# Algebraic Topology Homology Group

How would one go about calculating the first and second homology groups of a cube inscribed in a sphere?

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Is the cube solid? If so, I think what you're describing is homotopic to a sphere with 8 points identified, which I believe is homotopic to the wedge of a sphere and 7 circles. If not, the problem may be more difficult than that. – Jared Mar 18 '13 at 4:22
Sorry, its the surface of the cube. – dinky Mar 18 '13 at 4:25
You should be more precise about what space you want to study, really... – Mariano Suárez-Alvarez Mar 18 '13 at 6:49

I believe that my comment is still helpful in answering your question. If the cube were solid, then this topological space would be $S^2$ with eight points identified. This is homotopic to a wedge of $S^2$ with $7$ $S^1$'s. From this we see that the desired homology groups are $$H_2(X_{solid})=\mathbb{Z},\;\text{and}\;H_1(X_{solid})=\bigoplus_{i=1}^7\mathbb{Z}$$

Now if $X$ is the hollow cube in the sphere, then we can write $X_{solid}=X\cup B^3$ where we've filled in the cube with a $B^3$. Notice that $X\cap B^3$ is homotopic to $S^2$. An application of Mayer Vietoris will yield $$H_2(X)=\mathbb{Z}\oplus\mathbb{Z},\;\text{and}\;H_1(X)=\bigoplus_{i=1}^7\mathbb{Z}$$

It makes sense that hollowing out the cube will give us one more $2$-dimensional hole, but we won't get any additional $1$-dimensional holes because a loop around the cube is contractible (slide it to a face). To visualize the seven generators of the $H_1$ group, fix a vertex of the cube and consider the loops beginning at said vertex, traveling along the sphere to any of the other seven vertices, and then returning to the original vertex along the surface of the cube. The generators of the second homology group are the sphere are the cube.

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This is too long for a comment, so I'll post it as ananswer. The space ( call it $X$) is clearly homotopy equivalent to two concentric $2$-spheres and $8$ straight line segments connecting them. This space is homeomorphic to $2$ non concentric spheres with $8$ line segments connecting them, which in turn, upon moving the line segments so that they share their starting and ending points on the two spheres, and contracting one of the line segments to a point, is homotopy equivalent to the wedge of two $2$-spheres and seven circles, so that the space $\mathbf X$ is homotopy equivalent to the wedge sum of two spheres and seven circles. The conclusion is near, since (outside of degree $0$) $$H_{\bullet}(\bigvee X_i)\simeq\bigoplus H_{\bullet}(X_i).$$ This recovers Jared's result.

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Very nice approach. This was initially what I wanted to do, but didn't think of it appropriately. – Jared Mar 22 '13 at 1:10
@Jared thank you for your kind comment :) – Olivier Bégassat Mar 22 '13 at 1:20