# Fundamental group of 1-skeleton of cubes

• Oct 23rd 2010, 07:09 AM
math8
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?
• Oct 23rd 2010, 06:03 PM
Jose27
Quote:

Originally Posted by math8
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.
• Oct 27th 2010, 07:16 PM
Tinyboss
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.
• Oct 28th 2010, 06:58 AM
math8
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 $I^n$ .