Embedding homeomorphic manifolds

[Q-1]

Please forgive any confusion, I am not well acquainted with topological analysis and differential geometry, and I'm a novice with regards to this topic.

According to this theorem (I don't know the name for it), we cannot embed an n-dimensional space in an m-dimensional space, where n>m, without the former losing some of its structure.

So, does that mean that you can't render a higher n-dimensional space (for example, 4D), into a lower dimensional m-dimensional space (3D)? Is information lost, when trying to do so?

Edit: Does this relate or is applicable to holographic renderings?

 

[fresh_42]

How would you embed a full ball into a flat plane? You can only take perspectives, and these aren't bijective.

 

[Q-1]

You can only take perspectives, and these aren't bijective.

What do you mean by this?

Thanks.

 

[fresh_42]

If you have a 3D full ball laying on your desk, then you can take a photo, but all points on the line between the surface of the ball and the desk will be only one pixel of the photo. That is surjectivity and not injective, i.e. no embedding.

You can also say: A 3D object has 3 degrees of freedom, a plane only 2, so it cannot be embedded. That's basically how dimension is defined. The homeomorphism doesn't matter here, it only squeezes and stretches things.

If you want a detailed answer, then please define "dimension" first.

 

[TeethWhitener]

The homeomorphism doesn't matter here,

Is this true? Bijections exist between $\mathbb{R}^n, \mathbb{R}^m, n\neq m$, thanks to Cantor. I thought that dimension was a topological property (i.e., preserved under homeomorphisms).

 

[WWGD]

Yes, in a sense, you do lose information when you lower dimensions

Is this true? Bijections exist between $\mathbb{R}^n, \mathbb{R}^m, n\neq m$, thanks to Cantor. I thought that dimension was a topological property (i.e., preserved under homeomorphisms).

But none of these maps are even locally continuous, as this would violate " Invariance of dimension" . There are definitely surjections, but I don't think they are injective. But it is also a standard result ( though not too easy to prove/generalize) that for $m \neq n \mathbb R^n , \mathbb R^n$ are not homeomorphic.

 

[fresh_42]

Is this true? Bijections exist between $\mathbb{R}^n, \mathbb{R}^m, n\neq m$, thanks to Cantor. I thought that dimension was a topological property (i.e., preserved under homeomorphisms).

You are right, e.g. the Hilbert curve. I erroneously had in mind that it wasn't injective, i.e. I had an example in mind with crossing lines. In any case, it depends on the definition of dimension. I think there are a couple of them: Hausdorff, fraktal, locally Euclidean for manifolds. I had the latter in mind, but that was why I asked for the definition, because the term isn't nearly as obvious as it sounds. I guess Lebesgue's covering dimension is appropriate here if we talk about homeomorphisms and therewith about topologies. But unless stated which one to use, there will be room for confusion.

 

[WWGD]

If you use the Hausdorff ( On Golf?) dimension mentioned by Freshmeister, you will see that the Hausdorff dimension of $\mathbb R^n$ is $n$, which shows them to not be homeomorphic. We also use other types of dimension like Topological dimension: https://en.wikipedia.org/wiki/Lebesgue_covering_dimension
All I know in terms of proving this in other ways is still using heavy machinery, like Homology or Cohomology.

EDIT: Notice how we can go from lower to higher easily through $(x_1,..,x_n) \rightarrow (x_1, x_2,...,x_n,0)$, but we cannot do this in the opposite direction.

 

[mathwonk]

this is not so easy to prove. for the case of R^1 and R^2, we can use connectivity. I.e. R^1 can be separated into two disjoint open subsets by removing one point, but R^2 cannot, so they cannot be homeomorphic.

If we know the jordan curve theorem, then any closed simple curve separates R^2 into two open disjoint components, but there is obviously a circle that does not separate R^3, so R^2 cannot be homeomorphic to R^3.

To keep going we need some theorems on how to separate R^n by an embedded sphere, or some other roughly equivalent result.

Another related approach is to ask about continuous maps of spheres into these spaces. E.g. if R^n is homeomorphic to R^m, then also the one point compactifications of these spaces are homeomorphic. Thus it is sufficient to prove that the spheres S^n anmd S^m are not homeomorphic if n > m.

