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I was asked to find a CW complex for the Möbius strip with one 0-cell, two 1-cells, and one 2-cell.

I can find a CW complex for a Möbius strip with more cells (two 0-cells, three 1-cells and a single 2-cell), but this doesn't help me.

I was hinted that I need to attach the 2-cell's boundary to $ab^2$ where $a$,$b$ are the 1-cells. I don't see why this gives me a Möbius strip. Is there a way of coming up with it "automatically" from the fundamental polygon of the Möbius strip i.e. $I \times I$ under the identification $(x,0)\approx (1-x,1)$? There seems to be two 0-cells (two corners) and I'm asked for a single 0-cell so I'm guessing it's unrelated.. if so then what is the way to come up with this?

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Perhaps the CW complex that you found, the one with two 0-cells, three 1-cells, and a single 2-cell, might help you after all.

Look at your cells and ask yourself: Do I really need them all? Can I remove some of them, thereby decreasing the number of cells? Or can I merge several of them into fewer cells, thereby decreasing the number of cells?

Here's an example in dimension $1$. Suppose I take the following CW decomposition for the topological space $S^1$, meaning the unit circle in the plane. The $0$ skeleton $X^{(0)}$ has two 0-cells, $X^{(0)}=\{v_0,v_1\}$ where $v_0$ is the point $(1,0)$ and $v_1$ is the point $(-1,0)$; and the $1$-skeleton $X^{(1)}$ equals $X$ itself, having two open 1-cells, namely the open upper half semicircle and the open lower half semicircle of $S^1$

Now I look at the 0-cells, and I focus on the 0-cell $v_1$. I notice that if I continuously moved $v_1$ around the circle in the counterclockwise, making $e_0$ longer and $e_1$ shorter, eventually $v_1$ merges with $v_0$ making a single 0-cell, $e_0$ expands to become $S^1 - \{v_0\}$, and $e_1$ shrinks down into nothingness. In this manner I have visualized a new CW structure on $S^1$ with one 0-cell $v_0$ and one open 1-cell $S^1 - \{v_0\}$.

I suspect that if you carefully examine your CW decomposition of the Möbius band, something similar might happen.

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  • $\begingroup$ I don't see how to do it for the case of the Mobius band. When I imagine moving $v_1$ closer to $v_0$ along $e_0$ till it merges with it, eventually $e_0$ shrinks to nothing and I'm left with the right number of cells but I don't see how the disc can be attached now.. $\endgroup$ – Dan Apr 7 '18 at 14:52
  • $\begingroup$ Keep in mind: the open 2-cells are the path components of $X^{(2)} - X^{(1)}$; perhaps that will help. $\endgroup$ – Lee Mosher Apr 7 '18 at 22:29
  • $\begingroup$ I really don't see what you mean. Could you spell it out for me? $\endgroup$ – Dan Apr 19 '18 at 8:33
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Try "cutting and regluing" as shown in this picture. (My picture shows the steps in reverse, so read it from bottom to top!)

enter image description here

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