Contents

Contents

Idea

A topological manifold is a topological space (usually required to be Hausdorff and paracompact) which is locally homeomorphic to a Euclidean space $\mathbb{R}^n$ equipped with its metric topology.

Often one is interested in extra structure on topological manifolds, that make them for instance into differentiable manifolds or smooth manifolds or analytic manifolds or complex manifolds, etc. See at manifold for more on the general concept.

Definition

Locally Euclidean topological spaces

Definition

(locally Euclidean topological space)

A topological space $X$ is locally Euclidean if every point $x \in X$ has an open neighbourhood $U_x \supset \{x\}$ which is homeomorphic to the Euclidean space $\mathbb{R}^n$ with its metric topology:

$\mathbb{R}^n \overset{\phantom{AA} \simeq \phantom{AA}}{\longrightarrow} U_x \subset X \,.$

The “local” topological properties of Euclidean space are inherited by locally Euclidean spaces:

Proposition

(locally Euclidean spaces are $T_1$-separated, sober, locally connected, locally compact)

Let $X$ be a locally Euclidean space (def. ). Then

1. $X$ satisfies the $T_1$ separation axiom;

2. $X$ is sober;

3. $X$ is locally connected;

4. $X$ is locally compact in the sense that every open neighbourhood of a point contains a compact neighbourhood.

Proof

Regarding the first statement:

Let $x \neq y$ be two distinct points in the locally Euclidean space. We need to show that there is an open neighbourhood $U_x$ around $x$ that does not contain $y$.

By definition, there is a Euclidean open neighbourhood $\mathbb{R}^n \underoverset{\simeq}{\phi}{\to} U_x \subset X$ around $x$. If $U_x$ does not contain $y$, then it already is an open neighbourhood as required. If $U_x$ does contain $y$, then $\phi^{-1}(x) \neq \phi^{-1}(y)$ are equivalently two distinct points in $\mathbb{R}^n$. But Euclidean space, as every metric space, is $T_1$, and hence we may find an open neighbourhood $V_{\phi^{-1}(x)} \subset \mathbb{R}^n$ not containing $\phi^{-1}(y)$. By the nature of the subspace topology, $\phi(V_{\phi^{-1}(x)}) \subset X$ is an open neighbourhood as required.

Regarding the second statement:

We need to show that the map

$Cl(\{-\}) \;\colon\; X \to IrrClSub(X)$

that sends points to the topological closure of their singleton sets is a bijection with the set of irreducible closed subsets. By the first statement above the map is injective (via this lemma).

Hence it remains to see that every irreducible closed subset is the topological closure of a singleton. We will show something stronger: every irreducible closed subset is a singleton.

Let $P \subset X$ be an open proper subset such that if there are two open subsets $U_1, U_2 \subset X$ with $U_1 \cap U_2 \subset P$ then $U_1 \subset P$ or $U_2 \subset P$. By this prop. we need to show that there exists a point $x \in X$ such that $P = X \setminus \{x\}$ it its complement.

Now since $P \subset X$ is a proper subset, and since the locally Euclidean space $X$ is covered by Euclidean neighbourhoods, there exists a Euclidean neighbourhood $\mathbb{R}^n \underoverset{\simeq}{\phi}{\to} U \subset X$ such that $P \cap U \subset U$ is a proper subset. In fact this still satisfies the condition that for $U_1, U_2 \underset{\text{open}}{\subset} U$ then $U_1 \cap U_2 \subset P \cap U$ implies $U_1 \subset P \cap U$ or $U_2 \subset P \cap U$. Accordingly, by that prop. it follows that $\mathbb{R}^n \setminus \phi^{-1}(P \cap U)$ is an irreducible closed subset of Euclidean space. Sine metric spaces are sober topological space as well as $T_1$-separated, this means that there exists $x \in \mathbb{R}^n$ such that $\phi^{-1}(P \cap U) = \mathbb{R}^n \setminus \{x\}$.

In conclusion this means that the restriction of an irreducible closed subset in $X$ to any Euclidean chart is either empty or a singleton set. This means that the irreducible closed subset must be a disjoint union of singletons that are separated by Euclidean neighbourhoods. But by irreducibiliy, this union has to consist of just one point.

Regarding the third statement:

Let $x \in X$ be a point and $U_x \supset \{x\}$ a neighbourhood. We need to find a connected open neighbourhood $Cn_x \subset U_x$.

By local Euclideanness, there is also a Euclidean neighboruhood $\mathbb{R}^n \underoverset{\simeq}{\phi}{\to} V_x \subset X$. Since $\phi$ is a homeomorphism, and since $U_x \cap V_x$ is open, also $\phi^{-1}(U_x \cap V_x) \subset \mathbb{R}^n$ is open. This means that there exists an open ball $B_{\phi^{-1}(x)}^\circ(\epsilon) \subset \phi^{-1}(U_x \cap V_x)$. This is open and connected, and hence so is its homeomorphic image $\phi(B^\circ_{\phi^{-1}(x)}(\epsilon)) \subset X$. This is a connected open neighbourhood of $x$ as required.

