Differentiable Manifolds: Charts, Constructions, and Examples

1. Definition of Differentiable Manifolds

Calculus on a curved space begins with a local observation: a small piece of a sphere can be described by ordinary Euclidean coordinates. Different descriptions overlap, so we must also know how to change coordinates. A smooth manifold packages these descriptions together with the compatibility needed for differentiation.

We first make this compatibility precise, then examine maps, orientations, and quotient constructions. The examples test the definitions in several ways: spheres admit explicit charts, projective spaces identify points, and Grassmannians use linear maps as coordinates.

Definition 1. A topological manifold of dimension $d$ is a Hausdorff, second-countable space $M$ such that every $p\in M$ has an open neighbourhood $U$ homeomorphic to an open subset $U^{\prime}\subset\mathbb R^d$. We write the homeomorphism as $\varphi:U\to U^{\prime}$. No metric is assumed in this definition.

Definition 2. The neighbourhood $U$, together with homeomorphism $\varphi$, as defined in Definition 1, is called coordinate chart. The pullback of the coordinate functions from $\mathbb R^d$ is called local coordinates.

In other words, Definition 2 is saying, for all $p \in M$, there exists some neighbourhood $U \subset M$ and homeomorphism $h: U \to U' \subset \mathbb R^d$.

Definition 3. An atlas, denoted $\mathcal A$, is a collection of coordinate charts

\[ \mathcal A = \left\{ (U_\alpha, \varphi_\alpha) \right\}_{\alpha \in A}\tag{1}\]

where $\{U_\alpha\}_{\alpha \in A}$ form an open covering of $M$. $A$ is some set of indices.

Definition 4. An atlas $\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in A}$ is locally finite if every $p\in M$ has a neighbourhood that meets only finitely many chart domains $U_\alpha$. Requiring only that each point belong to finitely many domains is the weaker condition of point-finiteness.

Definition 5. Let $\mathcal A$ be a coordinate atlas for a topological manifold $M$. Assume that $(U_i,\varphi_i)$, $(U_j,\varphi_j)$ are two elements of $\mathcal A$. The map

\[\begin{aligned} \varphi_j \circ \varphi_i^{-1} : \varphi_i (U_i \cap U_j) \to \varphi_j (U_i \cap U_j) \end{aligned}\tag{2}\]

is a homemorphism between two open sets of $\mathbb R^d$. (2) is called the coordinate transition function of the charts.

Figure Differentiable Manifolds: Charts, Constructions, and Examples shows the coordinate transition function. To compare two descriptions of the same point, first return from coordinates to the manifold, then apply the second chart.

Coordinate transition between two overlapping charts. Both coordinate domains lie in Euclidean space of the manifold dimension.

Definition 6. An atlas $\mathcal A$ on $M$ is differentiable if all coordinate chart transition functions are differentiable.

Definition 7. The coordinate atlases $U = \{U_\alpha,\varphi_\alpha\}$, $V = \{V_\beta,\varphi_\beta\}$ are equivalent if

\[ \begin{aligned} \varphi_\beta \circ \varphi_\alpha^{-1} &: \varphi_\alpha (U_\alpha \cap V_\beta) \to \varphi_\beta (U_\alpha \cap V_\beta) \\ \varphi_\alpha \circ \varphi_\beta^{-1} &: \varphi_\beta (U_\alpha \cap V_\beta) \to \varphi_\alpha (U_\alpha \cap V_\beta) \end{aligned}\tag{3}\]

are diffeomorphisms.

Definition 8. A differentiable structure (or smooth structure) on a manifold $M$ is defined by an equivalence class of coordinate atlas, such that any coordinate transition function is a diffeomorphism.

Remark 9 (Maximal smooth atlas). Starting with a smooth atlas $\mathcal A$, collect every chart smoothly compatible with every chart of $\mathcal A$. This collection $\mathcal A_{\max}$ is itself a smooth atlas: near a point in the overlap of two new charts, insert an original chart and express their transition as a composition of two smooth transitions. It is maximal because any further compatible chart already belongs to the collection. Thus every smooth atlas determines a unique maximal smooth atlas. Maximality rules out a proper compatible enlargement; it does not rule out equality, and a compatible atlas $\mathcal B$ satisfies $\mathcal B\subseteq\mathcal A_{\max}$, not necessarily a strict inclusion.

Definition 10. A differentiable manifold (or smooth manifold) is a manifold $M$ equipped with differentiable structure.

