Meaning of isomorphism/diffeomorphism ## f: R^n\to M^m##

[shooride]

Can one define an isomorphism/diffeomorphism map ## f: R^n\to M^m## when ##n>m##? ##M## is a non-compact Riemannian manifold..

 

[andrewkirk]

Yes for diffeomorphism. The projection map is a diffeomorphism, eg ##f:\mathbb{R}^2\to\mathbb{R}## such that ##f((x,y))=x##.

What do you mean by isomorphism though? What is the algebraic structure that you are wanting to preserve?

If you define algebraic structures on domain and range, and find an isomorphism, it will not also be a diffeomorphism, because that would make it a homeomorphism, and there is a theorem that a n-dimensional manifold is not homeomorphic to a m-dimensional manifold if ##m\neq n##.

 

[WWGD]

Yes for diffeomorphism. The projection map is a diffeomorphism, eg ##f:\mathbb{R}^2\to\mathbb{R}## such that ##f((x,y))=x##.

What do you mean by isomorphism though? What is the algebraic structure that you are wanting to preserve?

If you define algebraic structures on domain and range, and find an isomorphism, it will not also be a diffeomorphism, because that would make it a homeomorphism, and there is a theorem that a n-dimensional manifold is not homeomorphic to a m-dimensional manifold if ##m\neq n##.

What do you mean, isn't a diffeomorphism invertible? ##p^{-1}(x_0)=(x_0, \mathbb R ) ##.

 

[WWGD]

Can one define an isomorphism/diffeomorphism map ## f: R^n\to M^m## when ##n>m##? ##M## is a non-compact Riemannian manifold..

No, dimension is a diffeomorphism invariant. Take, e.g., the Jacobian to see why this is not possible. Dimension ( I think this is the Lebesgue dimension) is a topological, i.e., homeomorphism - invariant

 

[andrewkirk]

What do you mean, isn't a diffeomorphism invertible?

https://cdn.mathjax.org/mathjax/latest/fonts/HTML-CSS/TeX/png/Main/Regular/085/2212.png?rev=2.5.3

Right you are. I forgot that part.

In that case there cannot be a diffeomorphism.

Edit: Ah, as I see you have noted in your next post.

 

[WWGD]

No problem, Andrew, I have made a few of those -- just hit the delete button, and I will delete my post; let me know.

 

[lavinia]

A key theorem that answers this question is Brouwer's Invariance of Domain which says that a continuous injective image of an open set of Rn into Rn is an open subset of Rn and that the mapping is a homeomorphism. It is not hard to show as a corollary that there is no continuous injection of Rn into Rn-m m>0. A differential map is continuous so the theorem applies to differential maps as well.

BTW: This question is really a question about topology not differential geometry.

 

[WWGD]

A key theorem that answers this question is Brouwer's Invariance of Domain which says that a continuous injective image of an open set of Rn into Rn is an open subset of Rn and that the mapping is a homeomorphism. It is not hard to show as a corollary that there is not continuous injection of Rn into Rn-m m>0. A differential map is continuous so the theorem applies to differential maps as well.

BTW: This question is really a question about topology not differential geometry.

But if you're dealing with diffeomorphisms, you can use tools not available with topology alone: a diffeomorphism gives you a vector space isomorphism between tangent spaces at a point and at its image. By properties of linear maps, both dimensions are the same. OF course, a homeomorphism does not exist either, by, e.g.,as you stated, invariance of domain. You can twist this using the result that continuous bijection between compact and Hausdorff is a homeo. Consider the closed unit ball B(0,1).Then f: B(0,1)--> f(B(0,1)) is a homeo. Now consider a small open disk in B(0,1), which is sent to an open set in R^n , contradicting Invariance of Domain.

 

[lavinia]

But if you're dealing with diffeomorphisms, you can use tools not available with topology alone: a diffeomorphism gives you a vector space isomorphism between tangent spaces at a point and at its image.

That is true.

By properties of linear maps, both dimensions are the same. en set in R^n , contradicting Invariance of Domain.

Not sure what you mean here.

 

[zinq]

Can one define an isomorphism/diffeomorphism map ## f: R^n\to M^m## when ##n>m##? ##M## is a non-compact Riemannian manifold..

Actually, any two manifolds that are topologically equivalent* must have the same dimension, so the answer here is No.

