A Is There a Simple Method to Analyze the Boundary of the 3-Sphere?

[jk22]

On wikipedia : https://en.m.wikipedia.org/wiki/3-sphere

We learn that this manifold is without boundary. Is there a simple analytical method to obtain out if its parametrization the fact that the boundary is empty ?

 

[PeroK]

On wikipedia : https://en.m.wikipedia.org/wiki/3-sphere

We learn that this manifold is without boundary. Is there a simple analytical method to obtain out if its parametrization the fact that the boundary is empty ?

What's the definition of boundary ?

 

[FactChecker]

I would refer to the hemispherical coordinates and point out the there is no path in terms of $\phi, \psi, \theta$ that does not remain in the 3-sphere.

 

[fresh_42]

I would refer to the hemispherical coordinates and point out the there is no path in terms of $\phi, \psi, \theta$ that does not remain in the 3-sphere.

But this is a tautology. A path is already within, or otherwise you need an embedding, an outside, which is not necessary to speak about manifolds with or without boundary.

 

[martinbn]

The sphere is the boundary of the ball. The boundary of a boundary is empty.

 

[mathwonk]

the 3 sphere is homogeneous, in the sense that every point has a nbhd that is homeomorphic to a nbhd of every other point. this not true on a manifold with non empty boundary, since a nbhd of a boundary point is not homeomorphic to a nbhd of an interior point.

 

[jk22]

The sphere is the boundary of the ball. The boundary of a boundary is empty.

In wikipedia it is said that boundary of boundary is included in the boundary : https://en.m.wikipedia.org/wiki/Boundary_(topology)
If a parametrization is given by let say :

$x=r\cos a\sin b$ $y=r\sin a\sin b$ $z=r\cos b$

With $r\in [0;1],a\in [0;2\pi],b\in [0;\pi]$

From the ball we reach the sphere by setting the parameter r at an extreme point 1. Is there a procedure like that to find boundaries.

 

[fresh_42]

From the ball we reach the sphere by setting the parameter r at an extreme point 1. Is there a procedure like that to find boundaries.

@PeroK posed the right question: What is the definition?

The definition is cum grano salis: Boundary points are those which are boundary points in the charts. So check all charts and you will find the boundary, or not.

 

[FactChecker]

I think that there is a difference between the 3-sphere embedded in 4-dimensional Euclidian space versus the stand-alone space with the inherited metric topology. As a stand-alone space, around any point, $(x_0,x_1,x_2,x_3)$, of the 3-sphere there is an open ball, $\{(y_0,y_2,y_3,y_4)\in S^3: (y_0-x_0)^2+(y_1-x_1)^2+(y_2-x_2)^2+(y_3-x_3)^2 \lt \epsilon\}$, completely contained in the 3-sphere.

EDIT: A boundary point of $S^3$ is a point in the closure of $S^3$ which is not in the interior of $S^3$. Since every point of the entire space is an interior point, there can be no boundary points.

 

[jk22]

I know there are several topological definitions of boundary : for example closure minus interior. But I don't know how to translate that in a manifold given by a parametrization, for example out of calculation with the metric.

 

[PeroK]

I know there are several topological definitions of boundary : for example closure minus interior. But I don't know how to translate that in a manifold given by a parametrization, for example out of calculation with the metric.

In other words, you don't know what definition of boundary to use?

 

[fresh_42]

I know there are several topological definitions of boundary : for example closure minus interior. But I don't know how to translate that in a manifold given by a parametrization, for example out of calculation with the metric.

You cannot see anything from a path within the manifold, because you are already in it (see post #4). The definition for manifolds is different from that of metric spaces. You need the charts for it, which are those metric spaces where it is defined in. It is always: go to the charts, do the job, and return to the manifold. Why should it be different here? Otherwise, if you consider it topologically, then you need an "outside" first, which is not automatically given for manifolds.

 

[FactChecker]

You cannot see anything from a path within the manifold, because you are already in it (see post #4). The definition for manifolds is different from that of metric spaces. You need the charts for it, which are those metric spaces where it is defined in. It is always: go to the charts, do the job, and return to the manifold. Why should it be different here? Otherwise, if you consider it topologically, then you need an "outside" first, which is not automatically given for manifolds.