1.1. Why Differentiable Manifolds?

On a smooth manifold $M$, we can do calculus for all functions $f: M \to \mathbb R$ that are smooth/differentiable. We speak about the class of functions $\mathscr C^{\infty}$ defined on $M$, and denote

\[\mathscr C^{\infty} (M) = \mathscr C^{\infty} (M,\mathbb R) = \left\{ f : M \to \mathbb R ~,~ \text{where $f$ is a smooth function} \right\} ~.\]

There is usually no ambiguity in the use of $\mathscr C^{\infty} (M)$, because the context generally lends itself to using a real-valued function defined on a differentiable manifold $M$.

Remark 11. For a chart $\varphi:U\to\varphi(U)\subset\mathbb R^d$, the coordinate expression of $f:M\to\mathbb R$ is $f\circ\varphi^{-1}:\varphi(U)\to\mathbb R$. We call $f$ smooth when all these expressions are $C^\infty$. If $\psi$ is a compatible chart, then on an overlap

\[f\circ\psi^{-1}=(f\circ\varphi^{-1})\circ(\varphi\circ\psi^{-1}).\]

The transition on the right is smooth. Smoothness is local, so this identity proves that equivalent atlases define the same class $C^\infty(M)$. Throughout these notes, “differentiable” in an atlas means smooth, rather than merely once differentiable.

Definition 12. An atlas of a differentiable manifold is oriented if all coordinate transition functions have positive functional determinants.

Remark 13. We denote the property of a coordinate transition function having positive functional determinant as $(\star)$.

Definition 14. A smooth manifold $M$ is said to be orientable if there exists an oriented atlas, as defined in Definition 12.

Definition 15. On a fixed smooth manifold of positive dimension, an orientation is an equivalence class of oriented atlases compatible with its smooth structure. Two oriented atlases are equivalent when every transition between them has positive Jacobian determinant. An oriented manifold is a manifold together with this choice.

Remark 16. Two oriented atlases represent the same orientation precisely when their union is an oriented atlas. An orientation is additional data on a smooth structure, not a different smooth structure.

Lemma 17. A connected orientable smooth manifold of positive dimension has exactly two orientations. The same assertion holds for a manifold with boundary, with the usual half-space charts.

Proof. Fix an oriented atlas. At any point, compare another oriented atlas with it by the sign of a coordinate-transition determinant. Changing either chart multiplies the determinant by positive factors, so this sign is well-defined. It is locally constant because a nonzero continuous determinant cannot change sign on a sufficiently small connected neighbourhood. Connectedness therefore makes the sign constant on all of $M$.

The positive choice gives the original orientation. Composing every original chart with a fixed reflection of $\mathbb R^d$ gives the negative choice: transitions are conjugated by reflections and still have positive determinant. These choices differ at every point, and the sign argument excludes a third orientation. For boundary charts, reflect a coordinate tangent to the boundary when possible; equivalently compare orientations of tangent frames, which also covers dimension one. ◻

Definition 18. A map $h:M\to N$ of smooth manifolds is smooth if it is continuous and, for every pair of charts $\varphi:U\to\mathbb R^d$, $\psi:V\to\mathbb R^e$, its coordinate expression

\[\psi\circ h\circ\varphi^{-1}:\varphi(U\cap h^{-1}(V))\longrightarrow\psi(V)\]

is smooth. Continuity ensures that the displayed domain is open. It suffices to check this in any pair of atlases, since other coordinate expressions are obtained by composing with smooth transition maps. Local finiteness is not required for this definition.

Remark 19. Definition 18 implies that smooth maps between differentiable manifolds depend on the smooth structures of $M$ and $N$, but not on the chosen representative coordinate atlases. This is because of the definition of equivalence relation in Definition 8.

Definition 20. Two differentiable manifolds are said to be diffeomorphic if there exists a smooth homeomorphism such that its inverse is smooth.

Lemma 21. Let $M$, $N$ be differentiable manifolds. Let $f : M \to N$ be a diffeomorphism. Then,

$M$ is orientable if and only if $N$ is orientable.
If $M$ and $N$ are orientable and connected, then $f$ induces either preserves or reverses orientation on $N$.

Proof. If $\{(U_\alpha,\varphi_\alpha)\}$ is an oriented atlas on $M$, transport it to $N$ as

\[\{(f(U_\alpha),\varphi_\alpha\circ f^{-1})\}.\]

The transition between two transported charts is exactly $\varphi_\beta\circ\varphi_\alpha^{-1}$, so its determinant is positive. This gives an orientation of $N$; applying the same construction to $f^{-1}$ proves the converse.

Now fix orientations on both connected manifolds. In oriented charts, $\det D(\psi\circ f\circ\varphi^{-1})$ never vanishes, since the derivative of the inverse is its inverse matrix. Its sign is independent of the oriented charts and is locally constant. The sets where this sign is positive and negative are disjoint open sets whose union is $M$; connectedness forces one to be empty. Thus $f$ either preserves the chosen orientations everywhere or reverses them everywhere. ◻

The definitions become more concrete when we construct charts. For an embedded surface, a parametrization supplies local coordinates; for a quotient, we must also check that the identifications preserve a local Euclidean description. The following examples illustrate both constructions.

