From the Poincaré and Galilei Algebras to the Spacetime Metric

Overview¶

We start from the abstract Lie algebras of the Poincaré and (bare) Galilei groups, defined purely by their commutation relations, and derive the invariant metric on the spacetime they act on. We do everything at the level of the algebra, using nothing but commutators and the Leibniz rule.

Conventions: 3+1 dimensions, no factors of $i$ in the structure constants, signature $(-, +, +, +)$, units $c = 1$. For Galilei we use the bare algebra (no central charge), which is sufficient for the spacetime metric.

The companion document Lie Groups I starts from explicit matrix generators and solves the same invariance condition. Here we derive everything from the commutation relations.

Strategy¶

The translation subspace as a representation¶

Let $V = \operatorname{span}\{P^0, P^1, P^2, P^3\}$ with $P^0 \equiv H$, the 4-dimensional subspace of translation generators inside the full Lie algebra.

The homogeneous generators $X \in \{J_i, K_i\}$ act on the whole algebra by the adjoint action

A direct inspection of the Poincaré or Galilei commutation relations (given in Parts I and II below) shows that for every generator $X \in \{J_i, K_i\}$ and every $\mu$,

so the adjoint action restricts to a linear representation of the homogeneous algebra on $V$.

Group invariance ⇒ Leibniz condition¶

Exponentiate. The group element $U = \exp(\theta X)$ acts on $V$ by conjugation:

A symmetric bilinear form $g$ on $V$ is invariant under the group action if

Differentiate at $\theta = 0$. Using $\dfrac{d}{d\theta}\big|_0 U P^\mu U^{-1} = [X, P^\mu]$ and the bilinearity of $g$,

Invariance forces this derivative to vanish:

for every generator $X \in \{J_i, K_i\}$ and every $\mu, \nu$. This is the Leibniz condition: invariance is the statement that $X$ acts as a derivation on $g$ with zero output.

It is a necessary condition. It is also sufficient: the infinitesimal Leibniz condition (at $\theta = 0$) implies the finite group invariance $g(U P^\mu U^{-1}, U P^\nu U^{-1}) = g(P^\mu, P^\nu)$ for all $\theta \in \mathbb{R}$. To see this, define

Differentiating at an arbitrary $\theta$:

By closure of $V$ under the adjoint action, $P^\mu_\theta \in V$, so the Leibniz condition (which holds for every pair of vectors in $V$, not just the basis) gives $f'(\theta) = 0$ identically. Hence $f$ is constant, $f(\theta) = f(0) = g(P^\mu, P^\nu)$, for all $\theta$. So the infinitesimal Leibniz condition is equivalent to finite group invariance along each one-parameter subgroup, and composing these recovers invariance under every element of the identity component of the group.

Writing $g(P^\mu, P^\nu) = \eta^{\mu\nu}$ with $\eta^{\mu\nu} = \eta^{\nu\mu}$, the Leibniz condition becomes a finite system of linear equations on the 10 components of $\eta$.

Spacetime as a homogeneous space¶

So far we have treated the translation subspace $V = \operatorname{span}\{P^0, P^1, P^2, P^3\}$ as if its basis carried a preferred geometric interpretation: $P^\mu$ generates translation along the $\mu$-th spacetime direction, and the parameters $a^\mu$ in $T_a = \exp(a^\mu P^\mu)$ are spacetime coordinates. This step needs justification, because the Lie algebra by itself does not announce which of its elements are “translations” and which are not.

The proper framework is that of a Klein geometry. Given a Lie group $G$ and a closed subgroup $H \subset G$, the coset space

is a smooth manifold on which $G$ acts transitively, with $H$ playing the role of the stabilizer of a chosen basepoint. At the algebra level this corresponds to a vector-space decomposition

where $\mathfrak{h}$ is the Lie subalgebra of $H$ (the isotropy at the basepoint) and $\mathfrak{m}$ is a complementary subspace identified with the tangent space $T_{eH}\mathcal{M}$. The pair $(\mathfrak{g}, \mathfrak{h})$ is what defines the geometry; the Lie algebra $\mathfrak{g}$ alone is not enough.

