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Forgive my basic understanding, my background is in physics/engineering, not mathematics. Please be gentle :)

As I understand it, k-forms are elements of a dual space that is multilinear and alternating, with k basis vectors (including some additional determinant-like properties - again, excuse my informalities).

Then, one can define a derivative on k-forms. For example, take the derivative of a 0-form (i.e., a smooth function) $\omega$, or, $d\omega = \sum_{j=1}^n\frac{\partial\omega}{\partial{x_{i}}}dx_j$.

This object now has a dual vector $dx_j$ waiting to eat a k-vector. Yet, we treat this as a differential as a function, which confuses me.

Maybe in other words, if we integrate a k-form $\omega = fdx_{I}$ over some region $\Omega$, we have $\int_{\Omega} \omega = \int_{\Omega}fdV$. Again, on the left-hand side, we have $\omega = fdx_{I}$, where $fdx_{I}$ is a multilinear map waiting to eat a k-vector and send it to $\mathbb{R}$. But when we integreate, we assume we have a number, without ever multiplying by a vector on the right-hand side.

How to reconcile this notation? One more apology for the likely naive question. Thanks!

Spinor
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3 Answers3

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In the 1d real case, the idea behind the integral $\int_{[a,b]}\mathrm d\omega=\int_{[a,b]}\omega'(x)\mathrm dx$ is to take little pieces of $[a,b]$ of length $\Delta x$ and weight them with $\omega'(x)$, then add them all up. Then in some sense take the limit $\Delta x\to0$. That's how you've probably learned it. But instead, consider the following idea: $\mathrm d\omega(x)$ is a function taking a small subinterval $[x,x+\Delta x]$ as input and giving its weighted length as output. That is, $\mathrm d\omega(x)([x,x+\Delta x]):=\omega'(x)\Delta x$. Then the integral means chopping up the domain $[a,b]$ into pieces, applying $\mathrm d\omega(x)$ to each piece and adding them all up (then take the limit).

In essence, $\mathrm d\omega(x)$ is a so-called measure. It tells us how much weight a subset of the domain has. This also works in 3d, but here we chop up our domain into parallelepipeds instead of intervals. And the measure $\mathrm d\omega$ can be viewed as taking the spanning vectors of the parallelepiped as input and giving the parallelepiped's weight as output. For small parallelepipeds, this weight should depend linearly on the spanning vectors, so it's fine to make $\mathrm d\omega$ a (multi)linear map. For instance, if $\mathrm d\omega=\mathrm dx\mathrm dy\mathrm dz$, then $\mathrm d\omega(u,v,w)$ gives the volume of the parallelepiped spanned by $u,v,w$.

This was for volume integrals. But it works similarly for line integrals. A line can be approximated by a series of vectors chained to each other. The integrand, say, $\vec{f}(\vec r)\cdot\mathrm d\vec r$, is a linear map at each $\vec r$ telling us the weight of the vector that starts at $\vec r$. If the vector is long, it has more weight. If it's aligned with $\vec f$, it has more weight due to the dot product. Then we add up all the weights and take the limit - voila, the line integral.

Now how does this tie back to the 1d case? Well, 1d vectors are just numbers and linear maps are just products of numbers, so in 1d it's fine to work with numbers alone instead of vectors and linear maps.

Vercassivelaunos
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without ever multiplying by a vector on the right-hand side

Oh, but we do, explicitly. That's a role of the integration operator: to supply the integrand form with infinitesimal tangent k-vectors (equiv., k tangent vectors). Quoting myself from https://math.stackexchange.com/a/4712357/73934 :

Integration intuitively is a procedure that 1) facets the manifold into infinitesimal n-vectors, 2) feeds those into the integrand (the n-form) and 3) sums up the results.

This is explained in more detail in the @Vercassivelaunos answer +1.

rych
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The easiest way to convince myself is the use of Stokes formula $\int_{\partial M}\omega=\int_Md\omega$. So if you already define the induced orientation of a oriented submanifold with smooth boundary, then the exterior differential can be unique determined by Stokes formula.

The most brutal way to answer this, is to resort of transition functions, that "this is just how the coordinate changes give you". If you run the arguments of change of variables you find the rules of chaning coordinate vector fields and integrations, "that is why they look like that".

In my prespective, in differential geometry, the definition of differential form is simply a model that fits in what we need as the true object (the forms) itself.

Basically, if we apply change of coordinates, the $\Omega$ may change but the number is not. So we can forget about $\omega$ as an object of pairing $k$-vectors, but simply as an object that splits out number from a $k$ dim set.

It's actually quiet hard to give a better picture as the answer varies between physics and analysis.

Liding Yao
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