Definition 22. $M^k \subset \mathbb R^d$ is a regular surface on $\mathbb R^d$ if, for all $p \in M^k$, there exists a neighbourhood $V$ of $p$ in $\mathbb R^d$ such that there is a map $\mathrm{x} : U \subset \mathbb R^k \to M \cap V$, where $U$ is an open set, where

$\mathrm{x}$ is a differentiable homeomorphism.
$(\dif x)_q : \mathbb R^k \to \mathbb R^d$ is injective for all $q \in U$.

Example 23 ($n$-sphere). Write $\mathbb S^n=\{x\in\mathbb R^{n+1}:|x|=1\}$. For $1\leq i\leq n+1$, put $U_i^\pm=\{x\in\mathbb S^n:\pm x_i>0\}$. Deleting coordinate $i$ defines a chart $h_i^\pm:U_i^\pm\to B^n(0,1)$. Its inverse inserts

\[x_i=\pm\sqrt{1-|u|^2}\]

into the vector $u\in B^n(0,1)$. Both directions are continuous, and the inverse is smooth because $1-|u|^2>0$.

For the parametrization $x_i^\pm=(h_i^\pm)^{-1}$, deleting coordinate $i$ from $D x_i^\pm(u)v$ gives $v$. Therefore $D x_i^\pm(u)v=0$ implies $v=0$, proving the required injectivity of the differential. Transition maps are compositions of coordinate deletion and these smooth inverse maps. The hemispheres cover the sphere, and the subspace topology is Hausdorff and second-countable. This verifies all manifold and regular-surface conditions.

Example 24 (Example of manifolds). $\mathbb R^d$ is a prototypical example of a manifold.

The general linear group $\operatorname{GL}_n (\mathbb R) = \left\{ A \in \mathcal M_n (\mathbb R) : \det A \neq 0 \right\}$ is a manifold.

The real projective space $\mathbb RP^n=\mathbb S^n/(x\sim -x)$ has the quotient topology. Its projection $\Pi(x)=[x]$ is two-to-one, so it is not a homeomorphism. For each $i$, let $U_i^+=\{x\in\mathbb S^n:x_i>0\}$ and $V_i=\{[x]:x_i\ne0\}$. Every point of $V_i$ has exactly one representative in $U_i^+$.

The projection is open, since $\Pi^{-1}(\Pi(O))=O\cup(-O)$ for every open $O\subset\mathbb S^n$. Thus $\Pi|_{U_i^+}:U_i^+\to V_i$ is a homeomorphism. Coordinates on $V_i$ are given by ratios:

\[\psi_i([x])=(x_1/x_i,\ldots,\widehat{x_i/x_i},\ldots,x_{n+1}/x_i)\in\mathbb R^n.\]

Its inverse inserts $1$ in the $i$th position, normalizes the resulting vector to unit length, and takes its equivalence class. On an overlap, the coordinate change divides each homogeneous coordinate by the nonzero coordinate chosen for the other chart. These rational functions are smooth on their domains, so $\{(V_i,\psi_i)\}_{i=1}^{n+1}$ is a smooth atlas.

For completeness, the quotient is Hausdorff: two distinct antipodal pairs on the sphere are disjoint finite sets at positive Euclidean distance, and sufficiently small antipodally invariant neighbourhoods of the pairs are disjoint. Their images are disjoint open quotient neighbourhoods. Images of a countable basis of the sphere form a countable basis of the quotient because $\Pi$ is open. This verifies the topological requirements as well as chart compatibility.

The sphere charts in Example 23 give the equivalent local-coordinate construction.

Example 25 (Free properly discontinuous actions). Let a discrete group $G$ act on a smooth manifold $M$ by diffeomorphisms, with $ep=p$ and $(gh)p=g(hp)$. Assume the action is free and proper: for every compact $K\subset M$, only finitely many $g\in G$ satisfy $gK\cap K\ne\varnothing$. Define $p\sim q$ when $q=gp$ for some $g\in G$, and give $M/G$ the quotient topology. Then the quotient map $\Pi:M\to M/G$ is a smooth covering map and a local diffeomorphism.

First, each $p$ has an open neighbourhood $U$ with $gU\cap U=\varnothing$ for all $g\ne e$. Choose a relatively compact neighbourhood of $p$; properness leaves only finitely many possible intersecting translates. Freeness gives $gp\ne p$ for each nonidentity element in this finite set, so shrink the neighbourhood to separate each such pair. The remaining group elements were already excluded by the original compact neighbourhood.

