Manifolds: doing all of it without coordinates

Idea. One identity does all of it. On the sphere \(x^2+y^2 = 1-z^2 = (1-z)(1+z)\), so dividing by \((1-z)^2\) gives \(r^2 = (1+z)/(1-z)\), which converts every \(z\) into an \(r^2\) and back.

1. The inverse. From \(r^2(1-z) = 1+z\) we get \(z = \dfrac{r^2-1}{r^2+1}\), hence \(1-z = \dfrac{2}{1+r^2}\) and

\[ x = u(1-z) = \frac{2u}{1+r^2},\qquad y = \frac{2v}{1+r^2},\qquad z = \frac{r^2-1}{r^2+1}. \]

On the sphere: \(4u^2+4v^2+(r^2-1)^2 = 4r^2 + r^4 - 2r^2 + 1 = (r^2+1)^2\), so the sum of squares is 1. ✓

2. The transition.

\[ \tilde u = \frac{x}{1+z} = \frac{x}{1-z}\cdot\frac{1-z}{1+z} = \frac{u}{r^2},\qquad \tilde v = \frac{v}{r^2}. \]

Its own inverse: if \(\tilde r^2 = \tilde u^2+\tilde v^2 = r^2/r^4 = 1/r^2\), then \(\tilde u/\tilde r^2 = (u/r^2)\cdot r^2 = u\). ✓

3. Domain. The overlap is \(U_N\cap U_S = S^2\setminus\{N,S\}\), whose image under \(\mathbf x\) is \(\mathbb R^2\setminus\{0\}\) — the south pole is the origin of the first chart and the north pole is missing from it altogether. Inversion is a quotient by \(r^2\), smooth wherever \(r^2\ne0\); adding the origin would put a zero in the denominator, and no continuous extension exists, since \(|\tilde{\mathbf x}|\to\infty\) there.

4. The Jacobian. \(\partial_u(u/r^2) = 1/r^2 - 2u^2/r^4 = (v^2-u^2)/r^4\) and \(\partial_v(u/r^2) = -2uv/r^4\); the second row swaps \(u\) and \(v\):

\[ A = \frac{1}{r^4}\begin{pmatrix} v^2-u^2 & -2uv \\ -2uv & u^2-v^2\end{pmatrix},\qquad \det A = \frac{-(u^2-v^2)^2 - 4u^2v^2}{r^8} = -\frac{1}{r^4}. \]

At \((3,4)\): \(r^2 = 25\), \(r^4 = 625\), \(v^2-u^2 = 7\), \(-2uv = -24\), so

\[ A\big|_{(3,4)} = \frac{1}{625}\begin{pmatrix}7 & -24\\ -24 & -7\end{pmatrix},\qquad \det = -\frac{1}{625}. \]

The point is \(\mathbf x^{-1}(3,4) = (6,8,24)/26 = \big(\tfrac3{13},\tfrac4{13},\tfrac{12}{13}\big)\), and its second-chart coordinates are \((3,4)/25 = \big(\tfrac3{25},\tfrac4{25}\big)\).

Check. Dimensions: everything here is dimensionless, the sphere having unit radius; if the radius were \(R\) then \(u\) would carry a length and \(\tilde u\) would be \(R^2/u\), so the transition is \(\tilde{\mathbf u} = R^2\mathbf u/r^2\) and \(\det A = -R^4/r^4\), still dimensionless as a Jacobian of coordinates with the same dimension must be. Special case: on the equator \(z=0\), \(r=1\) and the transition is the identity — both projections agree there, as the picture demands. Number: \(\big(\tfrac3{13}\big)^2+\big(\tfrac4{13}\big)^2+\big(\tfrac{12}{13}\big)^2 = (9+16+144)/169 = 1\). ✓

Another way

Complex notation collapses part 2. Write \(w = u+\mathrm iv\); then \(\tilde w = \tilde u + \mathrm i\tilde v = (u+\mathrm iv)/|w|^2 = w/(w\bar w) = 1/\bar w\), and \(1/\overline{(1/\bar w)} = w\) immediately. The map is anti-holomorphic, which is why \(\det A\lt0\); the reflected chart \((x,-y)/(1+z)\) gives \(\tilde w = 1/w\) and the Riemann sphere of unit 20.

Idea. A kinetic energy quadratic in the velocities is a metric on the configuration space, because velocities are tangent vectors and \(T\) is a symmetric bilinear form on them. Once that is seen, the only work is the atlas, and the atlas is the product of two circles'.

1. The atlas. Take \(U_1 = S^1\setminus\{(1,0)\}\) with \(\vartheta_1\in(0,2\pi)\) and \(U_2 = S^1\setminus\{(-1,0)\}\) with \(\vartheta_2\in(-\pi,\pi)\), both measuring the anticlockwise angle from \((1,0)\). Their union is \(S^1\) and their overlap is the two open half-circles. On the upper one \(\vartheta_2 = \vartheta_1\); on the lower one \(\vartheta_2 = \vartheta_1 - 2\pi\). Both are translations, hence \(C^\infty\) with \(C^\infty\) inverses, so this is a smooth atlas. (The transition is not one formula — it is locally constant on a disconnected overlap, which is perfectly allowed.)

For \(T^2\) take the four products \(U_\alpha\times U_\beta\) with charts \((\vartheta_\alpha,\vartheta_\beta)\), \(\alpha,\beta\in\{1,2\}\). A sample transition, from \((U_1\times U_1)\) to \((U_2\times U_1)\) on the part of the overlap where \(\theta_1\) is in the lower half-circle: \((\theta_1,\theta_2)\mapsto(\theta_1-2\pi,\ \theta_2)\), a translation of \(\mathbb R^2\). ✓

2. Products. Let \(M^m, N^n\) have atlases \(\{(U,\mathbf x)\}\), \(\{(V,\mathbf y)\}\). Give \(M\times N\) the product topology, which is Hausdorff and second countable because both factors are. The sets \(U\times V\) are open and cover, and \(\mathbf x\times\mathbf y: U\times V\to\mathbf x(U)\times\mathbf y(V)\subseteq\mathbb R^{m+n}\) is a homeomorphism onto an open set. Two such charts have transition \((\tilde{\mathbf x}\circ\mathbf x^{-1})\times(\tilde{\mathbf y}\circ\mathbf y^{-1})\), smooth because each factor is and smoothness of a map into a product is componentwise. Dimension \(m+n\). \(\square\)

3. The metric. Polarise: \(T = \tfrac12g_{ij}\dot\theta^i\dot\theta^j\) with

\[ g = \begin{pmatrix} (m_1+m_2)l_1^2 & m_2l_1l_2\cos\Delta \\ m_2l_1l_2\cos\Delta & m_2l_2^2\end{pmatrix},\qquad \Delta = \theta_1-\theta_2. \]

Symmetric by construction. Positive definite by Sylvester: \(g_{11} = (m_1+m_2)l_1^2\gt0\), and

\[ \det g = (m_1+m_2)m_2l_1^2l_2^2 - m_2^2l_1^2l_2^2\cos^2\Delta = m_2l_1^2l_2^2\big(m_1+m_2\sin^2\Delta\big) \gt 0 \]

for every \(\Delta\), since \(m_1\gt0\). It is a tensor field rather than an array because the components depend on position only through \(\cos\Delta\), and \(\cos\Delta\) is a well-defined function on \(T^2\) — unchanged by \(\theta_i\mapsto\theta_i+2\pi\), which \(\Delta\) itself is not. In a second chart of the atlas the coordinates differ from the first by a translation, so \(A = \mathbb 1\) and the components are the same functions of \(\Delta\); that is the agreement on the overlap, and it is why this metric is globally defined on the torus.

