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If ${\nabla }_X{e}_i={\sum }_j{\omega }_{ij}(X)e_j$ and ${\varphi }_i$ is the dual coframe, the first index of $\omega$ is the input frame vector. Expand the connection derivative of $Y={\sum }_i{\varphi }_i(Y)e_i$ and antisymmetrize in $X,Y$. The formula for the exterior derivative of a one-form evaluated on vector fields gives ${\tau }_j=d{\varphi }_j-{\sum }_i{\varphi }_i\wedge {\omega }_{ij}$. This is the first structure equation with the indicated index convention. Reversing the wedge order changes the sign.

On a $d$-dimensional Lie group, take the coframe dual to a basis of left-invariant vector fields. Its top wedge is nonvanishing and invariant under left translations, so its absolute density defines a left Haar measure. If the frame has coordinate matrix $A(x)$, the density is $|\det A(x){|}^{-1}d{x}^1\cdots d{x}^d$. This makes explicit computation of Haar measure a determinant calculation.

Let ${\eta }_{ab}=g(e_a,e_b)$ be the constant diagonal metric matrix in a local pseudo-orthonormal frame, and define ${\theta }_{ab}={\eta }_{ac}{\theta }^c{}_b$. The metric compatibility of the Levi-Civita connection givesThus the lowered connection matrix is antisymmetric, including in indefinite signature. Ordinary matrix antisymmetry of ${\theta }^a{}_b$ itself is not the corresponding assertion in indefinite signature.

Let ${\omega }^a$ be the dual coframe, and write a vector field as $Y=e_b{\omega }^b(Y)$. ThenSubtract the same expression with $X,Y$ exchanged and subtract ${\omega }^a([X,Y])$. By the definition of the exterior derivative, this gives the components of the torsion tensor:The Levi-Civita connection is torsion-free, so Cartan's first structure equation is

Use $R(X,Y)={\nabla }_X{\nabla }_Y-{\nabla }_Y{\nabla }_X-{\nabla }_{[X,Y]}$ and define its matrix of curvature two-forms by $R(X,Y)e_b=e_a{R}^a{}_b(X,Y)$. Applying two covariant derivatives to $e_b$ yieldsThe first three terms are $d{\theta }^a{}_b(X,Y)$ and the last two are $({\theta }^a{}_c\wedge {\theta }^c{}_b)(X,Y)$. Therefore Cartan's second structure equation isThe associated Riemann curvature tensor components satisfy $R^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{\omega }^c\wedge {\omega }^d$ with this convention.

For a position-dependent change of pseudo-orthonormal frame, write the frame row as $E^{\prime }=E\Lambda$, with ${\Lambda }^T\eta \Lambda =\eta$. The dual coframe obeys ${\omega }^{\prime }={\Lambda }^{-1}\omega$. Applying the Leibniz rule to $\nabla (E\Lambda )$ givesThe connection one-form has an inhomogeneous term because differentiating the moving basis produces $d\Lambda$. The curvature two-form instead transforms homogeneously. Indeed, curvature is tensorial in the vector to which it is applied, so $R(X,Y)(E\Lambda )=(R(X,Y)E)\Lambda =E^{\prime }{\Lambda }^{-1}R(X,Y)\Lambda$. ConsequentlyEquivalently, substituting the transformed connection into $d{\theta }^{\prime }+{\theta }^{\prime }\wedge {\theta }^{\prime }$ and using $d{\Lambda }^{-1}=-{\Lambda }^{-1}(d\Lambda ){\Lambda }^{-1}$ cancels all derivative terms and gives the same result. This transformation law holds for a general invertible frame change as well; the orthogonality condition ensures both frames remain pseudo-orthonormal.

Differentiate the product with the first factor close to the identity. A small displacement of its $x^{\prime }$ coordinate adds to $x$; a small $y^{\prime }$ displacement adds to $y$; and a small $z^{\prime }$ displacement changes $(x,y,z)$ by $(-x,y,1)$ times that displacement. The infinitesimal left actions therefore give the right-invariant vector fieldsFor right multiplication, regard $(x^{\prime },y^{\prime },z^{\prime })$ as the fixed point and the second factor as infinitesimal. Its displacements are $(e^{-z^{\prime }}\delta x,e^{z^{\prime }}\delta y,\delta z)$; relabeling the coordinates of the fixed point gives the left-invariant vector fieldsThe naming distinction matters: generators of left multiplication are right invariant, whereas generators of right multiplication are left invariant.