Reductive and symmetric Klein geometries¶

The vector-space splitting $\mathfrak{g} = \mathfrak{h} \oplus \mathfrak{m}$ needs a word of qualification, because the rest of this document silently relies on a non-trivial property of it.

A Klein geometry $(G, H)$ is called reductive if the splitting can be chosen so that $\mathfrak{m}$ is preserved by the adjoint action of $H$ — equivalently, at the algebra level,

Note that $\mathfrak{m}$ is not required to be a Lie subalgebra; only that $\mathrm{ad}_X$ for $X \in \mathfrak{h}$ maps $\mathfrak{m}$ back into itself.

If, in addition,

the geometry is called symmetric. Every symmetric Klein geometry is reductive; the converse is false.

Why reductivity matters here. Each step of our construction needs $\mathrm{Ad}(H)$ to act on $\mathfrak{m}$ separately from the rest of $\mathfrak{g}$:

• The Leibniz invariance condition asks for an $\mathrm{Ad}(H)$-invariant bilinear form on $\mathfrak{m}$. This presupposes that $\mathrm{ad}_X(Y) \in \mathfrak{m}$ for $X \in \mathfrak{h}$, $Y \in \mathfrak{m}$, i.e. exactly $[\mathfrak{h}, \mathfrak{m}] \subset \mathfrak{m}$.

• The Maurer–Cartan splitting $\omega = e + \omega_{\mathfrak{h}}$ into vielbein and spin connection, used to translate the algebraic metric on $\mathfrak{m}$ into a metric in coordinates (see the $S^2$ derivation in Part IV), needs an $\mathrm{Ad}(H)$-equivariant projection $\mathfrak{g} \to \mathfrak{m}$, which exists iff the geometry is reductive.

Without reductivity, “invariant form on $\mathfrak{m}$” is not even well-posed, because every choice of complement to $\mathfrak{h}$ in $\mathfrak{g}$ would lose pieces of $\mathrm{ad}_X(\mathfrak{m})$ back into $\mathfrak{h}$ under $\mathrm{Ad}(H)$. The whole machinery of this document is specifically a recipe for reductive Klein geometries.

Are there non-reductive Klein geometries? Yes, and they describe important geometries — they are simply not the ones where a metric is the primary invariant. The canonical examples are the parabolic geometries, where $H$ is a parabolic subgroup of a semisimple $G$:

In each case the parabolic $H$ has a Levi decomposition $H = L \ltimes N$ with $N$ unipotent, and the nilpotent pieces in $\mathfrak{h}$ prevent any choice of complement $\mathfrak{m}$ from being $\mathrm{Ad}(H)$-invariant. The right invariants on $G/H$ are no longer first-order tensors on $\mathfrak{m}$ (such as a metric) but live in filtrations of $\mathfrak{g}/\mathfrak{h}$ and capture geometric structure one or more derivative orders higher. This is the world of Tanaka–Morimoto–Čap–Slovák parabolic geometry, the modern home for conformal and projective geometry.

All six examples in this document. Poincaré, Galilei, $\mathfrak{e}(2)$, $\mathfrak{e}(3)$, $\mathfrak{so}(3)$, and $\mathfrak{so}(4)$ — together with their close cousins (de Sitter, anti-de Sitter, hyperbolic space) — are all not just reductive but symmetric. That is why each one admits a canonical metric (or pair of metrics) determined by an algebraic invariance condition, and why the construction works so cleanly. The bracket dichotomy

— flat vs. constant curvature, as summarised in the table after Part IV — is the symmetric-space dichotomy. The cases where it fails (e.g., neither inclusion holds, or $[\mathfrak{h}, \mathfrak{m}] \not\subset \mathfrak{m}$ in any complement) take us out of symmetric Klein geometry and, in the most interesting non-reductive case, into parabolic geometry.

Many familiar spaces are constructed this way: the sphere as $SO(3)/SO(2)$, Euclidean space as $E(n)/SO(n)$, hyperbolic space as $SO(n,1)/SO(n)$, and — the case at hand — Minkowski spacetime as $\text{Poincaré}/\text{Lorentz}$ and Galilean spacetime as $\text{Galilei}/\{J, K\}$. In each case the Lie group encodes the symmetries, and the chosen subgroup $H$ encodes “what fixes the origin”.

