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Math Help - Fundamental group of 1-skeleton of cubes

  1. #1
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    Fundamental group of 1-skeleton of cubes

    How do you compute the Fundamental group of the 1-skeleton of the 3-cube I^{3} = [0,1]^{3} ? What about the Fundamental group of the 1- skeleton of the 4-cube I^{4} ?

    I know the Fundamental group of a space X at a point x_{0} is the set of homotopy classes of loops of X based at x_{0} . And that the 1-skeleton of a space X is the union of all cells of the CW complex for X up to dimension 1. But how do you find the fundamental group of the 1 skeleton for those cubes?
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  2. #2
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    Quote Originally Posted by math8 View Post
    How do you compute the Fundamental group of the 1-skeleton of the 3-cube I^{3} = [0,1]^{3} ? What about the Fundamental group of the 1- skeleton of the 4-cube I^{4} ?

    I know the Fundamental group of a space X at a point x_{0} is the set of homotopy classes of loops of X based at x_{0} . And that the 1-skeleton of a space X is the union of all cells of the CW complex for X up to dimension 1. But how do you find the fundamental group of the 1 skeleton for those cubes?
    Assuming your CW complex has the same topology as I^3 when viewed inside euclidean space then your skeleton would be just the edges of the cube, and this is homeomorphic to the skeleton of I^2 times [0,1] now just use properties of the fundamental group. I guess you could do something similar for the other one, but my geometric intuition has failed me many times before, so you better check it carefully.
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  3. #3
    Senior Member Tinyboss's Avatar
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    1-skeletons of cubes are graphs, and to compute the fundamental group of a graph (which is always free) you just take a maximal tree, homotope it down to a point, and what you're left with is a wedge product of n circles. So your fundamental group is free on n generators.
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  4. #4
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    For the 1-skeleton of the 3-cube, I can visualize by drawing that there will be 5 edges not included in the maximal tree of the graph, so the fundamental group of the 1-skeleton of the 3-cube is the free product of 5 copies of Z .

    But what about for the 1-skeleton of the 4-cube I^4 ? Since I cannot visualize it, I am not sure how many edges won't be in the maximal tree(number of circuits) , so I am not sure how the fundamental group is going to look like, same thing with the n-cube I^n .
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