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In my algebraic topology class we showed how to calculate the homology groups of $S^n$, using the tools of singular homology, however we did not discuss other ways of doing it; my question - is there any relatively simple way of doing this, using simplicial homology? I tried thinking about this for a bit, but couldn't see any obvious direction.

Thanks.

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You could count the number of $n$-cycles that are not $n$-boundaries. –  PEV Jan 24 '11 at 22:17
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Yes, you can do it using simplicial homology. The idea is to compute the homology inductively. Start with $S^0$. Triangulate the suspension of $S^n$ as the union of two cones on $S^n$. –  Ryan Budney Jan 24 '11 at 22:24
    
Ryan - thanks, this way seems nice. One can also use the fact that $H_{n+1}(SX) = H_n(X)$ where $SX$ is the suspension of X (I proved this using tools of singular homology, though). –  shay Jan 25 '11 at 22:50
    
Yes, the Mayer-Vietoris sequence in reduced (singular) homology proves that $\tilde{H}_n(S(X))\cong \tilde{H}_{n-1}(X)$ for all positive integers $n$ (where $X$ is a topological space and $S(X)$ is the suspension of $X$). Once you note that the suspension of the $n$-sphere is homeomorphic to the $(n+1)$-sphere for all nonnegative integers $n$, you can apply Ryan's idea. –  Amitesh Datta Dec 15 '11 at 14:06

2 Answers 2

up vote 7 down vote accepted

Using simplicial homology you can triangulate the sphere as the boundary of an $(n+1)$-simplex, and work out the chain complex by hand.

With cellular homology it is even easier since $S^n$ is the union of an $n$-cell and a $0$-cell. The chain complex has a single $\mathbb Z$ in degree $0$ and a single $\mathbb Z$ in degree $n$. In all other degrees it is zero.

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Another rather quick way to compute the groups is with cellular homology. Here the $n$th chain group is generated by the $n$-cells of your CW-complex. The boundary map has to do with degree maps; but in your case it is simple. An $n$-sphere is a 0-cell with an $n$-cell attached by mapping the boundary $S^{n-1}$ to the 0-cell. If $n>1$ then all the maps in the chain complex must be 0 because the chain groups are trivial except for the $n$-chains and 0-chains. The homology of such a complex is easy to compute.

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Hi Joe, wouldn't this be sort of circular logic, though, since in order to show the equivalence between cellular and singular/simplicial homology (we had already shown that these two are equivalent) we use the homology of the sphere? –  shay Jan 25 '11 at 22:54
    
That's possible! –  Grumpy Parsnip Jan 27 '11 at 19:48

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