For the Poincaré algebra, the choice of subalgebra is essentially canonical. The Levi decomposition reads

with the radical $\mathfrak{t} = \operatorname{span}\{H, P^i\}$ being the unique maximal abelian ideal. Both $\mathfrak{t}$ and the Levi complement $\mathfrak{h} = \mathfrak{so}(3,1)$ are determined up to conjugation by the algebraic structure alone. So Minkowski spacetime $\mathbb{R}^{3,1}$ is intrinsically attached to the Poincaré algebra.

For the bare Galilei algebra, the situation is genuinely ambiguous. The maximal abelian ideal turns out to be $\operatorname{span}\{K_i, P^i\}$ — six-dimensional, since $[K_i, K_j] = [K_i, P^j] = [P^i, P^j] = 0$ and one verifies that bracketing this subspace with any generator stays inside it. The standard “spacetime translations” $\{H, P^i\}$ form a strictly smaller four-dimensional abelian ideal, and indeed there are two natural Klein pairs:

• $\mathfrak{h} = \{J_i, K_i\}$, $\mathfrak{m} = \{H, P^i\}$ — giving the four-dimensional Galilean spacetime;

• $\mathfrak{h}' = \{J_i, H\}$, $\mathfrak{m}' = \{K_i, P^i\}$ — giving the six-dimensional phase space (position × velocity).

Both are valid homogeneous spaces of the Galilei group. The Lie algebra alone does not single one of them out; the physical identification of spacetime as the space of events fixes the choice $\mathfrak{h} = \{J_i, K_i\}$ — that is, rotations and boosts are the transformations that leave a chosen event in place.

To see this more explicitly, note that the choice of Klein pair is fixed operationally, by specifying what kind of object the points of the homogeneous space are meant to be. An event in physics is a localised occurrence: a flash at a place at a moment. Two observers at different positions or with different velocities may label it by different coordinates, but it is the same physical thing. Asking which of the ten Galilei transformations does NOT move an event gives a clean answer:

So the transformations that fix an event are exactly $\{J^i, K^i\}$ and those that move events are exactly $\{H, P^i\}$. By the orbit–stabilizer theorem, the space of events is

with $\{H, P^i\}$ acting on it transitively.

Repeating the same procedure with a different physical primitive gives a different homogeneous space. Take inertial worldlines as the primitive (a worldline is a constant-velocity trajectory $x(t) = a + vt$):

The stabilizer of a worldline is thus $\{J^i, H\}$ and the resulting homogeneous space $G / \{J^i, H\}$ is the 6-dimensional space of inertial worldlines, parametrised by $(a, v)$. This is the underlying manifold of classical Galilean phase space, equivalently $T^* \mathbb{R}^3$.

One can take this further. Choosing as primitive “an event together with a velocity vector at it” gives a 7-dimensional homogeneous space $G / \{J^i\}$, with coordinates $(t, x, v)$ — Souriau’s evolution space of classical mechanics. Choosing the empty stabilizer recovers the full 10-dimensional group $G$ itself. Every choice of physical primitive thus gives a homogeneous space; the Lie algebra accommodates them all, and physics picks one by saying what kind of object the points are.

Two remarks are in order. First, the Klein construction returns only the smooth manifold — not the symplectic structure that makes the 6D space a Hamiltonian phase space. The symplectic form $\omega = dp_i \wedge dq^i$ involves the mass $m$ (since $p = m v$), which is not present in the bare Galilei algebra; it enters via the Bargmann central extension, where one of the commutators is modified to $[K_i, P^j] = m \delta_{ij}\, M$ with $M$ a central generator. So bare Galilei gives the 6D manifold; Bargmann gives the symplectic structure. More generally, phase spaces in physics arise as coadjoint orbits of (centrally extended) symmetry groups — the Kirillov–Kostant–Souriau picture — in which the Klein-pair construction supplies the manifold and the central extension supplies the symplectic geometry.

