Examples of 2-manifold homotopic but not homeomorphic

[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.

 

[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 ?

I thought you might like to think about this proof that π$_{3}$(S$^2$) is not zero. I will only give the idea of the proof.

The Hopf fibration maps S$^3$ into S$^2$ by sending entire circles to points on the 2 sphere. These circles are the intersections of the 3 sphere with complex lines in complex 2-space, C$^2$. A way to visualize these circles is to think of the unit length tangent circles to the 2 sphere. Each tangent circle projects to the point of tangency. Each of the circles in the Hopf vibration maps naturally onto one of the tangent circles by wrapping around it twice. (If you like Lie groups this two fold wrapping is the map from SU(2) onto SO(3).)

If one takes two points on the two sphere then there are two circles in the three sphere above them. One can show -e.g. using stereographic projection - that these two circles are linked like two linked rings. If the Hopf fibration were null homotopic, then these circles would not be linked. This was Hopf's original idea - I think.

Another proof requires some calculus on manifolds. This goes as follows.

if F:S$^3$ -> S$^2$ is any smooth map and ω is the volume element of the 2 sphere then
F*(ω) is an exact form because the three sphere has zero second real cohomology group.

So F*(ω) = dα for some 1 form, α. So α∧dα is a 3-form and thus can be integrated over the three sphere. One can show using Stokes theorem - actually quite easy - that the integral of α ∧dα is a smooth homotopy invariant. For the trivial map that sends the 3 sphere to a single point on the 2 sphere this integral is clearly zero. It is not hard to show that for the Hopf map, the integral is equal to 1. One can show that this integral is actually computing the linking of pairs of circles but the proof is messy.

 

[mathwonk]

removing a point from R makes it non connected but not for R^2. hence R and R^2 are not homeomorphic. My apologies if this was already noted.

 

[mathwonk]

for more detail on lavinia's beautiful answer, see Spivak Calculus on Manifolds, page 132-134, or Courant Differential and Integral Calculus vol. 2, page 409-411. (The linking number of two loops equals the oriented number of intersection points of one of the loops with any surface bordered by the other loop.)