I believe that a 3-sphere is defined as embeddable in the 4-dimensional Euclidean space. That would allow it to inherit a topology and a metric (although not the path-length metric along the surface).

 

[fresh_42]

I believe that a 3-sphere is defined as embeddable in the 4-dimensional Euclidean space. That would allow it to inherit a topology and a metric (although not the path-length metric along the surface).

The spheres can be defined multiple ways and most of them don't look like a ball. But even as a ball it sends the completely wrong signal to define the topology in the surrounding Euclidean space and speak of boundaries like subsets of that space. If you want to see it like this, never ever use the word manifold. Why should you? If you talk about manifolds and boundaries in the same context, then use the correct definition and do not mix two different contexts. This only creates misunderstandings, confusion and teach the wrong facts. A boundary of a manifold has a certain definition, the boundary of a subset of $(\mathbb{R}^n,\|,\|_p)$ has another. Even worse, you need the latter to define the former, but they are not the same.

Post #6 has an answer which doesn't use the definition. But it doesn't use embeddings or special cases either.

 

[FactChecker]

The spheres can be defined multiple ways and most of them don't look like a ball. But even as a ball it sends the completely wrong signal to define the topology in the surrounding Euclidean space and speak of boundaries like subsets of that space.

All 3-spheres are homeomorphic, so topologically we can consider it as being embedded in 4-dimensional Euclidian space. A question about the boundary is a topological question.

If you want to see it like this, never ever use the word manifold. Why should you? If you talk about manifolds and boundaries in the same context, then use the correct definition and do not mix two different contexts.

A 3-sphere is a manifold with specific properties, such as constant curvature. But this is a topological question about boundaries and specific topological properties of 3-spheres must be used to answer it.

This only creates misunderstandings, confusion and teach the wrong facts. A boundary of a manifold has a certain definition, the boundary of a subset of $(\mathbb{R}^n,\|,\|_p)$ has another. Even worse, you need the latter to define the former, but they are not the same.

Post #6 has an answer which doesn't use the definition. But it doesn't use embeddings or special cases either.

IMHO, that is "the tail wagging the dog". The 3-sphere has a very natural definition within 4-space and any topology should agree with that in all aspects because all 3-spheres are homeomorphic.

 

[mathwonk]

here is another answer: if p is a boundary point (in the sense of boundary of a manifold with boundary), then p has a contractible punctured open neighborhood. But for an interior point, no punctured neighborhood is contractible.

If you know about homology, another formulation of the difference is that the top dimensional relative homology of a nbhd of a point of a manifold, modulo the punctured nbhd, is zero if and only if the point is a boundary point. In the case of the 3 sphere, this group is Z at every point.

 

[FactChecker]

The boundary of a boundary is empty.

That sounds wrong. The boundary of a boundary is a subset of the boundary.
EDIT: I see that the definition of "boundary" as a manifold is quite different from the topological definition. In fact, the open disk, as a manifold, has no boundary. (see https://en.wikipedia.org/wiki/Boundary_(topology)#Boundary_of_a_boundary). With that in mind, I have to leave this discussion to the experts on manifolds.

 

[ChinleShale]

To summarize some of the above posts

Topological Boundary

No topological space by itself is a topological boundary since every point in it is an interior point.
The topological boundary of a subset of a topological space is those points which are in its closure that are not in its interior.

As a subset of Euclidean space a sphere is closed and has no interior. So it is its own topological boundary as a subset of Euclidean space.

Manifolds with Boundary

The topological boundary of a half space in Euclidean space, for instance the points whose x coordinate is non-negative, is those points whose x coordinate is equal to zero.

A manifold with boundary is defined as locally homeomorphic to a open set of a Euclidean half space of fixed dimension. Its boundary is those points that are mapped to the boundary of the Euclidean half space i.e. those points whose x coordinate is zero. The Euclidean half space is itself a manifold with boundary. So its topological boundary as a subset of Euclidean space and its "manifold boundary" are the same set.

A sphere is everywhere locally homeomorphic to an open subset of Euclidean space. @mathwonk in post #6 gives an elementary proof. So a sphere has no boundary as a manifold. Using that proof one can start with a parameterization around one point and others can be obtained by following the parameterization by a rotation of the sphere.

The manifold-boundary of a manifold is another manifold since each point on the boundary is surrounded by a open subset of the boundary of the Euclidean half space. This manifold does not have a boundary as @martinb says in post #5. This is easily seen from the definition of a manifold with boundary.

