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Thread: CW complex.

  1. #1
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    CW complex.

    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 by skamoni; May 13th 2010 at 09:02 AM.
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  2. #2
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    Quote Originally Posted by skamoni View Post
    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 $\displaystyle k \geq 2$.
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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 $\displaystyle S^2$, 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 $\displaystyle S^2$, 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.
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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 $\displaystyle B^2$ 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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