Local tools for bundles and transport

Written by GPT-6 Astra (OpenAI), Ultra effort, October 2026. Original exposition is dedicated under CC0 1.0. The attributed Brenner component and its marked completions in Section 3 retain CC BY-SA 4.0.

To construct a bundle or transport a vector, we need coordinates that solve equations, solutions that vary smoothly with their data, and smooth functions that join local constructions. We develop these tools in that order. The last construction applies them to a quotient of a Lie group.

Our manifolds are finite dimensional, Hausdorff and second countable. A smooth manifold is a space with charts whose transition functions are smooth. Its tangent vectors are coordinate velocities, identified by the derivatives of chart transitions. A Lie group is a smooth manifold whose multiplication and inversion are smooth. These are definitions; no subgroup or quotient theorem is included in them.

0. Analytic and linear foundations

We work over the ordered real field with the least-upper-bound property, using the usual natural numbers and induction. The free construction source for this section is Jiří Lebl's Basic Analysis, version 6.3, cited below. All analytic and linear arguments needed here are supplied in this lesson, including steps that the source assigns as exercises; the reference does not replace those proofs.

Lemma 0.0 (intermediate values and norm estimates). A continuous real function on an interval takes every value between any two of its values. The Euclidean length is a norm. In any normed space, the distance $d(x,y)=\|x-y\|$ is a metric and $\bigl|\|x\|-\|y\|\bigr|\leq\|x-y\|.$ The bounded linear maps between two normed spaces form a normed space with $\|A\|=\sup_{\|v\|\leq1}\|Av\|$, and $\|Av\|\leq\|A\|\|v\|$.

Proof. First suppose $f(a)<0<f(b)$. The nonempty set $S=\{x\in[a,b]:f(x)<0\}$ has a supremum $c$. Continuity and the strict endpoint signs give $a<c<b$: a small interval to the right of $a$ belongs to $S$, and a small interval to the left of $b$ misses $S$. If $f(c)<0$, continuity puts a point larger than $c$ in $S$, a contradiction. If $f(c)>0$, continuity gives an interval about $c$ disjoint from $S$, contradicting the definition of supremum. Thus $f(c)=0$. Subtract a target value, and reverse signs if necessary, to prove the general intermediate-value assertion.

Nonnegative square roots now exist by applying this assertion to $t^2$ between zero and a sufficiently large positive number. This polynomial is continuous because $(t+h)^2-t^2=2th+h^2\to0$. Roots are unique because $b^2-a^2=(b-a)(b+a)>0$ when $b>a\geq0$. Put $x\cdot y=\sum_i x_i y_i$ and $|x|=\sqrt{x\cdot x}$. If $y\ne0$, substitute $t=(x\cdot y)/|y|^2$ in $0\leq|x-ty|^2=|x|^2-2t(x\cdot y)+t^2|y|^2$. This gives $|x\cdot y|\leq|x||y|$; if $y=0$ the same inequality is immediate. Hence $|x+y|^2\leq|x|^2+2|x||y|+|y|^2$, which is the triangle inequality after taking nonnegative square roots. Positivity, definiteness and absolute homogeneity follow directly from the sum of squares. In any normed space, the three metric axioms follow from these norm axioms. The triangle inequality gives $\|x\|-\|y\|\leq\|x-y\|$; interchange $x,y$ for the reverse inequality. In particular the norm is continuous.

For a bounded linear map the displayed supremum is finite and nonnegative. Scaling a nonzero $v$ to a unit vector gives the stated bound; for $v=0$ it is immediate. This bound shows that $\|A\|=0$ forces $A=0$. Absolute homogeneity follows by moving the scalar outside the target norm. For every $\|v\|\leq1$, the target triangle inequality gives $\|(A+B)v\|\leq\|A\|+\|B\|$; taking the supremum proves the operator triangle inequality. These arguments also cover a zero-dimensional domain, whose closed unit ball is the singleton $\{0\}$. □

Lemma 0.1 (completeness and compactness). Euclidean space is complete. A closed subset of a complete metric space is complete. A metric space is compact exactly when every sequence has a subsequence converging in that space. Closed bounded subsets of a finite-dimensional Euclidean space are compact. A continuous image of a compact space is compact, a closed subset of a compact space is compact, finite unions of compact sets are compact, and compact subsets of Hausdorff spaces are closed. A continuous real function on a nonempty compact metric space attains a minimum and a maximum. A continuous map from a compact metric space to a metric space is uniformly continuous.

Proof. We first supply the real completeness and subsequence arguments in full, including the real-order steps sometimes assigned as exercises. The natural numbers are unbounded in $\mathbb R$: if they had a supremum $s$, some integer would exceed $s-1$, and its successor would exceed $s$. Thus $1/n\to0$.

A real Cauchy sequence $x_n$ is bounded: beyond some index it lies within one of a fixed term, and there are only finitely many earlier terms. Set $a_N=\inf_{n\geq N}x_n$ and $b_N=\sup_{n\geq N}x_n$; infima exist because $\inf S=-\sup(-S)$. The $a_N$ increase and the $b_N$ decrease. Put $a=\sup_N a_N$. For every $N$, $a_N\leq a\leq b_N$: every tail infimum is at most $b_N$, whether its index is before or after $N$. Given $\epsilon>0$, choose $N$ with $|x_j-x_k|<\epsilon$ for $j,k\geq N$. Fixing $k$, taking a supremum over $j$, and then an infimum over $k$, gives $b_N-a_N\leq\epsilon$. Therefore $|x_n-a|\leq\epsilon$ for $n\geq N$, proving real completeness.

For a bounded real sequence, start with a closed interval containing it and repeatedly choose a closed half-interval containing infinitely many terms. Write the nested intervals as $[a_j,b_j]$, of length $2^{-j}(b_0-a_0)$. Their lengths tend to zero since $2^j\geq j+1$. The number $a=\sup_j a_j$ lies in every interval: each left endpoint is at most every later right endpoint, and earlier left endpoints are smaller. Select successively an index greater than the preceding one whose term lies in the $j$-th interval. Its distance from $a$ is at most the interval length. This proves the real Bolzano–Weierstrass assertion without an unproved liminf or limsup step.

The zero-dimensional space consists of one point, so its completeness and compactness are immediate. For positive dimension, a Cauchy sequence in $\mathbb R^n$ is Cauchy in each coordinate. The just-proved real completeness gives coordinate limits. For $v\in\mathbb R^n$, $\max_i|v_i|\leq |v|\leq\sqrt n\max_i|v_i|,$ so coordinate convergence gives convergence in the Euclidean metric. If the sequence lies in a closed subset of a complete space, its ambient limit lies in that subset: otherwise an open ball about the limit disjoint from the subset contradicts convergence.

Here is the metric compactness argument, including the sequence steps. In a compact space, every sequence has a point every neighbourhood of which contains infinitely many sequence indices. Otherwise choose at each point a neighbourhood containing only finitely many indices; a finite subcover would account for only finitely many indices, a contradiction. Choose increasing indices in balls of radii $1/j$ about the resulting point. Their distances tend to zero, which is exactly convergence.

Conversely, suppose every sequence has a convergent subsequence, and fix an open cover. There is a $\delta>0$ such that every ball of radius $\delta$ is contained in a cover member. Otherwise choose $x_j$ whose radius-$1/j$ ball is contained in none. A subsequence converges to $x$. A cover member containing $x$ contains some ball $B(x,r)$; for large subsequence indices, $d(x_j,x)+1/j<r$, a contradiction. Now choose successively points at distance at least $\delta$ from all earlier choices whenever their radius-$\delta$ balls fail to cover the space. Infinitely many choices would have no convergent subsequence, since convergence makes pairwise distances in a tail less than $\delta$. The selection therefore stops. Each of the finitely many balls lies in a cover member, giving a finite subcover. The empty space satisfies both assertions directly.

