Examples of 2-manifold homotopic but not homeomorphic

[jk22]

that is what I am searching.

 

[wabbit]

This depends on what constraints you put on your surfaces. Since compact surfaces are classified by their homotopy (more specifically, by their orientability and Euler characteristic), you won't find an example there.
Going to open surfaces however, an example would be the cylinder and Möbius strip : both are homotopic to a circle, but they are not homeomorphic since one is orientable and the other is not.

 

[jk22]

thanks, and where could i find the ways of calculating the homotopy groups ?

It is then only in dimension 3 that closed manifolds can be homotopic but not homeomorphic ?

 

[wabbit]

thanks, and where could i find the ways of calculating the homotopy groups ?

Careful where you step in : ). In general, this is a very difficult question. But have a look at http://www.math.psu.edu/katok_a/TOPOLOGY/Chapter2.pdf for instance.
You may also want to look up "classification of two dimensional manifolds", this should give you some interesting links.

It is then only in dimension 3 that closed manifolds can be homotopic but not homeomorphic ?

The two-dimensional case is always much simpler in such questions. Also I should have mentionned this: it is very easy to find a manifold that is homotopic but not homeomorphic to a given compact 2-manifold: it just won't be two-dimensional. (Can you see how to do that?)
As to your question, have a look at https://en.wikipedia.org/wiki/Borel_conjecture.

 

[lavinia]

thanks, and where could i find the ways of calculating the homotopy groups ?

It is then only in dimension 3 that closed manifolds can be homotopic but not homeomorphic ?

Examples exist in higher dimensions than 3, maybe in every dimension higher than 3, not sure.

An interesting sub-case is that of homotopically equivalent compact manifolds without boundary that are not homeomorphic. There are examples of 3 manifolds.

All 2-manifolds that are compact without boundary are determined by their fundamental group.

One can also ask whether a homotopy equivalence between homeomorphic manifolds is homotopic to a homeomorphism.

 

[wabbit]

The thing with classification of manifolds, for the little I remember about it, is that it is a horribly complicated subject. Dimensions 4 and 3 tend to be the trickiest : one is trivial, two is easy, and 5 and above tend to fall into general rules (not that it's a trivial topic there by any means) - but the middle ones have entire books written about them, such as http://www.springer.com/mathematics/geometry/book/978-3-540-45898-2

 

[wabbit]

where could i find the ways of calculating the homotopy groups ?

Take a deep breath, and start with www.math.cornell.edu/~hatcher/AT/ATch4.pdf for instance.

 

[WWGD]

For homotopy 3-spheres, you deal with Poincare's conjecture (a theorem now, by
Perelman , etc.).

 

[WWGD]

Actually, even for 1-manifolds , if you include ones boundary, you have the open, half-open, closed intervals that are homotopy-equivalent but not homeomorphic. But maybe the most notable examples in any dimension are the contractible spaces, which are homotopically-equivalent to a point. So you can even have spaces of different dimensions being homotopic --but trivially not homeomorphic by cardinality reasons alone.

 

[jk22]

but trivially not homeomorphic by cardinality reasons alone.

Yes the dimensions have to be the same but i thought R and R2 for example have the same cardinality ?

 

[lavinia]

Yes the dimensions have to be the same but i thought R and R2 for example have the same cardinality ?

Yes. That is correct.

R and R2 are not homeomorphic but this is a deep theorem.

 

[lavinia]

In order to take away obvious cases such as the Moebius band and the circle, restrict to the case of compact manifolds without boundary. Such spaces are never contractible and never have the same homotopy type as compact manifolds without boundary in other dimensions.

As Wabbit pointed out, this gets extremely complicated. For three manifolds there are fundamental groups which determine the manifold up to homeomorphism and others that determine the homotopy type but not the homeomorphism type. The Poincare conjecture is true in all dimensions but the methods of proof are quite different for dimensions 5 and above than in dimension 4 and dimension3.

The homotopy groups of spheres is a whole subject by itself.

Here are some problems:

- Take $R^3$ minus both the z-axis and the unit circle in the xy-plane. Prove that this space is homotopically equivalent to a torus.

- A subspace A of a topological space W is called a strong deformation retract of W if there is a homotopy F: WxI -> W such that the following 3 conditions hold:
F(x,0) is the identity map
F(x,1) is completely contained in A
The composition A -> WxI ->A where the first arrow is the inclusions a -> (a,1) and the second is the homotopy,F, is the identity map on A.