For this we can define the concept of homotopic maps. Then we prove that the identity map of S^n to S^n is not homotopic to the constant map, but that for m < n, every continuous map from S^m to S^n is homotopic to a simplicial map which is not surjective, hence is homotopic to a constant map (since any proper subset of a sphere is contained in a contractible set). Thus when n>m, S^n, cannot be homotopic and thus not homeomorphic, to S^m. hence the same statements hold for R^n and R^m. (and thus also for all manifolds of different dimensions.)

I learned all this in a second year grad course on topology, but you can also prove it by homology theory, which was taught in first year topology in grad school. I just like homotopy theory better as more geometric and therefore to me more intuitive.

since i remember this over 50 years later, i wish to thank and compliment my teacher, Ed Brown Jr. of Brandeis.

Here is a link to one of his more beautiful results, that for good spaces, cohomology is "representable", i.e. equivalent to homotopy classes of maps into a specific space.

https://www.maths.ed.ac.uk/~v1ranick/papers/brown3.pdf

oops i am proving that R^m is not homeomorphic to R^n if n > m, and what is wanted is that R^n does not embed in R^m. If I think of how to do this variation I will post it later.

 

[WWGD]

this is not so easy to prove. for the case of R^1 and R^2, we can use connectivity. I.e. R^1 can be separated into two disjoint open subsets by removing one point, but R^2 cannot, so they cannot be homeomorphic.

If we know the jordan curve theorem, then any closed simple curve separates R^2 into two open disjoint components, but there is obviously a circle that does not separate R^3, so R^2 cannot be homeomorphic to R^3.

To keep going we need some theorems on how to separate R^n by an embedded sphere, or some other roughly equivalent result.

Another related approach is to ask about continuous maps of spheres into these spaces. E.g. if R^n is homeomorphic to R^m, then also the one point compactifications of these spaces are homeomorphic. Thus it is sufficient to prove that the spheres S^n anmd S^m are not homeomorphic if n > m.

For this we can define the concept of homotopic maps. Then we prove that the identity map of S^n to S^n is not homotopic to the constant map, but that for m < n, every continuous map from S^m to S^n is homotopic to a simplicial map which is not surjective, hence is homotopic to a constant map (since any proper subset of a sphere is contained in a contractible set). Thus when n>m, S^n, cannot be homotopic and thus not homeomorphic, to S^m. hence the same statements hold for R^n and R^m. (and thus also for all manifolds of different dimensions.)

I learned all this in a second year grad course on topology, but you can also prove it by homology theory, which was taught in first year topology in grad school. I just like homotopy theory better as more geometric and therefore to me more intuitive.

since i remember this over 50 years later, i wish to thank and compliment my teacher, Ed Brown Jr. of Brandeis.

Here is a link to one of his more beautiful results, that for good spaces, cohomology is "representable", i.e. equivalent to homotopy classes of maps into a specific space.

https://www.maths.ed.ac.uk/~v1ranick/papers/brown3.pdf

oops i am proving that R^m is not homeomorphic to R^n if n > m, and what is wanted is that R^n does not embed in R^m. If I think of how to do this variation I will post it later.

Don't we have something like borsuk ulam , that the n-sphere embeds in $\mathbb R^n$ or higher? Then, if m<n, the n-sphere does not embed in $\mathbb R^m$ so the two cannot be homeomorphic. Does that work?

 

[lavinia]

The Invariance of Domain Theorem says that a 1-1 continuous map $f:U→R^{n}$ of an open subset $U$ of $R^{n}$ into $R^{n}$ maps $U$ into an open subset of $R^{n}$.

If $g:U→R^{m}$ is an embedding with $m<n$ and $i:R^{m}→R^{n}$ is the inclusion map then $i \circ g: U→R^{n}$ is an injective continuous map so its image is open. But no subset of $i(R^{m})$ is open in $R^{n}$.

There is a confusing point here - confusing for me -which is why isn't it sufficient to use mathwonk's proof that $R^{n}$ and $R^{m}$ are not homeomorphic?

@mathwonk can you explain this?

 

[mathwonk]

it probably does suffice if you use it locally, since all open balls in R^n are homeomorphic to all of R^n. I was just trying to do it somewhat completely without quoting big theorems i did not prove.