4. Numbers. \(\cos(\pi/3) = \tfrac12\), \(\sin^2(\pi/3) = \tfrac34\):

\[ g_{11} = 0.8\times0.16 = 0.128,\quad g_{12} = 0.3\times0.4\times0.3\times0.5 = 0.018,\quad g_{22} = 0.3\times0.09 = 0.027 \]

all in kg m\(^2\), and \(\det g = 0.128\times0.027 - 0.018^2 = 0.003456 - 0.000324 = 0.003132\ \mathrm{kg^2m^4}\), which the formula confirms: \(0.3\times0.16\times0.09\times(0.5+0.3\times0.75) = 0.00432\times0.725 = 0.003132\). ✓ Then

\[ T = \tfrac12\big(0.128\times4 + 2\times0.018\times2\times(-3) + 0.027\times9\big) = \tfrac12(0.512-0.216+0.243) = 0.2695\ \mathrm J. \]

Check. Dimensions: \(g_{ij}\) is kg m\(^2\) and \(\dot\theta^i\) is s\(^{-1}\), so \(g_{ij}\dot\theta^i\dot\theta^j\) is kg m\(^2\) s\(^{-2}\) = J. ✓ Limit: as \(m_2\to0\) the metric degenerates to \(\mathrm{diag}(m_1l_1^2,0)\) with \(\det g\to0\) — the second bob is massless, its angle costs no energy, and the configuration space stops being metrically two-dimensional, which is exactly the degenerate Legendre transform unit 14 warned about. Special case \(\Delta = \pi/2\): \(g_{12} = 0\), the two angles decouple instantaneously, and \(\det g\) is largest, \(=m_2l_1^2l_2^2(m_1+m_2)\).

Idea. A global chart would make \(M\) homeomorphic to an open subset of \(\mathbb R^n\). Compactness then forces that subset to be closed as well, and the only non-empty clopen subset of the connected space \(\mathbb R^n\) is \(\mathbb R^n\) itself, which is not compact.

1. The proof. Suppose \((M,\mathbf x)\) is a chart with domain \(M\), so \(\mathbf x: M\to W\) is a homeomorphism onto an open \(W\subseteq\mathbb R^n\). Since \(M\) is compact and \(\mathbf x\) continuous, \(W\) is compact; compact subsets of \(\mathbb R^n\) are closed. So \(W\) is non-empty (as \(M\ne\emptyset\)), open and closed in \(\mathbb R^n\). But \(\mathbb R^n\) is connected, so \(W = \mathbb R^n\), which is not compact — contradiction. \(\square\)

2. The four. Each is a closed bounded subset of a Euclidean space, hence compact by Heine–Borel. \(S^1\subset\mathbb R^2\) and \(S^2\subset\mathbb R^3\): level sets of a continuous function, bounded by \(\|\cdot\| = 1\). \(T^2 = S^1\times S^1\): a product of compact spaces. \(SO(3)\subset\mathbb R^9\): closed because it is defined by the continuous equations \(A^{\mathsf T}A = \mathbb 1\) and \(\det A = 1\), and bounded because each column is a unit vector, so \(\|A\|_F = \sqrt3\). By part 1, none has a global chart, so each needs at least two.

3. \(S^1\), and the moral. The two-arc atlas of P2 works, so two suffice, and by part 1 one does not. Hence the minimum is exactly two.

For \(\mathbb R^2\setminus\{0\}\), identify it with \(\mathbb C^*\) and take \(F(w) = w/(1+|w|)\). Since \(|F(w)| = |w|/(1+|w|)\) increases strictly from \(0\) to \(1\) and \(F\) preserves direction, \(F\) is a bijection onto the punctured open unit disc \(\{0\lt|s|\lt1\}\), with inverse \(s\mapsto s/(1-|s|)\); both are smooth, \(|w|\) being smooth away from the origin. The image is open in \(\mathbb R^2\), so \(F\) is a global chart. The space has a hole and is covered by one chart; the sphere has none and is not. Compactness, not connectivity, is the operative property.

4. Where \(n\ge1\) is used. In the last step, "\(\mathbb R^n\) is not compact". For \(n = 0\), \(\mathbb R^0\) is a point, which is compact, and a zero-manifold consisting of a single point does have a global chart. The rest of the argument is unaffected.

Check. Dimensions: none — this is pure topology. Limit: the argument is sharp in the sense that dropping compactness destroys it, and part 3 exhibits the counterexample. Special case: \(M = \mathbb R^n\) itself is a non-compact manifold with a global chart, and the proof correctly produces no contradiction there, since \(W = \mathbb R^n\) is allowed when \(M\) is not compact.

Another way

Part 1 also follows from an extreme-value argument, which never mentions connectedness. Let \(x^1\) be the first coordinate function of the supposed global chart. It is continuous on the compact \(M\), so it attains a maximum, at some \(p\). But \(\mathbf x(M)\) is open, so it contains a ball around \(\mathbf x(p)\), and that ball contains points whose first coordinate exceeds \(x^1(p)\). Those points are \(\mathbf x(q)\) for some \(q\in M\), contradicting maximality.

Idea. A smooth structure is extra data on a topological space, and two different sets of data can still be carried into one another by a map. "Different structures" is a statement about the identity map; "diffeomorphic" allows any map.

1. \(\mathbf y(t) = t^3\) is a continuous bijection \(\mathbb R\to\mathbb R\) with continuous inverse \(s\mapsto s^{1/3}\), and \(\mathbb R\) is open. A single chart covering \(M\) is trivially an atlas — the only compatibility condition is the chart with itself, whose transition is the identity.

2. \(\mathbf y\circ\mathbf x^{-1}(s) = s^3\) is \(C^\infty\). But \(\mathbf x\circ\mathbf y^{-1}(s) = s^{1/3}\) is not differentiable at \(s = 0\): its difference quotient is \(h^{1/3}/h = h^{-2/3}\to\infty\). Compatibility requires both directions, so the charts are incompatible and \(\bar{\mathcal A}\ne\bar{\mathcal B}\).