Possibly the simplest proof of this uses homology, which is a way of assigning one abelian group H_k(X) — which can at times be the trivial group {0} — to each topological space and choice of positive integer k, in such a way that topologically equivalent spaces have isomorphic groups. A mild generalization of this idea assigns an abelian group, likewise, to every pair of topological spaces (X,A) where A is a subspace of X, and every positive integer k. (This is called a relative homology group.)

Then if M and N are topologically equivalent manifolds, there must be a homeomorphism h: M —> N. This implies that for any point p of M, the relative homology groups H_k(M,M-p) and H_k(N,N-h(p)) are isomorphic.

If the dimensions dim(M) and dim(N) were not equal, let's assume, without loss of generality, that d = dim(M) > dim(N). It can be shown that for any manifold M of dimension d, the relative group H_d(M,M-p) is isomorphic to the group ℤ of integers. It can also be shown that for any manifold N with d greater than the dimension of N, the homology group H_d(N,N-q) is the trivial group {0}, where q is any point of N.

But if M and N were homeomorphic they would have isomorphic relative groups. Since they do not, this shows that any two manifolds of unequal dimensions cannot be homeomorphic.
____________
* Note: Two smooth manifolds might be "homeomorphic" (topologically equivalent) or even diffeomorphic (having a smooth homeomorphism between them, with a smooth inverse) but the word "isomorphic" is not used for topological spaces.

 

[WWGD]

Actually, any two manifolds that are topologically equivalent* must have the same dimension, so the answer here is No.

Possibly the simplest proof of this uses homology, which is a way of assigning one abelian group H_k(X) — which can at times be the trivial group {0} — to each topological space and choice of positive integer k, in such a way that topologically equivalent spaces have isomorphic groups. A mild generalization of this idea assigns an abelian group, likewise, to every pair of topological spaces (X,A) where A is a subspace of X, and every positive integer k. (This is called a relative homology group.)

Then if M and N are topologically equivalent manifolds, there must be a homeomorphism h: M —> N. This implies that for any point p of M, the relative homology groups H_k(M,M-p) and H_k(N,N-h(p)) are isomorphic.

If the dimensions dim(M) and dim(N) were not equal, let's assume, without loss of generality, that d = dim(M) > dim(N). It can be shown that for any manifold M of dimension d, the relative group H_d(M,M-p) is isomorphic to the group ℤ of integers. It can also be shown that for any manifold N with d greater than the dimension of N, the homology group H_d(N,N-q) is the trivial group {0}, where q is any point of N.

But if M and N were homeomorphic they would have isomorphic relative groups. Since they do not, this shows that any two manifolds of unequal dimensions cannot be homeomorphic.
____________
* Note: Two smooth manifolds might be "homeomorphic" (topologically equivalent) or even diffeomorphic (having a smooth homeomorphism between them, with a smooth inverse) but the word "isomorphic" is not used for topological spaces.

I was trying to avoid (co)homology; I thought if the OP was familiar with them, s/he would not have asked the question. And I guess you could also use (c0)homology of the whole space , so that if M>N , then , if M is orientable, its highest homology, say m is, as in the case of the relative homology you used ## \mathbb Z ## , while, again as you said, the m-th homology (and anyone larger than Dim N) would be 0.

 

[andrewkirk]

I would also tend to avoid purely topological explanations. While they have wider application, they can be much more abstract and conceptually demanding. If the context is Riemannian manifolds, there's no need for them.

Indeed, my current feeling is that homology theory is the most abstract topic in mathematics I have studied. I love it, but to me it seems much less intuitive than diffeomorphisms between Riemannian manifolds, which are quite visual. One of the amazing things about homology theory is that one can achieve such practical, intuitive results, using such abstract mathematical structure, and that there is at present no easier known way to achieve such results. So it's not just abstruseness for the sake of abstruseness, as some branches of mathematics sometimes appear to be.

 

[mathwonk]

actually I can define a diffeomorphism betwen any two manifolds, it is a differentiable map with a differentiable inverse. I just can't prove one exists.

 

[zinq]

I would also tend to avoid purely topological explanations. While they have wider application, they can be much more abstract and conceptually demanding. If the context is Riemannian manifolds, there's no need for them.

Indeed, my current feeling is that homology theory is the most abstract topic in mathematics I have studied. I love it, but to me it seems much less intuitive than diffeomorphisms between Riemannian manifolds, which are quite visual. One of the amazing things about homology theory is that one can achieve such practical, intuitive results, using such abstract mathematical structure, and that there is at present no easier known way to achieve such results. So it's not just abstruseness for the sake of abstruseness, as some branches of mathematics sometimes appear to be.