For a bounded sequence in $\mathbb R^n$, apply the real subsequence argument to its first coordinate, then to the second coordinate along the chosen subsequence, and continue through all $n$ coordinates. The final subsequence converges in every coordinate and hence in norm. If the original sequence lies in a closed set, its limit remains there. The just-proved metric criterion gives compactness. This supplies all finite dimensions.

For a continuous image, pull back an open cover and push forward a finite subcover. For a closed subset, adjoin its open complement to the cover and then discard that extra member from a finite subcover. For a finite union, take a finite subcover on each constituent and unite these finitely many finite families. If $K$ is compact in a Hausdorff space and $x\notin K$, separate $x$ and each $y\in K$ by disjoint open sets $U_y,V_y$. Finitely many $V_y$ cover $K$; the intersection of the corresponding $U_y$ is a neighbourhood of $x$ disjoint from $K$. Thus the complement of $K$ is open.

For a continuous real function on nonempty compact $K$, the image is compact and therefore closed. The cover of its image by $(-j,j)$, $j\geq1$, has a finite subcover, proving boundedness. Its supremum $s$ is a limit of image points: choose one in $(s-1/j,s]$ for each $j$. Closedness puts $s$ in the image. Apply this to the negative function for the minimum. Finally, if a continuous map $f:K\to Y$ were not uniformly continuous, there would be $\epsilon>0$ and $x_j,y_j\in K$ with $d(x_j,y_j)<1/j$ but $d(f(x_j),f(y_j))\geq\epsilon$. A subsequence of $x_j$ converges to $x$; the triangle inequality makes the corresponding $y_j$ converge to the same $x$. Continuity and the target triangle inequality contradict the last bound. □

Lemma 0.2 (linear maps and complete function spaces). A subspace of a finite-dimensional vector space has a linear complement. Linear maps between finite-dimensional Euclidean spaces are bounded. If $E$ is a complete normed vector space, the bounded operators $E\to E$, with their operator norm, form a complete normed space, and $\|AB\|\leq\|A\|\|B\|$. Continuous functions from a compact interval to $\mathbb R^n$, with the supremum norm, form a complete normed space.

Proof. We recall the finite basis argument explicitly. An independent list has length at most a spanning list: starting with a spanning list, express the first independent vector in it and replace a vector with nonzero coefficient by that vector. The new list still spans, because the replaced vector can be solved for. At each subsequent step, the next independent vector has a nonzero coefficient on one of the as yet unreplaced vectors; otherwise it would be in the span of the preceding independent vectors. Replace that vector and continue. The process cannot continue past the length of the spanning list. In a subspace of $\mathbb R^n$, successively choose a vector outside the span of the previous choices whenever that span is not the whole subspace. Independence holds at each step, so the process stops after at most $n$ choices and gives a basis. Continue choosing vectors outside its span in the ambient space. The resulting ambient basis extends the subspace basis, and the span of the added vectors is a complement.

In standard coordinates, $(Av)_i=\sum_j a_{ij}v_j$, whence $|Av|\leq\sqrt m\left(\max_i\sum_j|a_{ij}|\right)|v|$ for an $m$-row matrix. Thus every such linear map has a finite operator norm. Directly from its definition, $|ABv|\leq\|A\|\|B\||v|$.

A supremum-norm Cauchy sequence $u_k$ has pointwise limits $u(t)$, by the first part of Lemma 0.1. Passing to the limit in $|u_k(t)-u_l(t)|\leq\epsilon$, uniformly in $t$, proves uniform convergence. Given $t_0$, choose $k$ with $\|u-u_k\|_\infty<\epsilon/3$ and then a neighbourhood on which $|u_k(t)-u_k(t_0)|<\epsilon/3$. The triangle inequality proves continuity of $u$. Each continuous function is bounded on the compact interval, by Lemma 0.1 applied to its norm. Thus the limit belongs to the asserted space. The norm axioms follow pointwise. □

Lemma 0.3 (integration and differentiation estimates). Continuous real functions on compact intervals are Riemann integrable. The integral of a continuous function has that function as its derivative in the interval interior, with the corresponding one-sided derivatives at endpoints, and integrating a continuous derivative gives the difference of endpoint values. Integration of continuous vector-valued functions is linear and satisfies $\left|\int_a^b v(t)\,dt\right|\leq\int_a^b |v(t)|\,dt \leq (b-a)\sup_{[a,b]}|v| \qquad(a\leq b).$ It commutes with uniform limits. For a continuously differentiable map on a neighbourhood of a segment, $f(x+h)-f(x)=\int_0^1 Df(x+th)h\,dt.$ The product and chain rules also hold for maps differentiable in norm between normed spaces, when their derivatives are bounded linear maps. Continuous bilinear maps, including operator composition, are smooth.

Proof. For a bounded real function $f$, let the lower and upper sums on a partition be the sums of the interval lengths multiplied by the infimum and supremum of $f$ on the respective intervals. Subdividing an interval raises its infima and lowers its suprema; adding the resulting inequalities shows that refinement increases lower sums and decreases upper sums. A common refinement of two partitions therefore shows that any lower sum is at most any upper sum. The supremum of the lower sums is consequently at most the infimum of the upper sums. These two finite numbers exist by the least-upper-bound property and boundedness of $f$.

For continuous $f$, Lemma 0.1 supplies boundedness and uniform continuity. If a partition has mesh at most $\delta$, its upper minus lower sum is at most $(b-a)\omega(\delta)$, where the oscillation bound $\omega(\delta)$ tends to zero. Thus the two numbers are equal; their common value defines the integral. Every tagged Riemann sum lies between the lower and upper sums and hence converges to that integral as the mesh tends to zero. Linearity, including negative scalar multiples and sums of two functions on the same partition, follows by passing the identities for tagged sums to the limit. Their additivity on adjoining intervals passes to the limit as well. The same argument proves monotonicity of integrals when the integrands are ordered. Define reversed integrals by changing the sign.

For $F(x)=\int_a^x f(t)\,dt$, additivity and the tagged-sum bound give $\left|\frac{F(x+h)-F(x)}h-f(x)\right| \leq\sup_{t\text{ between }x\text{ and }x+h}|f(t)-f(x)|\longrightarrow0.$ Thus $F'=f$. To prove the other direction explicitly, first note that a differentiable real function at an interior extremum has derivative zero: the left and right difference quotients have opposite weak signs and the same limit. A continuous function on a compact interval with equal endpoint values therefore has a zero derivative somewhere in the interior, unless it is constant; its minimum and maximum exist by Lemma 0.1, and any value unequal to the endpoints forces one of those extrema to occur inside. Subtracting the line joining the endpoints gives the mean value theorem for a continuous function differentiable in the interior. Apply it on every subinterval to a continuously differentiable $g$. Then $g(b)-g(a)$ is a tagged sum of $g'$, and letting the mesh tend to zero gives $g(b)-g(a)=\int_a^b g'(t)\,dt$. This also proves that a function with derivative zero on an interval is constant, by applying the mean value theorem to each compact subinterval.

For vectors, take the same tagged partition for all coordinates. The triangle inequality for each finite sum gives $\left|\sum v(t_i)\Delta t_i\right|\leq\sum |v(t_i)|\Delta t_i$. Passing to the limit proves the first bound; boundedness of the norm gives the second. Applying the bound to $v_k-v$ proves interchange with uniform limits; such a limit is continuous by the three-term estimate in Lemma 0.2.

For the norm chain rule write $f(x+h)=f(x)+Ah+r(h)$, with $r(h)=o(\|h\|)$, and $g(y+k)=g(y)+Bk+s(k)$, with $s(k)=o(\|k\|)$. Here $k=Ah+r(h)=O(\|h\|)$. Substitution leaves remainder $Br(h)+s(k)=o(\|h\|)$, proving derivative $BA$. Apply the just-proved scalar fundamental theorem coordinate by coordinate to $t\mapsto f(x+th)$, using this chain rule, to obtain the segment formula.