Show that A is homotopically equivalent to W

- Show that if the unit disk has a fixed point free mapping into itself then it is homotopically equivalent to the circle. In fact the circle would be a strong deformation retract of the disk. Conclude Brouwer's Fixed Point Theorem which says that every mapping of the disk into itself has a fixed point.

 

[jk22]

Thanks for the problems. I wanted to ask a question more : for pi1 we say that we take a loop and describe it by a continuous function f : [0;1]->Manifold such that f(0)=f(1)

But what if there is a knot in the loop we contract in the space ? Is a knot contractible to a point ?

I think a loop with a knot is homotopic to a circle but not homeomorphic this could be another simple example.

 

[wabbit]

R and R2 are not homeomorphic.

One thing to note is that the non homeomorphism of $R^n$ and $R^m$ is actually a nice early result of homotopy theory : once you remove one point from each, they are homotopic to the corresponding (n-1)-sphere, and the calculation of the lower* homotopy groups of spheres is one of the few easy ones you can make. If you are interested in the subject you should definitely look up this calculation which is found in pretty much any introduction to higher homotopy groups.

* Of course this being homotopy, nothing is easy, which is why I had to add "lower" to this statement : the calculation of $\pi_k(S^n)$ is easy only for $k\leq n$ - but this turns out to be sufficient for the purpose at hand.

 

[lavinia]

Thanks for the problems. I wanted to ask a question more : for pi1 we say that we take a loop and describe it by a continuous function f : [0;1]->Manifold such that f(0)=f(1)

But what if there is a knot in the loop we contract in the space ? Is a knot contractible to a point ?

In homotopy theory loops are allowed to self intersect and to pass through themselves. So a knot doesn't matter. For instance, a space filling loop that completely covers the sphere is homotopic to a point.

I suppose one could restrict the homotopies to disallow crossovers but that is not what homotopy theory assumes. I imagine in Knot theory one would make this restriction.

 

[lavinia]

One thing to note is that the non homeomorphism (and non homotopy equivalence) of $R^n$ and $R^m$ is actually a nice early result of homotopy theory : once you remove one point from each, they hare homotopic to the corresponding (n-1)-sphere, and the calculation of the lower* homotopy groups of spheres is one of the few easy ones you can make. If you are interested in the subject you should definitely look up this calculation which is found in pretty much any introduction to higher homotopy groups.

* Of course this being homotopy, nothing is easy, which is why I had to add "lower" to this statement : the calculation of $\pi_k(S^n)$ is easy only for $k\leq n$ - but this turns out to be sufficient for the purpose at hand.

Right!

 

[wabbit]

Right!

...except I had to remove the silly statement "non homotopy equivalence of $R^n$ and $R^m$ " from that post, but you have now captured my blooper for all posterity to contemplate : )
I'll leave it to the opening poster to see why it's such a silly statement.

 

[lavinia]

Here are some problems to give you an idea of what happens when you glue together manifolds with boundary.

- Show that the unit circle bundle of the 2 sphere is two solid tori glued together along their boundaries. Conclude that one can obtain real projective three space from two solid tori.

- Show that the 3 sphere can also be obtained from two solid tori glued together along their boundaries.

So you see that the gluing map matters. What other 3 manifolds can be obtained from gluing 2 solid tori together?

- Show that one can also obtain the three sphere by gluing two solid balls together along their boundaries,

 

[lavinia]

...except I had to remove the silly statement "non homotopy equivalence of $R^n$ and $R^m$ " from that post, but you have now captured my blooper for all posterity to contemplate : )
I'll leave it to the opening poster to see why it's such a silly statement.

Bloopers don't matter but I will be happy to delete if you want.

 

[wabbit]

Bloopers don't matter but I will be happy to delete if you want.

Not at all, I was just kidding, let this remain as a lesson in proof-reading before posting : )

 

[lavinia]

Not at all, I was just kidding, let this remain as a lesson in proof-reading before posting : )

BTW: Do you know the proof that the lens space L(7,1) andL(7,2) are homotopically equivalent but not homeomorphic?

 

[wabbit]

BTW: Do you know the proof that the lens space L(7,1) andL(7,2) are homotopically equivalent but not homeomorphic?

Oh dear ! I don't even remember what lens spaces are !

Edit : If I may briefly wander off-topic, a much more pedestrian question : how do you get the usual symbols for $N,Z,R,C$... in a post ? \mathbb R doesn't seem to work for me.