The map $\Pi$ is open, since $\Pi^{-1}(\Pi(O))=\bigcup_g gO$ is open for open $O$. If $p,q\in U$ have the same image, then $q=gp$; disjointness of translates forces $g=e$, hence $p=q$. Thus $\Pi|_U$ is a homeomorphism onto the open set $\Pi(U)$, and $\Pi^{-1}(\Pi(U))$ is the disjoint union of the translates $gU$.

To check Hausdorffness, take points $p,q$ in distinct orbits and compact neighbourhoods $K_p,K_q$. Only finitely many translates of $K_p$ meet $K_q$, by applying properness to their union. Since no such translate sends $p$ to $q$, shrink neighbourhoods $U\ni p$, $V\ni q$ so that $gU\cap V=\varnothing$ for all those finitely many elements. Then $\Pi(U)$ and $\Pi(V)$ are disjoint open neighbourhoods. Images of a countable basis of $M$ form a countable basis of the quotient.

Finally, transport charts through $\Pi|_U$. If two quotient charts overlap, their local lifts differ near any chosen point by one fixed group element $g$: shrink the second lift so that its image under $g$ lies in the first lift. Their transition is then $\varphi_1\circ g\circ\varphi_2^{-1}$, which is smooth. This proves the assertion. The weaker pointwise disjoint-neighbourhood condition alone does not establish Hausdorffness; the properness assumption supplies that missing step.

Example 26 (Orientability of $\mathbb RP^n$). For $n\geq1$, the space $\mathbb RP^n$ is orientable exactly when $n$ is odd. The exceptional space $\mathbb RP^0$ is a point and is orientable.

For a free properly discontinuous action, an orientation downstairs pulls back through the local diffeomorphism $\Pi$ to an orientation upstairs preserved by every element of $G$. Conversely, an invariant orientation upstairs defines an orientation downstairs: choose a lift of a tangent frame, declare its sign there, and observe that changing the lift applies an orientation-preserving group element. Quotient charts therefore agree in sign. Only the acting group elements need preserve the orientation, not all diffeomorphisms of $M$.

Orient $\mathbb S^n$ by its outward normal: a tangent frame $(v_1,\ldots,v_n)$ at $x$ is positive exactly when $(x,v_1,\ldots,v_n)$ is positive in $\mathbb R^{n+1}$. The antipodal map sends this ambient frame to $(-x,-v_1,\ldots,-v_n)$, multiplying its determinant by $(-1)^{n+1}$. It preserves sphere orientation exactly when $n$ is odd. For $n\geq1$, the sphere is connected and has only two orientations, and this sign is the same for either choice. Applying the quotient criterion to $G=\{1,-1\}$ proves both directions.

Example 27 (Grassmannian). For $n,m \in \mathbb N$, let $G(n,m)$ denote the collection of $n$-dimensional subspaces of $\mathbb R^{n+m}$. We will construct charts to show that $G(n,m)$ is an $nm$-dimensional manifold. The main observation is that a subspace transverse to a fixed complementary subspace is the graph of a unique linear map. The entries of that map will be our coordinates.

To start, endow $G(n,m)$ with a standard topology, such as the quotient topology from the Stiefel manifold of $n$-frames (or the metric topology induced by identifying subspaces with their orthogonal projection matrices in $\mathbb R^{(n+m) \times (n+m)}$). Pick $\Pi \in G(n,m)$. Denote by $\mathscr L(\Pi,\Pi^{\perp})$ the collection of linear maps between $\Pi$ and its orthogonal complement.

Construct a map from the space of linear maps $\Pi \to \Pi^{\perp}$ to the space of Grassmannians. For every $\Pi$, define

\[\begin{aligned} \varphi_{\Pi} : \mathscr L (\Pi,\Pi^{\perp}) \to G(n,m) : \alpha \mapsto \left( \mathbb 1_{\Pi} \oplus \alpha \right) (\Pi) ~. \end{aligned}\]

We understand $\left( \mathbb 1_{\Pi} \oplus \alpha \right)$ as a map $\Pi \to \Pi \oplus \Pi^{\perp} = \mathbb R^{n+m}$, by definition. In other words, the image of $\alpha$ under $\varphi_\Pi$ is its graph, i.e. $\{y + \alpha(y) : y \in \Pi\}$.