 

[WWGD]

it probably does suffice if you use it locally, since all open balls in R^n are homeomorphic to all of R^n. I was just trying to do it somewhat completely without quoting big theorems i did not prove.

How about the idea of Borsuk-Ulam, in that if n<m and S^{m-1} embeds in R^m , but not in R^n, then R^m, R^n cannot be isomorphic?

 

[mathwonk]

that sounds great! nice argument.

i confess i don't always remember the statement of the borsuk ulam theorem. It seems to me we had this discussion recently however of proving borsuk ulam, but i confess i have forgotten already how i proved it. oh yes it was a clever use of vector fields or something that i needed a hint for from chinn and steenrod. since i have a poor memory now i like to reduce everything to a few basic tools and try everytime to remind myself of how to use those tools to do the proofs. i.e. degree theory, homotopy, and homology which i know less well and find rather tediously complicated. i can visualize homotopy and path connectedness but not so much algebraic tools like homology.

my goal is to have a strong intuitive grasp of a few basic tools so that i can prove anything, but without quoting any theorems i do not know how to prove. for me this is all mental exercise to ward off alzheimer's and take pleasure in the contemplation of the concepts. thanks for the nice reminder of the value of the beautiful borsuk ulam theorem, i recall now appreciating earlier how clever it is to prove that more precise statement rather than the more general statement that the map cannot be injective. unfortunately i do not easily remember clever arguments, being less and less clever as i age.

so as to why i did not use some proof that to you youngsters is clearly sufficient, i just usually did not think of it. and i have this ideal that someday, if i understand a few basic principles well enough, then for every question i will be able to say: well if you use this principle, then it is easy. maybe i should include borsuk ulam among my fundamental principles! best wishes for stimulating responses as usual.

 

[lavinia]

Here is the statement of the Borsuk-Ulam Theorem on Wikipedia.

"In mathematics, the Borsuk–Ulam theorem states that every continuous function from an n-sphere into Euclidean n-space maps some pair of antipodal points to the same point. Here, two points on a sphere are called antipodal if they are in exactly opposite directions from the sphere's center."

 

[lavinia]

my goal is to have a strong intuitive grasp of a few basic tools so that i can prove anything, but without quoting any theorems

I once sat in on a basic course in algebraic topology. The teacher considered a valid proof to be an intuitive picture. If a student attempted a deductive argument he got angry.

It seems that mathematics is often thought of as mere deduction starting with axioms. Intuition is often overlooked.

 

[lavinia]

Is this true? Bijections exist between $\mathbb{R}^n, \mathbb{R}^m, n\neq m$, thanks to Cantor. I thought that dimension was a topological property (i.e., preserved under homeomorphisms).

Yes.

There are also continuous maps that are surjective. But there is no continuous bijection.

One can completely fill space of any finite dimension with a continuous curve.

 

[lavinia]

A Comment on the Invariance of Domain Theorem

Generally a continuous bijection $f:U→X$ onto a subset of a topological space $X$ is not a homeomorphism since the inverse function might not be continuous.

But in the special case of an open subset of $R^{n}$ mapped into $R^{n}$ it is true. This is why this theorem is so powerful.

 

[WWGD]

Yes.

There are also continuous maps that are surjective. But there is no continuous bijection.

One can completely fill space of any finite dimension with a continuous curve.

I think there isn't even a local bijection between the two, meaning between open sets. If there was , restrict to a closed (bounded, of course) ball in the sets in question. This gives then a continuous bijection between compact and Hausdorff, which is a homeomorphism. Then we have a homeo between a set in $\mathbb R^n$ and one in $\mathbb R^m$.

 

[mathwonk]

i will think further on invariance of domain now, thank you. in regard to my remarks about algebraic functors like homology versus geometric ones like homotopy, the theorem i linked to of edgar brown jr is remarkable: it turns out that for reasonable spaces (simplicial complexes, but not a topologists sine curve), the apparently algebraic cohomology functor is naturally equivalent to a homotopy functor! e.g. (I may get this wrong modulo spaces with base points versus not) H^1 is equivalent as a functor to the homotopy classes of maps to the circle! The point is that there is a duality, again for nice spaces, between maps of circles into a space, i.e. sort of 1st homology, and maps out of the space into a circle. I.e. given a 1-cocycle on a nice space X, there is a map of X into a circle that pulls back the standard generator of cohomology, namely dtheta, to that cocycle class. this generalizes in higher degrees to homotopy classes of maps into "eilenberg maclaine" spaces, spaces with homoopy groups in only one dimension.

