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Algebraic topology, groups and covering short, exact sequences

  1. Mar 26, 2009 #1

    KG1

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    Hi everyone!

    I would like to solve some questions:

    Classify up to isomorphism the four-sheeted normal coverings of a wedge of circles. describe them.

    i tried to to this and it is my understanding that such four sheeted normal coverings have four vertices and there are loops at each of the vertices. For this graph to be a normal covering the deck transformation group needs to be Z/2*Z/2

    second problem: show that a normal subgroup H of a free group L which has infinite index can't be generated by a finite subset. (use the deck transformation of an appropriate covering).
    I have no ideas for this one.

    Third problem: what is the fundamental group of RP²vRP²? can you find a normal cover of RP²vRP² which deck transformation group is Z/4Z?
    First of all the fundamental group of RP²vRP² is Z/2*Z/2.
    Second it is my understanding that this problem reduces to knowing whether we can have Z/4Z isomorphic to (Z/2*Z/2)/H with H a normal subgroup of Z/2*Z/2. Does such a normal subgroup exist?

    Fourth problem finally: we have p:E-->B a fibration. Analyse the exact sequence :
    pi_1(E,e)->pi_1(B,b)->pi_0(F,e)->pi_0(E,e)->pi_0(B,b)

    what does exact sequence mean at pi_0(F,e)?
    For this one I tried to say that there is an action of pi_0(B,b) on the fiber, an action which is transitive.In addtion pi_0(F,e) represent the connected components of the fiber. But I need to show that two points in the fiber F have same image by the induced inclusion i* of pi_0(F) in pi_0(E) if and only if they are in the same orbit under the action of the fundamental group of the base B

    Sorry for this long post and thx for any help or explanation!!!
     
  2. jcsd
  3. Mar 31, 2009 #2
    Third problem: what is the fundamental group of RP²vRP²? can you find a normal cover of RP²vRP² which deck transformation group is Z/4Z?
    First of all the fundamental group of RP²vRP² is Z/2*Z/2.
    Second it is my understanding that this problem reduces to knowing whether we can have Z/4Z isomorphic to (Z/2*Z/2)/H with H a normal subgroup of Z/2*Z/2. Does such a normal subgroup exist?

    Z/2*Z/2 has two generators of order two. Z/4 has one element of order 2. So if such an H were to exist then one or both of the generators would be zero mod H or they would be congruent mod H. None of these cases give you Z/4. If both generators are in H then the quotient group is trivial. If only one of them is in H then the quotient is Z/2.
    If they are equal mod H the quotient is Z/2.

    Interestingly, if you mod out by the normal subgroup generated by (ab)^4 where a and b are the two generators then the quotient has Z/4 as a subgroup of index 2. The entire quotient is the dihedral group of order 8.
     
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