A continuous bilinear map $P$ is bounded by a constant times the product of the norms. Indeed continuity at $(0,0)$ gives $\delta>0$ such that $\|P(u,v)\|\leq1$ when $\|u\|+\|v\|<\delta$. Scale nonzero $u,v$ separately to have norm $\delta/4$, and use bilinearity to get $\|P(u,v)\|\leq16\|u\|\|v\|/\delta^2$; a zero argument gives zero. Now expand $P(u+h,v+k)-P(u,v)=P(h,v)+P(u,k)+P(h,k).$ The last term is bounded by a constant times $\|h\|\|k\|$, and is a quadratic remainder. Its derivative is the displayed linear part, its second derivative is constant, and all higher derivatives vanish. The sum and scalar-multiple rules follow by adding or scaling the norm remainders. Repeated application gives the asserted smoothness and the higher product and chain rules. □

Lemma 0.4 (operator inverses). In a complete normed space, $I-A$ has a bounded two-sided inverse whenever $\|A\|<1$. The bounded operators possessing a bounded two-sided inverse form an open set, on which inversion is smooth, and $D(B\mapsto B^{-1})_B[H]=-B^{-1}HB^{-1}.$

Proof. The finite geometric identity is $(I-A)\sum_{j=0}^{N}A^j=I-A^{N+1}=\left(\sum_{j=0}^{N}A^j\right)(I-A)$. For $0<q<1$, write $q^{-1}=1+c$, with $c>0$. Induction gives $(1+c)^N\geq1+Nc$, so $q^N\to0$. The case $q=0$ is immediate. Scalar finite geometric sums then bound the operator tails by $q^{N+1}/(1-q)$ when $\|A\|\leq q<1$. Lemma 0.2 gives a limit, multiplication is continuous, and the finite identity proves that the limit is the two-sided inverse.

For $B$ with a bounded two-sided inverse and $\|B^{-1}H\|<1$, factor $B+H=B(I+B^{-1}H)$. The just-proved series gives invertibility and local boundedness of the inverse. The identity $(B+H)^{-1}-B^{-1}=-(B+H)^{-1}HB^{-1}$ first proves continuity. Substituting it once more leaves $(B+H)^{-1}-B^{-1}+B^{-1}HB^{-1} =(B+H)^{-1}HB^{-1}HB^{-1},$ a locally bounded quadratic remainder. This proves the derivative formula. It is continuous. If inversion is $C^r$, that formula and the smoothness of operator multiplication make its derivative $C^r$; induction proves smoothness. □

Lemma 0.5 (a flat smooth cutoff factor). The function $\eta(t)=e^{-1/t^2}$ for $t>0$, extended by zero for $t\leq0$, is smooth and all its derivatives vanish at zero.

Proof. We construct the scalar exponential using only the preceding lemmas. Put $L(x)=\int_1^x dt/t$ for $x>0$. The reciprocal is continuous there: $(x+h)^{-1}-x^{-1}=-h/(x(x+h))$ tends to zero. Lemma 0.3 gives $L'(x)=1/x>0$, so its mean value theorem makes $L$ strictly increasing. For fixed $y>0$, the chain rule shows that $L(xy)-L(x)$ has derivative zero in $x$. Its value at $x=1$ is $L(y)$; hence $L(xy)=L(x)+L(y)$. Moreover $L(2)\geq1/2$ by the integral bound. Thus $L(2^n)=nL(2)$ is unbounded above, and $L(2^{-n})=-nL(2)$ is unbounded below. The intermediate value theorem in Lemma 0.0 makes $L$ onto $\mathbb R$.

Let $E$ be its inverse. It is continuous: for $x=E(y)>0$ and $0<\epsilon<x$, the strict inequalities $L(x-\epsilon)<y<L(x+\epsilon)$ force $E(y+h)\in(x-\epsilon,x+\epsilon)$ when $|h|$ is smaller than both gaps. Write $\delta=E(y+h)-E(y)$. For $h\ne0$, strict monotonicity gives $\delta\ne0$, and $\frac{E(y+h)-E(y)}h =\left(\frac{L(x+\delta)-L(x)}\delta\right)^{-1} \longrightarrow\frac1{L'(x)}=x=E(y).$ Thus $E'=E$, and induction proves smoothness. We have $E(0)=1$, $E(y)>0$, and $E(s+t)=E(s)E(t)$ by the addition identity for $L$. In particular $E(-y)=1/E(y)$. We use the usual notation $e^y=E(y)$. For $x\geq0$, monotonicity gives $E(x)\geq1$. Repeated integration of $E'=E$, using Lemma 0.3, gives $E(x)\geq\sum_{j=0}^{N}x^j/j!$ for every integer $N\geq0$: integrate the preceding bound from zero and add $E(0)=1$. On $t>0$, the product and chain rules show by induction that every derivative of $e^{-1/t^2}$ is $p(1/t)e^{-1/t^2}$, with $p$ a polynomial. With $u=1/t$, the bound $E(u^2)\geq u^{2N}/N!$ shows that this expression tends to zero, even after division by $t$, on taking $2N$ greater than the relevant polynomial degree. Extend each derivative by zero. Its difference quotient at zero tends to zero, and its next derivative is continuous there. Induction proves the assertion. □

1. Contraction and local inversion

Lemma 1.1 (contraction). Let $(X,d)$ be nonempty and complete. If $T:X\to X$ satisfies $d(Tx,Ty)\leq qd(x,y)$ for a fixed $0\leq q<1$, then $T$ has exactly one fixed point.

Proof. Choose $x_0\in X$ and put $x_{j+1}=Tx_j$. Induction gives $d(x_{j+1},x_j)\leq q^j d(x_1,x_0)$. Thus, for $m>n$, $d(x_m,x_n)\leq\sum_{j=n}^{m-1}q^j d(x_1,x_0) \leq\frac{q^n}{1-q}d(x_1,x_0).$ The elementary geometric estimate in Lemma 0.4 proves this tends to zero. Completeness gives $x_j\to x$. Since $d(Tx,Tx_j)\leq qd(x,x_j)$, we have $Tx=\lim x_{j+1}=x$. Two fixed points satisfy $d(x,y)\leq qd(x,y)$, forcing their distance to be zero. □

Theorem 1.2 (smooth inverse function theorem). If $f:U\to\mathbb R^n$ is smooth on an open set and $Df(a)$ is invertible, then $f$ restricts to a smooth diffeomorphism between neighbourhoods of $a$ and $f(a)$.

Proof. Translate source and target and multiply the target by $Df(a)^{-1}$. It suffices to treat $a=0$, $f(0)=0$, $Df(0)=I$. Choose $r>0$ with $\overline B_r\subset U$ and $\|Df(x)-I\|\leq q<1$ on that closed ball. Continuity of the derivative permits this choice. By the segment formula of Lemma 0.3, $k(x)=x-f(x)$ is $q$-Lipschitz there. For $|y|<(1-q)r$, $T_y(x)=y+k(x),\qquad |T_y(x)|\leq |y|+q|x|<r$ maps the complete closed ball to itself. Lemma 1.1 supplies a unique fixed point $g(y)$, and that point is interior. It solves $f(g(y))=y$. Subtracting two fixed-point equations gives $|g(y)-g(z)|\leq (1-q)^{-1}|y-z|.$ Every $Df(x)$ on the ball is invertible by Lemma 0.4. Set $x=g(y)$ and $\delta=g(y+h)-g(y)$. Differentiability of $f$ gives $h=Df(x)\delta+o(|\delta|)$. The Lipschitz estimate makes the remainder $o(|h|)$; multiplying by $Df(x)^{-1}$ proves $Dg(y)=Df(g(y))^{-1}.$ This derivative is continuous, so $g$ is $C^1$. Lemmas 0.3–0.4 and the displayed identity imply inductively that $g\in C^{r+1}$ whenever $g\in C^r$. Hence $g$ is smooth. The open set $B_r\cap f^{-1}(B_{(1-q)r})$ maps bijectively to $B_{(1-q)r}$, since all its points satisfy the same unique fixed-point equation. These are the required neighbourhoods. Undo the initial changes. □

Corollary 1.3 (submersion coordinates). A smooth map with surjective differential at a point is locally a coordinate projection. Its level set is locally embedded, and it has a smooth local section through that point.