A Caveat: One might ask whether an open neighborhood of a boundary point on a manifold is itself homeomorphic to an open subset of Euclidean space in which case the definition of boundary would be redundant. The answer is no as @mathwonk proves in post #16.

A Picture: If one slices a n dimensional sphere along its equator, it falls apart into two n dimensional closed balls whose boundaries are the shared equatorial n-1 sphere. In the process a small, open ball around a point on the equator is split into two sets that look like small open subsets of Euclidean half space. When one glues the two large closed balls back together to get the sphere again, these two sets around the equator are pasted along their boundaries to make a small open ball. So all of the boundary points disappear and the sphere has no boundary. In the case of the 3 sphere its equator is a regular two dimensional sphere and the two pieces are ordinary solid balls. In 4 space these balls are puffed up into the fourth dimension but one can deflate them so that they reside in 3 space. A good visualization is then to imagine how the boundary spheres are glued back together to make the 3 sphere.

Footnote: One of the profound questions of 20'th century mathematics was 'When is a closed manifold without boundary the boundary of another manifold?'. The 3 sphere for instance has no boundary but is the boundary of a closed ball of one higher dimension. Incredibly, there are compact manifolds without boundary that cannot be made into the boundary of another manifold. The simplest example is the real projective plane. Mathematicians would say, "The projective plane is not a boundary". While this question was in some sense completely answered using Algebraic and Differential Topology, I know of no intuitive picture that tells the story.

 

[mathwonk]

i don't know how intuitive you will regard this, but think of euler characteristics, computed by a triangulation and counting vertices, edges faces, etc, in an alternating way. It seems obvious that the euler characteristic of a disjoint union of M and N is the sum of the euler characteristics of M and of N. Now if we had a 3 - manifold M with boundary equal to the projective plane P, then the euler characteristic of two disjoint copies of M, would thus be twice that of M, hence even. Now if we identify two disjoint copies of M along their common boundary P, we would get a 3-manifold W without boundary. Moreover, from its construction of gluing two copies of M along P, the euler characteristic of W is twice that of M minus the euler characteristic of P.

But every 3-manifold W without boundary has euler characteristic zero. (If W is orientable, by Poincare duality it has the same number of 1 cycles as 2 cycles, and if non orientable, it has an orientable double cover, hence has 1/2 the euler characteristic of an orientable 3-manifold, hence again zero.) Thus the euler characteristic of P must be even. But the euler characteristic of P is 1, since it has the sphere as double cover, so P has 1/2 the euler characteristic of the sphere.

I.e. if P were the boundary of M, then by forming the double of M, we see that the the euler characteristic of P would be twice that of M, but e(P) = 1.

 

[ChinleShale]

i don't know how intuitive you will regard this, but think of euler characteristics, computed by a triangulation and counting vertices, edges faces, etc, in an alternating way. It seems obvious that the euler characteristic of a disjoint union of M and N is the sum of the euler characteristics of M and of N. Now if we had a 3 - manifold M with boundary equal to the projective plane P, then the euler characteristic of two disjoint copies of M, would thus be twice that of M, hence even. Now if we identify two disjoint copies of M along their common boundary P, we would get a 3-manifold W without boundary. Moreover, from its construction of gluing two copies of M along P, the euler characteristic of W is twice that of M minus the euler characteristic of P.

But every 3-manifold W without boundary has euler characteristic zero. (If W is orientable, by Poincare duality it has the same number of 1 cycles as 2 cycles, and if non orientable, it has an orientable double cover, hence has 1/2 the euler characteristic of an orientable 3-manifold, hence again zero.) Thus the euler characteristic of P must be even. But the euler characteristic of P is 1, since it has the sphere as double cover, so P has 1/2 the euler characteristic of the sphere.

I.e. if P were the boundary of M, then by forming the double of M, we see that the the euler characteristic of P would be twice that of M, but e(P) = 1.

That is a nice argument. It works for any even dimensional manifold that is a boundary.

By intuitive I meant more a picture like the picture one has of filling up a sphere with water.

 

[mathwonk]

well picturing the triangulations, and the double of a manifold is as close as I could get. besides proving something is not possible does not allow a picture of doing it, it requires a condition that would hold, but does not. hopefully this let's you picture why the euler characteristic of the boundary of M, equals twice that of M, minus that of the double.

