# The Unapologetic Mathematician

## Cubic Singular Homology

Now that we’re armed with chains — formal sums — of singular cubes we can use them to come up with a homology theory. Since we will use singular cubes to build it, we call it “cubic singular homology”.

So, for each $k$ we have the abelian group $C_k$ of chains. For each one we want to define a map $\partial:C_k\to C_{k-1}$ with the property that $\partial^2=0$ so we can use this sequence of abelian groups as the setup for a homology theory. Now what would be an appropriate name for this sequence of groups of chains? I think “chain complex” fits nicely, no? It’s almost as if that’s where the name came from…

Etymologies aside, we need to pass from a $k$-dimensional chain to a $k-1$-dimensional one. And since every $k$-dimensional chain is a formal sum of singular $k$-cubes, we really just need to define it on $k$-cubes and extend by linearity.

The key idea here is that given a $k$-cube $c$ we want $\partial c$ to correspond to the boundary of $c$, and so it will be a certain combination of the $k-1$-dimensional “faces” of $c$. Heuristically, each face itself has faces, and each of these $k-2$-dimensional faces is part of two of the $k-1$-dimensional faces, and when we work everything out they will turn out to cancel each other off, leaving an “empty” second boundary.

So for each dimension we’re going to have two faces, one of which we get by setting that component to $0$ and the other of which we get by setting that component to $1$. Explicitly, we’ll define the following faces of the standard cube $I^k$:

\displaystyle\begin{aligned}I^k_{i,0}(a)&=(a_1,\dots,a_{i-1},0,a_i,\dots,a_{k-1})\\I^k_{i,1}(a)&=(a_1,\dots,a_{i-1},1,a_i,\dots,a_{k-1})\end{aligned}

for any $a\in[0,1]^{k-1}$. Then for any other singular cube $c$ we define the face $c_{i,j}=I^k_{i,j}$. Then we define the boundary operator by

$\displaystyle\partial c=\sum\limits_{i=1}^k\sum\limits_{j=0,1}(-1)^{i+j}c_{i,j}$

As an example, if $c$ is a $1$-cube then $\partial c=c(1)-c(0)$. We have to be careful here: this is not a “real” subtraction — $c(t)$ is a point in $M$, remember, which may not have any sense of subtraction at all. This is just a formal subtraction of one $0$-cube — one manifold point — from another. We will extend our definition to $0$-cubes by setting $\partial(c)=1\in\mathbb{Z}$ for all $0$-cubes $c$, and thus we automatically have $\partial\partial c=0$ for $1$-cubes $c$.

More generally for a singular $n$-cube $c$, we calculate

\displaystyle\begin{aligned}\partial\partial c&=\partial\left(\sum\limits_{i=1}^n\sum\limits_{k=0,1}(-1)^{i+k}c_{i,k}\right)\\&=\sum\limits_{i=1}^n\sum\limits_{k=0,1}\sum\limits_{j=1}^{n-1}\sum\limits_{l=0,1}(-1)^{i+k+j+l}(c_{i,k})_{j,l}\\&=\sum\limits_{i=1}^n\sum\limits_{k=0,1}\sum\limits_{j=1}^{n-1}\sum\limits_{l=0,1}(-1)^{i+k+j+l}c\circ I^n_{i,k}\circ I^{n-1}_{j,l}\end{aligned}

Now, if $i\leq j\leq n-1$ then it’s straightforward to check that $I^n_{i,k}\circ I^{n-1}_{j,l}=I^n_{j+1,l}\circ I^{n-1}_{i,k}$; the $+1$ creeps in because after we insert something between $a_{i-1}$ and $a_i$ we have an extra place to go to insert the other value. Thus we find that $(c_{i,k})_{j,l}=(c_{j+1,l})_{i,k}$, and so each term in the big sum above shows up twice. The thing is, one time it shows up with sign $(-1)^{i+k+j+l}$ and the other time it shows up with sign $(-1)^{j+1+l+i+k}$, which cancels off the first appearance. And so the whole sum collapses to $0$, just like we asserted.

And so we again define the group $Z_k$ of closed chains to be those chains $c\in C_k$ with $\partial c=0$ and the group $B_k$ of exact chains to be those $c\in C_k$ where there exists some $b\in C_{k+1}$ with $\partial b=c$. Again, $B_k\subseteq Z_k$, since $\partial\partial c=0$. And again, we define the homology group $H_k=Z_k/B_k$.

August 9, 2011