Proof. For its coordinate derivative $A$, choose a complement $C$ of $\ker A$, using Lemma 0.2. The restriction $A|_C$ is bijective onto the target: injectivity follows from the direct sum, and surjectivity follows by decomposing any preimage. Let $\ell$ be the projection onto $\ker A$. The derivative of $x\mapsto(f(x),\ell(x-a))$ is bijective. Theorem 1.2 makes this map a local coordinate system. There $f$ is the first projection; fixing the first coordinates gives its level set, and fixing the remaining coordinates at zero gives the section. □

Corollary 1.4 (constant rank). A smooth map of constant rank $r$ near a point has coordinate expression $(u,z)\mapsto(u,0)$, after restricting source and target neighbourhoods.

Proof. A rank-$r$ linear map has $r$ independent columns. Successively choose a nonzero coordinate of the first, eliminate that coordinate from the remaining columns, and repeat on the residual columns. These remain independent: a relation among the residuals would express a combination of the original remaining columns as a multiple of the pivot column, contradicting independence. This produces $r$ rows on which those columns form an invertible matrix. Thus its derivative has an invertible $r$-by-$r$ minor. Use the corresponding $r$ output functions and the unused input coordinates as new source coordinates, applying Theorem 1.2. The map becomes $(u,z)\mapsto(u,\psi(u,z))$. Its derivative has block form $\left(\begin{smallmatrix}I&0\\ *&D_z\psi\end{smallmatrix}\right)$. Any nonzero column of $D_z\psi$ would increase the rank past $r$; hence $D_z\psi=0$. On a product box, the segment formula makes $\psi$ independent of $z$. The target change $(u,w)\mapsto(u,w-\psi(u))$, whose explicit inverse adds $\psi(u)$, gives the stated expression. For $r=0$, the same segment argument makes the map constant. □

2. Differential equations and their parameters

Lemma 2.A (smooth parameter fixed points). Let $E$ be complete and normed, $C\subset E$ a closed ball, and $\Lambda\subset\mathbb R^m$ open. Let $T$ be smooth on an open neighbourhood of $C\times\Lambda$, take $C$ into itself for every parameter, and satisfy $\|T(u,z)-T(v,z)\|\leq q\|u-v\|,\qquad 0\leq q<1.$ If its fixed points are interior to $C$, their unique value $u(z)$ is smooth and $Du(z)=\bigl(I-D_uT(u(z),z)\bigr)^{-1}D_zT(u(z),z).$

Proof. Lemmas 0.1 and 1.1 give a fixed point for every $z$. Holding $z_0$ fixed and inserting $T(u(z_0),z)$ in the difference of the two fixed-point equations yields $(1-q)\|u(z)-u(z_0)\| \leq\|T(u(z_0),z)-T(u(z_0),z_0)\|.$ This proves continuity and, for $z=z_0+h$, the estimate $\delta=u(z_0+h)-u(z_0)=O(|h|)$. Differentiability of $T$ at the fixed point then gives $\delta=A\delta+Bh+o(\|\delta\|+|h|), \quad A=D_uT(u(z_0),z_0),\quad B=D_zT(u(z_0),z_0).$ Interior difference quotients and the contraction bound give $\|A\|\leq q$. By Lemma 0.4, $I-A$ has a bounded inverse. Multiplying proves the derivative formula with remainder $o(|h|)$. Its right side is continuous, so $u$ is $C^1$. If $u$ is $C^r$, the smoothness of $T$, the norm chain rule and smooth operator inversion make that right side $C^r$. Induction gives $u\in C^\infty$. □

Theorem 2.1 (local flow). Let $F(t,x,\lambda)$ be smooth on an open subset of $\mathbb R\times\mathbb R^n\times\mathbb R^m$. Every initial datum in its domain has a unique local solution of $x'=F(t,x,\lambda)$. Solutions depend smoothly on time, initial time, initial value and parameter. For fixed parameter $\lambda$, a solution extends across a finite endpoint if the graph $(t,x(t),\lambda)$ stays in a compact subset of the domain.

Proof. Around $(t_0,x_0,\lambda_0)$, choose a product of a closed time interval, a closed spatial ball and a closed parameter box, contained in the domain. Lemma 0.1 gives compactness. Lemma 0.1 bounds $|F|$ and $\|D_xF\|$ there by constants $B,L$. Choose a smaller spatial ball for the initial value and a sufficiently small $\epsilon>0$ so that $\epsilon L<1$ and $\epsilon B$ is less than half the remaining spatial margin. For data $z=(a,t_0,\lambda)$ in a smaller open neighbourhood, use the fixed interval $[-\epsilon,\epsilon]$ and set $(T(u,z))(s)=a+\int_0^s F(t_0+\tau,u(\tau),\lambda)\,d\tau.$ On a fixed closed supremum-norm ball about the constant curve $x_0$, the margin makes $T$ take values in the interior of that ball. Lemma 0.3 gives $\|T(u,z)-T(v,z)\|_\infty\leq\epsilon L\|u-v\|_\infty$. The space of curves is complete by Lemma 0.2. Lemma 1.1 gives a fixed point, and the coordinate fundamental theorem proves that it is a solution.

For uniqueness, any two solutions with the same datum remain in such a rectangle on a sufficiently small common interval and satisfy the same integral equation. Their difference is at most $\epsilon L$ times its supremum norm, so it vanishes. On any common connected time interval, the times where they agree form a closed set by continuity and an open set by this local argument. That set is the whole interval: if it stopped before any given time, its first stopping boundary would belong to it by closedness and extend it by openness. The backward argument is the same.

We verify the smoothness hypothesis of Lemma 2.A. The first curve derivative is $D_uT(u,z)[v](s)=\int_0^s D_xF(t_0+\tau,u(\tau),\lambda)v(\tau)\,d\tau.$ Derivatives in $t_0$ and $\lambda$ insert the respective derivatives of $F$; the derivative in $a$ is the constant curve. A derivative of any order is a finite sum of integrals of a corresponding multilinear derivative of $F$ applied to the increments. All coefficients are bounded on a slightly larger compact rectangle. For every derivative order, the segment formula applied to that derivative, and uniform continuity of its next derivative on this rectangle, give a Taylor remainder bounded by $\epsilon\omega(\|\Delta u\|_\infty+|\Delta z|)(\|\Delta u\|_\infty+|\Delta z|)$, where $\omega(r)\to0$. The same estimate proves continuity in the appropriate multilinear-operator norm. This proves the derivative assertions successively, rather than merely differentiating a prospective fixed point.

Lemma 2.A now gives smooth dependence on $z$ as a curve-valued map. Evaluation of a curve is bounded linear. Joint continuity in $(s,z)$ follows from $|v(z)(s)-v(z_0)(s_0)| \leq\|v(z)-v(z_0)\|_\infty+|v(z_0)(s)-v(z_0)(s_0)|.$ This applies to each parameter derivative. Parameter differentiation of the integral equation is justified by Lemma 0.3 and the uniform remainder estimates. The integrands are continuous, so the fundamental theorem gives the time derivative of each parameter derivative. Repetition gives all time and mixed derivatives. Substitute $s=t-t_0$ to recover the stated variables.

Finally, each point of a compact time-position set has a smaller rectangle with a positive time of existence valid for every starting point in that smaller rectangle. Finitely many cover the compact set; the minimum of their positive times is positive. Starting at an original solution time sufficiently near a finite endpoint therefore yields a solution interval crossing it. Uniqueness identifies the new solution with the old one on the overlap. The same finite-chart argument applies on a manifold: the chain rule transforms the differential equation under a chart change, and local uniqueness glues the solutions. □

Lemma 2.2 (invariant equations do not escape). Let $b:[a,c]\to T_eG$ be continuous. The right-invariant equation $g'(t)=(dR_{g(t)})_e b(t)$ has a unique $C^1$ solution on $[a,c]$ for every initial value at any time in that interval. For finitely piecewise continuous $b$, with finite one-sided limits at the division points, it has a unique continuous solution that is $C^1$ in the interior of each piece and satisfies the equation there.