3. Smoothness of \(f: M\to\mathbb R\) in the structure \(\mathcal B\) means \(f\circ\mathbf y^{-1}\) is \(C^\infty\). For \(f(t) = t\) that is \(s\mapsto s^{1/3}\), not differentiable at \(0\) — so the identity function is not smooth on \((\mathbb R,\mathcal B)\). For \(f(t) = t^3\) it is \(s\mapsto s\), which is smooth. ✓ In \(\mathcal A\) both are smooth, since \(f\circ\mathbf x^{-1}\) is \(s\) and \(s^3\) respectively, so \(C^\infty(\mathbb R,\mathcal A)\ne C^\infty(\mathbb R,\mathcal B)\) as sets of functions: the identity function belongs to the first and not to the second. In fact the containment is one-way: \(f\) is \(\mathcal B\)-smooth exactly when \(f(t) = h(t^3)\) for some smooth \(h\), and every such \(f\) is a composite of smooth maps, so \(C^\infty(\mathbb R,\mathcal B)\subsetneq C^\infty(\mathbb R,\mathcal A)\).

4. Take \(F: (\mathbb R,\mathcal A)\to(\mathbb R,\mathcal B)\), \(F(t) = t^{1/3}\). Its coordinate representation is \(\mathbf y\circ F\circ\mathbf x^{-1}(s) = \big(s^{1/3}\big)^3 = s\), the identity, which is smooth; the same computation for \(F^{-1}(t) = t^3\) gives \(\mathbf x\circ F^{-1}\circ\mathbf y^{-1}(s) = \big(s^{1/3}\big)^3 = s\). So \(F\) is a diffeomorphism.

No contradiction: part 2 says the identity map is not a diffeomorphism between the two structures, which is a statement about one particular map. Part 4 produces a different map that is. "Different smooth structures on the same set" and "non-diffeomorphic manifolds" are distinct notions, and only the second is of any interest.

Check. Dimensions: none. Limit: away from \(t=0\) everything in sight is smooth and the two structures agree locally — the incompatibility is a single point, and one point is enough. Special case worth knowing: \(\mathbb R^n\) carries exactly one smooth structure up to diffeomorphism for every \(n\ne4\); for \(n=4\) there are uncountably many, and there the distinction this problem trivialises becomes deep.

Idea. A rule assigning arrays to charts is a tensor field exactly when it is equivariant: one chart's array, pushed through the law, must reproduce the other chart's array. Two of these rules are built from tensors by tensorial operations and pass automatically; two are built by differentiating, and differentiation is where the law breaks.

Rule 1: not a tensor. \(S^{\mathsf T}\delta S = S^{\mathsf T}S\), and

\[ S^{\mathsf T}S = \begin{pmatrix}-3&-4\\-4&3\end{pmatrix}^2 = \begin{pmatrix}9+16 & 12-12\\ 12-12 & 16+9\end{pmatrix} = \begin{pmatrix}25&0\\0&25\end{pmatrix}. \]

The law predicts \(\tilde T_{11} = 25\); the rule says \(\tilde T_{11} = 1\). Convicted. (It would be a tensor only if every transition Jacobian were orthogonal, which inversion is not: it is \(5\) times an orthogonal matrix at \(P\), and its determinant is \(-1/25\) generally.)

Rule 2: a tensor — it is the round metric, and the deck verified \(\tilde g = g\) on the whole overlap. At \(P\): the array is \(\tfrac19\mathbb 1\), the law gives \(\tfrac19S^{\mathsf T}S = \tfrac{25}9\mathbb 1\), and the second chart's array is \(4/(1+\tfrac15)^2 = 4\cdot\tfrac{25}{36} = \tfrac{25}9\) on the diagonal. ✓

Rule 3: not a tensor. With \(\partial_uf = 4u/(1+r^2)^2\),

\[ \partial_u\partial_uf = \frac{4}{(1+r^2)^2} - \frac{16u^2}{(1+r^2)^3},\qquad \partial_u\partial_vf = -\frac{16uv}{(1+r^2)^3}. \]

At \(P\), \(r^2 = 5\), \((1+r^2)^2 = 36\), \((1+r^2)^3 = 216\):

\[ T = \frac1{27}\begin{pmatrix}-5&-4\\-4&1\end{pmatrix}, \qquad \text{law:}\quad S^{\mathsf T}TS = \frac1{27}\begin{pmatrix}-125&-100\\-100&25\end{pmatrix}. \]

Directly in the second chart, with \(f = (1-\tilde r^2)/(1+\tilde r^2)\), \(\partial_{\tilde u}f = -4\tilde u/(1+\tilde r^2)^2\) and \(\partial_{\tilde u}\partial_{\tilde u}f = -4/(1+\tilde r^2)^2 + 16\tilde u^2/(1+\tilde r^2)^3\); at \(\tilde r^2 = \tfrac15\), where \((1+\tilde r^2)^2 = \tfrac{36}{25}\) and \((1+\tilde r^2)^3 = \tfrac{216}{125}\),

\[ \tilde T = \frac1{27}\begin{pmatrix}-35&20\\20&-65\end{pmatrix} \ \ne\ \frac1{27}\begin{pmatrix}-125&-100\\-100&25\end{pmatrix}. \]

Convicted, decisively — even the signs disagree.

Rule 4: a tensor. It is \(\mathrm{d}f\otimes\mathrm{d}f\), a tensor product of two covector fields, and the tensor product of tensors is a tensor. Arithmetic confirms: at \(P\), \(\mathrm{d}f = \big(\tfrac29,\tfrac19\big)\), so \(T = \tfrac1{81}\begin{pmatrix}4&2\\2&1\end{pmatrix}\) and \(S^{\mathsf T}TS = \tfrac1{81}\begin{pmatrix}100&50\\50&25\end{pmatrix}\); directly, \(\mathrm{d}f = \big(-\tfrac{10}9,-\tfrac59\big)\) in the second chart, whose outer product is \(\tfrac1{81}\begin{pmatrix}100&50\\50&25\end{pmatrix}\). ✓

The distinction. Rule 4 applies a tensorial operation — a tensor product — to the tensor \(\mathrm{d}f\), whereas rule 3 applies \(\partial_i\) to the components of \(\mathrm{d}f\), and the transformation law of a derivative picks up the second derivatives of the transition map:

\[ \tilde\partial_a\tilde\partial_bf = S^i{}_aS^j{}_b\,\partial_i\partial_jf + \frac{\partial^2x^i}{\partial\tilde x^a\partial\tilde x^b}\,\partial_if . \]

The extra term vanishes exactly where \(\partial_if = 0\), that is at a critical point of \(f\). So rule 3 does give a well-defined tensor at the north and south poles — the Hessian at a critical point — and nowhere else. That is the fact Morse theory runs on.

