Take the 2-minute tour ×

In analysis course we encounter commonly the following definition of differentiable function: $f:U \rightarrow \mathbb{R^m}$, where $U \subset \mathbb{R^n}$ is differentiable when
$\exists \ T \in \mathcal{L}(R^m, R^n) $ such that $\forall a \in U$
$f(a + v) - f(a) = T(v) + r(v)$ where $\lim \frac{f(v)}{|v|} = 0$ when $v \rightarrow 0$

However, while studying in Spivak's Differential Geometry he defines differentiable structures for general manifolds, not considered as subsets of some $R^n$. He constructed maximal $C^\infty$-atlases and defined differentiability of a function by coordinated neighborhoods in the manifold.

Do these two definitions agree when the manifold is $R^n$ itself?

share|improve this question
Your definition of differentiable is wrong in several ways. $T$ should be from $\mathbb{R}^n$ to $\mathbb{R}^m$, the limit should be of $\frac{r(v)}{|v|}$, and (most importantly), $T$ should be allowed to depend on $a$. –  Chris Eagle Oct 19 '11 at 9:40

1 Answer 1

up vote 1 down vote accepted

Obviously the two concepts are the same when we regard $\mathbb R^n$ as a variety. Let's first consider the case of function of type $f \colon A \to \mathbb R$. We consider the atlas formed by all the immersions of opensets of $\mathbb R^n$ (identities if you considers charts to be homeomorphisms between opensets of variety and $\mathbb R^n$). With this atlas $f$ is differentiable if and only if for every chart $\varphi \colon U \to \mathbb R^n$ we have that $f \circ \varphi^{-1}$ is differentiable in classical sense (here $\varphi^{-1}$ is a section of $\varphi$); because of the choice of the atlas we have that $f \circ \varphi^{-1} = f|_V$ where $V= \varphi(U)$. So what we
get is that our function $f$ is differentiable in variety's sense if and only if for each openset $V \subset \mathbb R^n$ we have that $f|_V$ is differentiable in the classical sense, this clearly is equivalent to require that $f$ is differentiable in the classical sense. In the case of $f \colon A \to \mathbb R^m$ it suffices to see $f$ as a $m$-tuple of functions $\langle f_1,\dots,f_m \rangle$ in this way we can say that $f$ is differentiable just when each of the $f_i$ is differentiable.

Hope this may help.

share|improve this answer

Your Answer


