# 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

## Chains

We can integrate on the standard cube, and on singular cubes in arbitrary manifolds. What if it’s not very easy to describe a region as the image of a singular cube? This is where chains come in.

So a chain is actually pretty simple; it’s just a formal linear combination of singular $k$-cubes. That is, for each $k$ we build the free abelian group $C_k(M)$ generated by the singular $k$-cubes in $M$.

If we have a formal sum $c=a_1c_1+\dots+a_lc_l$ — the $c_i$ are all singular $k$-cubes and the $a_i$ are all integers — then we define integrals over the chain by linearity:

$\displaystyle\int\limits_c\omega=a_1\int\limits_{c_1}\omega+\dots+a_l\int\limits_{c_l}\omega$

And that’s all there is to it; just cover the $k$-dimensional region you’re interested in with singular $k$-cubes. If there are some overlaps, those areas will get counted twice, so you’ll have to cover them with their own singular $k$-cubes with negative multipliers to cancel them out. Take all the integrals — by translating each one back to the standard $k$-cube — and add (or subtract) them up to get the result!

August 5, 2011

## Integrals are Independent of Parameterization

If $c:[0,1]^k\to M$ is a singular $k$-cube and $\omega$ is a $k$-form on the image of $c$, then we know how to define the integral of $\omega$ over $c$:

$\displaystyle\int\limits_{c([0,1]^k)}\omega=\int\limits_{[0,1]^k}c^*\omega$

On its face, this formula depends on the function $c$ used to parameterize the region of integration. But does it really? What if we have a different function $d:[0,1]^k\to M$ with the same image? For convenience we’ll only consider singular $k$-cubes that are diffeomorphisms onto their images — any singular $k$-cube can be broken into pieces for which this is true, and we’ll soon deal with how to put these together.

Anyway, if $c([0,1]^k)=d([0,1]^k)$, then given our assumptions there is some diffeomorphism $f=c^{-1}\circ d:[0,1]^k\to[0,1]^k$ such that $d=c\circ f$. If $J_f$ is everywhere positive, then we say that $d$ is an “orientation-preserving reparameterization” of $c$. And I say that the integrals of $\omega$ over $c$ and $d$ are the same. Indeed, we calculate:

\displaystyle\begin{aligned}\int\limits_{c\circ f}\omega&=\int\limits_{[0,1]^k}(c\circ f)^*\omega\\&=\int\limits_{[0,1]^k}f^*c^*\omega\\&=\int\limits_fc^*\omega\\&=\int\limits_{f([0,1]^k)}\left[c^*\omega\right]\left(\frac{\partial}{\partial u^1},\dots,\frac{\partial}{\partial u^k}\right)\,d(u^1,\dots,u^k)\\&=\int\limits_{[0,1]^k}\left[c^*\omega\right]\left(\frac{\partial}{\partial u^1},\dots,\frac{\partial}{\partial u^k}\right)\,d(u^1,\dots,u^k)\\&=\int\limits_c\omega\end{aligned}

where we use our expression for the integral of $\omega$ over the image $f([0,1]^k)$ in passing from the third to the fourth line. Thus the integral is a geometric quantity, depending only on the image $c([0,1]^k)$ and the $k$-form $\omega$ rather than on any detail of the parameterization itself.

August 4, 2011

## Integration on Singular Cubes

The next step after standard cubes in our integration project is to define integration on “singular cubes”.

Given an $n$-dimensional manifold $M$, a singular $k$-cube in $M$ is a differentiable map $c:[0,1]^k\to M$ from the standard $k$-cube to the manifold. For example, a curve is a singular $1$-cube. We also consider $[0,1]^0=\{0\}$, so a singular $0$-cube in $M$ just picks out a single point of the manifold. We also will often abuse notation and write $c$ for the image $c([0,1]^k)\subseteq M$.

Now, let’s say we have some some subset $c([0,1]^k)\subseteq M$ described as the image of a singular $k$-cube $c$, and we have a $k$-form $\omega$ on the image of $c$. How shall we define the “integral” of $\omega$ over $c$? The most natural thing in the world is to pull back the form $\omega$ along $c$ to get a $k$-form $c^*(\omega)$ on $[0,1]^k$. Then we can define

$\displaystyle\int\limits_c\omega=\int\limits_{c([0,1]^k)}\omega=\int\limits_{[0,1]^k}c^*\omega$

Neat, huh?