 

[WWGD]

i will think further on invariance of domain now, thank you. in regard to my remarks about algebraic functors like homology versus geometric ones like homotopy, the theorem i linked to of edgar brown jr is remarkable: it turns out that for reasonable spaces (simplicial complexes, but not a topologists sine curve), the apparently algebraic cohomology functor is naturally equivalent to a homotopy functor! e.g. (I may get this wrong modulo spaces with base points versus not) H^1 is equivalent as a functor to the homotopy classes of maps to the circle! The point is that there is a duality, again for nice spaces, between maps of circles into a space, i.e. sort of 1st homology, and maps out of the space into a circle. I.e. given a 1-cocycle on a nice space X, there is a map of X into a circle that pulls back the standard generator pf cohomology, namely dtheta, to that cocycle class. this generalizes in higher degrees to homotopy classes of maps into "eilenberg maclaine" spaces, spaces with homoopy groups in only one dimension.

Isn't $H^1$ always the Abelianization of $\pi_1$ (or the other way around)?

 

[mathwonk]

yes indeed (the former), at least for path connected spaces (see p. 48, greenberg's lectures on algebraic topology). (Oops, not H^1 but H_1, since these are both covariant.) this is why i implied that homology is related to maps of circles into the space. this is indeed a nice relation of 1st homology with a geometric construct, although it is not entirely clear what the geometric content of "abelianization" is. this has no higher dimensional analog however. the only general relation between higher homotopy and higher homology i know of occurs in only one diomension, namely the first dimension in which a non zero homotopy group occurs. I.e. if all homotopy groups in dimensions less than n are zero, and n > 1, then the nth homotopy group is isomorphic to the nth homology group. But after that all bets are off.

Cohomology however which is often algebraically dual to homology is, for nice spaces (cw complexes), given in each dimension n, by homotopy classes of maps to a certain nth degree "classifying space". In dim one, that space is the circle, so that says that 1st homology, or essentially equivalently maps of the circle into the space, are the same as homotopy classes of maps of the space to the circle.

In dimension 2, although there is no general relation between 2nd homology and 2nd homotopy (i.e. the standard map from π2 to H2 is neither injective nor surjective in general), there is a degree 2 classifying space K(Z,2), (which happens to be infinite dimensional complex projective space), such that 2nd integral cohomology, i.e. H^2, is equivalent to homotopy classes of maps to the classifying space K(Z,2).

of course as stated above, in case π1 = 0, then H2 and π2 are isomorphic.

some of these general principles do not always hold for bad spaces. e.g. if one cuts apart a circle and then closes it up by a clasp formed by part of the topologists sine curve, one gets a space that admits no non trivial map of a circle, so π1 = 0. contrary to the representation theorem of brown mentioned above however, there are non trivial maps from this odd space to the circle. In particular this odd bracelet is simply connected, but its complement in the riemann sphere has 2 components.

this contradicts a principle i have had in mind from reading ahlfors book on complex analysis where he uses for certain subsets of the complex numbers, a definition of simple connectedness that equates it with the complement in the sphere being connected. It seems he used this definiion only for bounded open subsets.

 

[WWGD]

yes indeed (the former), at least for path connected spaces (see p. 48, greenberg's lectures on algebraic topology). (Oops, not H^1 but H_1, since these are both covariant.) this is why i implied that homology is related to maps of circles into the space. this is indeed a nice relation of 1st homology with a geometric construct, although it is not entirely clear what the geometric content of "abelianization" is. this has no higher dimensional analog however. the only general relation between higher homotopy and higher homology i know of occurs in only one diomension, namely the first dimension in which a non zero homotopy group occurs. I.e. if all homotopy groups in dimensions less than n are zero, and n > 1, then the nth homotopy group is isomorphic to the nth homology group. But after that all bets are off.

Cohomology however which is often algebraically dual to homology is, for nice spaces (cw complexes), given in each dimension n, by homotopy classes of maps to a certain nth degree "classifying space". In dim one, that space is the circle, so that says that 1st homology, or essentially equivalently maps of the circle into the space, are the same as homotopy classes of maps of the space to the circle.