Proof. In a fixed chart about $e$, restrict the position to a compact ball. The set of values of $b$ is compact, so the coordinate field and its spatial derivative have uniform bounds. The integral construction in Theorem 2.1 needs only continuity in time and a uniform spatial Lipschitz bound for existence and uniqueness. It therefore gives a common positive time $\epsilon$ for a solution $h$ starting at $e$, independently of the chosen starting time. At the endpoints of $[a,c]$, one can first extend $b$ constantly outside the interval.

To start at $g_0$, use $g(t)=h(t)g_0$. Indeed, $R_{g_0}\circ R_h=R_{hg_0}$, so the chain rule verifies the same equation. Restart after each time step smaller than $\epsilon$, to the right or left of the initial time. Finitely many steps cover $[a,c]$. Local uniqueness makes the pieces agree on overlaps. For piecewise continuous coefficients, apply the construction to each continuous piece, passing the endpoint value to the next one. □

Proposition 2.3 (group exponential). A Lie group has a smooth exponential map, its derivative at zero is the identity, it is a local diffeomorphism at zero, and it intertwines conjugation with the adjoint action.

Proof. For a constant $A\in T_eG$, the corresponding left-invariant equation is treated by left translation in the same proof. Write $g_A(t)$ for its identity-starting solution and define $\exp(A)=g_A(1)$. It is defined for all real $t$, by applying the finite-interval result on every bounded interval and using uniqueness. Differentiation shows that $s\mapsto g_A(ts)$ solves the equation for $tA$ from the identity; uniqueness gives $g_A(t)=\exp(tA)$. Translating a solution in time and then on the left gives $\exp((s+t)A)=\exp(sA)\exp(tA)$. The local smooth-dependence theorem and a finite number of such restarts prove that $A\mapsto\exp(A)$ is smooth near every $A$. The proved time-scaling identity gives $\left.\frac{d}{ds}\right|_{s=0}\exp(sA)=g_A'(0)=A$, so its derivative at zero is the identity map of $T_eG$. Theorem 1.2 consequently makes $\exp$ a local diffeomorphism. Finally, differentiating the group homomorphism $g\mapsto aga^{-1}$ along a solution gives another left-invariant equation, with initial velocity $\operatorname{Ad}(a)A$, where $\operatorname{Ad}(a)$ is that homomorphism's derivative at $e$. Uniqueness gives $a\exp(tA)a^{-1}=\exp(t\operatorname{Ad}(a)A).$

This proves each asserted property. □

For the countability arguments in the next section, the following elementary details will be useful. Pairs of nonnegative integers can be listed by increasing sum, listing the finitely many pairs of each sum in increasing first coordinate. Thus a countable union of listed countable families is countable, by listing the pairs of indices and discarding repetitions. A countable basis restricts to a countable basis on any subspace by intersecting its members with that subspace: every relatively open neighbourhood is the intersection of an ambient open set with the subspace, and a basis member can be chosen inside that ambient set at the chosen point. Finally, the exponential bound used in the attributed boundary completion below is already proved in Lemma 0.5, without assuming an exponential-series theorem.

3. Smooth weights with controlled support

The support of a function is the closure of the set on which it is nonzero. A compact exhaustion is a sequence of compact sets $A_n$ with $A_n\subset\operatorname{int}A_{n+1}$ and $\bigcup_{n\geq0}A_n=M$. We write $U(a,r)$ for an open Euclidean ball and $B(a,r)$ for its closed ball in the source passages below.

The following three complete statements and proofs are adapted from Holger Brenner and the Wikiversity contributors, Lecture 22 of Differentialgeometrie (Osnabrück 2023), Lemmas 22.6 and 22.9 and Theorem 22.10, through the complete English edition. They retain CC BY-SA 4.0. English translation, mathematical typography and local labels are adapted; the explicitly marked completions supply the countable-subcover argument, shell selection, local finiteness and bump-boundary derivatives. The source theorem asserts continuous differentiability; smoothness on our smooth manifolds follows from the same construction. Source images are not reused.

Lemma 3.A (compact exhaustion; Brenner 22.6). Let $M$ be a manifold with a countable basis for its topology. Then $M$ has a compact exhaustion.

Proof. For every point $P\in M$ there is an open coordinate neighborhood $P\in U$, $\alpha:U\to V$, and ball neighborhoods
$U(\alpha(P),\epsilon) \subset B(\alpha(P),\epsilon)\subset V.$
Since the coordinate map is a homeomorphism and closed balls are compact, $B_P=\alpha^{-1}(B(\alpha(P),\epsilon))$ is a compact subset of $M$ containing the open neighborhood $U_P=\alpha^{-1}(U(\alpha(P),\epsilon))$ of $P$. The sets $U_P$, $P\in M$, form an open cover of $M$, so by Exercise 2.8 (Measure and Integration Theory (Osnabrück 2022–2023)) there is a countable subcover.

Countable-subcover completion. Let $\{E_m\}_{m\geq0}$ be a countable basis. For every $E_m$ contained in some member of an open cover, choose one such member. These choices cover $M$: a point in a cover member lies in a basis element contained in that member. Thus every open cover has a countable subcover.

Denote it by $U_n$, $n\in\mathbb N$, (where the $U_n$ lie in the compact subsets $B_n$). We now define recursively a monotonically increasing map
$\mathbb N\longrightarrow\mathbb N,\qquad k\longmapsto n_k,$
such that
$A_k=\bigcup_{n=0}^{n_k}B_n$
is a compact exhaustion of $M$. As finite unions of compact sets, the $A_k$ are compact. We begin with $n_0=0$. Suppose $n_k$ has already been constructed. The set $A_k\cup B_{n_k+1}$ is compact and is therefore contained in a finite subunion $\bigcup_{n=0}^{n_{k+1}}U_n$, where the upper index is chosen so that $n_{k+1}\geq n_k+1$. With this choice,
$A_k\subset\bigcup_{n=0}^{n_{k+1}}U_n \subset\bigcup_{n=0}^{n_{k+1}}B_n=A_{k+1}.$
and this sequence is an exhaustion because the $U_n$, $n\in\mathbb N$, form an open cover of $M$.

Exhaustion completion. The middle union is open, so $A_k\subset\operatorname{int}A_{k+1}$; and $n_k\geq k$ ensures that every $U_n$ is eventually contained in an $A_k$. If the subcover is finite and $M$ is nonempty, repeat a member to index it by $\mathbb N$. For $M=\varnothing$, take $A_k=\varnothing$ throughout. □

Lemma 3.B (subordinate atlas; Brenner 22.9). Let $M$ be a manifold with a countable basis for its topology. Let $M=\bigcup_{i\in I}W_i$ be an open cover of $M$. Then there is a countable compatible atlas $(U_j,\alpha_j,V_j)$, $j\in J$, with ball neighborhoods
$U(0,\delta_j)\subset B(0,\epsilon_j)\subset V_j, \qquad 0<\delta_j<\epsilon_j,$
(where $0\in V_j\subset\mathbb R^d$ and $\delta_j<\epsilon_j$) such that for every $j\in J$ there is a $W_{i(j)}$ with $U_j\subset W_{i(j)}$ such that $M$ is covered by $\alpha_j^{-1}(U(0,\delta_j))$, $j\in J$, and every point $P\in M$ belongs to only finitely many of the sets $U_j$.

Proof. Let the open cover $W_i$, $i\in I$, be given. Furthermore, let $A_n$, $n\in\mathbb N$, be a compact exhaustion of $M$, which exists by Lemma 3.A. The open sets $\operatorname{int}A_{n+2}\setminus A_{n-1}$ also form an open cover, because every point $P\in M$ has a least $n\in\mathbb N$ such that $P\in A_n$ (with the convention $A_{-1}=\varnothing$). For this $n$, $P\notin A_{n-1}$ and $P\in A_n\subset\operatorname{int}A_{n+1}$.