Let’s look at what happens when $M=\mathbb{R}^n$ and $c$ is a singular $n$-cube. Since $c:[0,1]^n\to\mathbb{R}^n$ it has a Jacobian at each point in the unit cube, and we’ll keep things simple by assuming that it’s everywhere nonsingular.

Now if $\omega=f\,du^1\wedge\dots\wedge du^n$ is a top form on the image of $c$, then $c^*\omega$ is a top form on the unit cube, which we can again write in terms of a function and the canonical basis volume form on the cube. We can find this new function by plugging in the basic vector fields in order:

\displaystyle\begin{aligned}c^*\omega\left(\frac{\partial}{\partial u^1},\dots,\frac{\partial}{\partial u^n}\right)&=\left[\omega\circ c\right]\left(c_*\frac{\partial}{\partial u^1},\dots,c_*\frac{\partial}{\partial u^n}\right)\\&=\left[(f\circ c)du^1\wedge\dots\wedge du^n\right]\left(c_*\frac{\partial}{\partial u^1},\dots,c_*\frac{\partial}{\partial u^n}\right)\\&=(f\circ c)\det\left(du^i\left(c_*\frac{\partial}{\partial u^j}\right)\right)\\&=(f\circ c)\det\left(\frac{\partial}{\partial u^j}(u^i\circ c)\right)\end{aligned}

which is $f\circ c$ times the Jacobian determinant of $c$, which we will write as a function $J_c$. This, then, allows us to calculate

\displaystyle\begin{aligned}\int\limits_c\omega&=\int\limits_{[0,1]^n}(f\circ c)J_c\,du^1\wedge\dots\wedge du^n\\&=\pm\int\limits_{[0,1]^n}(f\circ c)\lvert J_c\rvert\,d(u^1,\dots,u^n)\end{aligned}

where we choose the positive sign if $J_c$ is everywhere nonnegative or the negative sign if it’s everywhere nonpositive on the unit cube. But now we can use the change of variables formula to see that

$\displaystyle\int\limits_c\omega=\pm\int\limits_{[0,1]^n}(f\circ c)\lvert J_c\rvert\,d(u^1,\dots,u^n)=\pm\int\limits_{c([0,1]^n)}f\,d(u^1,\dots,u^n)$

That is, we can evaluate the integral of $\omega$ over $c$ just by integrating the function $f$ over its image in the usual multivariable way. Now, in practice it may only really be feasible to use the change of variables formula to translate back to the unit cube and integrate $f\circ c$ (times the Jacobian!) there, but in principle this formula will come in handy.

August 3, 2011

## Integration on the Standard Cube

Sorry for the delay, I’ve had a packed weekend.

Anyway, we’re ready to start getting into integration on manifolds. And we start with a simple case that everything else will be built on top of.

We let $[0,1]^k\subseteq\mathbb{R}^k$ be the “standard $k$-cube”. We know that the space $\Omega^k([0,1]^k)$ of “top forms” — top because $k$ is the highest degree possible for a differential form on a differential form — has rank $1$ over the algebra $\mathcal{O}([0,1]^k)$ of smooth functions. That is, if $\omega$ is a top form then we can always write

$\displaystyle\omega=f\,du^1\wedge\dots\wedge du^k$

for some smooth function $f$ on the standard cube. Then we write

$\displaystyle\int\limits_{[0,1]^k}\omega=\int\limits_{[0,1]^k}f\,du^1\wedge\dots\wedge du^k=\int\limits_{[0,1]^k}f\,d(u^1\dots du^k)$

here we sorta pull a fast one, notationally speaking. On the left we’re defining the integral of a $k$-form $\omega$. In the middle we rewrite the form as above, in terms of a function and the canonical basis $k$-form $du^1\wedge\dots\wedge du^k$ made from wedging together the basic $1$-forms in order. And then on the right we suddenly switch to a $k$-dimensional Riemann integral over the standard $k$-cube. The canonical basis $k$-form $du^1\wedge\dots\wedge du^k$ corresponds to the volume element $d(u^1,\dots,u^k)$, and top forms are often also called “volume forms” because of this correspondence. In fact, it’s not hard to see that they’re related to signed volumes. This is the starting point from which all integration on manifolds emerges, and everything will ultimately come back to this definition.

August 2, 2011