All mixed Lie brackets of vector fields vanish. For example, $R_3(e^{-z})=-e^{-z}$ cancels $L_1(-x)=-e^{-z}$ in $[L_1,R_3]$, and $R_3(e^z)=e^z$ cancels $L_2(y)=e^z$ in $[L_2,R_3]$. The brackets with $R_1,R_2$ are zero because the coefficients of the $L_i$ have no $x$ or $y$ dependence; $[L_3,R_3]=0$ because the coefficients of $R_3$ have no $z$ dependence. HenceConceptually, the corresponding flows commute because $(ag)b=a(gb)$, by associativity of the Lie group multiplication.

The dual coframe iswhich satisfies ${\theta }^i(L_j)={\delta }_j^i$. Under left multiplication by $(x_0,y_0,z_0)$, the transformed coordinates are $X=x_0+e^{-z_0}x$, $Y=y_0+e^{z_0}y$, $Z=z_0+z$. Therefore $e^{Z}dX=e^{z}dx$, $e^{-Z}dY=e^{-z}dy$, and $dZ=dz$. Each one-form is a left-invariant differential form, so their sum of squares is the left-invariant metricThis is the standard metric of the Sol Lie group. As a sign check, $[L_3,L_1]=-L_1$, $[L_3,L_2]=L_2$, while $d{\theta }^1={\theta }^3\wedge {\theta }^1$ and $d{\theta }^2=-{\theta }^3\wedge {\theta }^2$, in agreement with the Maurer-Cartan equation.

Since the Riemannian orthonormal frame is positively oriented, its dual coframe satisfies ${\omega }_g={\omega }_1\wedge \cdots \wedge {\omega }_n$. The form $\nu$ is the interior product ${\iota }_X{\omega }_g$. The contraction rule for an exterior product givesOne can also obtain the sign by moving $E_i$ from the first argument into slot $i$ of the alternating form, using $i-1$ transpositions. Linearity in $X={\sum }_i{f}_i{E}_i$ proves the interior contraction of the Riemannian volume form formula:The exponents $i-1$ and $i+1$ have the same parity.

Use the standard convention that these compact smooth manifolds have no boundary. The interior product identity for a one-form gives $df\wedge {\iota }_X{\omega }_g=df(X){\omega }_g=X(f){\omega }_g$. For example, in the coframe from part (b), only the matching index survives in this wedge product, with the same pair of cancelling signs as in part (c). Thus the graded Leibniz rule and part (c) giveIntegrate and apply the Generalized Stokes theorem. Since $\partial M=\varnothing$, the integral of $d(f\nu )$ is zero. Rearranging proves integration by parts for Riemannian divergence:If a boundary were allowed, the exact general formula would instead have the additional term ${\int }_{\partial M}f\nu$ on the right; compactness alone does not remove that boundary contribution.

Let $v=(x,y{)}^T$ and $B(\rho )=\left(\begin{array}{cc}\cosh \rho & \sinh \rho \\ \sinh \rho & \cosh \rho \end{array}\right)$. Matrix multiplication givesThus $G$ is a semidirect product of the translation plane by the one-parameter Lorentz boost group. Differentiating left multiplication at the identity gives the left-invariant vector fieldsThey obey $[X_0,X_1]=X_2$, $[X_0,X_2]=X_1$ and $[X_1,X_2]=0$.

Their dual coframe is $d\rho$, $\cosh \rho dx-\sinh \rho dy$, and $-\sinh \rho dx+\cosh \rho dy$. The Haar measure from a left-invariant coframe is their absolute wedge product. Since $\det B(\rho )=1$,One can also check this directly: left multiplication has Jacobian determinant one. Right multiplication has a block triangular Jacobian with determinant one too, so the same measure is both left and right Haar measure.

The geodesic flow of a closed negatively curved surface is a topologically transitive Anosov flow. In particular, its Jacobi-field hyperbolicity and the usual negative-curvature transitivity theorem allow the smooth Livsic theorem to be applied. Set $b(x,v)={\beta }_x(v,v)$. The given closed-geodesic integrals are exactly its periodic-orbit obstructions on $SM$. Therefore

We make the differential operators and signs explicit. On the unit tangent bundle, let $V$ generate positive fibre rotation, $X$ be the geodesic vector field and $H=[V,X]$. For the canonical coframe of a surface unit tangent bundle, writeand let $\omega$ be the Levi-Civita connection form with $\omega (V)=1$. This uses $\theta$ for a one-form, keeping $\beta$ for the given tensor. The dual frame is $(X,H,V)$, and the structure equations areThese follow by writing the two horizontal forms as a rotated local orthonormal coframe and $\omega =d\vartheta +{\omega }_0$, with ${\omega }_0$ its connection 1-form. Equivalently, differentiating the sine and cosine components of $X$ twice in the fibre angle givesThe curvature equation also gives $[X,H]=KV$. The Liouville volume of a surface geodesic flow will be taken positive in the base/positive-fibre orientation, namely $\mu =\alpha \wedge \theta \wedge \omega$.

