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I am currently trying to calculate the homology groups for the 2 dimensional skeleton of the 4 dimensional cube.
My problem is that I have no idea what this looks like; I know that the 1 skeleton is the hypercube graph, but that's really it. Once I have the skeleton, I'm guessing I can just use $\Delta$-complexes to figure out its homology.
Any help is appreciated! Thanks.

TinaBelcher
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  • If you dip a cubical wire frame into soap solution (capturing a bubble in the center), the film makes the $2$-skeleton of the hypercube (minus the six faces of the frame cube itself). I couldn't find a photograph, but here is a computer-drawn image from wikipedia. – Andrew D. Hwang Mar 04 '15 at 02:09

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Let $I = [0,1]$. Let $X$ be the 2-skeleton of $I^4$. First off $H_1 X = H_1 I^4 = 0$ (use long exact sequence of the pair $(I^4, X)$. Now consider the cellular chain complex of $X$ with coefficients in a field $F$. We have $0 \to C_2(X) \to C_1(X) \to C_0(X) \to 0$ which can easily be seen to be $0 \to F^{24} \to F^{32} \to F^{16} \to 0$. Now $H_0(X) = F$ as $X$ is connected. So $ker(F^{32} \to F^{16}) = F^{32-15}= F^{17}$. By our $H_1 X$ computation, we have $im(F^{24} \to F^{32})$ must be 17 dimensional. So we can conclude $H_2(X) = F^7$.

Here's a tip on how to "see" what it looks like. The 0-cells of $I^4$ in a cubical decomposition are the points $x=(x_0,x_1,x_2,x_3)$ of $I^4$ where each $x_i$ is either $0$ or $1$. So there are $2^4$ of them. Each 1-cell is had by selecting a component to vary and the remaining components are either 0 or 1, e.g., the 2-nd component and a choice of $x_0,x_2,x_3$ in $\{0,1\}$, so $\{(x_0,t,x_2,x_3) \, : \, t \in [0,1]\}$. So we see that there are 4 ways to pick a component to have the parameter $t$, and there are $2^3$ ways to fill in the remaining components with 0's and 1's. Hence there are 32 1-cells in I^4. Similar combinatorics will show you what the 2-cells are.

Glen M Wilson
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  • Check out cubical sets for some other approaches to working with cubes: math.cmu.edu/~cnewstea/notes/cubicalsets.pdf – Glen M Wilson Mar 04 '15 at 03:19
  • Thank you for your response! I understand the shape now, and I get how you get the chain complex, but I'm unsure how you get that $ker(F^{32}\rightarrow F^{16})=F^{32-15}=F^{17}$ right away; isn't it possible some of the generators cancel out? – TinaBelcher Mar 04 '15 at 03:42
  • If you work over a field this is just the "rank-nullity" theorem of linear algebra. – Glen M Wilson Mar 04 '15 at 03:45
  • Oh sorry. It is because we know $im(F^{32} \to F^{16})$ must be 15-dimensional so that $H_0 X= F$. – Glen M Wilson Mar 04 '15 at 03:48
  • If you were to just use the long exact sequence of the pair $(I^4, X)$, you can use integer coefficients as well. – Glen M Wilson Mar 04 '15 at 03:53
  • Ahhhh, okay, I understand completely now. Thank you! – TinaBelcher Mar 04 '15 at 03:54
  • Sure thing. That was fun! – Glen M Wilson Mar 04 '15 at 03:55
  • @GlenMWilson I believe that $H_1=0$ but it seems to me that from the LES it just follows that $H_1(X)\simeq H_2(I^4, X)$, or not? – Peter Franek Mar 04 '15 at 04:34
  • $H_2(I^4,X) \cong H_2(I^4/X)$ and $I^4/X$ has no 2-cells. – Glen M Wilson Mar 04 '15 at 04:49
  • In this case, the LES is just a fancy way of saying that the part of the cellular chain complex for $X$ and $I^4$ are identical in the range $C_2 \to C_1 \to C_0 \to 0$. – Glen M Wilson Mar 04 '15 at 04:57
  • Hi, one last question - I was wondering if you could briefly clarify what you meant in your comment about integer coefficients - I tried looking at the LES, but I don't see how it follows that we can use integer coefficients. – TinaBelcher Mar 04 '15 at 05:08
  • The LES for the pair $(I^4, X)$ shows that $H_2 X \cong H_3 I^4/X$. The latter group can be computed directly from its cellular complex $\mathbb{Z} \to \mathbb{Z}^8 \to 0$ with $C_4(I^4/X)=\mathbb{Z}$ and $C_3(I^4/X)=\mathbb{Z}^8$. Just have to think about what the boundary map is! – Glen M Wilson Mar 04 '15 at 05:31