Clearly, $\varphi_\Pi$ is injective. Since $\dim \mathscr L(\Pi,\Pi^\perp) = nm$, the image $U_\Pi = \varphi_\Pi (\mathscr L(\Pi,\Pi^{\perp}))$ is an $nm$-dimensional coordinate neighborhood, and the collection of open sets $\{U_\Pi\}_{\Pi \in G(n,m)}$ covers $G(n,m)$.
It remains to check compatibility on overlaps. Once this is done, the dimension of the coordinate space gives $\dim G(n,m) = \dim \mathscr L(\Pi,\Pi^\perp) = nm$. Take any $\Pi,\Pi' \in G(n,m)$, and let $P'$ denote the orthogonal projection from $\mathbb R^{n+m}$ onto $\Pi'$. The idea is to study the transition maps between the charts $\varphi_\Pi$ and $\varphi_{\Pi'}$.

To do so, let $\alpha \in \mathscr L (\Pi, \Pi^{\perp})$ and $\beta \in \mathscr L (\Pi', (\Pi')^{\perp})$ be representations of the same subspace, such that $\varphi_\Pi(\alpha) = \varphi_{\Pi'}(\beta)$.

Claim 1: $F = \varphi_{\Pi'}^{-1} \circ \varphi_{\Pi}$ is smooth. This amounts to checking that $\beta$ depends smoothly on $\alpha$. Note that $F$ is a map between open subsets of spaces of linear maps. It is more convenient to write this down in terms of $\Pi'$ and $(\Pi')^{\perp}$. Let the following block matrix represent the identity map on $\mathbb R^{n+m}$ with respect to the direct sum decompositions $\mathbb R^{n+m} = \Pi \oplus \Pi^{\perp}$ and $\mathbb R^{n+m} = \Pi' \oplus (\Pi')^{\perp}$:

\[\begin{aligned} \begin{pmatrix} a & b \\ c & d \end{pmatrix} : \Pi \oplus \Pi^{\perp} \to \Pi' \oplus (\Pi')^{\perp} ~. \end{aligned}\]

It follows that the following maps

\[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \mathbb 1_\Pi \\ \alpha \end{pmatrix} ~~\text{and}~~ \begin{pmatrix} \mathbb 1_{\Pi'} \\ \beta \end{pmatrix}\]

have the same image in $\mathbb R^{n+m}$. As such, there exists some isomorphism $G: \Pi \to \Pi'$ such that

\[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \mathbb 1_\Pi \\ \alpha \end{pmatrix} = \begin{pmatrix} \mathbb 1_{\Pi'} \\ \beta \end{pmatrix} G ~.\]

Collecting equations, we have

\[\begin{aligned} a + b \alpha &= G \\ c + d \alpha &= \beta G \end{aligned}\]

which gives us the expression $\beta = (c + d \alpha)(a + b \alpha)^{-1}$. This expression shows $\beta$ depends on $\alpha$ smoothly, by Cramer’s rule for the inverse matrix. $\diamondsuit$

Claim 2: The map $f_\alpha : \Pi \to \Pi'$ defined by $f_\alpha(y) = P'(y + \alpha(y))$ is invertible. This justifies the inverse matrix in the transition formula. We use the following elementary fact: if a subspace $E \in G(n,m)$ satisfies $E \cap (\Pi')^\perp = \{0\}$, then the restriction of $P'$ to $E$ is an isomorphism onto $\Pi'$. Since $\dim E = n = \dim \Pi'$, we only need to show injectivity. If $e \in E$ and $P'(e) = 0$, then $e \in (\Pi')^\perp$. But $E \cap (\Pi')^\perp = \{0\}$, so $e = 0$. Thus $P'|_E : E \to \Pi'$ is an isomorphism.

Now we can argue $f_\alpha$ is invertible for any $\alpha$ in the domain of $F$. Let $E = \varphi_\Pi(\alpha)$. Since $\alpha$ is in the domain of $F$, $E$ is in the image of $\varphi_{\Pi'}$, which precisely means $E \cap (\Pi')^\perp = \{0\}$. By the property shown above, $P'|_E : E \to \Pi'$ is an isomorphism. Since $\mathbb 1_\Pi \oplus \alpha : \Pi \to E$ is also an isomorphism, their composition $f_\alpha = P' \circ (\mathbb 1_\Pi \oplus \alpha)$ is therefore invertible. Notice that the $\Pi'$ component of $y + \alpha(y)$ is exactly $a(y) + b(\alpha(y))$, so $f_\alpha = a + b\alpha = G$, which explicitly justifies the invertibility of $a + b\alpha$ in Claim 1. $\diamondsuit$
We can now show that for all $y \in \Pi$, $y + \alpha(y) = f_\alpha (y) + \beta(f_\alpha (y))$. By definition of $\varphi_{\Pi'}$, any element $e \in E = \varphi_\Pi(\alpha) = \varphi_{\Pi'}(\beta)$ can be written uniquely as $y' + \beta(y')$ for some $y' \in \Pi'$. For $e = y + \alpha(y)$, applying $P'$ gives $P'(e) = P'(y' + \beta(y')) = y'$ since $\beta(y') \in (\Pi')^\perp$. But by definition, $P'(e) = f_\alpha(y)$. Thus $y' = f_\alpha(y)$. Substituting this back into the decomposition of $e$, we get $y + \alpha(y) = f_\alpha(y) + \beta(f_\alpha(y))$. $\diamondsuit$

