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I am following an introductory course on algebraic topology. Here are two definitions we've made:

  • Given a smooth $n$-manifold $M$, a function $f : M \to \mathbb{R}$ is smooth iff, for each smooth chart $(U,\phi)$, the map $f \circ \phi^{-1} : \phi(U) \to \mathbb{R}$ is smooth.
  • A differential form on $U \subset \mathbb{R}^n$ is an element of $\mathcal{C}^{\infty}(U) \otimes_{\mathbb{R}} \Omega^*$, where $\Omega^* := \bigwedge \mathbb{R}^n$ is the exterior algebra on $\mathbb{R}^n$; in other words, it smoothly associates an element of the algebra $\Omega^*$ to each point of $U$.

However, we were told that the following definition for a differential form on a smooth manifold $M$ --- which is perhaps the straightforward way to synthesize the two above definitions --- is wrong:

  • A differential form on $M$ is a map $\omega : M \to \Omega^*$ such that in each smooth chart $(U,\phi)$, the map $\omega \circ \phi^{-1} : \phi(U) \to \Omega^*$ is a differential form on $\phi(U)$.

The reason given was that "the exterior derivatives in each smooth chart don't necessarily agree"; that is, it might be the case that for some smooth charts $(U,\phi),(V,\psi)$ and a point $y \in U \cap V$, we might have $(d(w \circ \phi^{-1}))(\phi(y)) \neq (d(w \circ \psi^{-1}))(\psi(y))$. However, I am unable to think of a "differential form" (according to this bad definition) on some manifold for which I can't come up with a "derivative form" which agrees with the actual derivative in each smooth chart. (For instance, it appears to work fine with the manifold being a circle and $\omega$ being the zero-form $\sin \theta$, with two charts giving the angle $\theta$; $d\omega$ would be $\cos \theta \; d\theta$). Does such a counterexample exist? Or is there something else about this "definition" that makes it wrong or unuseful?

(For completeness, the correct definition of a form on $M$ is a collection of forms $\{\omega_{(U,\phi)} \in \Omega^*(\phi(U))\}$ for each chart $(U,\phi)$, with agreement via pullbacks of transition maps.)

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    $\begingroup$ What is meant here by $\Omega^*$? Conventionally, there isn't a single exterior algebra, but the exterior algebra associated with a particular vector space. In the case of differential forms over a manifold $M$, it would be the exterior algebra of the cotangent space $T^*M$. Likewise, the exterior algebra of $T^*\mathbb{R}^n$ can be identified with the exterior algebra of $\mathbb{R}^n$ itself. $\endgroup$ – Kajelad Jul 11 at 2:24
  • $\begingroup$ I mean $\Omega^*(\mathbb{R}^n)$ for an $n$-manifold --- edited! $\endgroup$ – singerng Jul 11 at 2:33
  • $\begingroup$ Update: Taking one chart on the circle to give $2\theta$ and the other to give $\theta$ turns out to be an easy counterexample. :) Thanks @Elliot $\endgroup$ – singerng Jul 11 at 4:56
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It might be more clear here to start with the intrinsic characterization.

We say the exterior algebra of a manifold $\Omega^*M$ is the algebra generated by wedge products of elements of the cotangent bundle $\Omega^*M=\wedge^*T^*M$, or, equivalently, the set of alternating multilinear forms on $TM$. Differential forms are sections of this bundle.

More verbosely, a $k$-form $\omega$ associates each point $p\in M$ with an alternating map $\omega|_p:(T_pM)^k\to\mathbb{R}$ and this map varies smoothly as we vary $p$.

We can, of course, identify a fiber of $\Omega^*M$ with $\wedge^*\mathbb{R}^n$ (by choosing a basis for $T_p^*M$), but there is no canonical way of doing this even locally.

When choosing local coordinates $x^i$ o a neighborhood $U$, the differentials $dx^i$ induce just such a basis for $T^*M$, allowing us to identify $\Omega^*U$ with $\Omega^*\mathbb{R}^n$ locall, i.e. every $k$-form $\omega$ can be written as $$ \omega=\sum_{i_1<\dots<i_k}\omega_{i_1\dots i_k}dx^{i_1}\wedge\dots\wedge dx^{i_k} $$ For $\omega_{i_1\dots i_k}\in C^\infty U$. This frame, however, will depend on the choice of coordinates, subjest to the transformation law $dx^i=\sum_j\frac{\partial x^i}{\partial y^j}dy^j$.

The definition you give is essentially the set of all local coordinate descriptions of the differential form, plus a consistency condition to ensure that they agree.

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  • $\begingroup$ So in this sense, the definition isn't wrong, it's actually too strong? $\endgroup$ – singerng Jul 11 at 3:57
  • $\begingroup$ The first definition is incorrect, it gives us a map $M\to\wedge^*\mathbb{R}^n$, but there's no way to associate $\wedge^*\mathbb{R}^n$ with $\wedge^*(T_p^*M)$. The second definition in equivalent: If we know all the local coordinate descriptions (and they are consisten), we can uniquely identify the differential form. If we know the differential form, we can write down all of its local coordinate representations. $\endgroup$ – Kajelad Jul 11 at 4:01

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