Check. Dimensions: all four rules are dimensionless on the unit sphere. Limit: on the equator, \(r = 1\), the transition is the identity, \(S = \mathbb 1\), and every rule passes the test there — a reminder that equivariance must hold at every point, and that testing at a fixed point of the transition proves nothing. Number: the trace of rule 3's array in the first chart at \(P\) is \((-5+1)/27 = -\tfrac4{27}\); in the second it is \((-35-65)/27 = -\tfrac{100}{27}\). Neither is the other, and neither should be: the trace of a \((0,2)\)-array is not basis-independent, as unit 16 warned.

Idea. \(\mathrm{d}f\) is defined with no metric: it is just \(X\mapsto Xf\), so in coordinates it is the row of partial derivatives, and the only content is that the row transforms with \(S\) rather than with \(A\).

1. First chart: \(f = \dfrac{r^2-1}{r^2+1}\), so

\[ \mathrm{d}f = \frac{4u\,\mathrm{d}u + 4v\,\mathrm{d}v}{(1+u^2+v^2)^2}. \]

Second chart: \(\tilde r^2 = (x^2+y^2)/(1+z)^2 = (1-z)/(1+z)\), so \(f = \dfrac{1-\tilde r^2}{1+\tilde r^2}\) and

\[ \mathrm{d}f = -\frac{4\tilde u\,\mathrm{d}\tilde u + 4\tilde v\,\mathrm{d}\tilde v}{(1+\tilde u^2+\tilde v^2)^2}. \]

2. At \(P\), \(r^2 = 5\) and \((1+r^2)^2 = 36\), so \(\omega = \big(\tfrac{8}{36},\tfrac{4}{36}\big) = \big(\tfrac29,\tfrac19\big)\). And \(\tilde r^2 = \tfrac15\) with \((1+\tilde r^2)^2 = \tfrac{36}{25}\), so \(\tilde\omega = -\tfrac{25}{36}\big(\tfrac85,\tfrac45\big) = \big(-\tfrac{10}9,-\tfrac59\big)\). The law:

\[ \tilde\omega_1 = (-3)\tfrac29 + (-4)\tfrac19 = -\tfrac{10}9,\qquad \tilde\omega_2 = (-4)\tfrac29 + 3\cdot\tfrac19 = -\tfrac59 . \ \checkmark \]

3. In the first chart \(\mathrm{d}f = 0\) iff \(u = v = 0\), the origin, which is the south pole \((0,0,-1)\). In the second, \(\tilde u = \tilde v = 0\), the north pole \((0,0,1)\). Every other point of \(S^2\) lies in the first chart away from its origin, so there are exactly two critical points, the poles — which is what the geometry says: the height has one minimum and one maximum and no other stationary value.

4. \(\mathrm{d}f(R_z) = \omega_iR_z^i = \tfrac29(-1) + \tfrac19(2) = 0\). Inevitable: \(R_z\) generates rotation about the \(z\)-axis, which leaves \(z\) unchanged, so \(R_zf = 0\) everywhere — and \(\mathrm{d}f(R_z)\) is precisely \(R_zf\).

Check. Dimensions: on a unit sphere all quantities are dimensionless; with radius \(R\), \(f\) carries a length, \(u\) carries a length, and \(\mathrm{d}f\)'s components are dimensionless — consistent, since \(\mathrm{d}f(X)\) must have the dimension of \(f\) when \(X\) has that of \(\partial_u\), namely inverse length. Limit: at the equator \(r = 1\), \(\mathrm{d}f = (u\,\mathrm{d}u+v\,\mathrm{d}v)\), of magnitude 1 against the round metric — the height changes fastest there, as it must. Number: \(\|\mathrm{d}f\|^2_g = g^{ij}\omega_i\omega_j\) at \(P\) is \(9\big(\tfrac4{81}+\tfrac1{81}\big) = \tfrac59\), equal to \(1-z^2 = 1-\tfrac49\) at \(P\). ✓

Idea. The Leibniz rule sees only first-order information, and Hadamard's lemma is the statement that a smooth function can be split into a constant plus something first order with smooth coefficients. Put the two together and the derivation has nowhere to hide.

1. Hadamard. Fix \(s\in V\). Convexity puts the segment \(a+t(s-a)\), \(t\in[0,1]\), in \(V\). The map \(t\mapsto\hat f(a+t(s-a))\) is smooth on a neighbourhood of \([0,1]\), with derivative \((s^i-a^i)\partial_i\hat f(a+t(s-a))\) by the chain rule. The fundamental theorem of calculus gives

\[ \hat f(s)-\hat f(a) = (s^i-a^i)\underbrace{\int_0^1\partial_i\hat f\big(a+t(s-a)\big)\,\mathrm dt}_{=:\ h_i(s)} . \]

Each \(h_i\) is smooth: the integrand is smooth jointly in \((t,s)\) and \(t\) ranges over a compact interval, so differentiation under the integral sign is legitimate to all orders. And \(h_i(a) = \int_0^1\partial_i\hat f(a)\,\mathrm dt = \partial_i\hat f(a)\). \(\square\)

2. Derivations. First, \(D\) kills constants: \(D(1) = D(1\cdot1) = 1\cdot D1+1\cdot D1 = 2D1\), so \(D1 = 0\), and \(D(c) = cD(1) = 0\) by linearity. Now take a chart with \(a = \mathbf x(p)\) and a convex \(V\); as germs at \(p\),

\[ f = f(p) + (x^i-a^i)\,(h_i\circ\mathbf x). \]

Apply \(D\): the constant dies, and Leibniz on each product gives

\[ Df = \big(a^i-a^i\big)D(h_i\circ\mathbf x) + h_i(a)\,D(x^i) = \partial_i\hat f(a)\,\xi^i,\qquad \xi^i := D(x^i), \]

since \((x^i-a^i)(p) = 0\). So \(D = \xi^i\partial_i|_p\), and \(\{\partial_i|_p\}\) spans \(T_pM\). They are independent: if \(c^i\partial_i|_p = 0\), apply it to \(x^j\) to get \(c^j = 0\). Hence \(\dim T_pM = n\). \(\square\)

3. Locality. By linearity it suffices to show that \(Df = 0\) when \(f\) vanishes on a neighbourhood \(W\) of \(p\). Choose a smooth \(\chi\) with \(\chi(p) = 1\) and \(\operatorname{supp}\chi\subseteq W\) — a bump function, built from \(\mathrm e^{-1/t}\). Then \(\chi f\equiv0\) on all of \(M\), so

\[ 0 = D(\chi f) = \chi(p)\,Df + f(p)\,D\chi = Df . \]

It needs proof because \(D\) is defined by two algebraic axioms — linearity and Leibniz — neither of which mentions neighbourhoods. Nothing in the axioms forbids a functional from consulting the function far away; it is the Leibniz rule together with the existence of bump functions that forbids it. On germs the statement is automatic by construction, which is why the deck used germs.

