# 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.

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August 3, 2011 - Posted by | Differential Topology, Topology

## 12 Comments »

1. […] is a singular -cube and is a -form on the image of , then we know how to define the integral of over […]

Pingback by Integrals are Independent of Parameterization « The Unapologetic Mathematician | August 4, 2011 | Reply

2. […] 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 […]

Pingback by Chains « The Unapologetic Mathematician | August 5, 2011 | Reply

3. […] 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 […]

Pingback by Cubic Singular Homology « The Unapologetic Mathematician | August 9, 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 […]

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

5. […] the circle of radius will be a singular -cube. That is, it’s a curve in the plane that never touches the origin. Specifically, we’ll […]

Pingback by An Example (part 2) « The Unapologetic Mathematician | August 24, 2011 | Reply

6. […] we can take our differential form and our singular cube and put them together. That is, we can integrate the -form over the circle […]

Pingback by An Example (part 3) « The Unapologetic Mathematician | August 24, 2011 | Reply

7. […] defined how to integrate forms over chains made up of singular cubes, but we still haven’t really defined integration on manifolds. We’ve sort of waved our […]

Pingback by Integrals over Manifolds (part 1) « The Unapologetic Mathematician | September 5, 2011 | Reply

8. […] integrate forms as long as they’re supported within the image of an orientation-preserving singular cube. But what if the form is bigger than […]

Pingback by Integrals over Manifolds (part 2) « The Unapologetic Mathematician | September 7, 2011 | Reply

9. […] an oriented curve in the manifold . We know that this is a singular -cube, and so we can pair it off with a -form . Specifically, we pull back to on the interval […]

Pingback by Line Integrals « The Unapologetic Mathematician | October 21, 2011 | Reply

10. […] Given an -manifold we let be an oriented -dimensional “hypersurface”, which term we use because in the usual case of a hypersurface has two dimensions — it’s a regular surface. The orientation on a hypersurface consists of tangent vectors which are all in the image of the derivative of the local parameterization map, which is a singular cube. […]

Pingback by (Hyper-)Surface Integrals « The Unapologetic Mathematician | October 27, 2011 | Reply

11. […] we take to be an orientation-preserving embedding — a singular cube of top dimension. Then the pullback for some strictly-positive function . We conclude […]

Pingback by Compact Oriented Manifolds without Boundary have Nontrivial Homology « The Unapologetic Mathematician | November 24, 2011 | Reply

12. […] Armstrong: Integrals are Independent of Parametrization, Integration on Singular Cubes, Integration on the Standard Cube, Stokes’ Theorem […]

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