I recommend that you close your eyes when coming upon a topological explanation. Comparing diffeomorphisms with homology groups is like saying you like beautiful landscapes more than GPS devices. (Translation: You can't prove much with just a diffeomorphism. )

And as for "If the context is Riemannian manifolds, there's no need for them," just maybe you don't know all the ways that topological explanations can provide useful information in that context?

 

[lavinia]

I was trying to avoid (co)homology; I thought if the OP was familiar with them, s/he would not have asked the question. And I guess you could also use (c0)homology of the whole space , so that if M>N , then , if M is orientable, its highest homology, say m is, as in the case of the relative homology you used ## \mathbb Z ## , while, again as you said, the m-th homology (and anyone larger than Dim N) would be 0.

The manifolds may not be compact so highest homology may be zero For instance the homology of Rn is zero in all positive dimensions.
But Hn(M,M-p) = Z for any n manifold. This follows from excision.

Let A be an open neighborhood of p that is homeomorphic to an open ball in Rn. By excision,
Hq(M,M-p) = Hq(M-(M-A)),M-p-(M-A)) = Hq(A,A-p) for all q including n. A is contractible and A-p deforms onto a sphere so the exact sequence of the pair (A,A-p)

... Hq(A) -> Hq(A,A-p) -> Hq-1(A-p) -> Hq-1(A) ...

is

0 -> Hq(A,A-p) -> Hq-1(Sn-1) -> 0 except when q = 1.

 

[lavinia]

I would also tend to avoid purely topological explanations. While they have wider application, they can be much more abstract and conceptually demanding. If the context is Riemannian manifolds, there's no need for them.

Indeed, my current feeling is that homology theory is the most abstract topic in mathematics I have studied. I love it, but to me it seems much less intuitive than diffeomorphisms between Riemannian manifolds, which are quite visual. One of the amazing things about homology theory is that one can achieve such practical, intuitive results, using such abstract mathematical structure, and that there is at present no easier known way to achieve such results. So it's not just abstruseness for the sake of abstruseness, as some branches of mathematics sometimes appear to be.

To say that there is no need for topological proofs when one is dealing with smooth manifolds ignores much of the research in mathematics that deals with the interplay between algebraic topology, differential topology, and differential geometry.

BTW: A Riemannian manifold refers to a smooth manifold together with a Riemannian metric. Your question does not involve a Riemannian metric and is not a question about Riemannian geometry. It is a question about topology. WWGD's proof uses calculus on manifolds not Riemannian geometry. It is a differential topology proof. Zinq's proof is an algebraic topology proof.

 

[WWGD]

To say that there is no need for topological proofs when one is dealing with smooth manifolds ignores much of the research in mathematics that deals with the interplay between algebraic topology, differential topology, and differential geometry.

BTW: A Riemannian manifold refers to a smooth manifold together with a Riemannian metric. Your question does not involve a Riemannian metric and is not a question about Riemannian geometry. It is a question about topology. WWGD's proof uses calculus on manifolds not Riemannian geometry. It is a differential topology proof. Zinq's proof is an algebraic topology proof.

But if it is not possible to do it without a Riemannian metric, it is not possible to do it with one. So the Riemannian condition is somewhat superfluous. Is there some result that an n-dimensional Riemannian manifold cannot be equivalent to an m-dimensional one that only uses properties of the metric?

 

[lavinia]

One point that should be made here is that smooth manifolds may be topologically equivalent but not diffeomorphic. For such manifolds, as mathwonk pointed out, it is possible to define smooth equivalence but not possible find one.

The moral is that equivalence is relative to the category of mathematical objects. In the smooth category equivalence means diffeomorphic. In the topological category it mean homeomorphic. In the PL category it means piecewise linearly isomorphic.

 

[WWGD]

I guess the standard example of homeo but not diffeo is that of ## \mathbb R ## and ## (x,x^3): x \in \mathbb R ##.

 

[lavinia]

I guess the standard example of homeo but not diffeo is that of ## \mathbb R ## and ## (x,x^3): x \in \mathbb R ##.

all connected closed smooth 1 manifolds are either diffeomorphic to R or to the circle.

 

[WWGD]

But the map f: ##\mathbb R \to \mathbb R ## given by ## f(x)=x^3 ## is not a diffeomorphism; the inverse map is not differentiable at 0. I think that the differentiable structures are equivalent, but this is not a diffeomorphism.

