# The Unapologetic Mathematician

## 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