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Induced Orientation on Mfld. Boundary.

  1. Mar 16, 2009 #1

    WWGD

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    O.K, please let me see if I got it right:

    Let M be an orientable m-manifold with non-empty boundary B.

    Let p be a point in B , and let {del/delX^1,...,del/delX^(m-1) }_p

    be a basis for T_pB for every p in a boundary component .

    Let N be a unit normal field on B . Now, this is the induced orientation (is it?):


    We consider the collection {N_p, del/delX^1,...,delX^(m-1)}_p

    (with N_p normal to M at p.)

    AS IF p were a point in M, and not in the boundary B, (e.g., we can

    smooth out the boundary so that it disappears, or we can cap

    a disk or something, so that one boundary component disappears).

    Then, if this basis {N_p, del/delX^1,...,delX^(m-1)}_p for p in M

    of T_pM is oriented in agreement with the given orientation of M, (Jacobian of

    chart overlap is positive, etc. ) then the boundary is positively oriented,

    otherwise it is negatively oriented.

    Is this it?
    Thanks.
     
  2. jcsd
  3. Mar 16, 2009 #2

    WWGD

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    Then, if this basis {N_p, del/delX^1,...,delX^(m-1)}_p for p in M

    of T_pM is oriented in agreement with the given orientation of M, (Jacobian of

    chart overlap is positive, etc. ) then the boundary is positively oriented,

    I just realized we just need the change-of-basis matrix here must have

    positive determinant.


    otherwise it is negatively oriented.

    Is this it?
    Thanks.[/QUOTE]
     
  4. Mar 16, 2009 #3

    quasar987

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    Well, maybe you meant something else, but if N_p is normal to M, then N_p is not an element of T_pM (being normal to it!) and so it does not make sense to talk about {N_p, del/delX^1,...,delX^(m-1)} being a basis of T_pM.

    Instead, N_p is the unit vector that point outward of M. Meaning that given a (boundary) chart around p mapping a nbhd of p to [tex]\mathbb{R}^{m-1}\times \mathbb{R}_+[/tex], where R_+ denotes the closed half real line {x:x>=0}, N_p is [itex]-\partial/\partial x ^m[/itex].

    But modulo that change in the meaning of N_p, what you've written sounds good to me.
     
  5. Mar 16, 2009 #4
    N_p is not normal to M, but to the boundary of M. The tangent space of M can be defined in boundary points exactly as for interior points. There are three types of vectors in TpM, for a boundary point p:

    - vectors tangent to the boundary of M, these form a codimension 1 subvectorspace
    - outward pointing vectors
    - inward pointing vectors

    There are unfortunately several different conventions for the induced orientation. I think the most common one (also the one mentioned by WWGD) is the "outward first" convention, which is also compatible with Stoke's theorem. It goes as follows:

    Let [tex]p\in\partial M[/tex], [tex]w\in T_pM[/tex] be an outward-pointing vector. Then [tex](v_1,\ldots,v_{n-1})[/tex] is a positively oriented basis of [tex]T_p\partial M[/tex] iff [tex](w,v_1,\ldots,v_{n-1})[/tex] is a positively oriented basis of [tex]T_p M[/tex].
     
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