I am reading a paper on differential forms. They define it as follows:

A differential $k$-form on a domain $U$ is a continuous, infinitely differentiable map $\omega \colon U \to \Lambda^k (\Bbb{R}^{n \ast})$ where $\Bbb{R}^{n\ast}$ is the dual of $\Bbb{R}^n$ as a vector space.

In particular we may let $x_1,...x_n$ be a basis for $\Bbb{R}^n$ and let $dx_i\in \Bbb{R}^{n \ast}$ be the unique linear map $\Bbb{R}^n\to \Bbb{R}$ that satisfies $dx_i(x_j)=\delta_{ij}$

It then goes on to make a comment:

Intuitively, we may consider $dx_1 \wedge ... \wedge dx_k$ to be an oriented, $k$ dimensional volume element

I don't understand this remark. I know $dx$ from calculus is supposed to be an infinitesimal change and I also understand that wedge product is intuitively a way of turning vectors into higher dimensional objects.

But forget about wedge products. I don't understand how $dx$ defined as $\delta_{ij}$ has anything to do with an infinitesimal change.

What is going on here? Why is this abstract definition related to an infinitesimal change?


If you feed a list of $k$ vectors $v_1,\dots,v_k\in\Bbb R^n$ into the $k$-form $dx_1\wedge\dots\wedge dx_k$ you'll get the signed volume of the parallelepiped formed by $v_1',\dots,v_k'$, where $v_j'$ is the projection of $v_j$ into $x_1\dots x_k$-space.

  • $\begingroup$ Actually, I have one really stupid question. Where does the "infinitesimal" come in? You feed in $k$ vectors to the $k$ form and you get a volume, where does infinitesimal come in? $\endgroup$ – user223391 Jul 1 '17 at 0:54
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    $\begingroup$ "Infinitesimal" is a relic from the old-style, or physics-engineering, way of thinking of integrals. If you chop a $k$-dimensional surface into small pieces, the vectors $v_j$ you feed in here will be small vectors :) $\endgroup$ – Ted Shifrin Jul 1 '17 at 1:02

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