What is the intuition behind the definition of a differential of a function in differential geometry? i.e. $$df(p)(v_{p}) =v_{p} (f)(p) $$ where $v_{p} \in T_{p} M$ is a vector in the tangent space to the point $p\ M$.

Is it just that the notion of an infinitesimal change in a function is not mathematically rigorous (in the "traditional" sense that it's presented in elementary calculus). So in order to make the notion mathematically rigorous we note that a quantity that captures the notion of an infinitesimal change in a function should itself be a function of all the possible infinitesimal changes of the point that the function is evaluated at, i.e. it should be a function of the all possible tangent vectors in the vector space tangent to that point (as such quantities describe the possible directions that a function can pass through a point and all the possible "speeds"). In this sense the differential change in the function as it passes through the point in a given direction should be equal to the directional derivative of the function along the particular vector describing that direction?!


4 Answers 4


The most classical answer is probably: $df_p$ is the best linear approximation of $f$ at $p$, in a sense that can be made precise.

Personally I often like to think of it in the following way: while $f$ transforms positions, the differential of $f$ transforms velocities (or: while $f$ transforms points, $df$ transforms tangent vectors.)

u**strong text**

More precisely: let $t \mapsto x(t)$ be a curve in $M$ through $p$ at $t=0$. The image of this curve in $N$ is $t \mapsto y(t) = f(x(t))$. Let $v = x'(0) \in T_{p}M$ and $w = y'(0) \in T_{f(p)}N$. Then $w = df_p(v)$.

In short: $$\frac{d}{dt}_{|t=0}f(x(t)) = df_{x_0} (x'(0))$$

I find this interpretation of the differential useful both conceptually and also very often practically.

  • $\begingroup$ So is $df$ essentially quantifying how much $f$ changes as one "moves" along a particular tangent vector? Can the functions $dx^{i}$ be thought of as coordinate functions in the tangent space to a particular point? $\endgroup$
    – Will
    Mar 20, 2015 at 15:37
  • $\begingroup$ What do you mean by coordinate functions for the tangent space? It is true that if $v = \sum_{i = 1}^n v_i \frac{\partial}{\partial x_i}$, then $dx_i(v) = v_i$, if that's what you mean. $\endgroup$
    – A.P.
    Mar 20, 2015 at 16:39
  • $\begingroup$ yes, that's what I meant, that they give the coordinates of the tangent vectors (with respect to the coordinate basis $\frac{\partial}{\partial x^{i}})$ in the tangent space $\endgroup$
    – Will
    Mar 20, 2015 at 16:43
  • $\begingroup$ 1. Yes, $df(v)$ measures the (infinitesimal) variation of $f$ in the direction of $v$. Note that when $M$ is an open set in a vector space, $df_x(v)$ is just the infinitesimal variation $[f(x + tv) - f(x)]/t$ as $t \to 0$. What I wrote with curves is a generalization of that in the setting of manifolds. 2. Yes, $dx^i$ are linear coordinates in the tangent space. $\endgroup$
    – Seub
    Mar 20, 2015 at 16:46
  • 2
    $\begingroup$ That's right. The differential $dx_i$ is literally the differential of the coordinate function $x_i$. Don't overthink this though. If I wanted to be pedantic, I would say that $dx^i(v) = v^i$ is the definition of $\frac{\partial}{\partial x^i}$, not the other way around ($\frac{\partial}{\partial x^i}$ is the dual basis to $dx^i$ by definition) $\endgroup$
    – Seub
    Mar 20, 2015 at 17:14

I like to think it this way:

The "infinitesimal" perspective is not encoded in $df$ or in $v_p$, but in $f$ itself. This is because we define the derivations at a point $p$ not on smooth functions, but on the stalk of germs of smooth functions at $p$, i.e. on $$ \mathcal{C}^{\infty}_p := \left\{ (f,U) : p \in U, U \text{open}, f \in \mathcal{C}^{\infty}(U,\mathbb{R})\right\} / \sim $$ where $(f,U) \sim (g,V)$ iff $f \mid_{U \cap V} = g \mid_{U \cap V}$. In particular, this means that the class $f_p$ of $(f,U)$ captures the local behaviour of $f$ at $p$.