Regarding the fourth statement:

Let $x \in X$ be a point and let $U_x \supset \{x\}$ be an open neighbourhood. We need to find a compact neighbourhood $K_x \subset U_x$.

By assumption there exists a Euclidean open neighbourhood $\mathbb{R}^n \underoverset{\simeq}{\phi}{\to} V_x \subset X$. By definition of the subspace topology the intersection $U_x \cap V_x$ is still open as a subspace of $V_x$ and hence $\phi^{-1}(U_x \cap V_x)$ is an open neighbourhood of $\phi^{-1}(x) \in \mathbb{R}^n$.

Since Euclidean spaces are locally compact, there exists a compact neighbourhood $K_{\phi^{-1}(x)} \subset \mathbb{R}^n$ (for instance a sufficiently small closed ball around $x$, which is compact by the Heine-Borel theorem). Now since continuous images of compact spaces are compact, it follows that also $\phi(K) \subset X$ is a compact neighbourhood.

But the “global” topological properties of Euclidean space are not generally inherited by locally Euclidean spaces. This sounds obvious, but notice that also Hausdorff-ness is a “global property”:

Remark

(locally Euclidean spaces are not necessarily $T_2$)

It might superficially seem that every locally Euclidean space (def. ) is necessarily a Hausdorff topological space, since Euclidean space, like any metric space, is Hausdorff, and since by definition the neighbourhood of every point in a locally Euclidean spaces looks like Euclidean space.

But this is not so, Hausdorffness is a “non-local condition”.

Lemma

(connected locally Euclidean spaces are path-connected)

A locally Euclidean space which is connected is also path-connected.

Proof

Fix any $x \in X$. Write $PConn_x(X) \subset X$ for the subset of all those points of $x$ which are connected to $x$ by a path, hence

$PConn_x(X) \;\colon\; \left\{ y \in X \;\vert\; \underset{[0,1] \underoverset{cts}{\gamma}{\to} X }{\exists} \left( \left(\gamma(0) = x\right) \phantom{A} \text{and} \phantom{a} \left( \gamma(1) = y \right) \right) \right\} \,.$

Observe now that both $PConn_x(X) \subset X$ as well as its complement are open subsets:

To see this it is sufficient to find for every point $y \on PConn_x(X)$ an open neighbourhood $U_y \supset \{y\}$ such that $U_y \subset PConn_x(X)$, and similarly for the complement.

Now by assumption every point $y \in X$ has a Euclidean neighbourhood $\mathbb{R}^n \overset{\simeq}{\to} U_y \subset X$. Since Euclidean space is path connected, there is for every $z \in U_y$ a path $\tilde \gamma \colon [0,1] \to X$ connecting $y$ with $z$, i.e. with $\tilde \gamma(0) = y$ and $\tilde \gamma(1) = z$. Accordingly the composite path

$\array{ [0,1] &\overset{\tilde \gamma\cdot\gamma}{\longrightarrow}& X \\ t &\overset{\phantom{AAA}}{\mapsto}& \left\{ \array{ \gamma(2t) &\vert& t \leq 1/2 \\ \tilde(2t-1/2) &\vert& t \geq 1/2 } \right. }$

connects $x$ with $z \in U_y$. Hence $U_y \subset PConn_x(X)$.

Similarly, if $y$ is not connected to $x$ by a path, then also all point in $U_y$ cannot be connected to $x$ by a path, for if they were, then the analogous concatenation of paths would give a path from $x$ to $y$, contrary to the assumption.

It follows that

$X = PConn_x(C) \sqcup (X \setminus PConn_x(X))$

is a decomposition of $X$ as the disjoint union of two open subsets. By the assumption that $X$ is connected, exactly one of these open subsets is empty. Since $PConn_x(X)$ is not empty, as it contains $x$, it follows that its compement is empty, hence that $PConn_x(X) = X$, hence that $(X,\tau)$ is path connected.

Proposition

(equivalence of regularity conditions for Hausdorff locally Euclidean spaces)

Let $X$ be a locally Euclidean space (def. ) which is Hausdorff.

Then the following are equivalent:

1. $X$ is sigma-compact,

2. $X$ is second-countable,

3. $X$ is paracompact and has a countable set of connected components,

Proof

Generally, observe that $X$ is locally compact: By prop. every locally Euclidean space is locally compact in the sense that every point has a neighbourhood base of compact neighbourhoods, and since $X$ is assumed to be Hausdorff, this implies all the other variants of definition of local compactness, by this prop..

