This is related to a previous question of mine.

Basically, it has occurred to me that there are two distinct ways in which a $k-$dimensional smooth manifold looks locally like $\mathbb{R}^k$: (1) at each point it has a well-defined tangent space which is isomorphic in many ways to $\mathbb{R}^k$ (with the points of the tangent space, except for the origin, not being points of the manifold itself) (2) at each point it has a neighborhood which is homeomorphic to $\mathbb{R}^k$ (with the points of the neighborhood obviously being points of the manifold itself).

So then does the difference between differentiable manifolds and non-differentiable topological manifolds come down to the following:

Differentiable manifolds resemble $\mathbb{R}^k$ in the sense of both (1) and (2).

Non-differentiable topological manifolds resemble $\mathbb{R}^k$ only in the sense of (2).

And then an intuitive way of thinking of the implicit function/inverse function/rank theorems is that they prove that $(1) \implies (2)$?

Is the existence of exponential maps for Riemannian manifolds an alternative way to show that $(1) \implies (2)$?

  • 1
    $\begingroup$ I don't think so, the tangent space is something that is really about differentiability. For example the algebraic construction of $T_xM^*$, which looks like it doesn't talk about differentiability, breaks down here: Let $R$ being the germs of (here continuous!) functions near $x$, $\mathfrak m$ the ideal where the germs are $0$ at $x$ and then identifying $T_xM^*$ as the $R/\mathfrak m$ vector space $\mathfrak m/\mathfrak m^2$ yields $T_xM^*=0$, as every continuous function is a "signed root" of other continuous functions. $\endgroup$ – s.harp Oct 31 '16 at 12:08
  • 1
    $\begingroup$ I think definition of tangent space use the differentiable structure of manifold in an essential way.For example it is defined as the set of all derivations D at a point(say p).Now D acts on smooth function from a neighbourhood of p to R. The differentiable structure comes into play,to determine whether a function should be considered as smooth or not which is unambiguously defined using smooth chart transitions. $\endgroup$ – SAUVIK Oct 31 '16 at 15:18
  • 2
    $\begingroup$ No. They have something called a tangent microbundle, which is probably not worth worrying about until much, much later in your topological education. $\endgroup$ – user98602 Oct 31 '16 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.