What I have in mind is along the lines of this:

Let $M$ a topological space, $V$ a normed vector space, and

$$ \boxminus \colon M\times M \to V, $$ $$ \boxplus \colon M\times V \to M. $$

Then $M$ is called a manifold1 with tangent space $V$, if $$ p \boxplus (q \boxminus p) = q, $$ $$ p \boxplus \vec{0} = p $$ and, for sufficiently small2 vectors $v,w$, $$ (p \boxplus v) \boxplus w \approx p \boxplus (v+w). $$

Can this set of axioms be made equivalent to the standard definition of manifolds (with an atlas of ball-homeomorphic neighbourhoods)?

Note that rather than requiring the manifold to be locally homeomorphic to a ball in a vector space, I require it to be globally isomorphic to some subset of a vector space, in a way that's locally a homeomorphism.

It sure seems to work for simple examples. Christian Blatter suggests $S^1$. Well, if we interpret that as $[0,1]$ with the ends glued together, then we can just define $\theta \boxminus \phi$ as $\theta - \phi$ folded back into the interval $[-\tfrac12, \tfrac12[ \subset V \equiv \mathbb{R}$, and $\theta \boxplus \delta$ as $\theta + \delta$ confined to $[0,1[$. The axioms hold; in fact $\boxplus$ is associative.

The same works for tori of arbitrary dimension. For $S^2$ it immediately gets a bit messier, but I'd figure something Euler-angle-ish should do the trick in much the same way as in $S^1$. Where I'm uncertain is with multi-hole handlebodies, or Klein bottles etc. – and even if all those work, with the general case.

1 Evidently, this definition implies at least that the manifold will be path-connected. Likely also other things, but which exactly?

2 A precise formulation of this: let $\varepsilon>0$, $p\in M$, $v,w \in V$. Then there exists $\eta>0$, such that $$\left\|\bigl((p \boxplus \eta v) \boxplus \eta w\bigr) \boxminus \bigl(p \boxplus (\eta v + \eta w)\bigr) \right\| < \eta\cdot \varepsilon.$$

  • $\begingroup$ Try to realize a well known manifold, say $S^1$, along these lines! $\endgroup$ Jan 26, 2015 at 12:43
  • $\begingroup$ Trying to get intuition here, since I'm not sure I understand what $\boxplus$ and $\boxminus$ are "supposed to be": How do you define $\boxplus$ when $M = \mathbb{R}^2 \setminus \lbrace 0 \rbrace$? (What is your picture of a natural $V$ for this manifold?) $\endgroup$
    – mollyerin
    Jan 28, 2015 at 4:58
  • $\begingroup$ @mollyerin: $\mathbb{R}^2\setminus \{0\}$ is homeomorphic to $S^1 \times \mathbb{R}$, so I'd use $V = \mathbb{R}^2$, where one of the axes represents the tangent space of the angle and the other of the radius. I'd define, for instance, $(\phi,r) \boxplus (\delta_\phi, \delta_r) = ((\phi + \delta_\phi)_{[\text{mod}2\pi]}, \phi \cdot e^{\delta_\phi})$. $\endgroup$ Jan 28, 2015 at 12:16
  • $\begingroup$ Wait, I meant $(\phi,r) \boxplus (\delta_\phi, \delta_r) = \bigl((\phi + \delta_\phi)_{[\text{mod}2\pi]}, r \cdot e^{\delta_r}\bigr)$ of course. $\endgroup$ Jan 28, 2015 at 12:23
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    $\begingroup$ Also, in case it helps anyone's intuition, on a parallelizable Riemannian manifold $M$, one can use $\exp:M\times T_p M\rightarrow M$ as $\boxplus$. I'm still not sure what to make of $\boxminus$. $\endgroup$ Jan 28, 2015 at 14:14


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