Second, among all these candidates events are special because they are the most local and operational primitives: point-like, observable in principle by a single localised detection, and requiring no derived notion of velocity, mass, frame, or trajectory. Worldlines, phase-space points, and evolution-space points are all built out of events (equivalence classes of events under time translation, of worldlines under further equivalences, and so on). This operational primacy of events is what makes spacetime the natural arena for the laws of physics, and the other homogeneous spaces the natural arenas for derived constructions such as Hamiltonian dynamics or Lagrangian mechanics.

In the Poincaré case this whole subtlety disappears. The only abelian ideal is the spacetime translations $\{H, P^i\}$, so there is no analog of the 6-dimensional abelian ideal $\{K_i, P^i\}$ and no competing Klein pair. The “events as primitive” choice is canonical because it is the only choice. The Galilei algebra is ambiguous precisely because $[K_i, P^j] = 0$ — the same single algebraic difference that destroys non-degeneracy of the metric is what permits two distinct Klein pairs to exist in the first place.

The reason the translation parameters $a^\mu$ can be used directly as global coordinates on spacetime — in both the Poincaré and the standard Galilei case — is that in each case $\mathfrak{m}$ is an abelian ideal. Two consequences follow:

• Because $\mathfrak{m}$ is abelian, the exponential map $\mathfrak{m} \to T = \exp(\mathfrak{m})$ is a Lie-group isomorphism: $T$ is the additive group $(\mathbb{R}^4, +)$, with coordinates $a^\mu$.

• Because $\mathfrak{m}$ is an ideal complementary to $\mathfrak{h}$, every coset $gH$ has a unique representative in $T$, giving a global diffeomorphism $\mathcal{M} \cong T \cong \mathbb{R}^4$.

This is what makes Minkowski and Galilean spacetimes flat affine spaces — a special feature of the algebras at hand. For a non-abelian or non-ideal $\mathfrak{m}$ — for instance $SO(3)/SO(2) = S^2$ — the exponential map is not a global bijection and the resulting homogeneous space is curved. The further generalization to spaces that are not even homogeneous is the framework of Cartan geometry, where the homogeneous model varies smoothly from point to point; this is how curved spacetimes of general relativity are described, modelled point-wise on flat Minkowski space.

Finally, with the Klein-geometry picture in place, the connection between the abstract bilinear form $\eta^{\mu\nu}$ derived below and the spacetime metric is straightforward. The translation parameters $a^\mu$ transform under the homogeneous generators in exactly the same way as the basis $\{P^\mu\}$, since the conjugation $U T_a U^{-1} = T_{\Lambda a}$ at the group level differentiates to the same adjoint action $[X, P^\mu]$ at the algebra level. Hence the matrix $\eta^{\mu\nu}$ that solves the Leibniz invariance condition on $V$ is also the matrix of the invariant bilinear form on the spacetime coordinates $x^\mu = a^\mu$ — i.e., the metric.

Part I: The Poincaré algebra → Minkowski metric¶

Commutation relations¶

The Poincaré algebra has ten generators $\{J_i, K_i, P^i, H\}$ with

We use $P^0 = H$ and label $\eta^{\mu\nu}$ accordingly: $\eta^{00} = g(H, H)$, $\eta^{0i} = g(H, P^i)$, $\eta^{ij} = g(P^i, P^j)$.

Imposing invariance¶

Apply the Leibniz condition with $X = K_i$ and the indicated indices.

$(\mu, \nu) = (0, 0)$:

$(\mu, \nu) = (0, j)$:

So

$(\mu, \nu) = (j, k)$, both spatial:

which is automatic from $\eta^{0i} = 0$.

Rotations $J_i$ impose no further constraint: the diagonal isotropic form $\eta^{ij} \propto \delta^{ij}$ already commutes with them.

Result¶

Setting $\eta^{00} = -1$:

It is non-degenerate, $\det \eta = -1$, so the inverse $\eta_{\mu\nu} = \operatorname{diag}(-1, 1, 1, 1)$ is also invariant and is the same matrix.

Part II: The bare Galilei algebra → degenerate metrics¶

Commutation relations¶

The bare Galilei algebra has the same generators with two changes from Poincaré (boxed):

The two differences — $[K_i, K_j] = 0$ and $[K_i, P^j] = 0$ — are what change Minkowski into the Galilean structure.