Edit : Thanks jk22 , \mathbb{R} $\rightarrow \mathbb{R}$ works indeed ! Praise the Curly Braces !

 

[jk22]

$$\mathbb R $$

\mathbb{R} seems to work

I'm completely new to homotopy theory but I do not understand how pi3(s2) can be not trivial since we cannot put a 3-sphere in a 2-sphere ?

 

[lavinia]

$$\mathbb R $$

\mathbb{R} seems to work

I'm completely new to homotopy theory but I do not understand how pi3(s2) can be not trivial since we cannot put a 3-sphere in a 2-sphere ?

This shows that homotopy is different from embedding and immersion.

It is a good exercise to show that the Hopf fibration - which is a map of the 3 sphere onto the 2 sphere - is not null homotopic.

 

[wabbit]

This shows that homotopy is different from embedding and immersion

Also, jk22, generally you'll find that such "natural, as expected" results do hold in homology theory, which is in a way "homotopy without all the complications", but not in homotopy theory : the homology groups turn out to be more or less the homotopy groups where you remove all cyclical elements, and you might say that "the devil is in the cycles" here.
I don't know if you're familiar at all with homology, but if not that would certainly be something worth looking at as a complement to your interest in homotopy.

 

[jk22]

Yes, my misconception comes from my not knowing of the definition of homotopy : it is a continuous function from the unit interval cross X to Y, and I thought that X had to be in Y. But this is not the case.

 

[jk22]

btw what is the pi1 group of the double donut ? I arrive at at least z^10

 

[wabbit]

Wow how do you arrive at that ?

 

[jk22]

Well in the plane it is z^3 but i tried to visualize in dimension 3 but i lost myself in the counting.

 

[wabbit]

If I understand what you mean by "it's $\mathbb{Z}^3$ in the plane", then no, it isn't. What is $\pi_1(\mathbb{R}^n)$ ?
I could be wrong, but it sounds like you may not be using a textbook - if so, you need to get one (or an introduction available online, maybe start with one mentionned in previous posts), you won't get far without that.

 

[jk22]

If I understand what you mean by "it's $\mathbb{Z}^3$ in the plane", then no, it isn't. What is $\pi_1(\mathbb{R}^n)$ ?
.

This should be {0} since every loop is contractible to a point ?

Indeed I have no textbook but look online material.

By in the plane i meant a two holed plane. I reasoned like the loop can involve 1 point there are two possibilities or corcumscribed the two holes. What i haven't looked at is the group operation.

 

[wabbit]

This should be {0} since every loop is contractible to a point ?

Right.

By in the plane i meant a two holed plane.

OK, makes more sense. Still wrong, though : )

 

[jk22]

So It is the free group F2 for the 2 holed plane ?

 

[wabbit]

So It is the free group F2 for the 2 holed plane ?

Sounds right, going around either hole is a generator.

 

[lavinia]

A useful theorem for computing fundamental groups is Van Kampen's theorem. It follows immediately from it that the fundamental group of a figure 8 is the free group on two generators.

Exercise. Show that the plane with two holes in it is homotopy equivalent to a figure 8.
Show that a solid two handled torus is homotopy equivalent to a figues 8.

The fundamental group of the two handled torus (not solid) is not a free group. Try to compute its fundamental group.

 

[wabbit]

Hello again lavinia - I was wondering about something : to exhibit a homomorphism from the fundamental group of the two-holed plane to that of the torus, I was thinking one might use the map induced by $\mathbb{C}\backslash\{0,{1\over2}\}\rightarrow\mathbb{C}/\mathbb{Z}^2:z\rightarrow\frac{1}{z(z-{1\over2})}$.
Do you think that would work?

 

[lavinia]

Hello again lavinia - I was wondering about something : to exhibit a homomorphism from the fundamental group of the two-holed plane to that of the torus, I was thinking one might use the map induced by $\mathbb{C}\backslash\{0,{1\over2}\}\rightarrow\mathbb{C}/\mathbb{Z}^2:z\rightarrow\frac{1}{z(z-{1\over2})}$.
Do you think that would work?

The map will certainly give some homomorphism but probably not the one you want.

Its image is the plane minus the origin which has fundamental group,Z. So the homomorphism onto the fundamental group of the torus can not be surjective.
Thus is you were trying to abelianize the free group on two generators to get $Z^2$ that fails.