Now fix a local coordinate chart $((x_1,\dotsc,x_n),U)$ for some open neighbourhood $U$ of $p$, so that $\frac{\partial}{\partial x_1},\dotsc,\frac{\partial}{\partial x_n}$ is a basis for $T_p M$. Here $\frac{\partial}{\partial x_j}$ is by definition the (unique) derivation such that $$ \frac{\partial}{\partial x_j}(x_{i,p}) = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{otherwise} \end{cases} $$ This means that you can indeed think of $\frac{\partial}{\partial x_j}(f_p)$ as the local change of $f$ at $p$ in the "direction" $\frac{\partial}{\partial x_j}$, i.e. in the same direction in which the coordinate map $x_j$ increases.

So, what is the differential of $f_p$? I think of it a useful change of perspective: it allows you to think of $\frac{\partial}{\partial x_j}(f_p)$ not as a function of $f_p$, but as a function of $\frac{\partial}{\partial x_j}$. This is useful because it allows you to define a very natural map from smooth functions to the dual space $T^*_p M$. On one hand, with this map you can easily prove that $dx_1,\dotsc,dx_n$ is a basis for this space. On the other hand, it suggests that if we can reasonably map $0$-forms to $1$-forms, then it might not be too hard to extend this idea to a map from $n$-forms to $n+1$-forms. Indeed, this is possible and it is what we call the exterior derivative.

It might be useful to clarify that Seub is talking about a slightly different thing in his answer. Indeed, consider a morphism of differential manifolds $\phi\colon M \to N$ and fix a point $p \in M$. Then precomposition by $\phi$ gives a map from $\mathcal{C}^{\infty}_{\phi(p)}$ (on $N$) to $\mathcal{C}^{\infty}_p$ (on $M$), simply because if $f\colon N \to \mathbb{R}$ is smooth at $\phi(p)$, then $f \circ \phi$ is smooth at $p$. Then you can define a map $$ \phi_* \colon T_p(M) \to T_{\phi(p)}(N) $$ by putting $\big(\phi_*(v_p)\big)(f_{\phi(p)}) = v_p((f \circ \phi)_p)$ for every $f \in \mathcal{C}^{\infty}_{\phi(p)}$.

Why are these two things related? Suppose $N = \mathbb{R}$. Then the identity on $N$ has a germ $\mathbf{1}_{\phi(p)} \in \mathcal{C}^{\infty}_{\phi(p)}$ and we have $\big(\phi_*(v_p)\big)(\mathbf{1}_{\phi(p)}) = v_p((\mathbf{1} \circ \phi)_p) = v_p(\phi_p)$.

This perspective is very useful, because it allows you to "compare" the tangent spaces of two manifolds (in to corresponding points). I don't think that its motivation is measuring how $\phi$ changes, though; maybe historically, but not in this formalism. You may probably try to do so by passing through local charts, and indeed Boothby does something similar in examples 1.9 and 1.10, chapter 4, of his book (pp. 112-115 in my edition).

Note: Your question asked about the intuition (or, I think, motivation) of the definition of differential in differential geometry, and I hope I made this clear. On the other hand, from your comments it seems like the sources of your confusion are Spivak's book and the effort to reconcile this definition with the classical notion of differentials as "infinitesimal quantities". Now, to quote L. Ryder's book:

The $1$-form $\mathbf{d}x^{\mu}$ is not the same as the infinitesimal $dx^{\mu}$: it is not a 'number', but a member of the cotangent space $T_p^*$.

Just before the paragraph you quoted in your comments, Spivak writes (emphasis mine):

Classical differential geometers (and classical analysts) did not hesitate to talk about "infinitely small" changes $dx^i$ of the coordinates $x^i$, just as Leibnitz had. No one wanted to admit that this was nonsense, because true results were ob­tained when these infinitely small quantities were divided into each other (pro­vided one did it in the right way).