In an oriented orthonormal frame at $x$, put $v=(\cos \vartheta ,\sin \vartheta )$. The symmetric tensor evaluation isSince $V={\partial }_{\vartheta }$, this proves $(V^3+4V)b=0$.

The Lie bracket of vector fields identities give an operator identity that supplies the required equation. First,Using $VH=HV-X$ and $XV=VX-H$, differentiate this expression once more to obtainFor $\phi =V^{2}u+u$ and $Xu=b$, the first term vanishes by the explicit fibre calculation. Hence

A connection on a vector bundle $E\to M$ is an $\mathbb{R}$-linear mapsatisfying $\nabla (fs)=df\otimes s+f\nabla s$. Equivalently its directional version is real-bilinear and satisfiesFor a complex bundle it is complex-linear in the section variable and has the same Leibniz property.

Every smooth vector bundle over an ordinary Hausdorff, second-countable smooth manifold has such a connection. Choose trivializing open sets $U_a$, local frames $e_{a,i}$ and a subordinate locally finite partition of unity ${\rho }_a$. In each frame define the flat local connectionThen set $\nabla s={\sum }_a{\rho }_a{\nabla }^{(a)}(s{|}_{U_a})$, extending each weighted term by zero outside $U_a$. The supports lie inside the corresponding trivializing sets, making the extensions smooth, and local finiteness makes the sum smooth. Since ${\sum }_a{\rho }_a=1$, its Leibniz rule isThis is the construction of a vector bundle connection by a partition of unity.

The difference $A={\nabla }^{\prime }-\nabla$ of two connections is $C^{\infty }(M)$-linear in both its vector-field argument and its section argument. The derivative terms cancel, so $A_{X}s$ depends only on $X_p,s_p$ at each point. Smooth local coefficients therefore identify it withConversely, any such $A$ added to a connection preserves its axioms. Thus connections form an affine space modeled on ${\Omega }^1(M;\mathrm{End}E)$. Choosing a base connection ${\nabla }^0$ gives the noncanonical bijection $A\mapsto {\nabla }^0+A$; there is no distinguished zero connection and hence no canonical vector-space identification. This is the affine space of vector-bundle connections.

For a connection on $TM$, the torsion tensor is the alternating operation $\tau (X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$. The connection and bracket Leibniz rules giveAlternation then gives $\tau (X,fY)=f\tau (X,Y)$ as well. Hence it is tensorial, of type one contravariant and two covariant indices:It is equivalently a $TM$-valued two-form, a torsion form.

For the local frame $e_i$ with dual coframe ${\varphi }_j$, defineThe connection is $C^{\infty }$-linear in $X$, so these are smooth differential one-forms. Expansion in the frame gives ${\nabla }_X{e}_i={\sum }_j{\omega }_{ij}(X)e_j$, and uniqueness follows by applying each ${\varphi }_j$. Here the first index is the input frame vector and the second the output coefficient; this fixes the connection one-form convention.

Expand $Y={\sum }_i{\varphi }_i(Y)e_i$ and apply the connection Leibniz rule:Subtract the corresponding expression with $X,Y$ exchanged, and then subtract ${\varphi }_j([X,Y])$. By the exterior derivative of a one-form evaluated on vector fields,Since all pairs of tangent vectors can be tested, equality of the two-forms follows:This is the Cartan first structure equation with input-first indices. Its plus sign is correct with the stated index order and wedge order; the common output-first convention writes the same identity with a minus sign and reversed wedge factors.

The multiplication $(x^{\prime },y^{\prime },z^{\prime })(x,y,z)=(x^{\prime }+e^{-z^{\prime }}x,y^{\prime }+e^{z^{\prime }}y,z^{\prime }+z)$ defines the three-dimensional Sol Lie group. Its left-invariant vector fields $e^{-z}{\partial }_x,e^z{\partial }_y,{\partial }_z$ have dual coframe $e^{z}dx,e^{-z}dy,dz$. Their sum of squares is a left-invariant metric, $d{s}^2=e^{2z}d{x}^2+e^{-2z}d{y}^2+d{z}^2$.