4. A non-derivation. \(D(f) = f''(0)\) is linear and is not a derivation: \((fg)'' = f''g + 2f'g' + fg''\), so \(D(fg) = f(0)Dg + g(0)Df + 2f'(0)g'(0)\), and the last term is not zero in general — take \(f = g = t\), where \(D(fg) = 2\) while \(f(0)Dg+g(0)Df = 0\). It fails the Leibniz rule; linearity is fine. (Similarly \(D(f) = f(0)\) fails Leibniz, and \(D(f) = f'(0)^2\) fails linearity.)

Check. Dimensions: none; this is pure analysis. Limit: for \(n=1\) and \(\hat f(s) = s^2\) at \(a=0\), Hadamard gives \(h_1(s) = \int_0^12ts\,\mathrm dt = s\), so \(\hat f = 0 + s\cdot s\) ✓, and \(h_1(0) = 0 = \hat f'(0)\) ✓. Special case: the theorem fails for \(C^1\) germs — \(h_i\) is then only \(C^0\) and the space of derivations on \(C^1\) germs is in fact infinite-dimensional, which is exactly why manifolds are defined with \(C^\infty\).

Idea. Both expressions differentiate components, so both pick up derivatives of the transition Jacobian. In \(\mathcal L_X\omega\) the second term picks up the same derivative with the opposite sign, and the package is tensorial although neither half is.

1. \(\partial_i\omega_j\) fails. With \(\tilde\omega_b = S^j{}_b\omega_j\) and \(\tilde\partial_a = S^i{}_a\partial_i\),

\[ \tilde\partial_a\tilde\omega_b = S^i{}_aS^j{}_b\,\partial_i\omega_j \;+\; \big(\tilde\partial_aS^j{}_b\big)\,\omega_j , \]

and \(\tilde\partial_aS^j{}_b = \dfrac{\partial^2x^j}{\partial\tilde x^a\partial\tilde x^b}\): second derivatives of the transition map, symmetric in \(a,b\) by Clairaut. A tensor law has no such term, so the array is not tensorial. (It is tensorial for changes of chart that are affine, where the second derivatives vanish — which is why Cartesian-only treatments never notice.)

2. \(\mathcal L_X\omega\) survives. Compute the two pieces in the tilded chart.

\[ \tilde X^c\tilde\partial_c\tilde\omega_a = X^l\partial_l\big(S^i{}_a\omega_i\big) = S^i{}_a\,X^l\partial_l\omega_i + \big(X^l\partial_lS^i{}_a\big)\omega_i , \]

using \(\tilde X^c = A^c{}_kX^k\) and \(A^c{}_kS^l{}_c = \delta^l_k\). And

\[ \tilde\omega_c\,\tilde\partial_a\tilde X^c = S^l{}_a\,\omega_k\partial_lX^k + S^i{}_cS^l{}_a\big(\partial_lA^c{}_k\big)X^k\omega_i . \]

Differentiating \(S^i{}_cA^c{}_k = \delta^i_k\) along \(\partial_l\) gives \(S^i{}_c\partial_lA^c{}_k = -(\partial_lS^i{}_c)A^c{}_k\), which turns the second unwanted term into \(-S^l{}_a(\partial_lS^i{}_c)A^c{}_kX^k\omega_i = -\big(X^l\partial_lS^i{}_a\big)\omega_i\) — exactly the negative of the first. They cancel, leaving

\[ (\widetilde{\mathcal L_X\omega})_a = S^i{}_a\big(X^l\partial_l\omega_i + \omega_k\partial_iX^k\big), \]

the covector law. So it is the term \(\omega_k\partial_iX^k\) — the one repairing the covector slot — that cancels the offending piece generated by the transport term \(X^k\partial_k\omega_i\). Neither is tensorial alone.

3. On the sphere at \(P\). In the first chart \(\omega = \mathrm{d}f = \big(\tfrac{4u}{(1+r^2)^2},\tfrac{4v}{(1+r^2)^2}\big)\), so \(\partial_i\omega_j\) is the Hessian of P5:

\[ \partial_i\omega_j\big|_P = \frac1{27}\begin{pmatrix}-5&-4\\-4&1\end{pmatrix},\qquad \text{law predicts } \frac1{27}\begin{pmatrix}-125&-100\\-100&25\end{pmatrix}, \]

while the direct computation in the second chart gives \(\tfrac1{27}\begin{pmatrix}-35&20\\20&-65\end{pmatrix}\). Not tensorial. ✗

Now the Lie derivative. The flow of \(D\) is \(u\mapsto\mathrm e^tu\), so \(r^2\mapsto\mathrm e^{2t}r^2\) and \(\tilde u = u/r^2\mapsto\mathrm e^{-t}\tilde u\): hence \(\tilde D = -\tilde u\,\partial_{\tilde u}-\tilde v\,\partial_{\tilde v}\). Since \(\mathcal L_D\mathrm{d}f = \mathrm{d}(Df)\), compute the scalar \(Df\) in each chart:

\[ Df = u\,\partial_uf + v\,\partial_vf = \frac{4r^2}{(1+r^2)^2},\qquad \tilde Df = -\tilde u\,\partial_{\tilde u}f - \tilde v\,\partial_{\tilde v}f = \frac{4\tilde r^2}{(1+\tilde r^2)^2}. \]

At \(P\) both equal \(\tfrac{20}{36} = \tfrac59\) ✓ — a scalar, so they must. Differentiating,

\[ \mathcal L_D\omega\big|_P = \frac1{27}\big(-8,\ -4\big)\quad\text{and}\quad \widetilde{\mathcal L_D\omega}\big|_P = \frac1{27}\big(40,\ 20\big), \]

and the covector law reproduces the second from the first: \((-3)(-\tfrac8{27})+(-4)(-\tfrac4{27}) = \tfrac{40}{27}\) and \((-4)(-\tfrac8{27})+3(-\tfrac4{27}) = \tfrac{20}{27}\). ✓ Tensorial.

4. The offending term \(\big(\tilde\partial_aS^j{}_b\big)\omega_j\) is linear in \(\omega\) but its other factor comes from the chart change alone, so any repair must itself supply second derivatives of the transition map. Nothing built from \(\omega\) and \(\partial\omega\) does. There are only three sources in this course: a second vector field's derivative, giving the Lie derivative; an antisymmetrisation that kills the second derivatives outright, giving \(\mathrm{d}\omega\) in unit 34; or an extra field that transforms inhomogeneously, which is the connection of unit 32.

Check. Dimensions: all dimensionless on the unit sphere. Limit: if the transition map were affine, \(\partial^2x/\partial\tilde x\partial\tilde x = 0\) and \(\partial_i\omega_j\) would be tensorial — consistent with the fact that nothing in unit 16, which used only linear changes of basis, needed any of this. Number: the antisymmetric part of \(\partial_i\omega_j\) is zero in both charts, \(-4\) against \(-4\) and \(20\) against \(20\) off the diagonal, as it must be since \(\omega = \mathrm{d}f\) is exact — and that part is tensorial, being \(\mathrm{d}\omega\).