 

[ChinleShale]

well picturing the triangulations, and the double of a manifold is as close as I could get. besides proving something is not possible does not allow a picture of doing it, it requires a condition that would hold, but does not. hopefully this let's you picture why the euler characteristic of the boundary of M, equals twice that of M, minus that of the double.

Right. And your Euler characteristic proof is quite nice.

 

[ChinleShale]

well picturing the triangulations, and the double of a manifold is as close as I could get. besides proving something is not possible does not allow a picture of doing it, it requires a condition that would hold, but does not. hopefully this let's you picture why the euler characteristic of the boundary of M, equals twice that of M, minus that of the double.

A variant on the double argument: Assume that the projective plane is a boundary, $P^2=∂M$ as in your argument. $M$ is not orientable because $P^2$ is not orientable. Let $N$ be the oriented two fold covering space. It boundary is a single 2 sphere. Paste a solid 3 ball onto this sphere to make a 3 manifold without boundary $Y$. Its Euler characteristic is zero as with all 3 manifolds and one has the equation $0 = χ(Y) = χ(N)+χ(B^3)-χ(S^2) = χ(N)-1$ so $N$ has Euler characteristic 1. But $N$ has even Euler characteristic because it is a two fold covering space.

 

[ChinleShale]

On wikipedia : https://en.m.wikipedia.org/wiki/3-sphere

We learn that this manifold is without boundary. Is there a simple analytical method to obtain out if its parametrization the fact that the boundary is empty ?

You can parameterize the entire 3 sphere minus an arbitrary point using stereographic projection after rotating the point to the north pole.

 

[lavinia]

A variant on the double argument: Assume that the projective plane is a boundary, $P^2=∂M$ as in your argument. $M$ is not orientable because $P^2$ is not orientable. Let $N$ be the oriented two fold covering space. It boundary is a single 2 sphere. Paste a solid 3 ball onto this sphere to make a 3 manifold without boundary $Y$. Its Euler characteristic is zero as with all 3 manifolds and one has the equation $0 = χ(Y) = χ(N)+χ(B^3)-χ(S^2) = χ(N)-1$ so $N$ has Euler characteristic 1. But $N$ has even Euler characteristic because it is a two fold covering space.

I was browsing the Forums this morning and saw your post. I have tried and failed to find an intuitive picture of why the projective plane is not a boundary. Your post suggested a thought. It seems that your above proof using the two fold orientable cover can be restated as a three manifold with a solid ball removed has no fixed point free involution. The simplest case is the three ball itself since it is a three sphere with a solid ball removed. A fixed point free involution of a solid ball of any dimension that restricts to the antipodal map on the boundary would seem to violate the Intermediate Value Theorem. For instance, for the one ball, that is: the unit closed interval, the antipodal map maps 1 to 0 and 0 to 1 and the Intermediate Value Theorem rules out an extension to a fixed point free map of the interval into itself. This should generalize to higher dimensions.

Interestingly, if the orientable two fold covering space were simply connected then adding a three ball to the boundary would give a three sphere (by the Poincare Conjecture) and then by Alexander's Theorem, the bounding two sphere of the two fold covering space would bound a three ball and a generalized Intermediate Value Theorem would finish the theorem off.

But the orientable 2 fold cover might not be simply connected.

In that case, the universal cover, I would guess is a three sphere with a finite number of three balls removed, this because it is orientable and each boundary component covers the projective plane an so is a 2 sphere. Adding the three balls back in makes a simply connected closed three manifold and again by the Poincare Conjecture this must be a three sphere.

 

[r731]

What's the definition of boundary ?

A boundary point P is a point such that any neighberhood centered around P contains a point inside the set and a point outside the set.

 

[fresh_42]

A boundary point P is a point such that any neighberhood centered around P contains a point inside the set and a point outside the set.

The crucial question is what are in- and outside? Considered as a topological manifold in itself, you will have to use charts to define that. Considered as a subset of an Euclidean space things are different.

 

[PeroK]

A boundary point P is a point such that any neighberhood centered around P contains a point inside the set and a point outside the set.

Are there ever any points outside a manifold? Unless it's explicitly a sub-manifold.