This simply means that the classical formalism is not rigorous: you may think of "infinitesimal changes" if it helps you visualise what you're doing, but you can't trust any result you obtain in this way until you prove them using solid definitions.

He then goes on saying that while we cannot say how much is an "infinitely small" change (at least in the framework of standard analysis), we can still say "where our function is going", like with tangent vectors to a curve. Furthermore, classically the differential of a function is seen as its variation in front of an "infinitesimal" variation in its variables, so you could start formalising it as a function of this change. Since we said that we can still (intuitively) think of change as a tangent vector it then becomes a function on the tangent space.

In this sense, if you think of a smooth curve on a smooth manifold $M$ as a smooth function $c$ from $\mathbb{R}$ to $M$ (which you can, at least locally), then the "infinitesimal change" of the curve in front of an "infinitesimal change" of the parameter becomes a function from $T_p\mathbb{R}$ to $T_{c(p)}M$, which we call the differential of $c$.

  • $\begingroup$ So does the differential $dx_{p}^{i}$ quantify the local change in the coordinate map $x^{i}$ as one "moves" along a particular direction (specified by a tangent vector in the tangent space to the point $p\in M$)? Also, slightly off topic, but is the exterior part in the notion of an *exterior derivative" called as such, because when one operates it one is mapping to an object "outside" the original set that you started in (in a sense)? $\endgroup$
    – Will
    Mar 20, 2015 at 15:42
  • $\begingroup$ Yes, to both questions. $\endgroup$
    – A.P.
    Mar 20, 2015 at 16:32
  • $\begingroup$ Is there any way to relate the differential $dx^{i}_{p}$ to an "infinitesimal change in $x^{i}_{p}$"? I've read Spivak's notes in which he says that "the closest that we can come to describing the infinitesimal change in a function is to describe a direction in which this change is supposed to occur...", however, I found the explanation a little vague, and left me slightly unsure as to what was meant by it?! $\endgroup$
    – Will
    Mar 20, 2015 at 17:15
  • $\begingroup$ I think it means that while you can describe such change qualitatively, you cannot do so quantitatively (because in general your manifold is not "flat"). You may find helpful to read sections 3 and 4 of chapter 2 and section 1 of chapter 4 (especially the examples) of Boothby's "An Introduction to Differentiable Manifolds and Riemannian Geometry". $\endgroup$
    – A.P.
    Mar 20, 2015 at 17:54
  • $\begingroup$ Thanks for reference. I think what has confused me is whether $dx^{i}$ is considered as a function of a finite difference between to points (I understand that on a manifold, quantitatively this doesn't make sense), or whether the point is that as one can't add (or subtract) points on a manifold, quantitatively the best we can do is to describe instantaneous (i.e. infinitesimal) changes in coordinate functions (at a given point)... $\endgroup$
    – Will
    Mar 20, 2015 at 18:17

(The following is only about functions defined on some $\Omega\subset{\mathbb R}^n$)

If $p$ is a "generic" point for the function $f:\>{\mathbb R}^n\to{\mathbb R}$ then the rate of change of $f$ when walking away from $p$ depends on the chosen direction, but not in an arbitrary way: There is a direction of maximal increase, there is a plane through $p$ with virtually no change of $f$ when walking away staying in this plane, and there is a direction of maximal decrease. One can measure these various rates of change using directional derivatives: If $u$ is a unit vector one can define $$D_uf(p)=\lim_{t\to0+}{f(p+t u)-f(p)\over t}\ .$$ This idea does not tell us anything about how $D_uf(p)$ depends on $u$. In reality there is a certain vector $\nabla f(p)$, called the gradient of $f$ at $p$, such that $D_u f(p)$ can be computed for all $u$ by the formula $$D_u f(p)=\nabla f(p)\cdot u\ .\tag{1}$$ This shows that the "linear behavior" of $f$ when walking away from $p$ is encoded in this vector $\nabla f(p)$ once and for all.

