Understanding Differentials, Differential Forms, and Exterior Derivatives – Differential Geometry

differential-geometry

I don't get the relationship between differentials, differential forms, and exterior derivatives. (Too many $d$'s getting me down!) Here are the relevant (partial) definitions from Wikipedia; essentially the same definitions/terminology/notations are to be found in my notes.

  1. Pushforward. Let $\varphi : M → N$ be a smooth map of smooth manifolds. Given some $x \in M$, the differential of $\varphi$ at $x$ is a linear map $d\varphi_x : T_x M \rightarrow T_{f(x)}N$…

  2. Differential form. Let $M$ be a smooth manifold. A differential form of degree $k$ is a smooth section of the $k$th exterior power of the cotangent bundle of $M$. At any point $p \in M$, a $k$-form $\beta$ defines an alternating multilinear map $\beta_p : T_p M \times \cdots \times T_p M \rightarrow \mathbb{R}$…

  3. Exterior derivative. The exterior derivative is defined to be the unique $\mathbb{R}$-linear mapping $f \mapsto df$ from $k$-forms to $(k + 1)$-forms satisfying the following properties…

What I understand:

  • You apply $d$ to differential $k$-forms to get differential $(k+1)$-forms. Implicitly, this means "exterior derivative."

What I don't understand:

  • If $\varphi : M \rightarrow N$ is a smooth map of smooth manifolds, in what sense, if at all, is the differential of $\varphi$ a differential form? Is there any reason not to just call this the pushforward and consistently denote it $\varphi_*$?

  • If $f : M \rightarrow \mathbb{R}$ is a smooth map, does $df$ mean the differential of $f$, or does it mean the exterior derivative? Are these somehow miraculously the same? If so, why? It seems possible that they're the same, by identifying $T_x\mathbb{R}$ with $\mathbb{R}$. I don't understand the details.

  • What, if anything, is the connection between the differential of a smooth mapping and the exterior derivative of a differential form?

Best Answer

For a beginner just starting to come to grips with these ideas, I think the most useful answer is this:

Except in one special situation (described below), there is essentially no relationship between the exterior derivative of a differential form and the differential (or pushforward) of a smooth map between manifolds, other than the facts that they are both computed locally by taking derivatives and are both commonly denoted by the symbol $d$.

Differential geometry is loaded with notation, and sometimes we just run out of letters, so we have to overload a symbol by interpreting it in different ways in different situations. The fact that two things are represented by the same symbol doesn't always mean that they're "the same" in any deep sense.

The one situation in which the two concepts are directly related is for a smooth map $f\colon M\to\mathbb R$. In this case, we can consider $f$ either as a smooth map between manifolds or as a $0$-form. Considering it as a smooth map, for each $x\in M$, the pushforward is a linear map $df_x\colon T_xM\to T_{f(x)}\mathbb R$. Considering it as a $0$-form, its differential $df$ is a $1$-form, which means that for each $x\in M$ we have a linear functional $df_x\colon T_xM\to \mathbb R$. The link between the two is the fact that, because $\mathbb R$ is a vector space, there's a canonical identification $T_{f(x)}\mathbb R\cong\mathbb R$, and under that identification these two versions of $df_x$ are exactly the same map.

The excellent answer by @user86418 explains a sophisticated context in which both pushforwards and exterior derivatives can be viewed as special cases of a more general construction; but that's a context I wouldn't recommend that a beginner spend much time trying to come to terms with.

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