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Closed/Open Transition Function Question

  1. Jan 14, 2010 #1
    I am very early into my first look at topology (specifically, I am jumping to topology on smooth manifolds through the Baez/Muniain book - Gauge Fields, Knots and Gravity) and I have a few questions. Suppose we have overlapping sets on a manifold A and B like this

    http://img697.imageshack.us/img697/2991/venn.jpg [Broken]

    where A and B are both open. It can be shown that the union of A and B is also open, which sort of follows intuitively when looking at the boundaries. This means that, when you approach the 'boundary' from a point x contained in A union B, you can obtain a point in between x and the 'boundary' that is still contained in the union. But by applying the idea that a complement of an open set is a closed set, we see that when we approach from a point only within B, we can reach an edge where there is no other point inside B that is between the boundary and the original point. Same argument for A. The collection of these points would be the boundary. Check me if this reasoning isn't correct.

    My question is in regards to the use of the smooth transition function [tex]\phi_{A}\phi_{B}^{-1}:R^{n}\rightarrow R^{n}[/tex]. I am having trouble really nailing down what the transition function does. It seemed to me at first that


    But after thinking about it some more, that would lead to


    Which doesn't make any sense, if they were equal to each other then there would be no need for a transition function. So this implies to me that for each patch on the manifold there is an entirely new [tex]R^{n}[/tex] associated with it. So when there is a union between two patches A and B, it is possible/probable that the various regions within that union would be mapped to different places in each a different [tex]R^{n}[/tex]. What I mean is that there is a sort of indexing of the real spaces going on, for A there is [tex]R^{n}_{A}[/tex] and for B there is [tex]R^{n}_{B}[/tex], which is another nth dimensional space that is not associated with the others except by the transition functions.

    Is this all correct reasoning, if so I am set and question answered!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 16, 2010 #2
    In fact, we can say more: arbitrary unions of open sets are still open, and finite intersections of closed ones are still closed.

    This can be confusing, because, in Topology, open set don't have boundaries; I understand what you are trying to say, but it's put this way: if [tex]A[/tex] is an open set then, for any point [tex]x \in A[/tex], there is an open set [tex]V\left(a\right)[/tex] such that [tex]x\in V\left(a\right)[/tex] and [tex]V\left(a\right)\subseteq A[/tex]; [tex]V\left(a\right)[/tex] is called a neighbourhood of [tex]x[/tex]. Therefore, any point in an open set belongs to a neighbourhood contained in the set.

    Again, a boundary point of a set [tex]A[/tex] is best defined as being one such that, any neighbourhood of it, will contain points of [tex]A[/tex] and points not in [tex]A[/tex].

    Now for the transition functions; first, there is the mapping function [tex]\phi_A[/tex]; this function special in the sense that is an homeomorphism; this means that it's continuous, it maps an open set of the manifold to an open set in [tex]\mathbb R^{n}[/tex] bijectively and that is inverse is also continuous. The function's domain in the manifold is called a patch, and the patch's image in [tex]\mathbb R^{n}[/tex] is the set of local coordinates. The intuitive idea here is that, in a general manifold, it's not possible to have only one global coordinate system (this happens, for example, in a two-dimensional sphere; remember that spherical coordinates always leave a point out), so we have several local coordinates.

    When these local coordinate systems overlaps, it's necessary to have some way to pass from one to another (and yes, each mapping function goes to a "different" [tex]\mathbb R^{n}[/tex]); in other words, you need a coordinate transformation.
    In smooth manifolds, the transformation from B-coordinates to A-coordinates is done precisely by:

    \phi_{A}\phi_{B}^{-1}:R^{n}\rightarrow R^{n}

    And the one from A to B coordinates by the inverse of the above (remember that, as both function are bijective, their composition is also):


    In smooth manifolds, these transition maps are also required to be smooth functions, because this allows us to define useful objects, like vector and tensor fields, in the manifold and apply differential calculus to them without having to worry if the derivative exists.

    One last thing, you will need a companion text in Differential Geometry to go with that one.
  4. Jan 16, 2010 #3
    Thanks for the reply, you explanation helps a lot. What text do you recommend?. I really like the authors that are more of an intuitive chatty type, if there is author like this for a differential geometry text, that would be great. (I'm showing my physics-iness perhaps)
  5. Jan 16, 2010 #4
    There are several that I could recommend, but it's hard to find a chatty one in this field. If you were a mathematician, I would tell you to muddle through Spivak's five volums.

    But as you want this to follow the physics, try these ones (purely personal choices):

    (1) Topology for Physicists, by Albert Schwartz

    (2) A course in modern Mathematical Physics, by Peter Szekeres

    (3) Lectures on Differential Geometry, by Chern, Chen and Lam

    (4) Calculus on Manifolds, by Michael Spivak (this is only one volume)

    Also, many General Relativity textbooks have good introductions to Differential Geometry
  6. Jan 16, 2010 #5
    Thanks, I will hit up the library.
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