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Fundamental group of RP^n by recurrence?

  1. Apr 14, 2007 #1

    quasar987

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    Fundamental group of RP^n by recurrence!?

    1. The problem statement, all variables and given/known data
    That's it. Find the fundamental group of RP^n by recurrence.

    3. The attempt at a solution

    It's just obvious to me that it's Z/2 no matter n but what is this recurrence argument that I'm supposed to use?
     
  2. jcsd
  3. Apr 14, 2007 #2

    matt grime

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    Can you write RP^n as the union of copies of RP^m for m<n? Then you can use the Seifert Van Kampen theorem.

    EG you can show the fundamental group of R^n is trivial by saying R^n= R^{n-1}uR^{n-1}, picking the first and last n-1 coordinates, the intersection being R^{n-2} thus by induction the fundamental group is e amalgam e over e.

    Or more interestingly, you can show that the fundamental group of the bouquet of n copies of S^1 is F_n the free group on n generators.


    You can probably do the torus with n holes in it if you wanted to, though it requires a bit more thought.
     
  4. Apr 14, 2007 #3

    quasar987

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    I found [tex]\mathbb{R}P^n= \mathbb{R}P^{n-1}\cup B_n[/tex] (where B_n is the unit open ball) but this is only true as sets. The topology is wrong because in [tex]\mathbb{R}P^{n-1}\cup B_n[/tex], [tex]\mathbb{R}P^{n-1}[/tex] is open, but it's not supposed to be in [tex]\mathbb{R}P^n[/tex]. Plus, the intersection is void.

    Anyway, I got something else though. [tex]\mathbb{R}P^n\approx \mathbb{R}P^{n-1}\times \overline{B}_{n-1}[/tex]. And this, we've seen in class, implies that
    [tex]\pi_1(\mathbb{R}P^n)=\pi_1(\mathbb{R}P^{n-1})\times \pi_1(\overline{B}_{n-1})=\pi_1(\mathbb{R}P^{n-1})\times \{e\}=\pi_1(\mathbb{R}P^{n-1})[/tex]

    Tadam!
     
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