Temporal metric (bilinear form on $V$)¶

Apply the Leibniz condition with $X = K_i$:

$(\mu, \nu) = (0, 0)$:

$(\mu, \nu) = (0, j)$:

The spatial block is killed.

$(\mu, \nu) = (j, k)$: $\;0 + 0 = 0$ (automatic).

Only $\eta^{00}$ survives. Rotations leave it untouched. Setting $\eta^{00} = 1$:

This is the spacetime interval $\Delta s^2 = (\Delta t)^2$: absolute time.

Spatial metric (symmetric 2-tensor on $V$)¶

The temporal metric is degenerate ($\det = 0$), so it has no inverse. But the Galilei algebra admits a second, independent invariant: a symmetric contravariant 2-tensor

The Leibniz condition for $\xi$ is derived exactly as before. The group element $U = \exp(\theta X)$ acts on $V \otimes V$ on both factors:

Invariance $U \xi U^{-1} = \xi$ for all $\theta$, differentiated at $\theta = 0$, gives

for every generator $X \in \{J_i, K_i\}$. This is the Leibniz condition on a contravariant tensor — note that the same rule applies, but now contracted against $\xi^{\mu\nu}$ rather than evaluating $g$ on $P^\mu, P^\nu$.

Reading off coefficients of the basis $P^\alpha \otimes P^\beta$ with $[X, P^\mu] = L^\rho{}_\mu\, P^\rho$:

For Galilei $K_i$, the only nonzero commutator is $[K_i, H] = P^i$, i.e. $L^\rho{}_0 = \delta^\rho_i$. The invariance condition reduces to

Choosing $\alpha = i$, free $\beta$: $\xi^{0\beta} = 0$. Choosing $\beta = i$, free $\alpha$: $\xi^{\alpha 0} = 0$. Together: the entire time row and column of $\xi$ vanish.

The remaining spatial block $\xi^{ij}$ is rotation-invariant only if proportional to $\delta^{ij}$. Setting the proportionality to 1:

This is the Euclidean 3-metric on each slice of simultaneity.

Why two metrics for Galilei?¶

The two invariants live in different spaces:

Both are degenerate, so neither is the inverse of the other:

For Poincaré the corresponding $V^* \otimes V^*$ and $V \otimes V$ forms are both non-degenerate Minkowski — and they are mutual inverses, so the distinction collapses to a single metric.

That no non-degenerate $\eta$ exists for Galilei is visible directly from the calculation: the boost condition $\eta^{ij} = 0$ forces the entire spatial block to vanish, leaving $\det \eta = 0$.

Part III: Euclidean spaces — $\mathbb{R}^2$ and $\mathbb{R}^3$¶

The Poincaré and Galilei algebras are only two members of a much larger family. The Lie algebras $\mathfrak{e}(n)$ of Euclidean isometries — and the Lie algebras $\mathfrak{so}(n+1)$ of spherical isometries — admit the same Klein-pair + Leibniz-invariance analysis, and produce the standard Euclidean and spherical metrics. We work them out in 2 and 3 dimensions to show how generic the construction is.

Two-dimensional Euclidean space $\mathbb{R}^2$¶

The Euclidean algebra $\mathfrak{e}(2) = \mathfrak{so}(2) \ltimes \mathbb{R}^2$ has three generators $\{J, P^1, P^2\}$, with

Klein-pair candidates. The proper subalgebras of $\mathfrak{e}(2)$, up to conjugacy, are:

• $\{0\}$: $G/H$ is the whole 3-dim group manifold $E(2)$.

• $\{J\}$: 1-dim, generated by a rotation. Gives $\dim G/H = 2$.

• $\{aP^1 + bP^2\}$ for $(a, b) \neq 0$: 1-dim, generated by a translation. All such are conjugate under rotation. Gives $\dim G/H = 2$.

• $\{P^1, P^2\}$: 2-dim abelian (and an ideal). Gives $\dim G/H = 1$, the circle $S^1 \cong SO(2)$ of orientations.

