nLab topological manifold

Redirected from "locally Euclidean topological space".
Contents

Context

Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

Introduction

Basic concepts

Universal constructions

Extra stuff, structure, properties

Examples

Basic statements

Theorems

Analysis Theorems

topological homotopy theory

Manifolds and cobordisms

Contents

Idea

A topological manifold is a topological space (usually required to be Hausdorff and paracompact) which is locally homeomorphic to a Euclidean space n\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.

Topological manifolds form a category TopMfd.

Definition

Locally Euclidean topological spaces

Definition

(locally Euclidean topological space)

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

nAAAAU xX. \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 T_1 -separated, sober, locally connected, locally compact)

Let XX be a locally Euclidean space (def. ). Then

  1. XX satisfies the T 1T_1 separation axiom;

  2. XX is sober;

  3. XX is locally connected;

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

Proof

Regarding the first statement:

Let xyx \neq y be two distinct points in the locally Euclidean space. We need to show that there is an open neighbourhood U xU_x around xx that does not contain yy.

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

So let PXP \subset X be an open proper subset such that if there are two open subsets U 1,U 2XU_1, U_2 \subset X with U 1U 2PU_1 \cap U_2 \subset P then U 1PU_1 \subset P or U 2PU_2 \subset P. By this prop. we need to show that there exists a point xXx \in X such that P=X{x}P = X \setminus \{x\} it its complement.

Now since PXP \subset X is a proper subset, and since the locally Euclidean space XX is covered by Euclidean neighbourhoods, there exists a Euclidean neighbourhood nϕUX\mathbb{R}^n \underoverset{\simeq}{\phi}{\to} U \subset X such that PUUP \cap U \subset U is a proper subset. In fact this still satisfies the condition that for U 1,U 2openUU_1, U_2 \underset{\text{open}}{\subset} U then U 1U 2PUU_1 \cap U_2 \subset P \cap U implies U 1PUU_1 \subset P \cap U or U 2PUU_2 \subset P \cap U. Accordingly, by that prop. it follows that nϕ 1(PU)\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 1T_1-separated, this means that there exists x nx \in \mathbb{R}^n such that ϕ 1(PU)= n{x}\phi^{-1}(P \cap U) = \mathbb{R}^n \setminus \{x\}.

In conclusion this means that the restriction of an irreducible closed subset in XX 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 xXx \in X be a point and U x{x}U_x \supset \{x\} a neighbourhood. We need to find a connected open neighbourhood Cn xU xCn_x \subset U_x.

By local Euclideanness, there is also a Euclidean neighboruhood nϕV xX\mathbb{R}^n \underoverset{\simeq}{\phi}{\to} V_x \subset X. Since ϕ\phi is a homeomorphism, and since U xV xU_x \cap V_x is open, also ϕ 1(U xV x) n\phi^{-1}(U_x \cap V_x) \subset \mathbb{R}^n is open. This means that there exists an open ball B ϕ 1(x) (ϵ)ϕ 1(U xV x)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 ϕ(B ϕ 1(x) (ϵ))X\phi(B^\circ_{\phi^{-1}(x)}(\epsilon)) \subset X. This is a connected open neighbourhood of xx as required.

Regarding the fourth statement:

Let xXx \in X be a point and let U x{x}U_x \supset \{x\} be an open neighbourhood. We need to find a compact neighbourhood K xU xK_x \subset U_x.

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

Since Euclidean spaces are locally compact, there exists a compact neighbourhood K ϕ 1(x) nK_{\phi^{-1}(x)} \subset \mathbb{R}^n (for instance a sufficiently small closed ball around xx, which is compact by the Heine-Borel theorem). Now since continuous images of compact spaces are compact, it follows that also ϕ(K)X\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 2T_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”.

Example

counter-example: (non-Hausdorff locally Euclidean spaces)
An example of a locally Euclidean space (def. ) which is a non-Hausdorff topological space, is the line with two origins.

Lemma

(connected locally Euclidean spaces are path-connected)

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

Proof

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

PConn x(X):{yX|[0,1]ctsγX((γ(0)=x)Aanda(γ(1)=y))}. 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)XPConn_x(X) \subset X as well as its complement are open subsets:

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

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

[0,1] γ˜γ X t AAA {γ(2t) | t1/2 (˜2t1/2) | t1/2 \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 xx with zU yz \in U_y. Hence U yPConn x(X)U_y \subset PConn_x(X).

Similarly, if yy is not connected to xx by a path, then also all point in U yU_y cannot be connected to xx by a path, for if they were, then the analogous concatenation of paths would give a path from xx to yy, contrary to the assumption.

It follows that

X=PConn x(C)(XPConn x(X)) X = PConn_x(C) \sqcup (X \setminus PConn_x(X))

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

Proposition

(equivalence of regularity conditions for Hausdorff locally Euclidean spaces)

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

Then the following are equivalent:

  1. XX is sigma-compact,

  2. XX is second-countable,

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

Proof

Generally, observe that XX 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 XX is assumed to be Hausdorff, this implies all the other variants of definition of local compactness, by this prop..

1) \Rightarrow 2)

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

By sigma-compactness there exists a countable set {K iX} iI\{K_i \subset X\}_{i \in I} of compact subspaces. By XX being locally Euclidean, each admits an open cover by restrictions of Euclidean spaces. By their compactness, each of these has a subcover { nϕ i,jX} jJ i\{ \mathbb{R}^n \overset{\phi_{i,j}}{\to} X \}_{j \in J_i} with J iJ_i a finite set. Since countable unions of countable sets are countable, we have obtained a countable cover by Euclidean spaces { nϕ i,jX} iI,jJ i\{ \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 β n\beta_{\mathbb{R}^n} of base open sets. As a result the union iIjJ iϕ i,j(β n)\underset{{i \in I} \atop {j \in J_i}}{\cup} \phi_{i,j}(\beta_{\mathbb{R}^n}) is a base of opens for XX. 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 XX. This means that XX is second-countable.

