Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Two or more sources of study I use on Differential Geometry,most notably, Barrett O'Neill's book, defines an $ n $ dimensional manifold as follows: A set furnishes with a collection $ P $ of abstract "patches" (which are 1-1 and regular functions x $ :D \rightarrow M $, $D$ being an open set in $ \mathbb R^n $) satisfying : The covering property: Images of patches in $ P $ cover $M$.

The smooth overlap property:For an $x$,$y$ in $P$, functions $y^{-1}x$ and $ x^{-1}y $ are differentiable and defined on open sets in $\mathbb R^n$ .

The Hausdorff property.

I would like to know if this definition is any better than the manifold definition involving charts that one normally encounters in which the above properties are stated for $x^{-1}$ . It is to be noted that these discussions are all mainly concerned with $2-D$ surfaces. So is this definition merely well-suited because it helps the fresher realise that surfaces in $3-D$ space are stitched out of $2-D$ subsets of the plane??

One another thing is that regularity and 1-1 avoids self-intersecting surfaces and other problematic ones. So is this definition any good than the atlas-based one??

Is there anything more subtle that I am failing to grasp here, or is it merely a matter of convention, in which case this question would probably look foolish?

share|improve this question
3  
My 2 cents: It is merely a matter of convention. You switch to the usual definition by considering $x^{-1}$-functions and calling them "charts" instead of "patches". –  Giuseppe Negro Nov 11 '12 at 3:35
    
That's what i thought too, apart from the fact that it allows new entrants to visualise the construction of 2-D surfaces better. Thanks a lot anyways, I'll just a wait a little longer to see if anyone comes up with anything else. –  Vishesh Nov 11 '12 at 3:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.