 

[r731]

Are there ever any points outside a manifold? Unless it's explicitly a sub-manifold.

I don't think so. Are there any points outside the 2-D Euclidean plane (the coordinate system)?

 

[PeroK]

I don't think so. Are there any points outside the 2-D Euclidean plane (the coordinate system)?

By your definition, therefore, no manifold has a boundary.

 

[r731]

By your definition, therefore, no manifold has a boundary.

As far as I know topological spaces are preserved via homeomorphism. I doubt that a metric is required for a manifold.

 

[fresh_42]

As far as I know topological spaces are preserved via homeomorphism. I doubt that a metric is required for a manifold.

The term boundary of a manifold has a different meaning than the boundary in a topological space.

You gave the definition of the latter whereas I assume the former is meant.

 

[ChinleShale]

The topological definition of the boundary of a subset is its closure minus its interior. The interior of a subset is the largest open set contained within it. An entire topological space is open by definition and is also closed by definition so its closure minus its interior is the empty set. So one might say that the topological boundary of a topological space in itself is the empty set which is not nothing so to be super precise an entire topological space (rather than a subset)always has empty boundary.

In particular, the topological boundary of a manifold is empty.

Manifolds are defined by the condition that around each point there is an open subset that is homeomorphic to an open subset of Euclidean space of a fixed dimension. A manifold with boundary is defined to be locally homeomorphic to an open subset of the closed upper half space of Euclidean space. Its boundary is those points that are mapped to the boundary of the upper half space by some local homeomorphism.

If a local homeomorphism maps a point to the interior of this upper half space(interior when it is considered as a subset of Euclidean space) then no other can map it to the boundary of the half space. Conversely if a local homeomorphism maps a point to the boundary of the half space, then no other local homeomorphism can map it to the interior. This is a theorem that requires proof and it guarantees that the boundary of a manifold is well-defined. This is a theorem worth understanding. As an exercise try the proof for a smooth manifold.

If one considers the open subset of a manifold with boundary of those points that are mapped to the interior of the upper half space then the boundary of this subset is the boundary of the manifold. Perhaps this is what can cause some confusion.

 

[mathwonk]

" if a local homeomorphism maps a point to the boundary of the half space, then no other local homeomorphism can map it to the interior. This is a theorem that requires proof"

a point of the boundary has a neighborhood basis of contractible punctured sets, but no point of the interior does. I think that proves it. ( I may have said this before, since I thought of it so quickly.) does this scan?

added later: oh yes, one has to prove that a punctured ball is not contractible, and my proof of that uses some tools, like homology or homotopy, i.e. degree theory, e.g. via simplicial triangulation.

 

[ChinleShale]

@mathwonk

Some questions about your proof since you gave a sketch.

The punctured open ball has the homology of a sphere while a half ball with a boundary point removed is contractible. So an open ball and a half open ball are not homeomorphic. OK.

The argument about basis I think then wants to use this to say that if point $p$ is mapped to the boundary of the upper half space by homeomorphism $φ:U→R^{n}_{+}$ of an open set $U$ then $U$ can not contain any open subset that both contains $p$ and is homeomorphic to an open subset of $R^{n}$. Such a subset could not be punctured and still be contractible. Correct?

But what about the other way around? Suppose $p$ is mapped to the interior of the upper half space by a local homeomorphism of an open set $U$. How does your proof show that $U$ can not contain an open subset containing $p$ that can be mapped homeomorpically to an open subset of the boundary of $R^{n}_{+}$ and take $p$ to the boundary?

 

[jk22]

Btw : checking wikipedia about boundary :
https://en.m.wikipedia.org/wiki/Boundary_(topology)

I took $S=[0,1]^2\cap\mathbb{Q}^2$

Is the following correct :

$\partial S=[0,1]^2$
$\partial\partial S=([0,1]\times\{0,1\})\cup (\{0,1\}\times[0,1])$
$\partial\partial\partial S=\emptyset$ ?

 

[ChinleShale]

Btw : checking wikipedia about boundary :
https://en.m.wikipedia.org/wiki/Boundary_(topology)

I took $S=[0,1]^2\cap\mathbb{Q}^2$

Is the following correct :

$\partial S=[0,1]^2$
$\partial\partial S=([0,1]\times\{0,1\})\cup (\{0,1\}\times[0,1])$
$\partial\partial\partial S=\emptyset$ ?