The rule $$X\mapsto \nabla f(p)\cdot X$$ appearing in $(1)$ for the special case $X=u$, a unit vector, turns the vector $\nabla f(p)$ into a linear functional $\phi$ on the tangent space at $p$: For each $X\in T_p$ a value $\phi(X):=\nabla f(p)\cdot X$ is defined. This linear functional is called the differential of $f$ at $p$, and is denoted by $df(p)$. It is related to small changes of $f$ in the neighborhood of $p$ by the formula $$f(p+X)-f(p)=df(p).X+o\bigl(|X|\bigr)\qquad(X\to0)\ .\tag{2}$$ Note that in $(2)$ neither a scalar product, nor unit vectors appear.

  • $\begingroup$ How does $(2)$ translate for a smooth function $f \colon M \to \mathbb{R}$? I see that it should be possible by passing through a local chart at $p \in M$, but it would be nice if you could expand on this. $\endgroup$
    – A.P.
    Mar 21, 2015 at 13:39
  • $\begingroup$ @A.P.: This takes several pages in differential geometry books. Note that on a manifold $M$ we cannot write $p+X$. $\endgroup$ Mar 21, 2015 at 14:56
  • $\begingroup$ That was my point, exactly. From what I understand the OP is not concerned with the classical view of a differential as a way of measuring small changes of a function, but with how to translate this view in the context of (modern) differential geometry. $\endgroup$
    – A.P.
    Mar 21, 2015 at 15:11

Like always, the intuition comes from the Euclidean case — here even the one-dimensional case!

In the differential geometry of the line $\mathbb{R}$ with coordinate function $t$, we have

  • The standard vector field, usually denoted $\partial /\partial t$
  • The standard differential form, denoted $\mathrm{d} t$

The standard vector field captures the geometric aspect of one-variable calculus and the standard differential form captures the algebraic aspect.

These applications may be a little hard to tell apart at first, but it becomes more clear when you see how the line is used:

  • Maps $\mathbb{R} \to M$ describe curves on a manifold. Each curve pushes the standard vector field $\partial / \partial t$ forward to give the tangent vectors to the curve on $M$.
  • Maps $M \to \mathbb{R}$ describe scalar fields on a manifold. Each scalar field pulls the standard differential form $\mathrm{d} t$ back to give the differential of the scalar field on $M$.

Now, we know how to combine differential forms and vector fields on $\mathbb{R}$; e.g. the simplest case is that if $y = g(t)$, then $\mathrm{d} y \cdot \frac{\partial}{\partial t} = g'(t)$. This extends to the manifold: if we have a curve $\gamma$ and a scalar field $f$, then

$$ f^*(\mathrm{d} t) \cdot \gamma_*(\partial / \partial t) = (f \circ \gamma)'$$

This is consistent with the other two places we could do the combination:

$$ \mathrm{d} t \cdot (f_* \circ \gamma_*)(\partial / \partial t) = (f \circ \gamma)'$$ $$ (\gamma^* \circ f^*)(\mathrm{d} t) \cdot \partial/\partial t = (f \circ \gamma)'$$

As a side effect of my algebraic tendencies, I personally prefer a "differential forms first" approach to this sort of thing; IMO expressing things from the algebraic perspective tends to make calculation cleaner and more intuitive. In this approach, one might define the tangent vectors as linear functionals on the differentials.

e.g. $\mathrm{d}(x^2 y) = 2xy \mathrm{d}x + x^2 \mathrm{d} y$, and in $(x,y)$ coordinates, the vector $\partial / \partial x$ is the mapping that sends $\mathrm{d}x \to 1$ and $\mathrm{d} y \to 0$. I think this formulation is even a more direct description of what we mean by the notation $\partial / \partial x$: "the variation along with $x$ as $y$ is held constant".

I've never seen the foundations of differential geometry developed in this way, but I imagine you would start define the cotangent bundle in a manner analogous to however you would have defined the tangent bundle.

However, in other subjects where functions play more prominence at the foundational level, such as algebraic geometry, differential forms first is a standard approach.


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