The claim directly above implies that

\[\begin{aligned} (\mathbb 1_{\Pi'} \oplus \beta) \circ f_\alpha = \mathbb 1_\Pi \oplus \alpha ~, \end{aligned}\]

i.e. $\varphi_{\Pi'}(\beta) = \varphi_\Pi(\alpha)$, or, equivalently, that $F(\alpha) = \beta$. As such, we have an exact description of the smooth transition maps and their domains.

The graph charts therefore form the atlas

\[\left\{ (U_\Pi, \varphi_\Pi^{-1}) \right\}_{\Pi \in G(n,m)} ~,\]

but, in fact, we only need $\binom{n+m}{n}$ of such charts to cover $G(n,m)$ (for instance, by choosing $\Pi$ to be the coordinate subspaces of $\mathbb R^{n+m}$ spanned by subsets of the standard basis vectors).

For $n,m\geq1$, the Grassmannian $G(n,m)$ is orientable exactly when $n+m$ is even. Here is the sign calculation. At a plane $E$, its tangent space is $\operatorname{Hom}(E,E^\perp)$, as the graph chart shows. Changing orthonormal bases of $E$ and $E^\perp$ by matrices $A\in O(n)$, $B\in O(m)$ changes matrix coordinates by $H\mapsto B^{-1}HA$. The determinant of this linear map has sign $(\det A)^m(\det B)^n$. If the two bases together have the fixed ambient orientation, then $\det A\det B=1$; the sign is therefore $(\det A)^{n+m}$. When $n+m$ is even this gives a well-defined continuous tangent orientation.

When $n+m$ is odd, take $E_0=\operatorname{span}(e_1,\ldots,e_n)$ and the closed path of planes spanned by $\cos t\,e_1+\sin t\,e_{n+1},e_2,\ldots,e_n$, for $0\leq t\leq\pi$. Complete it by the rotating normal vector $-\sin t\,e_1+\cos t\,e_{n+1}$ and the other fixed normal vectors. At the endpoint, both basis-change determinants are $-1$; the tangent-frame determinant is $(-1)^{n+m}=-1$. An orientation cannot return with opposite sign along a closed path. This proves nonorientability. If either $n$ or $m$ is zero, the Grassmannian is a point instead.

For the topology omitted in the chart construction, write $L_\alpha=\binom{I}{\alpha}$. Orthogonal projection onto its graph is

\[P_{\operatorname{graph}\alpha}=L_\alpha(I+\alpha^*\alpha)^{-1}L_\alpha^*.\]

This is continuous in $\alpha$. In the block decomposition $\Pi\oplus\Pi^\perp$, its upper-left block is $(I+\alpha^*\alpha)^{-1}$ and lower-left block is $\alpha(I+\alpha^*\alpha)^{-1}$, so the inverse chart is the lower-left block times the inverse of the upper-left block. Invertibility is an open condition. Conversely, an invertible upper-left block implies that projection of the plane onto $\Pi$ is surjective and hence an isomorphism. Thus $U_\Pi$ is open and the graph parametrization is a homeomorphism. The projection-matrix topology is Hausdorff and second-countable as a subspace of a finite-dimensional matrix space. Finally, a full-rank matrix representing any plane has a nonzero $n$-row minor; the associated coordinate plane supplies one of the stated finite chart cover.

Example 28 (Stereographic projections). Let $N=(0,\ldots,0,1)$ and $S=-N$. The line $N+t(x-N)$ through $N$ and $x\ne N$ meets the hyperplane $x_{n+1}=0$ when $t=(1-x_{n+1})^{-1}$. This gives

\[\pi_N(x)=\frac{(x_1,\ldots,x_n)}{1-x_{n+1}},\qquad \pi_S(x)=\frac{(x_1,\ldots,x_n)}{1+x_{n+1}}.\]

The line lies in ambient Euclidean space, and its intersection with the hyperplane need not lie on the sphere. Solving for the inverse gives

\[\pi_N^{-1}(u)=\left(\frac{2u}{1+|u|^2},\frac{|u|^2-1}{1+|u|^2}\right),\qquad \pi_S^{-1}(u)=\left(\frac{2u}{1+|u|^2},\frac{1-|u|^2}{1+|u|^2}\right).\]

These formulas prove bijectivity and smoothness in both directions. On the overlap $u\ne0$, the transition is $u\mapsto u/|u|^2$; its derivative has one radial eigenvalue $-|u|^{-2}$ and $n-1$ tangential eigenvalues $|u|^{-2}$. Its determinant is $-|u|^{-2n}$. Composing one chart with a reflection makes the transition determinant positive, explicitly proving orientability, also for $n=1$ when the overlap is disconnected.

