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

[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.