CW complex.

Apr 2008
50
8
Let \(\displaystyle X\) be the topological space obtained as the quotient of the sphere \(\displaystyle S^2\) under the equivalence relation \(\displaystyle x \sim -x\) for \(\displaystyle x\) in the equatorial circle.

Describe a CW complex whose underlying space is \(\displaystyle X\), and compute \(\displaystyle H(X)\).

I know how to compute simplicial homology, I just can't describe the space in an adequate way, can anyone help me?
 
Last edited:
May 2010
95
38
Let \(\displaystyle X\) be the topological space obtained as the quotient of the sphere \(\displaystyle S^2\) under the equivalence relation \(\displaystyle x \sim -x\) for \(\displaystyle x\) in the equatorial circle.

Describe a CW complex whose underlying space is \(\displaystyle X\), and compute \(\displaystyle H(X)\).

I know how to compute simplicial homology, I just can't describe the space in an adequate way, can anyone help me?
The space X is the real projective plane, \(\displaystyle \mathbb{R}P^2\), which is the quotient space \(\displaystyle S^2/(x \sim -x)\), where -x is the antipodal point of x for each \(\displaystyle x \in S^2\).
The CW structure of \(\displaystyle \mathbb{R}P^2\) has one cell \(\displaystyle e^k\) in each dimension for k = 0, 1, and 2 such that \(\displaystyle e^0 \cup e^1 \cup e^2\).
The attaching map for \(\displaystyle e^k\) is the 2-sheeted covering projection \(\displaystyle \phi:S^{k-1} \rightarrow \mathbb{R}P^{k-1}\) for k=1,2, where \(\displaystyle \mathbb{R}P^0\) is a point and \(\displaystyle \mathbb{R}P^1\) is a circle.

Computing a boundary map is not trivial. A good reference for this is Hatcher's "Algebraic Topology" p 137-148, especially Example 2.42.

The resulting cellular chain complex is as follows:

\(\displaystyle \cdots \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{2} \mathbb{Z}\xrightarrow{0} \mathbb{Z} \rightarrow 0\)

Now the homology group of \(\displaystyle X = \mathbb{R}P^2\) is

\(\displaystyle H_0(X) = \mathbb{Z} \text{ }, H_1(X) = \frac{\mathbb{Z}}{2\mathbb{Z}}, \) and \(\displaystyle H_k(X) = 0 \) for \(\displaystyle k \geq 2\).
 
Apr 2008
50
8
Hi thanks for your reply. Are you sure that the space is the real projective plane? I thought the real projective plane was constructed by identifying all antipodal points on \(\displaystyle S^2\), whereas the space I'm looking at only identifies antipodal points on the equitorial circle.
 
May 2010
95
38
Hi thanks for your reply. Are you sure that the space is the real projective plane? I thought the real projective plane was constructed by identifying all antipodal points on \(\displaystyle S^2\), whereas the space I'm looking at only identifies antipodal points on the equitorial circle.
If you only identifies antipodal points on the single equatorial circle, then it is a totally different question. I'll try to put up the solution once it is ready.
 
May 2010
95
38
Let X be the space you referred.

Then X has the CW structure with a single 0-cell, one 1-cell, and two 2-cells.

The projective plane I referred is homeomorphic with the space obtained from closed ball \(\displaystyle B^2\) by identifying x with -x for each \(\displaystyle x \in S^1\). So we needed only one 2-cell.

I think the space you referred requires two 2-cells attaching the boundary of each 2-cell to \(\displaystyle \mathbb{R}P^1\).

The exact sequence looks like

\(\displaystyle 0 \rightarrow \mathbb{Z} \times \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \rightarrow 0\).

Now the resulting homology group is

\(\displaystyle H_0(X)=\mathbb{Z}\) and \(\displaystyle H_k(X)=0\) for \(\displaystyle k \geq 1\). Note that \(\displaystyle \mathbb{Z} \times \mathbb{Z} \xrightarrow{2} \mathbb{Z}\) in the above exact sequence is surjective.
 
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