Examples of 2-manifold homotopic but not homeomorphic

[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.)