# The Unapologetic Mathematician

## 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 - Posted by | Differential Topology, Topology

Comment by NICO VENGEANCE | August 2, 2011 | Reply

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