Induced Orientation on Mfld. Boundary.

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SUMMARY

The discussion centers on the concept of induced orientation on an orientable m-manifold M with a non-empty boundary B. It establishes that for a point p in B, the basis {N_p, ∂/∂X^1, ..., ∂/∂X^(m-1)}_p can be used to determine the orientation of the boundary. The orientation is positively defined if the change-of-basis matrix has a positive determinant, aligning with the orientation of M. The conversation also clarifies that N_p, being normal to the boundary, is not part of the tangent space T_pM, and discusses the "outward first" convention for induced orientation.

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  • Understanding of orientable m-manifolds
  • Familiarity with tangent spaces and normal vectors
  • Knowledge of change-of-basis matrices and determinants
  • Concept of Stokes' theorem in differential geometry
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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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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]
 
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 \mathbb{R}^{m-1}\times \mathbb{R}_+, where R_+ denotes the closed half real line {x:x>=0}, N_p is -\partial/\partial x ^m.

But modulo that change in the meaning of N_p, what you've written sounds good to me.
 
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 p\in\partial M, w\in T_pM be an outward-pointing vector. Then (v_1,\ldots,v_{n-1}) is a positively oriented basis of T_p\partial M iff (w,v_1,\ldots,v_{n-1}) is a positively oriented basis of T_p M.
 

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