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?


3 Answers 3


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.


$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\renewcommand{\phi}{\varphi}$As you may know, differential forms on $M$ can be "vector-valued", i.e., they can take values in some vector bundle $E \to M$. In the simplest case, $E$ is the trivial real line bundle, and "$E$-valued differential forms" are just "differential forms". In general, if $(\Basis_{j})_{j=1}^{\ell}$ is a local frame for $E$ in some trivializing coordinate neighborhood $U$ (i.e., both $E$ and $TM$ are trivial over $U$), an $E$-valued $k$-form looks like $$ \sum_{j=1}^{\ell} \omega_{j} \Basis_{j} $$ with the $\omega_{j}$ ordinary $k$-forms in $U$.[1]

If $\phi:M \to N$ is a smooth map, and if $p:E \to N$ is a vector bundle, there is a pullback bundle $\phi^{*}E \to M$ whose total space is $$ \{(x, v) \text{ in } M \times E: \phi(x) = p(v)\} $$ and whose projection map is projection to the first factor. Intuitively, put a copy of the fibre $E_{\phi(x)}$ over $x$ for each $x$ in $M$.

The push-forward is sometimes introduced informally as a mapping $\phi_{*}:TM \to TN$, but technically that's not right. Actually, $\phi_{*}$ takes values in $\phi^{*}TN$ and is a mapping between vector bundles over $M$.

  • The differential $d\phi$ may be viewed as a $1$-form on $M$ taking values in $\phi^{*}TN$, the pullback of the tangent bundle by $\phi$. That's a fancy way of saying that if $v$ is a tangent vector to $M$ at a point $x$, then $d\phi(x)(v)$ is an element of $T_{\phi(x)}N$, and is linear in $v$. (Of course, $d\phi(x)(v)$ actually measures something useful, the "rate of change" of $\phi$ at $x$ in the direction $v$.)

  • For a real-valued function $f$, the differential $df$ is indeed the exterior derivative, by the usual definition of the exterior derivative. To mesh this with the preceding item, note that $T\Reals$ is a trivialized vector bundle: We know what "$1$" means when we speak of real-valued functions and forms, so $T\Reals = \Reals \times \Reals$ and consequently $f^{*}T\Reals = M \times \Reals$. That means we can view the differential $df$ as a $1$-form with values in $M \times \Reals$, i.e., as a real-valued $1$-form.

  • Your third bullet point may be answered by the first bullet point above, but it's probably worth mentioning that a smooth map $\phi:M \to N$ could be termed an "$N$-valued $0$-form", whose differential is $d\phi$.

Notationally, I think of $d\phi(x)$ as the linear fibre map sending a tangent vector $v$ in $T_{x}M$ to $d\phi(x)(v)$ in $T_{\phi(x)}N$, while the push-forward $\phi_{*}$ is the bundle map sending $(x, v)$ in $TM$ to $\bigl(\phi(x), d\phi(x)(v)\bigr)$ in $TN$. So, they're not quite identical if you need to split hairs, but in practice you can usually talk only about the push-forward.

  1. Technically, an $E$-valued differential $k$-form is a smooth section of the vector bundle $\bigwedge^{k}T^{*}M \otimes E$. If you change trivializations, the form "coefficients" transform like ordinary differential forms, and the "vector parts" transform like sections of $E$. In general, there's no natural exterior derivative on $E$-valued forms; you have to know how to differentiate sections of $E$, as well. If $E$ has locally constant transition functions, however, then there is a natural exterior derivative, given by $$ d\left(\sum \omega_{j} \Basis_{j}\right) = \sum (d\omega_{j}) \Basis_{j}. $$ One can't mention this without making a plug for holomorphic vector bundles: Holomorphic functions act like constants with respect to the Dolbeault operator $\bar{\partial}$ (i.e., $f$ is holomorphic if and only if $\bar{\partial}f = 0$), so there's a well-defined $\bar{\partial}$ operator $$ \bar{\partial}\left(\sum \omega_{j} \Basis_{j}\right) = \sum (\bar{\partial}\omega_{j}) \Basis_{j} $$ for differential forms with values in a holomorphic vector bundle $E$. (!)
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    $\begingroup$ What's informal about getting from $\varphi : X \rightarrow Y$ to $\varphi_* : TX \rightarrow TY$? Isn't tangent space functorial, with $T\varphi = \varphi_*$? $\endgroup$ Commented Oct 30, 2016 at 4:26
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    $\begingroup$ If $X$ is a vector field on $M$ (i.e., a section of $TM$), then $\varphi_{*}X$ is generally not (in fact, "usually not") a section of $TN$, but is a section of $\varphi^{*}TN$. (First, $\varphi_{*}X$ is not defined off the image of $\varphi$. Second, if $\varphi$ is not injective, $\varphi_{*}X$ may be multiple-valued.) By "informal", I meant to convey that "to a first approximation" one may think of $\varphi_{*}:TM \to TN$, but generally it's preferable to view the push-forward taking values in the pullback bundle $\varphi^{*}TN$. $\endgroup$ Commented Oct 30, 2016 at 11:35
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    $\begingroup$ I'm not sure I understand. Under the point of view I was taught, the pushforward of $\varphi : M \rightarrow N$ doesn't act on vector fields on $M$, rather it acts on based vectors in $M$. In other words, it acts on the elements of $TM$. Are you saying that a better point of view is to regard $\varphi_*$ as acting on sections of the aforementioned bundles, instead of elements? In other words, are you saying that given a smooth map $\varphi : M \rightarrow N$, we should think of $\varphi_*$ as a mapping $\Gamma(TM) \rightarrow \Gamma(\varphi^*TN)$ instead? If so, what are the benefits... $\endgroup$ Commented Feb 7, 2020 at 2:37
  • $\begingroup$ ... of this change in point of view? (Given a total space $E$ equipped with an arrow down to a base space $M$, I'm writing $\Gamma(E)$ for the set of sections of the arrow to the base space.) $\endgroup$ Commented Feb 7, 2020 at 2:38

They are not equal! Pushforward gives you the value of function in the given point ,if you set $N=IR$ but in exterior case,it gives you the value of the derivative of the function in the given point. An easier case is this: $M=IR$ & $N=IR$ and let a f be differentiable function : $f:M \to N$. It can be $f(x)=\sin(x)$ Then the pushforward of f,maps vector $\frac{d}{dx}$ at $x=1$ to $\frac{d}{dx}$ at $x=\sin(1)$. But exterior derivative of $f, df=\cos(x)dx$ maps $\frac{d}{dx}$ at $x=1$ to $\cos(1)$.even in the vector-value on form case,it maps $\frac{d}{dx}$ to $\cos(1) \frac{d}{dx}$ at $x=\sin(1).$

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    $\begingroup$ This is incorrect; the pushforward $f_*$ also maps $\frac d{dx}$ at $1$ to $\cos(1)\frac d{dx}$ at $\sin(1)$. If we view tangent vectors as derivations on the germs of smooth functions, then the definition is $f_*(v)[g]:=v[g\circ f]$, and $\frac d{dx}[g]=g'(x)$. So at $x=1$, $f_*(\frac d{dx})[g]=\frac d{dx}(g\circ f)=(g\circ f)'(1)=g'(f(1))f'(1)=g'(\sin(1)){\cos(1)}$; while at $x=\sin(1)$, $\cos(1)\frac d{dx}[g]=\cos(1)g'(x)=\cos(1)g'(\sin(1))$, which is the same. In general, $f_*(\frac d{dx})=\cos(x)\frac d{dx}$ for this function $f$. $\endgroup$ Commented Oct 24, 2022 at 15:34

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