An explicit computation would invoke computing the winding numbers of the images of small positively oriented circles centered at the origin and at 1/2 .

I didn't check it carefully but their images of seem to wind around the origin once counter clockwise .

For instance the circle of radius 1/4 centered at the origin is mapped to

$1/4e^{-iθ}/(1/4e^{iθ} -1/2)$

so $dw/w = - i - {1/4e^{iθ}/(1/4e^{iθ} - 1/2)}$

 

[wabbit]

Thanks for your reply, indeed I was trying to exhibit an explicit morphism showing abelization at work. But after reading your comments I think it was just a poor attempt (just one detail, though, the map is in fact surjective onto the torus $\mathbb{C}/\mathbb{Z}^2$, but otherwise it just isn't right). What we'd want is a map sending say a small circle around 0 to a horizontal line, and a circle around 1/2 to a vertical line, but I don't see an easy way to do that.

 

[lavinia]

Thanks for your reply, indeed I was trying to exhibit an explicit morphism showing abelization at work. But after reading your comments I think it was just a poor attempt (just one detail, though, the map is in fact surjective onto the torus $\mathbb{C}/\mathbb{Z}^2$, but otherwise it just isn't right). What we'd want is a map sending say a small circle around 0 to a horizontal line, and a circle around 1/2 to a vertical line, but I don't see an easy way to do that.

Yes it is surjective onto the torus but not onto the plane. The homomorphism of fundamental groups factors through the fundamental group of the plane minus the origin.

I apologize though for some muddled thinking.

Any mapping of any space into the torus that factors through the plane must induce the zero homomorphism on fundamental groups because the plane is simply connected. Here factoring through the plane means throughout the covering of the torus by the plane.

 

[wabbit]

Not so nice as an analytical function perhaps, but a way to exhibit this abelization morphism is as follow :

Take a figure eight made of two loops of string. Now embed it in (tie it around) a doughnut so that one loop goes around a "small circle" and the other goes around a "great circle". There you get the morphism I was talking about. Interesting to think about how eactly this makes the original free generators commute.

 

[wabbit]

Yes it is surjective onto the torus but not onto the plane. The homomorphism of fundamental groups factors through the fundamental group of the plane minus the origin.

Indeed ! I wasn't seeing why I couldn't get it right - of course you are right, this is the fundamental obstacle - thanks.

 

[lavinia]

Not so nice as an analytical function perhaps, but a way to exhibit this abelization morphism is as follow :

Take a figure eight made of two loops of string. Now embed it in a doughnut so that one loop goes around a "small circle" and the other goes around a "great circle". There you get the morphism I was talking about. Interesting to think about how eactly this makes the original free generators commute.

That works.

If the two loops are denoted a and b, then by definition the torus is constructed by attaching a 2 disk along its boundary circle to the loop $aba^{-1}b^{-1}$ in the figure eight. So this commutator must be zero in the torus.

 

[lavinia]

Oh I wanted this left as an exercise for op : )

yikes. Sorry.

Try the two handled torus.

 

[wabbit]

You are diabolical : )

 

[jk22]

I have a question i consider the homotopy of the 8 and the plane with two holes. If i consider a loop around both points in the plane this cannot be mapped continuously to 2 circle around each point. So i thought there should be 3 generators for the plane with 2 holes ?

 

[lavinia]

I have a question i consider the homotopy of the 8 and the plane with two holes. If i consider a loop around both points in the plane this cannot be mapped continuously to 2 circle around each point. So i thought there should be 3 generators for the plane with 2 holes ?

Two generators only. The loop that encloses both holes deforms to the figure eight.

 

[jk22]

This deformation is not inversible but it does not need to be ?

 

[wabbit]

What do you mean by inversible here?

 

[jk22]

That the loop 8 cannot be transformed continuously to a circle. (Since one point should be transformed in two points)

In fact the relation deforms to is not commutative.

 

[wabbit]

That the loop 8 cannot be transformed continuously to a circle.

Right, in the sense that the figure 8 is not homotopy equivalent to the circle.
But this is not what we need here, only that a certain path going around the figure 8 can be deformed continuously to a path going around a circle enclosing both holes - "path" here does not designate a space, but a map from the circle to a space.

In fact the relation deforms to is not commutative.

Ah but it is commutative in homotopy, whether you consider the relation "space X is homotopy equivalent to space Y", or the (different) one used here, "path 1 is homotopic to path 2"
Other notions of deformation may very well not be commutative, but they are not at at play here.