First, recall a general definition of a differentiable function as follows. Suppose $f :D\subset \mathbb R^3 \rightarrow \mathbb R$. Then $f$ is differentiable at $\mathbf a \in D$ if there is a linear function $L:\mathbb R^3 \rightarrow \mathbb R$ such that

$$\lim_{\mathbf x\rightarrow\mathbf a}\frac{\|f(\mathbf x)-[f(\mathbf a)+L(\mathbf x-\mathbf a)]\|}{\|\mathbf x-\mathbf a\|} = 0.$$

Now consider a definition of a differentiable function defined on a regular surface given in most differential geometry textbook. Let $f:V\subset S\subset \mathbb R^3\rightarrow \mathbb R$ be a function defined on an open subset $V$ of a regular surface $S$ (which itself is a subset of $\mathbb R^3$). Then $f$ is differentiable at $\mathbf p\in V$ if for some parameterization $\mathbf x: U\subset\mathbb R^2\rightarrow S$ with $\mathbf p\in \mathbf x(U)\subset V$, the composition $f\circ\mathbf x:U\subset \mathbb R^2\rightarrow \mathbb R$ is differentiable at $\mathbf x^{-1}(\mathbf p)$.

My question is whether the two definitions are consistent with each other. To be more specific, if the two definitions are consistent, then

  1. Why do we need a new definition for differentiable functions defined on a regular surface which is in fact a subset of $\mathbb R^3$ anyway?
  2. What is the corresponding linear function $L$ in the second definition?

Thank you!


I think your questions will be answered most clearly if we try to identify the assumptions under which the definitions you wrote for differentiability make sense. The first thing to note is, like you wrote, that the first definition tells you when a map is differentiable and what is its differential while the second definition only tells you when a map is differentiable. A differentiable map from a surface also has a differential, but it is usually discussed separately from the definition of differentiability.

The first definition you have given makes sense for functions $f \colon D \rightarrow \mathbb{R}$ defined on an open subset $D \subseteq V$ of some normed vector space $(V, ||\,||)$. We write $x - a$, and we require from the map $L \colon V \rightarrow \mathbb{R}$ to be linear, so $V$ better have the structure of a vector space and we use norm to make sense of the limit and estimate the size of $x - a$. The interpretation of $L$ we have in mind is that given $v \in V$ with $||v|| = 1$, $Lv = \frac{d}{dt} f(a + tv)|_{t =0}$ is directional derivative of $f$ in the direction of $v$.

This definition makes sense even if $D$ is not open if we take the limit in $D$, treating $D$ for example as a metric space with a metric induced from the norm $||\,||$. However, this raises many problems. For example, if $D = \{ (x,y,z) \, | \, z = 0 \} \subset \mathbb{R}^3$, then the derivative $L$ won't be unique. The limit $x \to a$ is taken in $D$, so $x - a$ always lie in $D$ and thus $L(x-a)$ that appears in the limit depends only on how $L$ acts on the subspace $D$ and not on the whole $\mathbb{R}^3$. The problem is that in $D$, one can approach a point $a$ using only directions that lie in the xy plane and so it doesn't make sense to require a priori from $L$ to be defined on the whole vector space $\mathbb{R}^3$ but to be defined only on the directions that are relevant to the limit. Of course, we can take $D$ to be something like $\{ (x,|x|,0) \, | \, x \in \mathbb{R} \}$ and then there are only two directions from which we can approach $a = (0,0,0)$ inside $D$ and so it doesn't make sense to encode the directional derivative in an operator that is defined on a vector space.

That is why $D$ is taken be an open set and so each point $a \in D$ can be approached inside $D$ through all possible directions and we say that $f$ is differentiable at $x = a$ if all possible directional derivatives can be "encoded uniformly in a linear operator" $L \colon \mathbb{R}^3 \rightarrow \mathbb{R}$.

Now, if $S \subseteq \mathbb{R}^3$ is a regular surface, it is never open in $\mathbb{R}^3$ so we need a different approach. We think of a regular surface as a two-dimensional object and so to check the differentiability of $f$ at $p$, we intuitively know that we need to check it only with respect to the directions the point $p$ can be approached inside $f$. The easiest definition is then to compose $f$ with a coordinate chart around $x = p$ (which turns $f$ into a function of two variables) and then to say $f$ is differentiable if the composition is differentiable.

Connecting to what I wrote before, if you read further, you'll see that the fact that $S$ is a regular surface and not an arbitrary subset of $\mathbb{R}^3$ guarantees that at each $p \in S$ there is a two-dimensional affine vector subspace of $\mathbb{R}^3$ called that tangent plane that consists of the velocities ("directions") of all the curves that pass through $p$ and live on $S$. This tangent plane depends on the point $p \in S$ and changes when we move $p$ around. The differential of $f$ at $p \in S$ will be defined as a linear map $df_p \colon T_pS \rightarrow \mathbb{R}$ and if $f$ is the restriction of a differentiable map $\tilde{f} \colon \mathbb{R}^3 \rightarrow \mathbb{R}$, then under appropriate identifications, $df_p$ will be the restriction of $d\tilde{f}_p$ to the two dimensional subspace of "relevant" directions tangent to $S$.

  • $\begingroup$ Thank you for your very detailed response. I do not understand the following comments you made: "if $S\subset \mathbb ℝ^3$ is a regular surface, it is never open in $\mathbb ℝ^3$." Why is this the case, please? $\endgroup$ – LaTeXFan Apr 17 '14 at 8:32
  • $\begingroup$ A set $U \subseteq \mathbb{R}^3$ is open if for every point $p \in U$ exists an open ball $B(p,\varepsilon_p)$ of $\mathbb{R}^3$ such that $B(p, \varepsilon_p) \subseteq U$. A regular surface is something "two dimensional" (like a sphere, a torus, a paraboloid) sitting in $\mathbb{R}^3$. It will never contain an open ball of $\mathbb{R}^3$. This follows from example from the fact that a regular surface is locally a graph of a function of two variables and a graph of a function is never open. $\endgroup$ – levap Apr 19 '14 at 15:33

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