The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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August 9, 2011 - Posted by | Differential Topology, Topology

8 Comments »

  1. [...] want to show that the cubic singular homology we’ve constructed is actually a functor. That is, given a smooth map we want a chain map , [...]

    Pingback by Functoriality of Cubic Singular Homology « The Unapologetic Mathematician | August 10, 2011 | Reply

  2. [...] -form over a -chain . We also have a differential operator on differential forms and a boundary operator on chains. We can put these together in two ways: either we start with a -form , take its [...]

    Pingback by Stokes’ Theorem (statement) « The Unapologetic Mathematician | August 17, 2011 | Reply

  3. [...] start by considering the case where is the standard cube . Whipping out the definition of the boundary operator , the integral on the right proceeds as [...]

    Pingback by Stokes’ Theorem (proof part 1) « The Unapologetic Mathematician | August 18, 2011 | Reply

  4. [...] course, we only have to handle the case of a general singular cube, since we defined the boundary operator to be additive; if is a general chain — a formal sum of singular cubes — then is the [...]

    Pingback by Stokes’ Theorem (proof part 2) « The Unapologetic Mathematician | August 20, 2011 | Reply

  5. [...] the edge — but on the boundary subspace it’s a different story. Just like we wrote the boundary of a singular cubic chain, we write for this boundary. Any point that gets sent to by a [...]

    Pingback by Manifolds with Boundary « The Unapologetic Mathematician | September 13, 2011 | Reply

  6. [...] the boundary. In the latter case, without loss of generality, we can assume that is exactly the face of where the th coordinate is [...]

    Pingback by Stokes’ Theorem on Manifolds « The Unapologetic Mathematician | September 16, 2011 | Reply

  7. [...] what does this mean for homology? Well, for cubic singular homology it means that is exact if is simply-connected. Indeed, if is a closed -chain, then it must be [...]

    Pingback by Simply-Connected Spaces « The Unapologetic Mathematician | December 14, 2011 | Reply

  8. [...] seen that if a manifold is simply-connected then the first degree of cubic singular homology is trivial. I say that the same is true of the first degree of de Rham [...]

    Pingback by Simply-Connected Spaces and Cohomology « The Unapologetic Mathematician | December 17, 2011 | Reply


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