In dimension 2, although there is no general relation between 2nd homology and 2nd homotopy (i.e. the standard map from π2 to H2 is neither injective nor surjective in general), there is a degree 2 classifying space K(Z,2), (which happens to be infinite dimensional complex projective space), such that 2nd integral cohomology, i.e. H^2, is equivalent to homotopy classes of maps to the classifying space K(Z,2).

of course as stated above, in case π1 = 0, then H2 and π2 are isomorphic.

some of these general principles do not always hold for bad spaces. e.g. if one cuts apart a circle and then closes it up by a clasp formed by part of the topologists sine curve, one gets a space that admits no non trivial map of a circle, so π1 = 0. contrary to the representation theorem of brown mentioned above however, there are non trivial maps from this odd space to the circle. In particular this odd bracelet is simply connected, but its complement in the riemann sphere has 2 components.

this contradicts a principle i have had in mind from reading ahlfors book on complex analysis where he uses for certain subsets of the complex numbers, a definition of simple connectedness that equates it with the complement in the sphere being connected. It seems he used this definiion only for bounded open subsets.

There is something here about K(Z, $\infty$) being Complex Projective space or something. Is that a sort of direct limit? You know how that works?

 

[Infrared]

There is something here about K(Z, $\infty$) being Complex Projective space or something. Is that a sort of direct limit? You know how that works?

There is a fibration (even a fiber bundle) S^1\to S^\infty\to\mathbb{C}P^\infty. The long exact sequence for a fibration then tells you that \mathbb{C}P^\infty is K(\mathbb{Z},2) since S^\infty is contractible.

 

[mathwonk]

thank you! ( I just repeated something i read and have not gone through the calculation. Thanks for reminding of the value of the Hopf fibration!)

as to the invariance of domain theorem lavinia emphasized, i have been thinking about it and realized it follows for the jordan curve theorem. so here is another case of a beautiful theorem that is an easy corollary of another harder theorem. i.e. a basic (fairly easy) principle in topology is that a bijective continuous map of compact hausdorff spaces is a homeomorphism, i.e. is an open map. (for compact hausdorff spaces, closed is equivalent to compact, and compact sets are preserved by continuous maps so all continuous maps are also closed.) so for (weak) invariance of domain we want to prove that a continuous bijection of R^n is a homeomorphism, i.e. is open.

So consider the image of a closed sphere. It is a closed subset homeomorphic to a sphere, hence by jordan, separates R^n into two connected components, both open and one bounded. then ask for the image of the interior of the sphere. It must be connected and bounded since the image of the closed ball is compact hence bounded. Moreover the image of the exterior of the sphere must be connected hence must be the unbounded component. Hence the map sends the interior of the closed sphere onto the interior of the image of the sphere.

Now since a closed ball in R^n is compact hausdorff, this restricted map is a homeomorphism. it follows that the whole map is a homeomorphism of R^n. Thus the fact that a bijective continuous map of R^n is a homeomorphism, is a fairly easy corollary of the jordan curve theorem. so again we have a situation where the more basic and underlying theorem is the jordan curve theorem, dealing with connectivity.

 

[lavinia]

thank you! ( I just repeated something i read and have not gone through the calculation.)

as to the invariance of domain theorem lavinia emphasized, i have been thinking about it and realized it follows for the jordan curve theorem. so here is another case of a beautiful theorem that is an easy corollary of another harder theorem. i.e. a basic (fairly easy) principle in topology is that a bijective continuous map of compact hausdorff spaces is a homeomorphism, i.e. is an open map. (for compact hausdorff spaces, closed is equivalent to compact, and compact sets are preserved by continuous maps so all continuous maps are also closed.) so for (weak) invariance of domain we want to prove that a continuous bijection of R^n is a homeomorphism, i.e. is open.

So consider the image of a closed sphere. It is a closed subset homeomorphic to a sphere, hence by jordan, separates R^n into two connected components, both open and one bounded. then ask for the image of the interior of the sphere. It must be connected and bounded since the image of the closed ball is compact hence bounded. Moreover the image of the exterior of the sphere must be connected hence must be the unbounded component. Hence the map sends the interior of the closed sphere onto the interior of the image of the sphere.

