4
$\begingroup$

I'm asked to prove that the following is a CW structure for the 3-sphere, (as a part of an exercise involving defining the Cw structure of the Lens Spaces) I'm asked to prove that the following is a CW decomposition

enter image description here

But in order to visualise it I started computing an homeomorphism between the interior of any $2$-cell and a $2$-disk and guessing where the $2$-cells are attached. Without much results. Same problem for the $3$-cells.

I tried looking in some other alg. top. books like Hatcher's in order to see if there are some explicit computations and I found a more geometric interpretation. Which should help clarify, but I cannot work with it properly:

enter image description here

With Hatcher's description is not clear to me where the $2$-cells are attached and how, moreover it seems that they are attached to an $S^1$ which is not part of the CW structure and by an interior point to one of the $(0,p_j)$, where $p_j$ is the j-th roots of unity. In other words, it seems that it's not a $2$-cell because the identification is not along the boundary and it doesn't attach to the right skeleton.

I want to work with the cell description provided at the beginning of the question, because then it would be easy to formalise it and to show that the rotation defining a lens space is a cellular map. So, someone can explain (or give an hint) about why the sets $e_r^i$'s are cells and how are they attached to the $i-1$-skeleton (e.g. helping on the computations explained above)?

$\endgroup$
3
$\begingroup$

Let me use $X^{(i)}$ to denote the $i$-skeleton, the union of all open cells $e^d_r$ with $d \le i$. From the wording of your question it looks like you already understand the 1-skeleton $X^{(1)}$, so I will make that assumption.

The quick summary is that the open cells are given implicitly, and from this one needs to write down those cells parametrically.

So you need a parameterization of $e^2_r$. To get this, first notice that for each $(z_0,z_1) \in e^2_r$ we have $|z_0| < 1$, because $z_1 \ne 0$. Next notice that $z_0$ uniquely determines $z_1$ by the requirements that $|z_1|^2 = 1 - |z_0|^2$ and $\text{arg}(z_1) = 2 \pi r / p$. It follows that $e^2_r$ is parameterized by the set of all $z_0$ such that $|z_0|<1$, which is a good start because that set is an open 2-cell, namely the interior $D^o$ of the unit disc $D \subset \mathbb{C}$. So, let me denote the open cell parameterization map by $f^o_r : D^o \to e^2_r$, which is given by the formula $$f^o_r(z_0) = \bigl(z_0, \sqrt{1 - |z_0|^2} \text{exp}(2 \pi i r / p) \bigr) $$

The next step is to show that $f^o_r : D^o \to e^2_r$ extends to a continuous map $f_r : D \to X$ taking $\partial D$ to the 1-skeleton; once that is done it follows that $f_r$ is a characteristic map for closed 2-cell with interior $e^2_r$. But the formula for $f_r$ is staring us in the face: it is the same as the formula for $f^o_r$ but extended to the whole disc $D$. And then, when you plug in a point $z_0 \in \partial D = S^1$, you get $f^o_r(z_0)=(z_0,0)$ which is clearly contained in $X^{(1)}$.

Can you take it from here with the 3-cells?

$\endgroup$
  • $\begingroup$ Perfect! This map (it was so easy to find argh! :( ) explains everything! Actually I was trying to define a map which does the job of your map but I wasn't able to do that! Thank you very much! $\endgroup$ – Luigi M Dec 3 '14 at 13:49
  • 1
    $\begingroup$ @LuigiM: By the way, although I did this purely with formulas, in the back of my mind is a true visualization, based on picturing these objects in $\mathbb{R}^3$, thinking of $S^3$ as the one-point compactification of $\mathbb{R}^3$. It's worthwhile trying to think of it that way. $\endgroup$ – Lee Mosher Dec 3 '14 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.