in the three equations cancels the terms $4[(C-B)+(A-C)+(B-A)]x^1{x}^2{x}^3$. The Hamiltonian flow therefore lies on both an energy level and a sphere, a symplectic leaf of this signed rotational bracket.

For the Lie derivative of a tensor field, the scalar rule follows because a vector field $X$ differentiates products of smooth functions. On vector fields the Lie bracket of vector fields satisfiesThis follows by applying both sides to an arbitrary smooth function and expanding the two compositions of derivations. Both the scalar and vector-field operators are real-linear, so the extension theorem applies and gives the unique contraction-compatible tensor operator ${\mathcal{L}}_X$.

For a type-$(1,1)$ tensor $A$, regard it as the pointwise endomorphism of $TM$ obtained by contracting its covector slot with a vector. The endomorphism-induced tensor derivation starts withIt is real-linear and obeys $D_A(fY)=fA(Y)=f{D}_{A}Y+(D_{A}f)Y$. The scalar operator zero is a derivation in every dimension, so there is no zero-dimensional obstruction here. Consequently it also extends uniquely. On a differential one-form, the two operators areTheir actions on arbitrary tensor fields follow by the tensor-product rule; $D_A$ adds $A$ in every contravariant slot and subtracts its dual action in every covariant slot.

Functions in this calculation belong to $C^{\infty }(M)$: the printed $C^{\infty }(X)$ in Q2(c) is a typographical error. The contraction $C_2^1$ pairs the sole covector with the new vector $Y$.

The tensor denoted $X\otimes df$ acts as the rank-one endomorphism $Y\mapsto df(Y)X=Y(f)X$. In the paper's covector-first ordering it is written $df\otimes X$, using the canonical interchange of tensor factors; this is the same endomorphism.

On any smooth function $h$,On a vector field $Y$, the Lie bracket of vector fields givessoMultiplication by $f$ preserves the tensor-product Leibniz rule and commutation with tensor contractions, since the latter are linear over smooth functions. Hence the difference is a real-linear contraction-compatible tensor derivation whose scalar operator is zero. Its agreement with $D_{X\otimes df}$ on functions and vector fields determines its action on every tensor by uniqueness:For example, on a differential one-form this gives $-\omega (X)df$. Equivalently ${\mathcal{L}}_{fX}\omega =f{\mathcal{L}}_X\omega +\omega (X)df$, which checks the sign independently. This is the scaled Lie derivative defect identity.

The product Riemannian metric isIt is smooth and symmetric. If $(u_1,u_2)\ne 0$, at least one component is nonzero and contributes a strictly positive squared norm; the other contributes a nonnegative term. Thus $g$ is positive definite.

Let the factor connections be Levi-Civita connections. The torsion form of the product affine connection vanishes on two lifted fields from one factor because it equals the factor torsion; on opposite-factor fields it vanishes because both covariant derivatives and the Lie bracket of vector fields vanish. By tensoriality, torsion vanishes for all fields.

Check metric compatibility on lifted fields as well. When the two paired fields are from one factor, their metric pairing depends only on that factor. Differentiation along that factor gives the usual factor compatibility identity, and differentiation along the other factor gives zero on both sides. When the paired fields are from different factors, their pairing is identically zero and their derivatives stay in their original summands, so both sides are again zero. Since $\nabla g$ has tensoriality, these checks imply $\nabla g=0$ globally.

The existence and uniqueness of the Levi-Civita connection therefore givesEquivalently, applying the Christoffel symbol formula to the block metric gives the factor coefficients and zero mixed coefficients, because each factor metric is independent of the other coordinates.

The tangent bundle is $TM={⨆}_{p\in M}{T}_{p}M$, with its smooth vector-bundle structure from coordinate charts. A vector field is a smooth section of this bundle. The Lie bracket of vector fields is their commutator as derivations:In coordinates, $[X,Y{]}^k=X^j{\partial }_j{Y}^k-Y^j{\partial }_j{X}^k$.

The four numbered conditions here are hypotheses of one argument, so the following sections assemble its proof. Let $j:\partial M\hookrightarrow M$. Because all $X_i$ are tangent to the boundary, evaluating $\eta ={\iota }_{X_1}\cdots {\iota }_{X_n}\omega$ on a tangent vector $Y$ there evaluates the volume form on $n+1$ vectors in the $n$-dimensional boundary tangent space. Alternation gives $\boxed{j^{\ast }\eta =0}$. This uses the interior product of a differential form and will make the boundary restriction of the de Rham cohomology class vanish.