Now since a closed ball in R^n is compact hausdorff, this restricted map is a homeomorphism. it follows that the whole map is a homeomorphism of R^n. Thus the fact that a bijective continuous map of R^n is a homeomorphism, is a fairly easy corollary of the jordan curve theorem. so again we have a situation where the more basic and underlying theorem is the jordan curve theorem, dealing with connectivity.

@mathwonk Nice proof. The idea is that the image of the interior of a closed ball around any point must be open by the Jordan Curve Theorem.

I am unsure about the last paragraph. It seems unnecessary.

 

[mathwonk]

I think you are right that openness has already been proved, hence homeomorphism.

One can also extend this proof to the stronger invariance theorem you mention, that an injective continuous map is open, by using a slightly stronger statement of jordan, that not only is the complement of the embedded sphere consisting of two open components, but the sphere actually wraps around each point of the bounded component, hence the map on the closed ball must be onto every such interior point, so again, without bijectiveness, we get the interior of the ball mapping onto the open bounded component. Maybe this is also unnecessary, but I did not see it.

Anyway, I like your emphasis on what this proves, as it is in general interesting to know in what categories a bijective morphism is an isomorphism. It holds e.g. in compact hausdorff spaces, and in banach spaces, and groups of course, but your observation seems to show it is also true for n dimensional manifolds. It fails in the presence of certain singularities, since in algebraic geometry a smooth curve can map bijectively onto a curve with a cusp, but not isomorphically. In algebraic geometry, "normality" is sufficient, but smoothness, i.e. non singularity is of course ok too.

 

[lavinia]

Cohomology however which is often algebraically dual to homology is, for nice spaces (cw complexes), given in each dimension n, by homotopy classes of maps to a certain nth degree "classifying space". In dim one, that space is the circle, so that says that 1st homology, or essentially equivalently maps of the circle into the space, are the same as homotopy classes of maps of the space to the circle.

For the case of closed orientable smooth $n$-manifolds, most of the duality follows from the Implicit Function Theorem together with Sard's Theorem. If $x$ is a regular value of a smooth map $f:M→S^1$ then the Implicit Function Theorem says that $f^{-1}(x)$ is a smooth closed codimension 1 submanifold. This hypersurface is orientable since its normal bundle is trivial and $M$ is orientable. The Poincare dual of $f^{-1}(x)$ is an element of $H^{1}(M;Z)$.

Sard's Theorem says that the critical values of a smooth function are closed and have measure zero so regular values always exist and one always has these hypersurfaces. The homology classes of these hypersurfaces are the same for any regular value of $f$ and are the same for any function $g$ that is smoothly homotopic to $f$. Conversely, if two functions yield the same oriented hypersurface, then they are smoothly homotopic. The proofs again just use the Implicit Function Theorem. One gets an injective map from homotopy classes of maps of $M$ into the circle into the first integer cohomology group of $M$.

Using the Tubular Neighborhood Theorem, it can also be shown that any closed orientable hypersurface is the inverse image of a regular value of some function into the circle. This implies that the map is also surjective since every codimension 1 integer homology class of an orientable smooth manifold is represented by an orientable closed hypersurface.

Note:
One can use the naturality of Poincare duality to show that these hypersurfaces are Poincare dual to the pull back of the fundamental integer cohomology class of the circle. Any regular value is Poincare dual to the fundamental class of the circle and the induced hypersurface is dual to the pull back of the fundamental class. This guarantees homotopy invariance and independence of the regular value.

 

[Infrared]

Anyway, I like your emphasis on what this proves, as it is in general interesting to know in what categories a bijective morphism is an isomorphism. It holds e.g. in compact hausdorff spaces, and in banach spaces, and groups of course, but your observation seems to show it is also true for n dimensional manifolds.

It's not true for manifolds. Map [0,1) to the unit circle using the complex exponential. Any smooth bijection with a critical point is also a counterexample in the smooth case (like the cubic function on \mathbb{R}).

 

[lavinia]

It's not true for manifolds. Map [0,1) to the unit circle using the complex exponential. Any smooth bijection with a critical point is also a counterexample in the smooth case (like the cubic function on \mathbb{R}).

I think an $n$ dimensional manifold is meant to be without a boundary. In that case, the theorem is true.