The antipodal map $A(x)=-x$ is a diffeomorphism because $A^2=\operatorname{id}$. Its orientation sign is $(-1)^{n+1}$, by the outward-normal calculation in Example 26.

Example 29 (Complex manifolds). A complex manifold of complex dimension $d$ has charts in $\mathbb C^d$ whose transition maps are holomorphic with holomorphic inverses. For a transition $F$, holomorphicity means $\partial F^j/\partial\overline z^k=0$; it does not mean $\partial F^j/\partial z^k=0$.

Write its complex derivative as $A=B+iC$. Its real derivative, in real and imaginary coordinates, is

\[\begin{pmatrix}B&-C\\ C&B\end{pmatrix}.\]

After complexifying this real matrix and using the variables $z=x+iy$, $\overline z=x-iy$, it becomes block diagonal with blocks $A$ and $\overline A$. Consequently its real determinant is $\det_{\mathbb C}A\,\overline{\det_{\mathbb C}A}=|\det_{\mathbb C}A|^2>0$; nonvanishing follows from the holomorphic inverse. Thus complex coordinate charts determine an orientation of the underlying real manifold.

Example 30 ($\mathbb CP^n$). The complex projective space $\mathbb CP^n$ is the set

\[\begin{aligned} \mathbb CP^n \triangleq \{ \ell \subset \mathbb C^{n+1} : \ell \text{ is a $1$-dimensional complex subspace} \} ~. \end{aligned}\]

Let $\mathbb C^*$ denote the multiplicative group of nonzero complex numbers, i.e. $\mathbb C^* = \mathbb C \setminus \{0\}$ is endowed with the multiplication operation. In this notation, $\mathbb CP^n \cong (\mathbb C^{n+1} \setminus \{0\}) / \mathbb C^*$. The equivalence class of $z \in \mathbb C^{n+1} \setminus \{0\}$ is given by

\[\begin{aligned} [z] \triangleq \{\lambda z : \lambda \in \mathbb C^*\} ~, \end{aligned}\]

so the associated line is given by $\ell = \mathbb C z$. It follows from inspection that an atlas on $\mathbb CP^n$ is given by the open cover $U_i \triangleq \{[z] = [z_0 : \cdots : z_n] : z_i \neq 0\}$ for all $i \in \{0,1,\cdots,n\}$. The coordinate charts are $\varphi_i : U_i \to \mathbb C^n$ given by

\[\begin{aligned} \varphi_i ([z]) \triangleq \left( \frac{z_0}{z_i} , \cdots, \frac{z_{i-1}}{z_i}, \frac{z_{i+1}}{z_i}, \cdots, \frac{z_n}{z_i} \right) ~. \end{aligned}\]

Each $\varphi_i$ is a homeomorphism for the quotient topology. The ratio map on $\{z_i\ne0\}$ is continuous and constant on complex lines, so it descends continuously through the quotient; the quotient projection is open because the saturation of an open set is the union of its nonzero scalar multiples. The inverse inserts $1$ in coordinate $i$ and then takes the equivalence class, hence is continuous. On an overlap with chart $j$, the denominator $z_i$ is nonzero, and the transition formula is

\[\begin{gathered} \left( z_0, \cdots, z_{j-1}, z_{j+1}, \cdots, z_n \right) \\ \overset{\varphi_j^{-1}}{\mapsto} [z_0 : \cdots : z_{j-1} : 1 : z_{j+1} : \cdots : z_n] \overset{\varphi_i}{\mapsto} \left( \frac{z_0}{z_i} , \cdots, \frac{z_{i-1}}{z_i}, \frac{z_{i+1}}{z_i}, \cdots, \frac{z_{j-1}}{z_i}, \frac{1}{z_i}, \frac{z_{j+1}}{z_i}, \cdots, \frac{z_n}{z_i} \right) \end{gathered}\]

The displayed ratios are holomorphic on the overlap $z_i\ne0$, and exchanging $i$ and $j$ supplies the holomorphic inverse. Holomorphicity follows from this formula, not from being a smooth diffeomorphism. These charts therefore give a complex manifold once the topological separation requirement is checked. One direct check sends $[z]$ to the Hermitian projection matrix $zz^*/|z|^2$: this is a continuous injective map from the quotient into a Hausdorff matrix space, so distinct points have disjoint open neighbourhoods. The finite chart cover gives second countability. The longer quotient-topology discussion appears in the note Why Complex Projective Space Is Hausdorff.