By considering the intersections $W_i\cap(\operatorname{int}A_{n+2}\setminus A_{n-1})$, we may assume that every set in the cover lies in some $\operatorname{int}A_{n+2}\setminus A_{n-1}$. For every point $P\in M$ there is an open (compatible) coordinate neighborhood $P\in U_P$, contained in one of the $W_i$ and having ball neighborhoods
$U(0,\delta_P)\subset B(0,\epsilon_P)\subset V_P.$
such that $P\in\alpha_P^{-1}(U(0,\delta_P))$ and $\delta_P<\epsilon_P$. These sets $\alpha_P^{-1}(U(0,\delta_P))$, $P\in M$, also form an open cover of $M$. By Exercise 2.8 (Measure and Integration Theory (Osnabrück 2022–2023)) we may pass to a countable subcover. Thus we may assume that a system of charts $U_j$, $j\in\mathbb N$, together with ball neighborhoods $U(0,\delta_j)\subset B(0,\epsilon_j)\subset V_j$ is given such that $\alpha_j^{-1}(U(0,\delta_j))$, $j\in\mathbb N$, is also an open cover of $M$, each $U_j$ is contained in some $W_{i(j)}$, and the relationship to the compact exhaustion described above holds. We will define a subset $J\subset\mathbb N$ such that the family $U_j$, $j\in J$, also satisfies the finiteness property.

Shell-selection completion. Make these choices separately for each compact layer $A_{n+1}\setminus\operatorname{int}A_n$, requiring its charts to lie in $\operatorname{int}A_{n+2}\setminus A_{n-1}$. This is possible because that entire layer is contained in that open shell. Also choose charts covering $A_0$ inside $\operatorname{int}A_1$. Each layer has a countable subcover by the same basis argument (applied to its subspace topology). Their countable union is the initial chart system. Thus the initial system contains enough charts of each specified shell for the following finite selections; an arbitrary countable subcover of all shells need not have this property.

First, compactness of $A_0$ gives a finite index set $J_{-1}$ such that the sets $\alpha_j^{-1}(U(0,\delta_j))$, $j\in J_{-1}$, cover $A_0$. For $n\in\mathbb N$ consider the compact set $A_{n+1}\setminus\operatorname{int}A_n$. It is covered by finitely many of the sets $\alpha_j^{-1}(U(0,\delta_j))$, $j\in\mathbb N$, and only indices $j$ for which $U_j$ is contained in $\operatorname{int}A_{n+2}\setminus A_{n-1}$ are needed. Denote the corresponding finite index set by $J_n$, and set
$J=J_{-1}\cup\bigcup_{n\geq0}J_n.$
Then each $A_k$ meets only finitely many of the sets $U_j$, $j\in J$.

Cover and local-finiteness completion. The smaller balls cover $M$: if a point first enters the exhaustion at $A_r$, it lies in $A_0$ when $r=0$, and in $A_r\setminus\operatorname{int}A_{r-1}$ otherwise. A chart selected from $J_n$ with $n\geq k+1$ misses $A_k$, since it misses $A_{n-1}\supset A_k$. Only the finitely many selections $J_{-1},J_0,\ldots,J_k$ can meet $A_k$. For a point $P$, choose $k$ with $P\in\operatorname{int}A_k$. That interior is a neighbourhood meeting only finitely many charts. Consequently the selected atlas is locally finite, which is stronger than the pointwise finiteness in the source statement. □

Definition 3.C (partition of unity; Brenner 22.7). Let $X$ be a topological space. A family of functions $h_j:X\to\mathbb R$ indexed by $j\in J$ is called a partition of unity if the following properties hold.

We have $h_j(X)\subset[0,1]$ for every $j\in J$.

Every point $P\in X$ has an open neighborhood $P\in U$ such that the restricted functions $h_j|_U$ are identically zero, with only finitely many exceptions.

We have $\sum_{j\in J}h_j=1$.

If all $h_j$ are continuous, this is called a continuous partition of unity.

The second property ensures that the sum in (3) is defined, since for every point $P\in X$ and all but finitely many $j\in J$ the equality $h_j(P)=0$ holds. On a manifold such a partition is called differentiable if all $h_j$ are differentiable functions.

Definition 3.D (subordinate partition; Brenner 22.8). Let $X=\bigcup_{i\in I}W_i$ be an open cover of a topological space $X$. A partition of unity $h_j:X\to\mathbb R$ indexed by $j\in J$ is called a partition of unity subordinate to the cover if for every $j\in J$ there is an open set $W_{i(j)}$ in the cover such that the support of $h_j$ is contained in $W_{i(j)}$.

Theorem 3.E (partition of unity; Brenner 22.10). Let $M$ be a differentiable manifold with a countable basis for its topology. Then every open cover admits a continuously differentiable partition of unity subordinate to that cover.

Proof. By Lemma 3.B we may assume that an open cover consists of coordinate domains $U_j$, $j\in J$, ($J$ countable) with $\alpha_j:U_j\to V_j$ and with ball neighborhoods
$U(0,\delta_j)\subset B(0,\epsilon_j)\subset V_j, \qquad \delta_j<\epsilon_j,$
(with $\delta_j<\epsilon_j$) such that the sets $\alpha_j^{-1}(U(0,\delta_j))$ also cover $M$ and every point $P\in M$ belongs to only finitely many of the $U_j$, and in particular to only finitely many of these sets $\alpha_j^{-1}(U(0,\delta_j))$. On $V_j$ consider the function $g_j$ defined by the following formula. Notation for the formula: $s_j(v)=\delta_j^2-\|v\|^2$.
$g_j(v)= \begin{cases} e^{-1/s_j(v)^2}, &\|v\|<\delta_j,\\ 0, &\|v\|\geq\delta_j. \end{cases}$
This function is positive precisely on $U(0,\delta_j)$ and its support is $B(0,\delta_j)$. Considering the two open subsets (which cover $V_j$) $U(0,\epsilon_j)$ and $V_j\setminus B(0,\delta_j)$ shows that $g_j$ is infinitely differentiable.

Boundary-derivative completion. Put $\eta(t)=e^{-1/t^2}$ for $t>0$, and $\eta(t)=0$ for $t\leq0$. On $t>0$, every derivative is $p(1/t)e^{-1/t^2}$ for a polynomial $p$, by induction using the product and chain rules. Every such expression, even after division by $t$, tends to zero as $t\downarrow0$: with $u=1/t$, the exponential-series bound $e^{u^2}\geq u^{2N}/N!$ dominates any specified polynomial by taking $2N$ larger than its degree. Inductively, the difference quotient at zero of each extended derivative is zero, and the next derivative is continuous there. Thus $\eta$ is smooth with all derivatives zero at zero. Since $g_j(v)=\eta(\delta_j^2-\|v\|^2)$, the claimed smoothness includes the boundary sphere.

Define a function $\widetilde g_j:M\to\mathbb R$ by
$\widetilde g_j(x)= \begin{cases} g_j(\alpha_j(x)), &x\in\alpha_j^{-1}(U(0,\delta_j)),\\ 0, &\text{otherwise}. \end{cases}$
This function is continuously differentiable on $M$, because the “annular strip” $B(0,\epsilon_j)\setminus U(0,\delta_j)$ allows a smooth transition.

Extension completion. The compact support $\alpha_j^{-1}(B(0,\delta_j))$ lies inside $U_j$; every point outside $U_j$ has a neighbourhood missing that support. Inside $U_j$, the boundary-derivative completion applies. Thus $\widetilde g_j$ is continuously differentiable, and is smooth when the charts are smooth.

Set
$\widetilde g(x):=\sum_{j\in J}\widetilde g_j(x).$
This is a finite sum at every point, since the support of $\widetilde g_j$ is contained in
$\alpha_j^{-1}(B(0,\delta_j))\subset U_j.$
This function is continuously differentiable on $M$ and positive everywhere, because the $\widetilde g_j(x)$ are positive on the covering sets $\alpha_j^{-1}(U(0,\delta_j))$. Then the functions
$h_j=\frac{\widetilde g_j}{\widetilde g}$
form the required partition of unity.

