Edit: I have a proof here but when I spoke last with my professor, she told me something was close, but not quite. Can someone help me patch this proof? I've been trying to get this down for quite a while and am starting to feel a bit frustrated with myself.

I am trying to show that the second (simplicial) homology group or a compact, connected orientable surface without boundary is ismormophic to $\mathbb Z$.

I can show that this group is non-trivial by triangulating the surface, and considering the chain consisting of all the 2-simplices of the triangulation. Since the surface is orientable, and the simplices are oriented compatibly, each edge of each two simplex will appear twice pointing in opposite directions (a fact which wasn't hard to prove, but I won't type up because LaTeX), and so they cancel. That means the chain consisting of all two-simplices is a cycle. Since there are no three-simplices when dealing with surfaces, there is nothing to be the boundary of. So the two-homology group is non-trivial at least.

Now, to show that $H_2(S)$ for some orientable $S$ is not larger than $\mathbb Z,$ we proceed by contradiction. Suppose that $H_2(S)$ had a second generator. Let's call the above chain of all 2-simplices $K=\sum_{i=1}^n \sigma_i.$ If there is another generator for $H_2(S)$, it is also a cycle, so we can write $L=\sum_{j=1}^k a_j \sigma_j$, with the property that $\partial L=\sum a_j \partial \sigma_j =0.$

Now suppose that some positively oriented 1-simplex $(+)v$ appears in $\partial \sigma_x$, and $(-)v$ appears in $\partial \sigma_y$, then the coefficients $a_x$ and $a_y$ must be equal. Since $v$ was arbitrary, we have $a=a_1=...=a_k.$ So now we can say that $\partial L=\sum_{j=1}^n \sigma_j=0,$ by removing the coefficients and cancelling. This means that $L$ is only a distinct generator for $H_2(S)$ when $k\neq n.$

If this is true then, we can take the difference of the two chains, and produce a third chain so that $K$ is a linear combination of these two chains, producing a basis for $H_2(S)$ isomorphic to $\mathbb Z \bigoplus \mathbb Z$. These two chains are $K-L$ and $L$. However, since $S$ is a connected surface, it is not the disjoint union of two other surfaces, and moreover, since it is connected, it is path-connected. This means that there is a path from any point belonging to a simplex in $K-L$ to any point belonging to a simplex appearing in $L$. However, since there are no 2-simplices between them, the path must go through only 1-simplices of $S$. But then any point in this 1-simplex does not have neighborhoods homeomorphic to 2-disks, so that $S$ is not a surface. A contradiction. So this proves that $k=n,$ so that $L$ is not a distinct generator for $H_2(S).$ Since $H_2(S)$ has only one generator, and $S$ is orientable, $H_2(S)$ is isomorphic to $\mathbb Z$.

What's going wrong here?

  • $\begingroup$ You are working too hard. Try just proving it cyclic. Given another cycle, use connectedness to prove its a multiple of the first. Since being torsion free forces sub modules to be torsion fee you know it's $\mathbb{Z}$. $\endgroup$ – Charlie Frohman Jul 29 '16 at 2:59

Now suppose that some positively oriented 1-simplex $(+)v$ appears in $\partial \sigma_x$, and $(-)v$ appears in $\partial \sigma_y$, then the coefficients $a_x$ and $a_y$ must be equal. Since $v$ was arbitrary, we have $a=a_1=...=a_k.$

This doesn't follow immediately. You've shown that if $\sigma_x$ and $\sigma_y$ share a boundary edge, then $a_x=a_y$ (to be clear, you should also mention the fact that $\sigma_x$ and $\sigma_y$ are the only two triangles that contain that boundary edge, so that you can't get a $v$ term from $\partial a_j$ for $j\neq x,y$). But why does this imply that all of the $a_j$ are equal?

The reason is that $S$ is connected. You know that if $\sigma_x$ and $\sigma_y$ share an edge, then $a_x=a_y$. But since $S$ is connected, every triangle in $S$ can be reached by starting at $\sigma_x$ and repeatedly "taking the triangle on the other side of a boundary edge".

To be precise, let $\sim$ be the equivalence relation on triangles generated by the relation "shares a boundary edge". The union of the triangles in any equivalence class is a clopen subset of $S$, and so there can only be one equivalence class by connectedness. Now the relation $\sim'$ defined by $\sigma_x\sim' \sigma_y$ if $a_x=a_y$ must contain $\sim$, since $a_x=a_y$ whenever $\sigma_x$ and $\sigma_y$ share a boundary edge. Thus $\sim'$ also has only one equivalence class.

But this actually proves something stronger than you claimed: it proves that you must have $a_1=a_2=\dots=a_n$, not just $a_1=a_2=\dots=a_k$, because nothing in this argument says you have to restrict to $\sigma_x$ such that $a_x\neq 0$. If you write $a$ for the common value of all the coefficients, you then have $L=aK$. So every cycle is an integer multiple of $K$.

  • $\begingroup$ Although it's been nearly two years since I asked this question, I really do appreciate this answer. This justification was wrapped up in the "since $v$ was arbitrary" phrase, and probably did warrant at the time this kind of clarification. I've asked questions on this site in the past about how to deal with rigor, something I have mostly, but probably still not completely, overcome. While this question is not so hard to me in hindsight, somehow a little closure goes a long way. $\endgroup$ – Alfred Yerger Dec 20 '16 at 4:00

You wrote already everything down precisely then! You know that a homomorphism between free abelian groups can be represented by a matrix. Write it down and you will see the kernel (i.e. $H_2$) which is $\langle (1,\cdots,1) \rangle$.

  • $\begingroup$ I didn't use a matrix in my computation. I wrote down a list of simplices and computed their boundaries with a double sum. It's quite ugly. $\endgroup$ – Alfred Yerger Jan 28 '15 at 7:10
  • $\begingroup$ then do the same for every simplex and you have your matrix. $\endgroup$ – Daniel Valenzuela Jan 28 '15 at 7:25

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