 

[zinq]

As for examples of smooth manifolds that are homeomorphic but not diffeomorphic:

The history of this is very interesting. It was not suspected that homeomorphic smooth manifolds might not always be diffeomorphic until 1956, when quite by accident John Milnor discovered that the 7-sphere S⁷ (the unit sphere in 8-dimensional euclidean space R⁸) as a topological space can be given a non-standard differentiable structure — also known as a smooth structure — for which there is no diffeomorphism with the standard S⁷.

(In fact it turns out that there are exactly 15 distinct smooth structures on S⁷, where two structures are called "distinct" if there is no diffeomorphism between them. If all smooth structures on a given topological manifold are diffeomorphic, the structure is called "unique".)

It is known that there is a unique smooth structure on S¹, S², S³, S⁵, S⁶, S¹², and S⁶¹. At present, no other-dimensional spheres are known to have a unique smooth structure — so having "exotic" smooth structures is overwhelmingly more the rule than the exception for spheres.

The most pressing problem in differential topology today is whether the 4-sphere S⁴ has any exotic smooth structure. Surprisingly, this has turned out to be the most difficult dimension to deal with.

 

[WWGD]

As for examples of smooth manifolds that are homeomorphic but not diffeomorphic:

The history of this is very interesting. It was not suspected that homeomorphic smooth manifolds might not always be diffeomorphic until 1956, when quite by accident John Milnor discovered that the 7-sphere S⁷ (the unit sphere in 8-dimensional euclidean space R⁸) as a topological space can be given a non-standard differentiable structure — also known as a smooth structure — for which there is no diffeomorphism with the standard S⁷. (In fact it turns out that there are exactly 15 distinct smooth structures on S⁷, where two structures are called "distinct" if there is no diffeomorphism between them.)

It is known that there is a unique smooth structure on S¹, S², S³, S⁵, S⁶, S¹², S⁶¹. At present, no other-dimensional spheres are known to have a unique smooth structure — so having "exotic" smooth structures is overwhelmingly more the rule than the exception for spheres.

The most pressing problem in differential topology today is whether the 4-sphere S⁴ has any exotic smooth structure. Surprisingly, this has turned out to be the most difficult dimension to deal with.

I think 4 is special because it is large enough to allow for bizarreness but not so large so that bizarreness can be done away with, e.g., Whitney trick.

 

[mathwonk]

WWGD, your statement in post #21, amounts to saying that projection of the graph {(x,x^3), all real x} onto the y-axis is not a diffeomorphism. Projection onto the x-axis however is a diffeomorphism.

 

[WWGD]

WWGD, your statement in post #21, amounts to saying that projection of the graph {(x,x^3), all real x} onto the y-axis is not a diffeomorphism. Projection onto the x-axis however is a diffeomorphism.

But don't we need both to be diffeomorphisms to have the map from ## \mathbb R ## to itself to be a diffeomorphism?

 

[lavinia]

But don't we need both to be diffeomorphisms to have the map from ## \mathbb R ## to itself to be a diffeomorphism?

Here is a hint.

You have given the real line two differentiable structures. The first is the maximal atlas of charts compatible with the identity map - viewed as a chart.
The second is the maximal atlas of charts compatible with x -> x^1/3.

- Show that these two structures are different. This means that the two atlases are not compatible with each other.
- Show as mathwonk said that the two structures are diffeomorphic and that x -> x^3 is in fact a diffeomorpism from the first structure to the second.

So two different differentiable structures can still be diffeomorphic. On the 7 sphere they are not only different but they are also not diffeomorphic.

 

[mathwonk]

yes the map from R to itself defined by x-->x^3 is not a difeomeorphism, or equivalently the map from R to itself defined by composing the maps x-->(x,x^3)-->x^3 is not a diffeomorphism. Thus this is indeed an example of a homeomorphism that is not a diffeomorphism. The first map x-->(x,x^3) however is a difeomorphism.

Thus the set of points {(x,x^3), all real x} is not an example of a manifold that is homeomorphic but not diffeomorphic to R, i.e. it is diffeomeorphic to R. That means not that there cannot be a map between them which is a homeomorphism and not a diffeomorphism, rather it means that there is some map which is a diffeomorphism. So even if there is a map which is a homeomorphism and not a difeomeorphism, as long as there is another map which is a diffeomeorphism, then the two manifolds are diffeomeorphic. So it is possible for two diffeomorphic manifolds to admit one map which is a siffeomorphism, and also admit another map which is a homeomorphism and not a diffeomorphism.