How do you justify each these three steps?

 

[mathwonk]

"But what about the other way around? Suppose p is mapped to the interior of the upper half space by a local homeomorphism of an open set U. How does your proof show that U can not contain an open subset containing p that can be mapped homeomorpically to an open subset of the boundary of R+n and take p to the boundary?"

By shrinking around the boundary point, and composing, it seems this would give a homeomorphism from a half disc H around the boundary point y, to a bounded connected open set W in the interior, taking y to some point x in W. Then the complement of y in the half disc, i.e. the punctured half disc H - {y}, would be homeomorphic to the punctured set W-{x}. But one of them is contractible and the other, namely W - {x}, is not. I.e. a small sphere centered at x wraps once around x, and hence cannot be deformed off it, as a contraction would do.

I.e. as stated originally, y has arbitrarily small contractible punctured neighborhoods in H, while x has no contractible punctured nbhds in W. So H cannot be homeomorphic to W, with y going to x.
Perhaps I should say (H,y) and (W,x) have different local homology groups, i.e. the local homology of H at y is zero, unlike that of W at x.

believe that? sorry, I am a bit rusty on topology. But I am using excision here to compute the local homology of W at x, by replacing W by a ball centered at x. So even though originally I am trying to compare a punctured half ball to a punctured open set in R^n, excision let's me just compare a punctured half ball to a punctured ball.

 

[trees and plants]

Could the theorem for boundary of A, $\partial{A}=\overline{A}-{A}^\circ$ where the closure and interior are to be computed, help? By using the definitions of closure and interior, with spherical neighbourhood, could it help?

 

[fresh_42]

Could the theorem for boundary of A, $\partial{A}=\overline{A}-{A}^\circ$ where the closure and interior are to be computed, help? By using the definitions of closure and interior, with spherical neighbourhood, could it help?

No, because ...

The term boundary of a manifold has a different meaning than the boundary in a topological space.

You gave the definition of the latter whereas I assume the former is meant.

 

[ChinleShale]

Could the theorem for boundary of A, ∂A=A¯−A∘ where the closure and interior are to be computed, help? By using the definitions of closure and interior, with spherical neighbourhood, could it help?

As @fresh_42 says in post #42 the two definitions of boundary, the topological definition and the definition of the boundary of a manifold are not the same.

Here is an example: Start with the closed unit ball in $R^3$. This is a three dimensional manifold with boundary equal to the unit sphere.

As a subset of $R^3$ its interior is the open unit ball so its closure minus its interior is the unit sphere. This is the same as its boundary as a manifold rather than as a subset of $R^3$.

As a subset of itself though, it is both open and closed so its boundary is empty.

Suppose $R^3$ is thought of as the points in $R^4$ whose fourth coordinate is zero. Then the unit ball in $R^3$ becomes a subset of $R^4$. As a subset of $R^4$ it is closed and has empty interior. So its boundary is the entire unit ball.

 

[ChinleShale]

@infinitely small

Continuing with post #41

Notice that the only case in post #41 where the boundary of the solid ball was both its topological and manifold boundary was when it was a subset of 3 dimensional space. This is because open sets in the manifold 3 ball are also open sets in $R^3$. One might wonder whether this would work for an arbitrary manifold with boundary. Sadly, this is not possible in general. Even a three dimensional manifold with boundary may not be realizable as a subset of $R^3$. For instance, a 3 dimensional manifold whose boundary is a Klein bottle can not be a subset of $R^3$. More generally an n-dimensional manifold with boundary may only be realizable as a subset of a higher dimensional Euclidean space.

 

[trees and plants]

The boundary of a manifold is the complement of its interior and its interior is the set of all points in the manifold which have neighborhoods homeomorphic to an open subset of $R^n$ read if you want this https://en.wikipedia.org/wiki/Manifold where it says about manifold with boundary, boundary and interior.

Perhaps this helps? If we show that the complement of the interior of the 3-sphere is the empty set then the work is done.

 

[WWGD]

Cant we just use some homological/topological argument?

The 3-sphere is the boundary of the 3 -ball. The boundary of a boundary is empty. EDIT: We can use, e.g., Simplicial homology, then it follows dod =d^2 =0.(empty). Don't know if this is too over the top , i.e., unnecessarily complicated.