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In exponential scale coordinates, the real affine group has matrices $\left(\begin{array}{cc}{e}^{\alpha }& \beta \\ 0& 1\end{array}\right)$. The left and right Maurer-Cartan forms are respectively $Dd\alpha +T{e}^{-\alpha }d\beta$ and $Dd\alpha +T(d\beta -\beta d\alpha )$. Their dual vector fields obey $[D_L,T_L]=T_L$ and $[D_R,T_R]=-T_R$, displaying the opposite bracket sign for right-invariant vector fields.

The Maurer-Cartan form of the translation-dilation group of the plane has these coefficients in its dilation–translation basis, with $a=1,2$. Left multiplication rescales each coordinate differential and $\rho$ equally, proving they are left-invariant differential forms. Their dual left-invariant vector fields are $\rho {\partial }_{\rho }$ and $\rho {\partial }_{x^a}$. The Maurer-Cartan equation is $d{\lambda }^0=0$, $d{\lambda }^a=-{\lambda }^0\wedge {\lambda }^a$. The tensor ${\sum }_j{\lambda }^j\otimes {\lambda }^j$ is a positive definite left-invariant metric.

In the faithful matrices $g=\left(\begin{array}{cc}a& b\\ 0& 1\end{array}\right)$ with $a>0$, the Maurer-Cartan form has entries $da/a$ and $db/a$. Thus the left-invariant differential forms ${\sigma }^D=da/a$, ${\sigma }^T=db/a$ obey $d{\sigma }^D=0$ and $d{\sigma }^T=-{\sigma }^D\wedge {\sigma }^T$. The Lie algebra generators satisfy $[D,T]=T$.

For massless up and down quarks, the Quantum chromodynamics kinetic term splits into two independent chirality sectors,The colour covariant derivative acts identically on both flavours and commutes with constant flavour matrices. Independent transformations $q_R\mapsto A{q}_R$, $q_L\mapsto B{q}_L$, with $A,B\in SU(2)$, therefore leave the action invariant. This is chiral symmetry. The non-singlet axial transformations have traceless flavour generators, so they are not broken by the colour axial anomaly; the anomalous axial $U(1)$ is not being included in this symmetry.

For $V_{ij}={\bar{q}}_{Lj}{q}_{Ri}$, transforming each index givesThe nonzero condensate $⟨V⟩=-vI$ is unchanged precisely when $A=B$. Thus chiral symmetry breaking leaves the diagonal vector subgroup $SU(2{)}_V$. Three of the six generators are broken, producing three massless Goldstone bosons in the exact massless limit. They are the pions, represented at low energy by the coset coordinate $U\in SU(2)$.

The uniqueness of the two-derivative two-flavour chiral action follows from the symmetry, rather than merely from guessing an invariant. The chiral group acts transitively on $U$, so every derivative-free invariant is constant. At $U=I$, the stabilizer acts on the three tangent directions by conjugation, the irreducible rotation representation. Its invariant symmetric quadratic form is a single multiple of ${\delta }_{ab}$. Transporting that form around the group gives a unique invariant two-derivative metric. Equivalently, the Maurer-Cartan form $U^{†}{\partial }_{\mu }U$ is traceless, so an independent double-trace term vanishes; the remaining contracted single trace is the same metric. Lorentz invariance supplies the spacetime contraction. Up to total derivatives, a constant term and its overall coefficient, the massless action is thereforeThe positive coefficient is conveniently expressed through the pion decay constant $F$.

Use the Pauli matrices with $\mathrm{tr}({\tau }_a{\tau }_b)=2{\delta }_{ab}$. Expanding the matrix exponential gives ${\partial }_{\mu }U=i{\tau }_a{\partial }_{\mu }{\pi }_a/F+\cdots$. Henceso the three fields have canonical kinetic terms. There is no nonconstant invariant potential in the exact chiral theory, consistent with their masslessness.

For real diagonal $M$, the two terms in the mass trace areTheir sum therefore gives the usual Dirac mass term,These terms explicitly break chiral symmetry. In the low-energy chiral perturbation theory prescription, $V\mapsto -vU$ yieldsThe Pauli matrix algebra implies $(\mathbf{\pi }\cdot \mathbf{\tau }{)}^2=|\mathbf{\pi }{|}^{2}I$. ThusandComparing with the canonically normalized scalar mass term gives the leading pion mass from a two-flavour condensate:Positive quark masses therefore give positive pion squared masses. Even if $m_u\ne m_d$, this leading action depends only on their sum and gives equal masses for all three pions. Strong isospin-breaking effects beyond this order and electromagnetic effects can split them. In the condensate convention used here, $⟨\bar{u}u⟩=⟨\bar{d}d⟩=-2v$, which also explains the factor of two in the mass relation. The massive pions are pseudo-Goldstone bosons, becoming massless as both quark masses tend to zero.

For a principal bundle $\pi :E\to B$, the vertical space is $V_e=\ker d{\pi }_e$. A principal connection specifies smooth complements $T_{e}E=H_e\oplus V_e$ with $(R_g{)}_{\ast }{H}_e=H_{eg}$. Equivalently its Lie-algebra-valued connection one-form $\mathcal{A}$ satisfies $\mathcal{A}({\xi }_E)=\xi$ and $R_g^{\ast }\mathcal{A}={\mathrm{Ad}}_{g^{-1}}\mathcal{A}$. Every base vector has a unique horizontal lift because $d\pi :H_e\to T_{\pi (e)}B$ is an isomorphism. A base curve lifts from a chosen initial fiber point by solving $\mathcal{A}(\dot{e})=0$; this defines parallel transport. Closed curves can give nontrivial holonomy, and failure of horizontal fields to close under brackets is measured by the curvature of a principal connection.

For a smooth strictly convex rigid body, assume contact with the plane is unique and varies smoothly with orientation. Write a configuration as $(R,c_{\parallel })\in SO(3)\times {\mathbb{R}}^2$. Contact determines the reference point's height $h(R)$, so the full reference position is $c=(c_{\parallel },h(R))$. Let $r(R)$ be the spatial vector from that point to the contacting material point and let $\hat{\Omega }=\dot{R}{R}^{-1}$ encode the spatial angular velocity. The instantaneous contact velocity isThe no-slip condition is its vanishing. Its vertical component follows already from differentiating contact: the normal component of the changing contact point along the body surface is zero, giving $\dot{h}=-(\Omega \times r{)}_z$. The two remaining equations areTranslations of $c_{\parallel }$ define a free right ${\mathbb{R}}^2$ action, with quotient $SO(3)$. The translational connection for a convex body rolling on a plane isHere $\Omega$ denotes the vector-valued right Maurer-Cartan form on $SO(3)$. This one-form is invariant under translations and returns a translation vector on a vertical translation, so it satisfies both principal connection axioms. A prescribed orientation path has the unique rolling horizontal liftThis is a kinematic connection; inertia and gravity determine which of the allowed paths is dynamically realized.

For a sphere of radius $a$, $r=-an$, where $n$ is the upward unit normal. The same no-slip condition gives ${\dot{c}}_{\parallel }=a\Omega \times n$. It leaves the component $\Omega \cdot n$ arbitrary. If one additionally forbids twisting about $n$, then $\Omega =n\times {\dot{c}}_{\parallel }/a$ and a plane path uniquely lifts to an orientation path by $\dot{R}=\hat{\Omega }R$. This is the familiar $SO(3)$ principal connection over the plane; its horizontal lifts have noncommuting rotation generators and hence nonzero curvature. No-slip alone does not supply that latter two-dimensional horizontal distribution. Corners or nonsmooth changes of contact require piecewise treatment rather than the smooth bundle construction.

A Lie algebra homomorphism $\theta :{\mathfrak{g}}_1\to {\mathfrak{g}}_2$ always defines a local group homomorphism near the identities by the exponential charts:for sufficiently small $X$. Preservation of the BCH multiplication law makes this a local homomorphism. The issue is extending it consistently around all paths and periods in the source.

Suppose first that $G_1$ is connected and simply connected. For a piecewise smooth path $\gamma :[0,1]\to G_1$ from $e_1$ to $g$, letbe its left logarithmic velocity. Solve in $G_2$ the differential equationThe equation exists throughout the finite path interval: bounded smooth body velocity gives a uniform local existence interval at the identity, and left translation repeatedly continues the solution. Define the proposed map by the endpoint $\Phi (g)=\eta (1)$.

Bracket preservation makes this endpoint invariant under fixed-endpoint homotopy. To see the mechanism, for a two-parameter path write $A={\gamma }^{-1}{\partial }_t\gamma$ and $B={\gamma }^{-1}{\partial }_s\gamma$, using the Maurer-Cartan form rather than literal multiplication for nonmatrix groups. The Maurer-Cartan equation givesAfter applying $\theta$, the same identity holds for $a=\theta A$ and $b=\theta B$. If $c={\eta }^{-1}{\partial }_s\eta$, the transport equation impliesBoth have zero initial value because the path starts at a fixed identity. Uniqueness gives $c=b$. At the fixed endpoint, $B(s,1)=0$, hence ${\partial }_s\eta (s,1)=0$. The endpoint is therefore unchanged by the homotopy. Simple connectedness then makes it independent of the chosen path.

For paths to $g$ and $h$, concatenate the first with the left translate by $g$ of the second. Its transported endpoint is $\Phi (g)\Phi (h)$, because left logarithmic velocity is unchanged by left translation. It follows that $\Phi (gh)=\Phi (g)\Phi (h)$. Near the identity, choose $\gamma (t)=\exp (tX)$; transport yields $\Phi (\exp X)=\exp (\theta X)$. Thus $\Phi$ is smooth near the identity and, by translation, everywhere, and its differential is $\theta$. The uniqueness argument from the root solution givesThis is integration of a Lie algebra homomorphism.

For a connected source that is not simply connected, first integrate $\theta$ on its universal cover $p:{\tilde{G}}_1\to G_1$. The resulting map $\tilde{\Phi }:{\tilde{G}}_1\to G_2$ descends exactly whenIndeed two lifts of one element differ by a kernel element, so this condition is both necessary and sufficient for their images to agree. It is the period obstruction to integration of a Lie algebra homomorphism. For $G_1=G_2=S^1$, identify the Lie algebras with $\mathbb{R}$ and take $\theta (t)=ct$. The integrated map on the cover is $t\mapsto e^{ict}$, which respects the source period $2\pi$ precisely when $c\in \mathbb{Z}$. Therefore half-scaling is a valid Lie algebra homomorphism but does not integrate to a circle homomorphism. If both groups are written as simply connected covers modulo central subgroups, the equivalent descent condition is that the integrated cover map send ${\Gamma }_1$ into ${\Gamma }_2$.

For disconnected $G_1$, first solve this problem on $G_1^{\circ }$, obtaining ${\varphi }_0$. One must then assign images $a_{\gamma }$ to representatives $s_{\gamma }$ of the component group. These assignments must satisfy the conjugation and multiplication relations. Explicitly, with $s_e=e$ and $s_{\gamma }{s}_{\delta }=s_{\gamma \delta }{c}_{\gamma ,\delta }$ for $c_{\gamma ,\delta }\in G_1^{\circ }$, the conditions areWhen these can be solved, $\varphi (s_{\gamma }h)=a_{\gamma }{\varphi }_0(h)$ is a smooth global homomorphism. They may prevent existence or allow several extensions with the same differential. Thus a bracket-preserving linear map gives local data; global integration requires the source's periods and, when present, its component relations to be respected.

The Maurer-Cartan form is $g^{-1}dg={\lambda }^j{X}_j$. Direct matrix multiplication givesThese are left-invariant differential forms: for a constant group element $a$, $(ag{)}^{-1}d(ag)=g^{-1}dg$. Equivalently, left translation sends $(\rho ,x)$ to $(a\rho ,b+ax)$, so all three coordinate differentials and the denominator acquire the same factor $a>0$.

The left-invariant coframe of the translation-dilation group is pointwise linearly independent because $\rho >0$. Consequentlyis positive definite and unchanged under every left translation. It therefore defines the required left-invariant metric, with precisely the coordinate expression obtained from this sum. The notation $d{\rho }^2$ in that metric means $d\rho \otimes d\rho$, not the differential of ${\rho }^2$.

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.

Multiplying the matrices gives the Heisenberg group lawDifferentiating at the identity gives all strictly upper-triangular matrices. With $X=E_{12}$, $Y=E_{23}$ and $Z=E_{13}$, the commutator givesThis is the Heisenberg Lie algebra.

The right Maurer-Cartan form isThus the right-invariant coframe of the real Heisenberg group is ${\sigma }^1=dx$, ${\sigma }^2=dy$, ${\sigma }^3=dz-ydx$. Under a right translation,we have $d{x}^{\prime }=dx$, $d{y}^{\prime }=dy$, and $d{z}^{\prime }-y^{\prime }d{x}^{\prime }=dz-ydx$. All three right-invariant differential forms are preserved.

Every right-invariant Riemannian metric is determined by an arbitrary inner product at the identity. In this coframe its most general expression isEquivalently,Products here are symmetric products. If a pseudo-Riemannian metric tensor is intended, replace positive definiteness by nondegeneracy. Since every right translation preserves the coframe, it is an isometry. The action is faithful, and $a\mapsto R_{a^{-1}}$ is a group homomorphism, embedding a copy of $G$ into the isometry group.

The one-parameter right translations produce the Killing frame for a right-invariant Heisenberg metric:Their flows are respectively $(x+t,y,z)$, $(x,y+t,z+xt)$ and $(x,y,z+t)$. Each preserves $h$, and their Lie brackets of vector fields are $[K_X,K_Y]=K_Z$, with the other two zero. This explicitly realizes the Heisenberg Lie algebra. These Killing vector fields are left-invariant vector fields; the vector fields dual to the right-invariant coframe are instead ${\partial }_x+y{\partial }_z,{\partial }_y,{\partial }_z$, and should not be substituted for these generators.

For the diagonal case, putThe Kaluza-Klein decomposition along a Killing field uses $V=h(K_X,K_X)=\alpha +\chi y^2$. Completing the square in $dx$ gives the Heisenberg metric Kaluza-Klein reduction:These depend only on the quotient coordinates $(y,z)$ and satisfy $h=V(dx+A{)}^2+\gamma$. For a diagonal indefinite metric tensor, the same expressions hold wherever $V\ne 0$; a null Killing vector field cannot be treated with this completed-square decomposition.

For a Matrix Lie group $G$, the left Maurer-Cartan form identifies tangent vectors with the Lie algebra by translating them to the identity:It is a Lie algebra-valued differential form of degree one. For a fixed $h\in G$, replacing $g$ by $hg$ gives $(hg{)}^{-1}d(hg)=g^{-1}dg$, so it is a left-invariant differential form.

Differentiate $g^{-1}g=I$ to get $d(g^{-1})=-g^{-1}(dg)g^{-1}$. The exterior derivative then givesHere the exterior product of matrix-valued differential forms includes matrix multiplication; the order of the matrices matters. Hence the Maurer-Cartan equation is

Write $\rho ={\sigma }^{\alpha }{T}_{\alpha }$. Antisymmetry of the exterior product impliesComparison of the Lie algebra components yields the Maurer-Cartan equation in a Lie-algebra basis:Thus the canonical antisymmetric choice is $f^{\gamma }{}_{\alpha \beta }=-\frac{1}{2}{c}^{\gamma }{}_{\alpha \beta }$. The factor one half is required because the sum includes both ordered pairs $(\alpha ,\beta )$ and $(\beta ,\alpha )$. If only $\alpha <\beta$ is summed, its coefficient is $-c^{\gamma }{}_{\alpha \beta }$. Strictly, the equality of differential forms determines only the antisymmetric part of $f$: one may add any tensor symmetric in $\alpha ,\beta$ without changing it. This is the symmetric-part ambiguity in Maurer-Cartan coefficients.

A faithful matrix representation of the orientation-preserving real affine group isIts action on $(x,1{)}^T$ has first component $ax+b$. The group operation and inverse areDifferent transformations have different matrix entries, proving faithfulness. With the Lie algebra basisthe left Maurer-Cartan form isThe requested left-invariant differential forms therefore form the Maurer-Cartan coframe of the real affine group:For a direct invariance check, left translation by $(a_0,b_0)$ sends $(a,b)$ to $(a_{0}a,b_0+a_{0}b)$, and the pullbacks of the two displayed differential forms are unchanged. Their dual left-invariant vector fields are $a{\partial }_a$ and $a{\partial }_b$, whose bracket is $a{\partial }_b$, in agreement with $[D,T]=T$.

Direct differentiation gives, for example,Differentiating ${\sigma }^1,{\sigma }^2$ similarly and substituting the definitions yields $d{\sigma }^1={\sigma }^2\wedge {\sigma }^3$ and $d{\sigma }^2={\sigma }^3\wedge {\sigma }^1$. Hence these Maurer-Cartan forms on the three-sphere obey

For $U=e^{2i\pi /f_{\pi }}$, the Maurer-Cartan form expands asSubstituting, using cyclicity of the trace, and collecting even powers givesOdd terms cancel because the sigma-model metric is invariant under $U\leftrightarrow U^{†}$.

For a matrix Lie group, the left-invariant Maurer-Cartan form isDifferentiating $g^{-1}g=I$ gives $d(g^{-1})=-g^{-1}(dg)g^{-1}$. Henceand therefore

Represent the real affine group byMatrix multiplication reproducesThe Maurer-Cartan form isso a basis of left-invariant one-forms isThe dual left-invariant vector fields areIndeed ${\sigma }^i(E_j)={\delta }_j^i$, and left translation preserves the one-forms and vector fields.

For the Heisenberg group law $(x,y,z)(a,b,c)=(x+a,y+b,z+c+xb)$, the right Maurer-Cartan form has coefficients $dx,dy,dz-ydx$. They satisfy $d{\sigma }^1=d{\sigma }^2=0$ and $d{\sigma }^3={\sigma }^1\wedge {\sigma }^2$. A general right-invariant Riemannian metric is a constant positive-definite symmetric matrix in this coframe.

A differential form $\alpha$ on a Lie group is right-invariant if $R_g^{\ast }\alpha =\alpha$ for every right translation $R_g(p)=pg$. A right Maurer-Cartan form is represented in a matrix group by $dg{g}^{-1}$, whereas the left version is $g^{-1}dg$. Its components supply a right-invariant coframe.