Of these, the choice that gives “the standard $\mathbb{R}^2$ of points” is $\mathfrak{h} = \{J\}$. The operational reason is the orbit–stabilizer analysis of which transformations fix a point: translations $P^1, P^2$ move points to other points, while a rotation $J$ centered on the point itself leaves the point fixed. So $\mathfrak{h} = \{J\}$, $\mathfrak{m} = \{P^1, P^2\}$, and

Metric derivation. Define the symmetric bilinear form $g(P^i, P^j) = \eta^{ij}$ on $\mathfrak{m}$. Three independent components: $\eta^{11}, \eta^{22}, \eta^{12}$. The Leibniz condition is imposed for every $X \in \mathfrak{h}$; here only $X = J$:

Plugging in the brackets:

• $i = j = 1$: $g(P^2, P^1) + g(P^1, P^2) = 2 \eta^{12} = 0$ $\Rightarrow \eta^{12} = 0$.

• $i = 1, j = 2$: $g(P^2, P^2) + g(P^1, -P^1) = \eta^{22} - \eta^{11} = 0$ $\Rightarrow \eta^{11} = \eta^{22}$.

Hence $\eta^{ij} = \lambda\, \delta^{ij}$ for an overall positive scale $\lambda$ (an arbitrary unit of length squared). Choosing $\lambda = 1$:

This is the standard Euclidean metric on $\mathbb{R}^2$. Because $\mathfrak{m}$ is an abelian ideal, the exponential map gives a global diffeomorphism $\mathfrak{m} \to \mathbb{R}^2$, and $\eta^{ij}$ on $\mathfrak{m}$ becomes $ds^2 = (dx^1)^2 + (dx^2)^2$ on $\mathbb{R}^2$.

Translating to polar coordinates. Polar coordinates make the same point that was sharper on $S^2$: the coordinate form of the metric depends on the chart, while the underlying bilinear form does not. We run the five-step Maurer–Cartan procedure used for $S^2$, now on $\mathfrak{e}(2)$ instead of $\mathfrak{so}(3)$. The output is the familiar polar-coordinate metric of flat $\mathbb{R}^2$ — and the connection $\omega_{\mathfrak{h}}$ comes out flat by direct calculation, confirming that the space is flat despite the coordinate-dependent metric components.

Step 1: Coordinate-defining section. Pick the basepoint at the origin and define polar coordinates $(r, \phi)$ algebraically by the section

$P^1$ translates the basepoint by $r$ along the 1-direction (giving point $(r, 0)$); $J$ then rotates by $\phi$ around the origin (giving $(r\cos\phi, r\sin\phi)$). These are standard polar coordinates by construction. The section is regular on $r > 0$; at $r = 0$ the $\phi$-coordinate is ill-defined, exactly as for the polar chart on $\mathbb{R}^2$.

Step 2: Maurer–Cartan form. Compute $\sigma^{-1}\, d\sigma = \omega_r\, dr + \omega_\phi\, d\phi$.

$\partial_r \sigma = e^{\phi J} P^1 e^{r P^1}$ gives, after multiplying on the left by $\sigma^{-1}$,

(since $[P^1, P^1] = 0$). Next, $\partial_\phi \sigma = J\, \sigma$ gives

The Lie-algebra calculation: $[-rP^1,\, J] = r\,[J, P^1] = r P^2$; the next bracket $[-rP^1,\, r P^2] = -r^2 [P^1, P^2] = 0$ closes the series. Hence

Step 3: Read off the vielbein. With $\mathfrak{m} = \{P^1, P^2\}$ and $\mathfrak{h} = \{J\}$, split $\omega = e + \omega_{\mathfrak{h}}$:

Reading off vielbein components $e_\mu = e^a{}_\mu\, P_a$:

equivalently the coframe

Step 4: Killing vector fields. Using $Y_\xi^\mu = (e^{-1})^\mu{}_a\, [\mathrm{Ad}_{\sigma^{-1}}\xi]^a$, i.e. $\dot r =$ ($P^1$-coefficient of $\sigma^{-1} \xi\, \sigma$) and $\dot\phi =$ ($P^2$-coefficient) $/\, r$:

• $\xi = J$: $\sigma^{-1} J\, \sigma = J + r P^2$ (computed above), so $\dot r = 0$, $\dot\phi = 1$, giving $J = \partial_\phi$ as expected.

