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On page $100$ of Spivak's Calculus on Manifolds the following definition is made:

If $\omega$ is a $k$-form on $\mathbb{R}^k$, then $\omega = f\ dx_1\land\ldots\land dx_k$ for a unique function $f:\mathbb{R}^k\to\mathbb{R}$. We define $$\int_{[0,1]^k}\omega := \int_{[0,1]^k}f.$$

I wish to get a better grasp of the above definition.

From my understanding of this post one may interpret $\omega$ as a function that, at each point $p\in\mathbb{R}^k$, takes in $k$ vectors $v^1,\ldots,v^k$ representing a $k$-dimensional parallelotope $P$ and spits out a number proportional to its hypervolume. Such number being $$ f(p) \ A(v^1,\ldots,v^k)$$ for a point $p\in P$, a function $f:\mathbb{R}^k\to\mathbb{R}$, and the alternating $k$-tensor $A:{(\mathbb{R}^k)}^k\to\mathbb{R}$ defined by $$A = x_1\land\ldots\land x_k = \text{Alt}(x_1\otimes\ldots\otimes x_k) = \sum_{\sigma\in\mathbb{S}_k}\text{sgn}(\sigma) \prod_{j=1}^k v^{\sigma(j)}_j$$ although the explicit computation of $A$ seems secondary to the fact it is multilinear and alternating.

How should one interpret the numbers $f(p)$ and $A(v_1,\ldots,v_k)$ at the moment of computing the integral?

It would also probably help if someone could provide an example where concrete values are given to the numbers above.

Sam
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1 Answers1

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Well, you can think of the space $M=[0,1]^k$ (WLOG, consider $k=3$ for a better intuition) as having a natural notion of "volume", given by your $A$. It is natural, because the vectors $(1,0,0),(0,1,0),(0,0,1)$, attached at any $p\in M$, span an infinitesimal volume of $$ A((1,0,0),(0,1,0),(0,0,1))= $$ $$ =dx_1\wedge dx_2 \wedge dx_3 ((1,0,0),(0,1,0),(0,0,1))=1. $$ With this "sense of volume", the total volume (i.e., adding together the infinitesimal volumes) of $M$ is 1.

But you can wonder: what if I have another "sense of volume"? An alternative sense of volume is mathematically formalized through your $k$-form $\omega$. In every point $p\in M$, $$ \omega(p)=f(p)dx_1\wedge dx_2 \wedge dx_3 $$ is an externally prescribed way of measuring volumes at the tangent space $T_pM$. This can appears because you are considering another metric on $M$, or because $M$ is a "perfect Euclidean space" but expressed in other coordinates, or whatever other reason. With this new way of measuring volumes (or technically, volume form) the vectors $(1,0,0),(0,1,0),(0,0,1)$ attached at $p$ span an infinitesimal volume $f(p)$. Now, the total volume of $M$ is no longer 1, but the sum of all the infinitesimal volumes $f(p)$, i.e., $$ \int_M \omega=\int_M f. $$

A. J. Pan-Collantes
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  • +1 for the good answer. Could you provide an example of how a different metric on $M$ gives rise to a different "sense of volume" $\omega(p)$? – Sam Apr 07 '23 at 12:31
  • Once you have a metric in every tangent space (Riemannian metric) you can compute an orthonormal basis (in every point) for that metric. Then you can prescribe that the volume spanned by this basis is 1. I have the details written in my personal notes – A. J. Pan-Collantes Apr 07 '23 at 14:33