Idea. A one-parameter group of diffeomorphisms of the sphere descends to a flow in each chart; differentiating the chart coordinates along it at \(t = 0\) is the vector field, by definition A of a tangent vector.

0. The identity. \(r^2 = (1+z)/(1-z)\) gives \(\dfrac{z}{1-z} = \dfrac{(1+z)-(1-z)}{2(1-z)} = \dfrac{r^2-1}{2}\). ✓

1. The three fields. For \(R_z\): \(\dot x = -y,\ \dot y = x,\ \dot z = 0\), so \(\dot u = -y/(1-z) = -v\) and \(\dot v = u\). For \(R_x\): \(\dot x = 0,\ \dot y = -z,\ \dot z = y\), so

\[ \dot u = \frac{x\dot z}{(1-z)^2} = \frac{xy}{(1-z)^2} = uv,\qquad \dot v = \frac{-z(1-z)+y^2}{(1-z)^2} = -\frac{z}{1-z}+v^2 = \frac{1-r^2}{2}+v^2 . \]

For \(R_y\): \(\dot x = z,\ \dot z = -x\), so \(\dot u = \dfrac{z}{1-z} - u^2 = \dfrac{r^2-1}{2}-u^2\) and \(\dot v = -xy/(1-z)^2 = -uv\). Collecting:

\[ R_z = -v\,\partial_u+u\,\partial_v,\quad R_x = uv\,\partial_u+\tfrac12(1-u^2+v^2)\,\partial_v,\quad R_y = -\tfrac12(1+u^2-v^2)\,\partial_u - uv\,\partial_v . \]

2. The brackets. Using \([X,Y]^i = X^j\partial_jY^i-Y^j\partial_jX^i\). For \([R_z,R_x]\), first component:

\[ (-v)(v) + (u)(u) - \Big[uv\cdot0 + \tfrac12(1-u^2+v^2)(-1)\Big] = u^2-v^2+\tfrac12(1-u^2+v^2) = \tfrac12(1+u^2-v^2), \]

second component: \((-v)(-u)+u\,v - \big[uv\cdot1 + \tfrac12(1-u^2+v^2)\cdot0\big] = uv\). Together that is \(-R_y\). The same computation gives

\[ [R_x,R_y] = -R_z,\qquad [R_y,R_z] = -R_x,\qquad [R_z,R_x] = -R_y . \]

3. The sign. The generators of a left action form an anti-homomorphic image of the Lie algebra: \([R_A,R_B] = -R_{[A,B]}\), because the flow of \(R_A\) is left multiplication by \(\exp(tA)\) and composing left multiplications reverses the order of the corresponding operators on functions. The map \(\varphi(J_k) = -R_k\) fixes it:

\[ [\varphi(J_x),\varphi(J_y)] = [-R_x,-R_y] = [R_x,R_y] = -R_z = \varphi(J_z), \]

and cyclically. \(\varphi\) is linear, bijective onto the span, and bracket-preserving: a Lie algebra isomorphism \(\mathfrak{so}(3)\to\mathrm{span}\{R_x,R_y,R_z\}\).

4. The zeros. \(R_z = (-v,u)\) vanishes only at \(u=v=0\), the south pole; and in the second chart \(\tilde R_z = (-\tilde v,\tilde u)\) vanishes only at the north pole. Two zeros. The hairy ball theorem says no smooth vector field on \(S^2\) is nowhere-zero, so at least one was compulsory; Poincaré–Hopf sharpens this: the indices must sum to \(\chi(S^2) = 2\), and each of these zeros is a rotation, of index \(+1\), so two is the minimum and \(R_z\) attains it.

Check. Dimensions: none. Limit: at the origin of the first chart (the south pole) \(R_x = (0,\tfrac12)\) and \(R_y = (-\tfrac12,0)\) — non-zero and orthogonal, as rotations about two horizontal axes should be at the bottom of the sphere, and the factor \(\tfrac12\) is the stereographic scale there, where \(g_{ij} = 4\delta_{ij}\) and so \(\|R_x\|^2 = 4\cdot\tfrac14 = 1\): unit speed at unit distance from the axis. ✓ Special case: on the equator \(r = 1\), \(R_x = (uv,\ v^2)= v(u,v)\) — radial, and vanishing at \((\pm1,0)\), which are the two ends of the \(x\)-axis, where rotation about \(x\) indeed moves nothing.

Idea. \(\mathcal L_Xg = 0\) says the flow of \(X\) is a family of isometries. For a conformally flat metric \(\Omega^2\delta\) the equation splits into "how much the flow stretches" plus "how much \(\Omega\) changes along it", and the two must cancel.

1. \(R_x\) is Killing. Write \(X = R_x\), \(X^1 = uv\), \(X^2 = \tfrac12(1-u^2+v^2)\). Then

\[ \partial_1X^1 = v,\quad \partial_2X^1 = u,\quad \partial_1X^2 = -u,\quad \partial_2X^2 = v. \]

Transport: \(X^k\partial_kr^2 = 2u^2v + v(1-u^2+v^2) = v(1+u^2+v^2) = v(1+r^2)\), so

\[ X^k\partial_k\Omega^2 = \frac{-8}{(1+r^2)^3}\cdot v(1+r^2) = \frac{-8v}{(1+r^2)^2} = -2v\,\Omega^2 . \]

Now assemble each component.

\[ (\mathcal L_Xg)_{11} = -2v\Omega^2 + 2\Omega^2\partial_1X^1 = \Omega^2(-2v+2v) = 0, \] \[ (\mathcal L_Xg)_{22} = -2v\Omega^2 + 2\Omega^2\partial_2X^2 = \Omega^2(-2v+2v) = 0, \] \[ (\mathcal L_Xg)_{12} = 0 + \Omega^2\big(\partial_1X^2+\partial_2X^1\big) = \Omega^2(-u+u) = 0 . \]

All three vanish. ✓ The off-diagonal cancellation is the statement that the flow is conformal, and the diagonal one that the conformal factor exactly compensates the change in \(\Omega\).

2. The dilation. \(\partial_iD^j = \delta_i^j\) and \(D^k\partial_kr^2 = 2r^2\), so

\[ (\mathcal L_Dg)_{ij} = \frac{-8}{(1+r^2)^3}2r^2\delta_{ij} + 2\Omega^2\delta_{ij} = \Big[\frac{-16r^2}{(1+r^2)^3}+\frac{8}{(1+r^2)^2}\Big]\delta_{ij} = \frac{8(1-r^2)}{(1+r^2)^3}\delta_{ij}, \] \[ \text{so}\qquad \mathcal L_Dg = \lambda g,\qquad \lambda = \frac{2(1-r^2)}{1+r^2} . \]

Zero only on \(r = 1\), the equator. At \(P\), \(r^2 = 5\): \(\lambda = 2(-4)/6 = -\tfrac43\). So \(D\) is conformal Killing and not Killing: its flow slides points along meridians, preserving angles and shrinking lengths near \(P\).

