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Math Help - CW complex.

  1. #1
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    CW complex.

    Let X be the topological space obtained as the quotient of the sphere S^2 under the equivalence relation x \sim -x for x in the equatorial circle.

    Describe a CW complex whose underlying space is X, and compute 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 by skamoni; May 13th 2010 at 10:02 AM.
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  2. #2
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    Quote Originally Posted by skamoni View Post
    Let X be the topological space obtained as the quotient of the sphere S^2 under the equivalence relation x \sim -x for x in the equatorial circle.

    Describe a CW complex whose underlying space is X, and compute 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, \mathbb{R}P^2, which is the quotient space S^2/(x \sim -x), where -x is the antipodal point of x for each x \in S^2.
    The CW structure of \mathbb{R}P^2 has one cell e^k in each dimension for k = 0, 1, and 2 such that e^0 \cup e^1 \cup e^2.
    The attaching map for e^k is the 2-sheeted covering projection \phi:S^{k-1} \rightarrow \mathbb{R}P^{k-1} for k=1,2, where \mathbb{R}P^0 is a point and \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:

    \cdots \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{2} \mathbb{Z}\xrightarrow{0} \mathbb{Z} \rightarrow 0

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

    H_0(X) = \mathbb{Z} \text{ }, H_1(X) = \frac{\mathbb{Z}}{2\mathbb{Z}}, and H_k(X) = 0 for k \geq 2.
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  3. #3
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    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 S^2, whereas the space I'm looking at only identifies antipodal points on the equitorial circle.
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  4. #4
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    Quote Originally Posted by skamoni View Post
    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 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.
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  5. #5
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    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 B^2 by identifying x with -x for each 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 \mathbb{R}P^1.

    The exact sequence looks like

    0 \rightarrow \mathbb{Z} \times \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \rightarrow 0.

    Now the resulting homology group is

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