Sum completion. The local finiteness proved in Lemma 3.B makes $\widetilde g$ a finite sum on a neighbourhood of each point; pointwise finiteness alone would not suffice for differentiability. Its positivity makes division legitimate. The functions $h_j$ take values in $[0,1]$, have the required supports and local finiteness, and sum to one. When the charts are smooth, every step is smooth. On the empty manifold the empty family satisfies the assertion. □

Theorem 3.1 (smooth partition and cutoff). Every open cover of a smooth manifold admits a locally finite smooth partition of unity subordinate to it. Its supports can be chosen compact and contained in coordinate neighbourhoods lying in members of the cover. If $K$ is closed and $U$ is an open neighbourhood of $K$, there is a smooth function $\chi:M\to[0,1]$ equal to one on a neighbourhood of $K$, with support contained in $U$.

Proof. Apply Theorem 3.E using smooth compatible charts. Its construction gives the asserted smoothness and compact supports. For the last assertion apply it to $\{U,M\setminus K\}$, and let $\chi$ be the sum of the functions assigned to $U$. The sum is smooth by local finiteness. Its support is contained in the union of their supports, which is closed and contained in $U$. To see closedness, any point outside the union has a neighbourhood meeting only finitely many supports; intersect that neighbourhood with the complements of those finitely many closed sets. The resulting open neighbourhood misses the union. Near a point of $K$, local finiteness similarly permits a neighbourhood missing every support assigned to $M\setminus K$. There $\chi=1$, since the whole partition sums to one. □

4. Quotients by a closed Lie subgroup

An embedded Lie subgroup $H\subset G$ is a subgroup that is an embedded submanifold, with the restricted smooth group operations. We assume both embeddedness and closedness.

Theorem 4.1 (smooth homogeneous quotient). If $H$ is an embedded closed Lie subgroup of $G$, the coset space $G/H$ is a Hausdorff second-countable smooth manifold. The map $q:G\to G/H$ is a smooth submersion with smooth local sections, and $G\to G/H$ is a principal $H$-bundle.

Proof. Put $\mathfrak h=T_eH$, choose a linear complement $\mathfrak m$ in $T_eG$, and consider $F:\mathfrak m\times H\longrightarrow G, \qquad F(X,h)=\exp(X)h.$ The derivative at $(0,e)$ is $(X,Y)\mapsto X+Y$: this follows by differentiating multiplication on each factor at the identity. It is invertible. Theorem 1.2 gives neighbourhoods $W\subset\mathfrak m$, $V\subset H$ for which $F|_{W\times V}$ is a diffeomorphism onto an open subset. Shrink $W$ so that its derivative stays invertible at every $(X,e)$. Because $H$ is embedded, there is an ambient neighbourhood $N$ of $e$ with $N\cap H\subset V$. By continuity of $(X,Y)\mapsto\exp(X)^{-1}\exp(Y)$, shrink $W$ again so that every such product, for $X,Y\in W$, belongs to $N$.

If $\exp(X)h=\exp(Y)k$, then $hk^{-1}=\exp(X)^{-1}\exp(Y)\in V$. Consequently $F(X,hk^{-1})=F(Y,e)$, and local injectivity gives $X=Y$, then $h=k$. Thus $F:W\times H\to G$ is injective. At $(X,h)$, its derivative is obtained from that at $(X,e)$ by right translation in the subgroup and in the group, and is therefore invertible. It is a local diffeomorphism everywhere, so its image $O=\exp(W)H$ is open and its local smooth inverses agree. Hence $F:W\times H\to O$ is a diffeomorphism.

Give $G/H$ the quotient topology. The map $q$ is open because $q^{-1}(q(U))=UH=\bigcup_{h\in H}Uh$ is open for open $U$. The coordinate map $W\to q(O)$, $X\mapsto q(\exp X)$, is a continuous bijection. It is open: an open $W'\subset W$ has open $F(W'\times H)\subset G$, whose image under $q$ is exactly the corresponding coordinate image. We thus have a homeomorphism. Its left translates cover $G/H$.

On overlapping translated charts, a representative $b\exp X$ is sent to the first component of $F^{-1}(a^{-1}b\exp X)$. This expression is smooth wherever it is needed; membership of the same coset in the $a$-chart means precisely that $a^{-1}b\exp X\in O$. These are smooth chart transitions. In the corresponding product charts $q$ is the projection $(X,h)\mapsto X$, and $X\mapsto a\exp X$ is a smooth section.

If $gH\ne g'H$, then $g^{-1}g'\notin H$. Closedness of $H$ and continuity of $(v,v')\mapsto v^{-1}v'$ give neighbourhoods $U,U'$ of $g,g'$ with $u^{-1}u'\notin H$ for every pair. Therefore the open sets $q(U),q(U')$ are disjoint, proving Hausdorffness. The images of a countable basis of $G$ form a countable basis of $G/H$, since $q$ is open. The right action of $H$ on $G$ is smooth, free, and has exactly the fibres of $q$ as orbits. The product maps just constructed intertwine this action with $(X,h)k=(X,hk)$. These are precisely principal-bundle trivializations. □

5. Worked examples

Example 5.1 (inner products as a quotient). Let $n\geq1$, and let $\operatorname{Sym}_n$ be the real symmetric matrices. Lemma 0.4 makes $\mathrm{GL}(n,\mathbb R)$ open in the matrix space and gives its smooth inversion; multiplication is a bilinear map, smooth by Lemma 0.3. It is consequently a Lie group. The orthogonal group is an embedded closed subgroup. To prove embeddedness, set $P(g)=g^Tg$. For $g^Tg=I$, $DP_g[V]=V^Tg+g^TV.$ Every $S\in\operatorname{Sym}_n$ is the image of $V=\tfrac12gS$. Thus Corollary 1.3 makes $P^{-1}(I)$ an embedded submanifold. It is closed in $\mathrm{GL}(n,\mathbb R)$ by continuity. Multiplication and inversion restrict to it; their smoothness in submanifold coordinates follows from the local coordinate projection defining an embedded submanifold. Its tangent space at $I$ consists of skew-symmetric matrices. Every matrix is the unique sum of its symmetric and skew-symmetric parts, $(A+A^T)/2$ and $(A-A^T)/2$.

The quotient coordinate is $g\mathrm O(n)\longmapsto g^{-T}g^{-1}.$ It is unchanged by right multiplication by an orthogonal matrix. Conversely, if the expressions for $g,k$ agree, then $Q=g^{-T}g^{-1}=k^{-T}k^{-1}$ gives $(g^{-1}k)^T(g^{-1}k)=I$, so the cosets agree. Every positive-definite symmetric $Q$ occurs. Here is the complete construction. Starting with the standard basis $e_j$, define $v_j=e_j-\sum_{i<j}(s_i^TQe_j)s_i, \qquad s_j=\frac{v_j}{\sqrt{v_j^TQv_j}}.$ Inductively the earlier $s_i$ span the earlier standard basis vectors, so $v_j\ne0$; otherwise $e_j$ would lie in their span. Its $Q$-inner product with every earlier $s_i$ vanishes by the formula, and normalization gives $s_j^TQs_j=1$. Hence the matrix $S$ with columns $s_j$ is invertible and satisfies $S^TQS=I$, so $Q=S^{-T}S^{-1}$. Positive square roots exist by Lemma 0.0 applied to $t^2$ and are unique by strict monotonicity on $(0,\infty)$; they are smooth there by Theorem 1.2 applied to $t\mapsto t^2$. The construction is consequently smooth in $Q$. Positive-definite matrices form an open set: on the compact Euclidean unit sphere the positive function $v\mapsto v^TQv$ has positive minimum $m$, and a matrix perturbation of operator norm less than $m$ preserves positivity. Thus $Q\mapsto S(Q)$ supplies a smooth section on this whole cone. The map from the quotient to the cone is smooth in the local sections of Theorem 4.1, and its inverse is the smooth map $Q\mapsto q(S(Q))$. It is therefore a diffeomorphism.

