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)?


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?

  • $\begingroup$ 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

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