The projection map (x,x^3)-->x^3 is a homeomorphism and not a difeomorphism (to the y axis), but the projection map (x,x^3)-->x, is a diffeomorphism from the graph {(x,x^3) all x in R} to the x axis, with inverse diffeomorphism the map x-->(x,x^3). I.e. the map x-->(x,x^3) has two coordinate functions, both differentiable, hence it is differentiable, and the projection (x,x^3)-->x is the restriction of a linear map hence also differentiable.

Lavinia's point is different, that there is a diffeomorphism from the usual structure on R to the one given by composing with the homeomorphism x-->x^3. This is even more interesting.

 

[WWGD]

Here is a hint.

You have given the real line two differentiable structures. The first is the maximal atlas of charts compatible with the identity map - viewed as a chart.
The second is the maximal atlas of charts compatible with x -> x^1/3.

- Show that these two structures are different. This means that the two atlases are not compatible with each other.
- Show as mathwonk said that the two structures are diffeomorphic and that x -> x^3 is in fact a diffeomorpism from the first structure to the second.

EDIT: So two different differentiable structures can still be diffeomorphic. On the 7 sphere they are not only different but they are also not diffeomorphic.

I understand, I am aware of the existence of a unique maximal differentiable structure for the Reals, that the two structures are diffeomorphic .I was mentioning that while the differentiable structures are diffeomorphic, I believe, this map is not a diffeomorphism. This is all I meant.

 

[mathwonk]

your post "I guess the standard example of homeo but not diffeo is that of R and (x,x3):x∈R." does not complete the words homeo and diffeo to morphic or morphism, but since you gave examples of manifolds and not maps it seems natural to assume you meant homeomorphic and not diffeomeorphic. and that is incorrect.

 

[WWGD]

your post "I guess the standard example of homeo but not diffeo is that of R and (x,x3):x∈R." does not complete the words homeo and diffeo to morphic or morphism, but since you gave examples of manifolds and not maps it seems natural to assume you meant homeomorphic and not diffeomeorphic. and that is incorrect.

How so? I never claimed the two were not homeomorphic. I only said, or t least meant to say that the map is not a diffeomorphism. This is correct.

 

[lavinia]

I understand, I am aware of the existence of a unique maximal differentiable structure for the Reals, that the two structures are diffeomorphic .I was mentioning that while the differentiable structures are diffeomorphic, I believe, this map is not a diffeomorphism. This is all I meant.

Right. I misread your statement. I thought you were saying that if one takes x^1/3 as a chart then one would obtain a different smooth version of R. My mistake. I thought you were responding to my statement about smooth manifolds that are topologically equivalent but not diffeomorphic.

Interestingly, in the case of the two different differentiable structures on R, the identity map is not a diffeomorphism but x^3 is while in your case the identity is a diffeomorphism but x^3 is not.

 

[WWGD]

No problem, I have made plenty of mistakes myself .

 

[mathwonk]

WWGD: I am just saying we were left to fill in your statement, since you used only partial words. And since you did not present a map at all, we were not inclined to realize you were thinking of a particular map.

I.e. you said:

"I guess the standard example of homeo but not diffeo is that of R and (x,x3):x∈R."

That can be filled in several ways. Since you presented examples of two manifolds, but no map between them, I thought you meant to say:

" "I guess the standard example of homeo[morphic] but not diffeo[morphic manifolds] is that of R and (x,x3):x∈R." which is false.

You however apparently meant to say:

"I guess the standard example of [a] homeo[morphism] but not [a] diffeo[morphism] is that of [the horizontal projection between the y axis, i.e.] R and [the graph] (x,x3):x∈R." which is true.

So I made the mistake of assuming I understood what you meant.

 

[WWGD]

OK, yes, I was sloppy in this post; I will be more careful from now on, sorry.

 

[zinq]

I can't help but mention a case of too-low differentiability that the graph of y = x^3 is related to:

Consider this graph and its copies after being translated by each possible amount to the right or left. In other words, the set of curves

G_c = {((x, (x-c)³ | x ∈ ℝ}

for each real number c.

It's clear that these curve represent the trajectories of some vector field. But in too many ways! That is, one trajectory through (0,0) could be G_0 (the graph of y = x³). But another trajectory containing (0,0) of the same vector field would have to be the x-axis (since all vectors on the x-axis have slope 0).

The existence and uniqueness theorem for ordinary differential equations tells us that a C¹ vector field — one that is (at least) once-continuously differentiable — always has a unique solution for any given initial conditions.

Conclusion: Although the set of curves {G_c | c ∈ ℝ} seem as smooth as can be, in fact there is no vector field tangent to them that is continuously differentiable.