Induced Orientation on Mfld. Boundary.

In summary: If p\in\partial M, w\in T_pM be an inward-pointing vector, then (v_1,\ldots,v_{n-1}) is a negatively oriented basis of T_p\partial M iff (w,v_1,\ldots,v_{n-1}) is a negatively oriented basis of T_p M.In summary, if an outward-pointing vector is in the T_pM and has a positive orientation, then the boundary of M is positively oriented. If an inward-pointing vector is in the T_pM and has a negative orientation, then the boundary of M is negatively oriented.
  • #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.
 
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  • #2
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]
 
  • #3
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.
 
  • #4
quasar987 said:
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.

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].
 

FAQ: Induced Orientation on Mfld. Boundary.

1. What is "Induced Orientation on Mfld. Boundary."?

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