Example 31 (Lie Group). A Lie group combines a smooth manifold with a group operation compatible with its smooth structure. We return to this compatibility in Section Lie Groups and Their Lie Algebras. Here are the definition and some examples.

A Lie group $G$ is a smooth manifold equipped with a group structure so that the group multiplication

\[ m : G \times G \to G : (g_1,g_2) \mapsto g_1 g_2\tag{4}\]

is a smooth map. Here are some basic examples of Lie groups:

$\mathbb R^n$ with addition operation.
$\mathbb R^* = \mathbb R \setminus \{0\}$ with multiplication operation.
$\mathbb S^1$ with multiplication operation.
$\operatorname{GL}_n(\mathbb R)$ with matrix multiplication. So are $\operatorname{SL}_n(\mathbb R)$, $\operatorname{SO}(n)$, $\operatorname{O}(n)$, and so on, under matrix multiplication.

The existence of a smooth structure does not by itself supply a compatible group operation. For example, the only spheres that admit a Lie group structure are $\mathbb S^0$, $\mathbb S^1$, and $\mathbb S^3$; among all the compact connected 2-dimensional surfaces, the only one admits a Lie group structure is $\mathbb T^2 = \mathbb S^1 \times \mathbb S^1$. There are many constraints for a manifold to be a Lie group. For example, a Lie group must be analytic manifold, and the tangent bundle of a Lie group is always trivial: $TG \cong G \times \mathbb R^n$. In particular, any Lie group is orientable.

The usual definition also asks that inversion $g\mapsto g^{-1}$ be smooth. It follows here from smooth multiplication and the inverse function theorem: the derivative in $h$ of $gh$ is the invertible derivative of a left translation, so the equation $gh=e$ can locally be solved smoothly for $h$. These local solutions agree with the group inverse.

To see why the tangent bundle is trivial, choose a basis $v_1,\ldots,v_d$ of $T_eG$ and set $X_i(g)=(dL_g)_e v_i$. Multiplication is smooth, so these are smooth vector fields. Each $L_g$ is a diffeomorphism, so they form a basis at every point. The map $(g,\sum a_iv_i)\mapsto\sum a_iX_i(g)$ is a smooth vector-bundle isomorphism, with inverse induced by $dL_{g^{-1}}$. Declaring this frame positive gives an orientation on every component. The classification statements about spheres and compact connected surfaces above are background results, rather than consequences of this frame construction.

For the background assertions, Merry, Differential Geometry (2019), Remark 12.14, gives the sphere restriction: a compact connected abelian Lie group is a torus, while a compact connected nonabelian Lie group has nonzero third real cohomology. The cohomology of a sphere therefore leaves only $S^1$ or $S^3$ among connected positive-dimensional spheres; $S^0=\{1,-1\}$, complex unit numbers, and unit quaternions give the three examples. These cohomology facts are external inputs to the classification.

For the surface claim, average an inner product on the two-dimensional Lie algebra over the adjoint action of the compact group. The result is invariant, so every $\operatorname{ad}_X$ is skew-adjoint. Hence $[X,Y]$ is perpendicular to both $X$ and $Y$; for a basis $X,Y$ this forces $[X,Y]=0$. The connected group is therefore abelian. Its exponential map is a local diffeomorphism and a homomorphism from $\mathbb R^2$, so its image is an open subgroup and hence the whole connected group. Its discrete kernel is a full lattice because the quotient is compact. Thus the group is a two-torus. Compactness and connectedness are essential to this statement.

The compatible real-analytic structure is a separate regularity theorem; see Kirillov, Introduction to Lie Groups and Lie Algebras, Remark 2.2. It means that suitable analytic charts exist, not that every chart of the original smooth atlas is analytic.

Carmo, Manfredo Perdigão do. 1992. Riemannian Geometry. Mathematics. Theory & Applications. Boston: Birkhäuser.

Hatcher, Allen. 2002. Algebraic Topology. Cambridge ; New York: Cambridge University Press.

All errors are mine. The main reference for this collection of notes is (Carmo 1992). This collection of notes describes basic concepts in the study of differential geometry. Unless otherwise stated, this collection focuses on Riemannian geometry; throughout, “differential manifold” is used in place of “Riemannian manifold”. This collection is meant to be an introduction to Riemannian geometry, with emphases only on the fundamental building blocks of the study of Riemannian manifolds.↩︎

Reading: Carmo (1992), Riemannian geometry.