3. Surfaces of revolution. General fact: if \(g_{ij}\) is independent of the coordinate \(x^m\), then \(X = \partial_m\) has \(X^k = \delta^k_m\) constant, so both correction terms vanish and \(\mathcal L_Xg_{ij} = \partial_mg_{ij} = 0\). Here \(g = \mathrm{diag}(1,F(\rho)^2)\) has no \(\phi\)-dependence, so \(\partial_\phi\) is Killing for every \(F\) — the surface is rotationally symmetric by construction. For \(\partial_\rho\), the same criterion demands \(\partial_\rho g_{ij} = 0\), that is \(2FF' = 0\), so \(F\) must be a non-zero constant: the cylinder, and no other surface of revolution. (Indeed a cylinder is flat and translation-invariant; a sphere, where \(F = \sin\rho\), is not.)

4. Rindler. The shift is \(X_{\text{old}} = c^2/a + X\), which leaves \(\mathrm{d}X^2\) alone and turns the lapse into \((a/c)^2\big(c^2/a+X\big)^2 = c^2\big(1+aX/c^2\big)^2\) — the same metric, with the reference observer now at the origin. Its components \(-(1+aX/c^2)^2c^2\) and \(1\) depend on \(X\) only, so by the same criterion \(\partial_T\) is Killing. Physically: the Rindler metric is static, an accelerating observer's description of flat spacetime does not change with their own time coordinate, and the conserved quantity \(g(\partial_T,\dot\gamma)\) is what unit 30's redshift argument was implicitly using — the ratio of clock rates between two heights is constant in time, which is why the Pound–Rebka experiment has a steady answer.

Check. Dimensions: in part 4, \(aX/c^2\) is \(\mathrm{m\,s^{-2}\cdot m/(m^2s^{-2})} = 1\), dimensionless ✓, and both metric terms carry m\(^2\). Limit: in part 2, \(\lambda\to-2\) as \(r\to\infty\), the north pole, where the dilation's flow is fastest in the stereographic picture and slowest on the sphere; and \(\lambda\to+2\) at the south pole. Number: the three rotations give \(3 = n(n+1)/2\) Killing fields for \(n = 2\), so \(S^2\) is maximally symmetric and no fourth Killing field exists — \(D\) could not have been one.

Idea. With constant metric components the transport term dies and the Killing equation becomes a statement that the derivative matrix of \(X\) is antisymmetric. Antisymmetric matrices plus constants is exactly the Euclidean group's Lie algebra, and the counting comes out on its own.

1. Solving. With \(g_{ij} = \delta_{ij}\) constant,

\[ (\mathcal L_Xg)_{ij} = \delta_{kj}\partial_iX^k + \delta_{ik}\partial_jX^k = \partial_iX^j+\partial_jX^i = 0 . \]

Taking \(ij = 11\) and \(22\): \(\partial_xX^1 = 0\) and \(\partial_yX^2 = 0\), so \(X^1\) depends only on \(y\) and \(X^2\) only on \(x\). Taking \(ij = 12\): \(\partial_xX^2 = -\partial_yX^1\). The left side is a function of \(x\) alone, the right of \(y\) alone, so both equal a constant \(\omega\). Integrating, \(X^2 = \omega x + b\) and \(X^1 = -\omega y + a\):

\[ X = (a-\omega y)\,\partial_x + (b+\omega x)\,\partial_y, \]

three free constants \(a, b, \omega\).

2. The algebra. \(T_1 = \partial_x\) and \(T_2 = \partial_y\) are translations; \(R = -y\,\partial_x+x\,\partial_y\) is rotation about the origin. Brackets:

\[ [T_1,T_2] = 0,\qquad [R,T_1] = -T_2,\qquad [R,T_2] = T_1 . \]

(For instance \([R,T_1]^i = R^j\partial_jT_1^i - T_1^j\partial_jR^i = 0 - \partial_x(-y,x) = -(0,1)\).) This is \(\mathfrak{e}(2)\), the Lie algebra of the Euclidean group: an abelian ideal of translations with the rotation acting on it by a \(90^\circ\) turn.

3. The bound. \(n(n+1)/2 = 3\) for \(n = 2\), and we found three: attained, so the plane is maximally symmetric. The bound exists because a Killing field is determined by \(X_m\) and the antisymmetric matrix \(\partial_{[i}X_{j]}\) at any single point \(m\) — here \((a,b)\) and \(\omega\) — which is \(n + n(n-1)/2 = n(n+1)/2\) numbers. The symmetric part of \(\partial X\) is forced to vanish by the Killing equation itself, and the second derivatives are then determined (on a general manifold, by the curvature), so no further freedom survives.

4. Minkowski. With \(g = \mathrm{diag}(-1,1)\) and coordinates \((t,x)\), the Killing equations are \(-\partial_tX^0 \cdot 2 = 0\) in the \(00\) slot, \(2\partial_xX^1 = 0\) in the \(11\) slot, and \(-\partial_xX^0+\partial_tX^1 = 0\) in the \(01\) slot. So \(X^0 = X^0(x)\), \(X^1 = X^1(t)\) and \(\partial_xX^0 = \partial_tX^1 = \kappa\), constant, giving

\[ X = (a+\kappa x)\,\partial_t + (b+\kappa t)\,\partial_x . \]

Again three constants. The rotation is replaced by the boost \(B = x\,\partial_t+t\,\partial_x\) — note the relative plus sign, the hyperbolic counterpart of the rotation's minus — whose flow is the Lorentz boost of unit 26, and whose integral curves are the hyperbolae of unit 27's uniformly accelerated worldline.

Check. Dimensions: none in part 1; in part 4, \(\kappa\) is dimensionless with \(x^0 = ct\). Limit: setting \(\omega = 0\) leaves only translations, the symmetries of a plane with a marked direction but no marked point; setting \(a = b = 0\) leaves rotations about the origin. Number: the brackets close, so the three-dimensional space really is a subalgebra — computing \([R,[T_1,T_2]] = 0\) and checking the Jacobi identity on \((R,T_1,T_2)\) gives \([R,[T_1,T_2]]+[T_1,[T_2,R]]+[T_2,[R,T_1]] = 0 + [T_1,-T_1]+[T_2,-T_2] = 0\). ✓

Idea. The commutator square is the definition of the bracket made visible. Both orders of travel reach the same point to first order; the bracket is the whole of the second-order discrepancy.

1. Second-order flow. Along an integral curve \(\dot\gamma^i = X^i(\gamma)\), so

\[ \ddot\gamma^i = \frac{\mathrm d}{\mathrm dt}X^i(\gamma) = \partial_jX^i(\gamma)\,\dot\gamma^j = \big(X^j\partial_jX^i\big)(\gamma). \]

Taylor at \(t = 0\) with \(\gamma(0) = s\) gives the stated expansion.