• $\xi = P^1$: $\sigma^{-1} P^1 \sigma = \mathrm{Ad}_{e^{-\phi J}}\, P^1 = \cos\phi\, P^1 - \sin\phi\, P^2$ (rotation by $-\phi$ in the $(P^1, P^2)$-plane, since $\mathrm{ad}_J$ acts on $\{P^1, P^2\}$ as a 90° rotation), giving

  $$P^1 \;=\; \cos\phi\, \partial_r - \frac{\sin\phi}{r}\, \partial_\phi.$$

  (46)

• $\xi = P^2$: similarly $\sigma^{-1} P^2 \sigma = \sin\phi\, P^1 + \cos\phi\, P^2$, giving

  $$P^2 \;=\; \sin\phi\, \partial_r + \frac{\cos\phi}{r}\, \partial_\phi.$$

  (47)

These are the standard polar-coordinate expressions for the Cartesian translation Killing fields $\partial_x, \partial_y$.

Step 5: The metric in coordinates. The bridge from form on $\mathfrak{m}$ to tensor field on $\mathbb{R}^2$ is identical to Step 5a of the $S^2$ case: $g = \lambda\, \delta_{ab}\, e^a \otimes e^b$. Expanding,

Choosing the conventional unit $\lambda = 1$:

This is the flat Euclidean metric of $\mathbb{R}^2$, written in polar coordinates. The coordinate components $g_{\mu\nu} = \operatorname{diag}(1, r^2)$ do depend on $r$, even though the space is flat — the $r^2$ is the squared length of $\partial_\phi$ at radius $r$, in exact analogy with the $\sin^2\theta$ of $S^2$.

Why this is flat — verified at the algebra level. The Cartan curvature 2-form of the connection $\omega_{\mathfrak{h}}$ on the model space is

Here $\omega_{\mathfrak{h}} = J\, d\phi$, so $d\omega_{\mathfrak{h}} = dJ \wedge d\phi = 0$ (since $J$ is a constant element of the algebra), and $[\omega_{\mathfrak{h}}, \omega_{\mathfrak{h}}] = [J, J]\, d\phi \wedge d\phi = 0$ trivially. Hence

i.e., zero sectional curvature everywhere. By contrast, on $S^2$ one gets $\omega_{\mathfrak{h}} = \cos\theta\, J^3\, d\phi$, whose exterior derivative $-\sin\theta\, J^3\, d\theta \wedge d\phi$ is non-zero — precisely the constant positive curvature $1/R^2$.

So the polar-coordinate $\mathbb{R}^2$ exhibits the moral lesson sharply: the coordinate components of the metric vary, but the curvature (read off the connection 1-form, an algebraic Lie-algebra object) vanishes. The bracket-level distinction $[\mathfrak{m}, \mathfrak{m}] = 0$ is doing exactly the work that makes flat space flat.

Cartesian derivation: $\sigma = \exp(x P^1 + y P^2)$¶

The derivation above was the traditional one — it used a section $\sigma = e^{\phi J}\, e^{r P^1}$ with the rotation generator $J$ inside (as a “fiber direction”) and a translation generator radially. The same machinery accepts another natural choice: skip the rotation entirely and use only the translations,

Geometrically, this is a single translation by the vector $(x, y)$ from the basepoint. Because $[P^1, P^2] = 0$, this is the same as $e^{xP^1}\, e^{yP^2}$ in either order — the two translations commute, so no Baker–Campbell–Hausdorff correction arises. Let us run the four-step algorithm with this section.

Step 1 — structure constants¶

The Klein pair is unchanged: $\mathfrak{h} = \mathrm{span}(J)$, $\mathfrak{m} = \mathrm{span}(P^1, P^2)$. The non-zero brackets are

The crucial novelty is the $\mathfrak{m}$–$\mathfrak{m}$ bracket: $[P^1, P^2] = 0$. We will see exactly what this triviality does to the algorithm.

Step 2 — Maurer–Cartan form¶

We need $\omega = \sigma^{-1}\, d\sigma$. Use the standard $3 \times 3$ affine representation of $\mathfrak{e}(2)$: