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Let M be an n-dimensional real manifold.
Axiom A: there is a nonvanishing n-form on M.
Axiom B: M has an atlas such that for every two charts (V,f), (W,g) in this atlas and for every a in the intersection of f(V) and g(W), the determinant of the derivative of gf-1 at a is strictly positive.
It is well known that A implies B. If M is covered by a countable atlas, then B implies A. I looked through a few standard textbooks but could not find any answer to the following
Question: Where can I find a counter example that B does not imply A?
Thank you in advance. Because it is so fundamental, I must overlook some exercises or examples in the standard textbooks. Thanks to CiaPan so that I have the following new signiture. twma 03:36, 4 August 2007 (UTC)
The following is what I understand although it may not be standard. Let M be a nonempty set. An n-patch on M is a pair (V, f) where V is a subset of M and f:V--> Rn is an injection. Two n-patches (V,f),(W,g) on M are compatible if both are open in Rn and the coordinate transformation gf-1 : is a diffeomorphism, i.e. infinitely differentiable with invertible derivative. A cover AA of M by n-patches is called an n-atlas if every two members in AA are compatible. For every (V,f) in AA , since (V,f) is compatible with itself, the set f(V) is open in Rn as part of the definition.
Let AA be an atlas on a set M. Let TT denote the family of subsets B of M such that for every (V,f) in AA , the set is open in Rn. Then TT is a topology on M. It is called the manifold topology induced by AA. A manifold is a nonempty set with an atlas so that the manifold topology is separated, i.e. distinct points have disjoint neighborhoods.
An n-patch on M is called a chart if it is compatible with every patch in AA. An atlas may contain only a few patches but we have plenty of charts to play around. In particular, a subset B of M is open iff for every m in B , there is a chart (V,f) at m with . This is frequently used as the defintion of manifold topology.
By the way, being separated is not the same as being disconnected in my mind. Paracompactness is not part of the definition of a manifold in this article. Because every manifold M is locally homeomorphic to an open subset of Rn, is there any counter example to show that M as defined here is not metrizable?
Hope this would be acceptable. Thank you in advance. twma 21:12, 5 August 2007 (UTC)
The long line is a manifold but not metrizable. My first original question still need help. Thank you in advance. twma 09:43, 6 August 2007 (UTC)
Suppose that there are two maps of some geographic area, to different scales. If the maps represent exactly the same area, the smaller-scale one will be correspondingly smaller than the other. If the smaller is laid on the larger, completely within it, it is intuitively obvious that there will be exactly one point where a pin can be pushed through both maps, with the marks on the maps representing the same place on the ground. But how can this point be found?…217.43.210.194 19:40, 4 August 2007 (UTC)
To be as general as possible, if the smaller map's lower left corner is at (x,y) on the larger map and it has been rotated by an angle θ clockwise, but still wholly within the other, is it possible to get the co-ordinates of the point explicitly?…217.43.210.194 00:03, 5 August 2007 (UTC)