Example 5.2 (a noncompact group). For the multiplicative group $\mathbb R_{>0}$, the equation in Lemma 2.2 is $g'=b(t)g$. Its solution from $g(a)=g_0>0$ is $g(t)=g_0\exp\left(\int_a^t b(s)\,ds\right).$ The fundamental theorem and chain rule in Lemma 0.3 verify both the equation and the initial value; positivity of the exponential keeps it in the group. Uniqueness comes from Lemma 2.2. Noncompactness of the group does not obstruct existence on the compact time interval.

6. Exercises and complete solutions

Exercise 6.1 (easy). Solve $x'=x^2$, $x(0)=1$, and explain why this does not contradict Lemma 2.2.

Solution. The function $x(t)=(1-t)^{-1}$ satisfies the equation and the initial value on $(-\infty,1)$, by differentiation. Theorem 2.1 gives uniqueness there. It cannot extend across $1$ as a real continuous solution, since it tends to infinity as $t\uparrow1$, so this interval is maximal. The additive group has $dR_x=I$; a right-invariant equation has spatially constant velocity, whereas $x^2$ depends on position. □

Exercise 6.2 (medium). A smooth vector bundle is locally a product $U\times\mathbb R^r$, with changes of fibre coordinates given by smoothly varying invertible linear maps. A smooth section is a map to its fibres with smooth coordinate functions. Show that a section given on a neighbourhood of a closed set extends globally with agreement on a smaller neighbourhood.

Solution. Let the original section be $s$ on $U\supset K$. Choose $\chi$ from Theorem 3.1. Define $\widetilde s=\chi s$ on $U$ and the zero vector in each fibre outside $U$. The definition is independent of local linear fibre coordinates. Inside $U$ it is smooth by the product rule. At every point outside $U$, the complement of the closed support of $\chi$ is a neighbourhood on which it is zero. Thus it is smooth everywhere and agrees with $s$ near $K$. □

Exercise 6.3 (medium). Construct the smooth local sections of $\mathbb R/\mathbb Z$ explicitly.

Solution. The integers are a closed discrete subgroup of the additive Lie group $\mathbb R$: the complement is the union of the intervals $(n,n+1)$, and $(n-1/2,n+1/2)$ meets the subgroup only at $n$. Its point charts make it an embedded zero-dimensional Lie subgroup; the group operations in those charts are constant maps between zero-dimensional spaces. Theorem 4.1 therefore gives a Hausdorff second-countable quotient. Give it the quotient topology. The quotient map is open, because the full preimage of the image of an open interval is the union of its integer translates. On an interval of length less than one it is injective, hence is a homeomorphism onto its open image. Inverse maps from such images to intervals give an atlas: a transition is $x\mapsto x+k$, where the integer $k$ is locally constant, since two continuously varying representatives have integer-valued difference. These transitions are smooth. The inverse interval maps are smooth local sections. Addition and negation are well-defined on the quotient because integer representatives differ by integers. Locally these operations are addition and negation of interval representatives followed by the quotient map, so they are smooth. □

Exercise 6.4 (hard). In $\mathbb T^2=(\mathbb R/\mathbb Z)^2$, let $H$ be the image of $t\mapsto([t],[\alpha t])$, where $\alpha$ is irrational. Prove that $H$ is a proper dense subgroup and $\mathbb T^2/H$ is not Hausdorff. Explain the relevance of the hypotheses in Theorem 4.1.

Solution. The product circle charts from Exercise 6.3 give the torus its smooth structure. It is Hausdorff because points differing in one coordinate can be separated in that coordinate, and second countable because products of two countable bases form a countable basis. Divide $[0,1)$ into $N$ intervals of length $1/N$ and place the $N+1$ fractional parts of $0,\alpha,\ldots,N\alpha$ in them. Two lie in the same interval: otherwise the $N$ intervals would contain at most $N$ of these $N+1$ points. Subtracting, and changing sign if necessary, produces integers $k,m$ with $0<k\alpha-m<1/N$; the strict lower bound follows from irrationality. Put $\delta=k\alpha-m$. For any $\beta\in[0,1)$, take $j=\lfloor\beta/\delta\rfloor$. Then $0\leq\beta-j\delta<1/N$, while $[j\delta]=[jk\alpha]$. Thus integer multiples of $\alpha$ are dense modulo one. Given first coordinate $[a]$, use $t=a+l$, $l\in\mathbb Z$, to fix it and approximate any second coordinate. Every product of open arcs therefore meets $H$, proving density.

The fibre of the first coordinate over $[0]$ meets $H$ in the countable set $\{([0],[\alpha l]):l\in\mathbb Z\}$. The circle is uncountable. The series with terms $a_j3^{-j}$, $a_j\in\{0,1\}$, converge: their tails are at most $3^{-n}/2$, by the geometric bound of Lemma 0.4, so real completeness applies. The numbers $\sum_{j\geq1}a_j3^{-j}$, $a_j\in\{0,1\}$, lie in $[0,1/2]$, and different sequences give different numbers, since the first differing term exceeds the sum of the later possible differences. The set of binary sequences cannot be listed as $b^{(1)},b^{(2)},\ldots$: the sequence $a_j=1-b^{(j)}_j$ differs from the $j$-th listed sequence in its $j$-th entry for every $j$. This proves uncountability. Hence this fibre is not exhausted by $H$, and $H$ is proper.

If an open nonempty subset of the torus is saturated under $H$, its complement is closed and also $H$-invariant. A point in that complement would bring with it its dense $H$-orbit, forcing the complement to be the whole torus. Thus the only open sets of the quotient are the empty set and the whole quotient, which has more than one point. It is not Hausdorff.

The parametrization has trivial kernel: a kernel element has $t\in\mathbb Z$ and $\alpha t\in\mathbb Z$, so irrationality forces $t=0$. Transport the smooth structure of $\mathbb R$ to its image $H$ through this bijection. In the local interval charts of Exercise 6.3, the inclusion into the torus has derivative $(1,\alpha)$, which is injective. Thus it is an injective immersion and defines an immersed Lie subgroup. It is not embedded. An embedded submanifold is locally the zero set of its transverse coordinates, so it is locally closed near the identity. If this dense proper subgroup were locally closed, its intersection with a sufficiently small open neighbourhood would be both dense and closed there, hence equal to the neighbourhood. It would then be an open subgroup and therefore also closed (its other cosets are open), contrary to density and properness. Both the closedness and embeddedness assumptions of Theorem 4.1 have therefore been checked against this example. □

References

[Lebl] Jiří Lebl, Basic Analysis: Introduction to Real Analysis, volumes I and II, free version 6.3, 15 May 2026. Author's free edition. Section 0 supplies the complete analytic and linear arguments used here. The text is available under CC BY-SA 4.0 and, alternatively, CC BY-NC-SA 4.0.

[Teschl] Gerald Teschl, Ordinary Differential Equations and Dynamical Systems, free author preliminary version, April 2012, Sections 2.1–2.2, 2.4 and 2.6. Used for mathematical source comparison; no source prose, images or PDF is reproduced. The fixed-point parameter argument and every group-equation argument used here are proved in this lesson.

[Mrowka] Tomasz Mrowka, Geometry of Manifolds, MIT 18.965, Fall 2004, free Lecture 4, Theorems 5.1–5.2. The finite-dimensional proofs here use only the explicitly proved linear and analytic steps; no Banach open-mapping theorem is invoked.

[Brenner] Holger Brenner and the Wikiversity contributors, Differentialgeometrie (Osnabrück 2023), Lecture 22, revision 1052940, page histories. English translation and adaptation: Smooth Manifolds and Differential Geometry, complete edition v2026.09.01-complete. The attributed Section 3 and its marked completions retain CC BY-SA 4.0.

[Michor] Peter W. Michor, Topics in Differential Geometry, freely available author version, Section 5.11. The quotient construction is proved in Section 4 with its embeddedness hypothesis explicit; no closed-subgroup theorem is assumed.