Results 1 to 3 of 3

Math Help - Covering space action

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    82

    Covering space action

    I am having problems with 1.3.24 in Hatcher's Algebraic Topology:

    24. Given a covering space action of a group G on a path-connected, locally path-connected space X, then each subgroup H in G determines a composition of covering spaces X -> X/H -> G. Show:

    a. Every path-connected covering space between X and X/G is isomorphic to X/H for some subgroup H in G

    The best I can do here is say that since we have a covering space action then pi_1(X/H1) = pi_1(X/H2); not sure how to proceed.

    b. Two such covering spaces X/H1 and H/H2 of X/G are isomorphic iff H1 and H2 are conjugate subgroups of G.

    c. The covering space X/H -> X/G is normal iff H is a normal subgroup of G, in which case the group of deck transformations of this cover is G/H.

    (definition of covering space actionat (*) on page 72 in this pdf http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by harbottle View Post
    I am having problems with 1.3.24 in Hatcher's Algebraic Topology:

    24. Given a covering space action of a group G on a path-connected, locally path-connected space X, then each subgroup H in G determines a composition of covering spaces X -> X/H -> G. Show:

    a. Every path-connected covering space between X and X/G is isomorphic to X/H for some subgroup H in G

    The best I can do here is say that since we have a covering space action then pi_1(X/H1) = pi_1(X/H2); not sure how to proceed.

    b. Two such covering spaces X/H1 and X/H2 of X/G are isomorphic iff H1 and H2 are conjugate subgroups of G.

    c. The covering space X/H -> X/G is normal iff H is a normal subgroup of G, in which case the group of deck transformations of this cover is G/H.

    (definition of covering space actionat (*) on page 72 in this pdf http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf)
    For (a), Since X is a path-connected, locally path-connected space and the action of G is a covering space action, \pi_1(X) is a subgroup of \pi_1(X/G). By using a Galois's correspondence, there exists a covering space X/H of X/G such that \pi_1(X/H) is a subgroup of \pi_1(X/G). In a similar vein, X is the covering space of X/H.

    For (b), an isomorphism between covering spaces p_1:\tilde{X_1} \rightarrow X and p_2:\tilde{X_2} \rightarrow X is a homeomorphism f:\tilde{X_1} \rightarrow \tilde{X_2} such that p_1 = p_2f. This implies that for each x \in X, f preserves the covering space structure, taking p_1^{-1}(x) to p_2^{-1}(x).

    If H_1 and H_2 are conjugate subgroups of G, then so are p_{1*}\pi_1(X/H_1, x_1) and  p_{2*}\pi_1(X/H_2, x_2) in \pi_1(X/G, x_0), where x_1 \in p_1^{-1}(x_0) and x_2 \in p_2^{-1}(x_0). The remaining step is to show that there exists a bijection between each point x_1=p_1^{-1}(x_0) and x_2=p_2^{-1}(x_0) for each point x_0 \in X/G.

    For (c), if X/H \rightarrow X/G is normal, then L=p_*\pi_1(X/H, x_1) is a normal subgroup of M=\pi_1(X/G, x_0). The group of deck transformation is N(L)/L (Hatcher p71, proposition 1.39b). Since L is a normal subgroup of M=\pi_1(X/G, x_0), we have N(L)/L=M/L. It remains to show that M/L=G/H. This equation intuitively makes sense, but I can't think of the rigorous proof for now. Can you proceed from here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    82
    Quote Originally Posted by aliceinwonderland View Post
    For (a), Since X is a path-connected, locally path-connected space and the action of G is a covering space action, \pi_1(X) is a subgroup of \pi_1(X/G). By using a Galois's correspondence, there exists a covering space X/H of X/G such that \pi_1(X/H) is a subgroup of \pi_1(X/G). In a similar vein, X is the covering space of X/H.

    For (b), an isomorphism between covering spaces p_1:\tilde{X_1} \rightarrow X and p_2:\tilde{X_2} \rightarrow X is a homeomorphism f:\tilde{X_1} \rightarrow \tilde{X_2} such that p_1 = p_2f. This implies that for each x \in X, f preserves the covering space structure, taking p_1^{-1}(x) to p_2^{-1}(x).

    If H_1 and H_2 are conjugate subgroups of G, then so are p_{1*}\pi_1(X/H_1, x_1) and  p_{2*}\pi_1(X/H_2, x_2) in \pi_1(X/G, x_0), where x_1 \in p_1^{-1}(x_0) and x_2 \in p_2^{-1}(x_0). The remaining step is to show that there exists a bijection between each point x_1=p_1^{-1}(x_0) and x_2=p_2^{-1}(x_0) for each point x_0 \in X/G.

    For (c), if X/H \rightarrow X/G is normal, then L=p_*\pi_1(X/H, x_1) is a normal subgroup of M=\pi_1(X/G, x_0). The group of deck transformation is N(L)/L (Hatcher p71, proposition 1.39b). Since L is a normal subgroup of M=\pi_1(X/G, x_0), we have N(L)/L=M/L. It remains to show that M/L=G/H. This equation intuitively makes sense, but I can't think of the rigorous proof for now. Can you proceed from here?
    Thank you very much for this. Some of the concepts are still new to me so I need to take a little while to work out what everything means; I'll work through your solution and see what I can do.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Covering space
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: December 1st 2011, 10:42 PM
  2. [SOLVED] Transitive action, blocks, primitive action, maximal subgroups.
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: August 6th 2011, 08:39 PM
  3. Group Action On Space of Multilinear Forms
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: November 11th 2010, 03:10 PM
  4. Covering space
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: June 30th 2009, 08:22 AM
  5. Covering space
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: February 23rd 2009, 04:26 AM

Search Tags


/mathhelpforum @mathhelpforum