1) \Rightarrow 3)

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

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

2)\Rightarrow 1) Let XX 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 XX be paracompact with countably many connected components. We show that XX is sigma-compact.

Since XX is locally compact, there exists a cover {K i=Cl(U i)X} iI\{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

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

as well as a locally finite cover {Cl(V j)X} jJ\{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)Cl(V_j) are contained in one of the compact subspaces K iK_i. Since subsets are closed in a closed subspace precisely if they are closed in the ambient space, the Cl(V j)Cl(V_j) are also closed as subsets of the K iK_i. Since closed subsets of compact spaces are compact it follows that the Cl(V j)Cl(V_j) are themselves compact and hence form a locally finite cover by compact subspaces.

Now fix any j 0Jj_0 \in J.

We claim that for every jJj \in J there is a finite sequence of indices (j 0,j 1,,j n=j)(j_0, j_1, \cdots, j_n = j) with the property that V j kV j k+1V_{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 XX is connected. From this the general statement follows since countable unions of countable sets are countable. Hence assume that XX is connected. It follows from lemma that XX is path-connected.

Hence for any xV j 0x \in V_{j_0} and yV jy \in V_{j} there is a path γ:[0,1]X\gamma \colon [0,1] \to X connecting xx with yy. 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 iV_i that covers the image of this path. This proves the claim.

It follows that there is a function

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

which sends each V jV_j to the minimum natural number as above.

We claim now that for all nn \in \mathbb{N} the preimage of {0,1,,n}\{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)X} jJ\{ Cl(V_j) \subset X\}_{j \in J} is a countable cover of XX by compact subspaces, hence that XX is sigma-compact.

We prove this last claim by induction. It is true for n=0n = 0 by construction. Assume it is true for some nn \in \mathbb{N}, hence that f 1({0,1,,n})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 nVf 1({0,,n})V K_n \coloneqq \underset{V \in f^{-1}(\{0,\cdots, n\})}{\cup} V

is compact. By local finiteness of the {V j} jJ\{V_j\}_{j \in J}, every point xK nx \in K_n has an open neighbourhood W xW_x that intersects only a finite set of the V jV_j. By compactness of K nK_n, the cover {W xX} xK n\{W_x \subset X\}_{x \in K_n} has a finite subcover. In conclusion this implies that only a finite number of the V jV_j intersect K nK_n.

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

Topological manifold

Definition

(topological manifold)

A topological manifold is a topological space which is

  1. locally Euclidean (def. ),

  2. paracompact Hausdorff.

If the local Euclidean neighbourhoods nUX\mathbb{R}^n \overset{\simeq}{\to} U \subset X are all of dimension nn for a fixed nn \in \mathbb{N}, then the topological manifold is said to be a nn-dimensional manifold or nn-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 nn-dimensional topological manifold XX (def. ), then

  1. an open subset UXU \subset X and a homeomorphism ϕ: nAAU\phi \colon \mathbb{R}^n \overset{\phantom{A}\simeq\phantom{A}}{\to} U is also called a local coordinate chart of XX.

  2. an open cover of XX by local charts { nϕ iUX} iI\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,jIi,j \in I the intersection of the iith chart with the jjth chart in such an atlas by

    U ijU iU j U_{i j} \coloneqq U_i \cap U_j

    then the induced homeomorphism

    nAAϕ i 1(U ij)Aϕ iAU ijAϕ j 1Aϕ j 1(U ij)AA n \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 ii to chart jj.

graphics grabbed from Frankel

Definition

(differentiable and smooth manifolds)

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

  1. a topological manifold XX (def. );

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

A pp-fold differentiable function between pp-fold differentiable manifolds

(X,{ nϕ iU iX} iI)AAfAA(Y,{ nψ jV jY} jJ) (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 iIi \in I and jJj \in J then

    nAA(fϕ i) 1(V j)ϕ if 1(V j)fV jψ j 1 n \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 pp-fold differentiable function between open subsets of Euclidean space.

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

Properties

Proposition

Let XX be a kk-fold differentiable manifold and let SXS \subset X be an open subset of the underlying topological space (X,τ)(X,\tau).

Then SS carries the structure of a kk-fold differentiable manifold such that the inclusion map SXS \hookrightarrow X is an open embedding of differentiable manifolds.

Proof

Since the underlying topological space of XX 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 XX has a single connected component. By this prop this implies that XX 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 SS is still locally Euclidean: since XX is locally Euclidean every point xSXx \in S \subset X has a Euclidean neighbourhood in XX and since SS is open there exists an open ball in that (itself homeomorphic to Euclidean space) which is a Euclidean neighbourhood of xx contained in SS.

For the differentiable structure we pick these Euclidean neighbourhoods from the given atlas. Then the gluing functions for the Euclidean charts on SS are kk-fold differentiable follows since these are restrictions of the gluing functions for the atlas of XX.

Examples

See the examples at differentiable manifold.

References

Historical articles:

Textbook accounts:

  • John M. Lee, Introduction to topological manifolds, Graduate Texts in Mathematics 202, Springer (2000) [ISBN: 0-387-98759-2, 0-387-95026-5]

    Second edition: Springer (2011) [ISBN:978-1-4419-7939-1, doi:10.1007/978-1-4419-7940-7, errata pdf]

See also:

Last revised on June 11, 2024 at 17:08:56. See the history of this page for a list of all contributions to it.