1) $\Rightarrow$ 2)

Let $X$ be sigma-compact. We show that then $X$ is second-countable:

By sigma-compactness there exists a countable set $\{K_i \subset X\}_{i \in I}$ of compact subspaces. By $X$ being locally Euclidean, each admits an open cover by restrictions of Euclidean spaces. By their compactness, each of these has a subcover $\{ \mathbb{R}^n \overset{\phi_{i,j}}{\to} X \}_{j \in J_i}$ with $J_i$ a finite set. Since countable unions of countable sets are countable, we have obtained a countable cover by Euclidean spaces $\{ \mathbb{R}^n \overset{\phi_{i,j}}{\to} X\}_{i \in I, j \in J_i}$. Now Euclidean space itself is second countable (by this example), hence admits a countable set $\beta_{\mathbb{R}^n}$ of base open sets. As a result the union $\underset{{i \in I} \atop {j \in J_i}}{\cup} \phi_{i,j}(\beta_{\mathbb{R}^n})$ is a base of opens for $X$. But this is a countable union of countable sets, and since countable unions of countable sets are countable we have obtained a countable base for the topology of $X$. This means that $X$ is second-countable.

1) $\Rightarrow$ 3)

Let $X$ be sigma-compact. We show that then $X$ is paracompact with a countable set of connected components:

Since locally compact and sigma-compact spaces are paracompact, it follows that $X$ is paracompact. By local connectivity (prop. ) $X$ is the disjoint union space of its connected components (this prop.). Since, by the previous statement, $X$ is also second-countable it cannot have an uncountable set of connected components.

2)$\Rightarrow$ 1) Let $X$ be second-countable, we need to show that it is sigma-compact.

This follows since locally compact and second-countable spaces are sigma-compact.

3) $\Rightarrow$ 1)

Now let $X$ be paracompact with countably many connected components. We show that $X$ is sigma-compact.

Since $X$ is locally compact, there exists a cover $\{K_i = Cl(U_i) \subset X\}_{i \in I}$ by compact subspaces. By paracompactness there is a locally finite refinement of this cover. Since paracompact Hausdorff spaces are normal, the shrinking lemma applies to this refinement and yields a locally finite open cover

$\mathcal{V} \coloneqq \{V_j \subset X \}_{j \in J}$

as well as a locally finite cover $\{Cl(V_j) \subset X\}_{j \in J}$ by closed subsets. Since this is a refinement of the orignal cover, all the $Cl(V_j)$ are contained in one of the compact subspaces $K_i$. Since subsets are closed in a closed subspace precisely if they are closed in the ambient space, the $Cl(V_j)$ are also closed as subsets of the $K_i$. Since closed subsets of compact spaces are compact it follows that the $Cl(V_j)$ are themselves compact and hence form a locally finite cover by compact subspaces.

Now fix any $j_0 \in J$.

We claim that for every $j \in J$ there is a finite sequence of indices $(j_0, j_1, \cdots, j_n = j)$ with the property that $V_{j_k} \cap V_{j_{k+1}} \neq \emptyset$.

To see this, first observe that it is sufficient to show sigma-compactness for the case that $X$ is connected. From this the general statement follows since countable unions of countable sets are countable. Hence assume that $X$ is connected. It follows from lemma that $X$ is path-connected.

Hence for any $x \in V_{j_0}$ and $y \in V_{j}$ there is a path $\gamma \colon [0,1] \to X$ connecting $x$ with $y$. Since the closed interval is compact and since continuous images of compact spaces are compact, it follows that there is a finite subset of the $V_i$ that covers the image of this path. This proves the claim.

It follows that there is a function

$f \;\colon\; \mathcal{V} \longrightarrow \mathbb{N}$

which sends each $V_j$ to the minimum natural number as above.

We claim now that for all $n \in \mathbb{N}$ the preimage of $\{0,1, \cdots, n\}$ under this function is a finite set. Since countable unions of countable sets are countable this implies that $\{ Cl(V_j) \subset X\}_{j \in J}$ is a countable cover of $X$ by compact subspaces, hence that $X$ is sigma-compact.

We prove this last claim by induction. It is true for $n = 0$ by construction. Assume it is true for some $n \in \mathbb{N}$, hence that $f^{-1}(\{0,1, \cdots, n\})$ is a finite set. Since finite unions of compact subspaces are again compact (this prop.) it follows that

$K_n \coloneqq \underset{V \in f^{-1}(\{0,\cdots, n\})}{\cup} V$

is compact. By local finiteness of the $\{V_j\}_{j \in J}$, every point $x \in K_n$ has an open neighbourhood $W_x$ that intersects only a finite set of the $V_j$. By compactness of $K_n$, the cover $\{W_x \subset X\}_{x \in K_n}$ has a finite subcover. In conclusion this implies that only a finite number of the $V_j$ intersect $K_n$.