2. The composition. Write \(\varphi^X_t(s) = s+tX+\tfrac{t^2}2(X\!\cdot\!\partial)X\), all fields evaluated at \(s\), and compose the four maps keeping terms of total degree \(\le2\) in \((t,s)\). Terms in \(t\) alone come only from the first and third factors, which are \(\varphi^X_t\) and \(\varphi^X_{-t}\); their composition is the identity exactly, so every pure-\(t\) term cancels — and identically for pure-\(s\) with \(\varphi^Y_{\pm s}\). What is left at degree two is the cross terms, which are

\[ ts\Big[\big(X^j\partial_jY^i\big) - \big(Y^j\partial_jX^i\big)\Big] = ts\,[X,Y]^i , \]

the two coming from "differentiate \(Y\) after having been displaced by \(tX\)" and its mirror image. So the square closes short by \(ts[X,Y]+O(3)\).

3. Prediction at \(P\). From P9, \([R_z,R_x] = -R_y\), whose components are \(\big(\tfrac12(1+u^2-v^2),\ uv\big)\). At \((u,v) = (2,1)\) that is \((2,2)\). Hence for \(t = s = 0.02\),

\[ ts\,[X,Y]_P = 4\times10^{-4}\times(2,2) = (8,\ 8)\times10^{-4}. \]

4. Is the error \(O(t^3)\)? The relative errors are

\[ t = 0.02:\quad \frac{7.8804-8}{8} = -1.50\%,\qquad \frac{8.1238-8}{8} = +1.55\% , \] \[ t = 0.05:\quad \frac{4.81334-5}{5} = -3.73\%,\qquad \frac{5.20252-5}{5} = +4.05\% . \]

If the absolute error is \(O(t^3)\) and the leading term is \(O(t^2)\), the relative error is \(O(t)\), so it should scale by \(0.02/0.05 = 0.4\) between the two rows. The measured ratios are \(1.50/3.73 = 0.402\) and \(1.55/4.05 = 0.383\) — both close to \(0.4\). Consistent. ✓

Check. Dimensions: the chart coordinates are dimensionless, \(t\) and \(s\) are angles in radians, and \([X,Y]\) has the dimension of a coordinate per radian, so \(ts[X,Y]\) is a coordinate displacement. ✓ Limit: if \(Y = X\) the four flows compose to the identity exactly and the gap is zero — matching \([X,X] = 0\). Special case: at the south pole \((u,v) = (0,0)\), \([R_z,R_x] = (\tfrac12, 0)\), non-zero, so even at a fixed point of one of the two flows the square fails to close — the bracket is about the fields near the point, not their values at it.

Another way

On the sphere both flows are rotation matrices, so the square is \(R_x(-s)R_z(-t)R_x(s)R_z(t)\), and the Baker–Campbell–Hausdorff formula — named in unit 07, and here only its first correction term, which you can also get by multiplying the four exponential series out to second order — gives it as \(\exp\big(-ts[J_z,J_x]+O(3)\big) = \exp\big(-tsJ_y+O(3)\big)\) with unit 16's matrices, whose displacement of \(P\) is \(-ts\,R_y\). The bracket route gives \(ts[R_z,R_x] = -ts\,R_y\) as well. The two agree, and the sign that reconciles \([J_z,J_x] = +J_y\) with \([R_z,R_x] = -R_y\) is P9's anti-homomorphism, appearing once more.

Idea. Two directions you can move in do not make a surface you can stay on. What decides is whether the bracket escapes the span: if it does, the two directions generate a third, and the "surface" would have to be three-dimensional.

1. \(X = (1,0,0)\) and \(Y = (0,1,x)\) are linearly independent at every point — their first two components already are — so \(\mathcal D\) is a smooth rank-2 distribution. The bracket:

\[ [X,Y]^i = X^j\partial_jY^i - Y^j\partial_jX^i = \partial_x(0,1,x) - 0 = (0,0,1) = \partial_z . \]

Is \(\partial_z\in\mathrm{span}\{X,Y\}\)? A combination \(\alpha X+\beta Y = (\alpha,\beta,\beta x)\) equals \((0,0,1)\) only if \(\alpha = \beta = 0\) and \(\beta x = 1\) — impossible. So \(\mathcal D\) is not involutive.

2. No surface. Suppose \(z = h(x,y)\) had \(\mathcal D\) as its tangent planes on some open set. A vector \((a,b,c)\) is tangent iff \(c = a\,\partial_xh+b\,\partial_yh\). Applying this to \(X = (1,0,0)\) gives \(\partial_xh = 0\); applying it to \(Y = (0,1,x)\) gives \(\partial_yh = x\). Then

\[ \partial_y\partial_xh = \partial_y(0) = 0,\qquad \partial_x\partial_yh = \partial_x(x) = 1 . \]

Clairaut's theorem (unit 03) says the mixed partials of a \(C^2\) function are equal. They are not. Contradiction. \(\square\) The same argument works for a surface given implicitly, by the implicit function theorem.

3. Reachability. Flowing along \(X\) for time \(a\) changes \(x\) by \(a\) and nothing else. Flowing along \(Y\) for time \(b\) changes \(y\) by \(b\) and \(z\) by \(bx\), with \(x\) held fixed. So: move to the target \(y\) with \(Y\), move to the target \(x\) with \(X\), and then adjust \(z\) with the commutator square — go \(X\) by \(a\), \(Y\) by \(b\), back \(X\) by \(a\), back \(Y\) by \(b\), which returns \(x\) and \(y\) to their values and changes \(z\) by \(ab\) exactly (not merely to leading order: \(z\) gains \(b(x+a)\) then loses \(bx\)). Choosing \(ab\) freely reaches any \(z\). Hence the set reachable from a point is all of \(\mathbb R^3\). If the leaves existed they would contain every reachable point, hence be three-dimensional, contradicting rank 2. This is Chow's theorem in its simplest instance, and it is the exact opposite of Frobenius: brackets that escape make the distribution more useful, not less.

4. The car. With \(A = \cos\theta\,\partial_x+\sin\theta\,\partial_y\) and \(B = \partial_\theta\),

\[ [B,A] = \partial_\theta(\cos\theta,\sin\theta,0) = (-\sin\theta,\cos\theta,0), \]

which is the direction perpendicular to the car's heading — exactly the sideways motion the car cannot perform directly. Drive, steer, reverse, straighten: that is parallel parking, and it works because the bracket of the two controls points sideways.

Check. Dimensions: in part 4, \(\theta\) is dimensionless and \(A\) carries inverse length, so \([B,A]\) does too and is comparable with \(A\). ✓ Limit: replacing \(Y\) by \(\partial_y\) (drop the \(x\partial_z\)) gives \([X,Y] = 0\), involutive, and the leaves are the planes \(z = \mathrm{const}\) — the integrable case, recovered by deleting exactly the term that caused the trouble. Number: the commutator square with \(a = b = 0.1\) changes \(z\) by \(ab = 0.01\), exactly, with no higher-order error, because these flows are affine in the coordinates.