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Definition of Manifold

  1. Aug 6, 2011 #1
    I have a question about the definition of a manifold given in my analysis book. Here is the definition:

    Let [itex] 0 < k \le n [/itex]. A k-manifold in [itex] \mathbb{R}^n [/itex] of class [itex] C^r [/itex] is a set [itex] M \subset \mathbb{R}^n [/itex] having the following property: For each p in M, there is an open set V of M containing p, a set U that is open in either [itex] \mathbb{R}^k [/itex] or [itex] \mathbb{H}^k [/itex], and a continuous bijection [itex] \alpha: U \rightarrow V [/itex] such that:

    1) [itex] \alpha [/itex] is of class [itex] C^r [/itex]
    2) [itex] \alpha^{-1} [/itex] is continuous
    3) [itex] D\alpha(x) [/itex] has rank k for each x in U

    The map [itex] \alpha [/itex] is called a coordinate patch on M about p.

    Note that the set [itex] \mathbb{H}^k [/itex] is upper half-space. That is, it is the set [itex] \{ x = (x_1, ..., x_k) \in \mathbb{R}^k : x_k \ge 0 \} [/itex]

    My question is, why do require that the set U at least be open in [itex] \mathbb{H}^k [/itex]? What is so special about [itex] \mathbb{H}^k [/itex]?
     
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  3. Aug 6, 2011 #2

    HallsofIvy

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    We don't. Your quote says "open in either [itex]\mathbb{R}^k[/itex] or [itex]\mathbb{H}^k[/itex]" (emphasis mine).

    If the p is on a surface of the manifold then we have to use [itex]\mathbb{H}^k[/itex], mapping the surface of the manifold to the surface of [itex]\mathbb{H}^k[/itex], the points with [itex]x_k= 0[/itex]. If p is not on the surface, we can use [itex]\mathbb{R}^k[/itex] which, of course, does not have a surface.
     
  4. Aug 6, 2011 #3

    lavinia

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    The upper half plane allows the manifold to have a boundary. Points that map to the x axis on a coordinate chart are on the boundary. If the manifold has no boundary then R^n would be used instead of the half plane.
     
  5. Aug 7, 2011 #4
    Perhaps my question wasn't worded well. Let me try again:

    Suppose M is our k-manifold and [itex] p \in \partial M [/itex], that is, p is in the boundary of M. If [itex] \alpha: U \rightarrow V [/itex] is a coordinate patch on M about p, then by my definition U is open in [itex] \mathbb{H}^k [/itex], and if x is the point in [itex] \mathbb{H}^k [/itex] such that [itex] \alpha(x) = p [/itex] then [itex] x \in \mathbb{R}^{k-1} \times 0 [/itex]. So lower half space should work as well, right? That is, we can find a coordinate patch [itex] \alpha*: U* \rightarrow V* [/itex] such that [itex] \alpha*(x) = p [/itex] and [itex] U* [/itex] is open in lower half space. So is it just a preference amongst mathematicians that we use upper half space instead of lower half space? Or are there properties of upper half space that we require that lower half space doesn't possess?
     
  6. Aug 7, 2011 #5

    Hurkyl

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    The upper and lower half-spaces are equivalent for most intents and purposes.
     
  7. Aug 7, 2011 #6

    lavinia

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    reread that answer I gave.
     
  8. Aug 7, 2011 #7
    I reread it but I don't see the point that you are driving at? Based on your previous post, lower half-space should work just as I explained, no?
     
  9. Aug 8, 2011 #8

    HallsofIvy

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    The "upper half plane" and "lower half plane" can be used interchangebly (or, for that matter, "left half plane" and "right half plane" or many more in higher dimensions). The point is to get a boundary which [itex]R^n[/itex] does not have.
     
  10. Aug 8, 2011 #9

    lavinia

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    the x axis provides for a boundary. Any homeomorphism from a neighborhood to a neighborhood of a point on the x axis that lies in the lower half plane can be followed by reflection around the x axis to get a neighborhood in the upper half plane.

    So the lower half plane is a manifold with boundary.
     
  11. Aug 8, 2011 #10
    Got'cha. Thanks for the replies guys!
     
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