Now by definition $f^{-1}(\{0,1,\cdots, n+1\})$ is a subset of those $V_j$ which intersect $K_n$, and hence itself finite.

Topological manifold

Definition

(topological manifold)

A topological manifold is a topological space which is

1. locally Euclidean (def. ),

If the local Euclidean neighbourhoods $\mathbb{R}^n \overset{\simeq}{\to} U \subset X$ are all of dimension $n$ for a fixed $n \in \mathbb{N}$, then the topological manifold is said to be a $n$-dimensional manifold or $n$-fold. This is usually assumed to be the case.

Remark

(varying terminology)

Often a topological manifold (def. ) is required to be sigma-compact. But by prop. this is not an extra condition as long as there is a countable set of connected components.

Differentiable manifolds

Definition

(local chart and atlas and gluing function)

Given an $n$-dimensional topological manifold $X$ (def. ), then 1. an open subset $U \subset X$ and a homeomorphism $\phi \colon \mathbb{R}^n \overset{\phantom{A}\simeq\phantom{A}}{\to} U$ is also called a local coordinate chart of $X$.

2. an open cover of $X$ by local charts $\left\{ \mathbb{R}^n \overset{\phi_i}{\to} U \subset X \right\}_{i \in I}$ is called an atlas of the topological manifold.

3. denoting for each $i,j \in I$ the intersection of the $i$th chart with the $j$th chart in such an atlas by

$U_{i j} \coloneqq U_i \cap U_j$

then the induced homeomorphism

$\mathbb{R}^n \supset \phantom{AA} \phi_i^{-1}(U_{i j}) \overset{\phantom{A}\phi_i\phantom{A}}{\longrightarrow} U_{i j} \overset{\phantom{A}\phi_j^{-1}\phantom{A}}{\longrightarrow} \phi_j^{-1}(U_{i j}) \phantom{AA} \subset \mathbb{R}^n$

is called the gluing function from chart $i$ to chart $j$.

graphics grabbed from Frankel

Definition

(differentiable and smooth manifolds)

For $p \in \mathbb{N} \cup \{\infty\}$ then a $p$-fold differentiable manifold is

1. a topological manifold $X$ (def. );

2. an atlas $\{\mathbb{R}^n \overset{\phi_i}{\to} X\}$ (def. ) all whose gluing functions are $p$ times continuously differentiable.

A $p$-fold differentiable function between $p$-fold differentiable manifolds

$(X, \{\mathbb{R}^{n} \overset{\phi_i}{\to} U_i \subset X\}_{i \in I}) \overset{\phantom{AA}f\phantom{AA}}{\longrightarrow} (Y, \{\mathbb{R}^{n'} \overset{\psi_j}{\to} V_j \subset Y\}_{j \in J})$

is

such that

• for all $i \in I$ and $j \in J$ then

$\mathbb{R}^n \supset \phantom{AA} (f\circ \phi_i)^{-1}(V_j) \overset{\phi_i}{\longrightarrow} f^{-1}(V_j) \overset{f}{\longrightarrow} V_j \overset{\psi_j^{-1}}{\longrightarrow} \mathbb{R}^{n'}$

is a $p$-fold differentiable function between open subsets of Euclidean space.

Notice that this in in general a non-trivial condition even if $X = Y$ and $f$ is the identity function. In this case the above exhibits a passage to a different, but equivalent, differentiable atlas.

Properties

Proposition

Let $X$ be a $k$-fold differentiable manifold and let $S \subset X$ be an open subset of the underlying topological space $(X,\tau)$.

Then $S$ carries the structure of a $k$-fold differentiable manifold such that the inclusion map $S \hookrightarrow X$ is an open embedding of differentiable manifolds.

Proof

Since the underlying topological space of $X$ is locally connected (this prop.) it is the disjoint union space of its connected components (this prop.).

Therefore we are reduced to showing the statement for the case that $X$ has a single connected component. By this prop this implies that $X$ is second-countable topological space.

Now a subspace of a second-countable Hausdorff space is clearly itself second countable and Hausdorff.

Similarly it is immediate that $S$ is still locally Euclidean: since $X$ is locally Euclidean every point $x \in S \subset X$ has a Euclidean neighbourhood in $X$ and since $S$ is open there exists an open ball in that (itself homeomorphic to Euclidean space) which is a Euclidean neighbourhood of $x$ contained in $S$.

For the differentiable structure we pick these Euclidean neighbourhoods from the given atlas. Then the gluing functions for the Euclidean charts on $S$ are $k$-fold differentiable follows since these are restrictions of the gluing functions for the atlas of $X$.

Examples

See the examples at differentiable manifold.

Textbook accounts include

• John Lee, Introduction to topological manifolds, Graduate Texts in Mathematics, Springer (2000) (errata pdf)