
#1
Oct1512, 06:41 PM

P: 391

Hi, Everyone:
A way of defining an orientation form when given a codimension1 , orientable nmanifold N embedded in R^{n+1} , in which the gradient ( of the parametrized image ) is nonzero (I think n(x) being nonzero is equivalent to N being orientable), is to consider the nowherezero normal vector n(x), and to define the form w(v)_x : =  n(x) v1 , v2 ,...,v_n1 (**) Where {vi}_i=1,..,n1 is an orthogonal basis for T_x N , written as column vectors, and n(x) is the vector normal to N at x , so that we write:  n_1(x) v_11 v_21..... v_n1  n_2(x) v_12 v_22......v_n2 ................................ ................................ n_n(x) v_1n v_2n.......v_nn For vi= (vi1, vi2,....,vin ) Then the vectors in (##) are pairwise orthogonal, and so are Linearlyindependent. *QUESTION* : How do we define a form for a curve of codimension1, and, in general, for orientable manifolds of codimension larger than 1 ? I have seen the expression t(x).v , meaning <t(x),v> ,for the curve. But the tangent space of a curve is 1dimensional, so, how is this a dot product? Also, for codimension larger than one: do we use some sort of tensor contraction? Thanks. Thanks. 



#2
Nov3012, 07:09 AM

Sci Advisor
P: 1,716

If the normal bundle to the submanifold is trivial then you can get an orientation form by contracting the orientation form of the ambient manifold by a smooth set of lineally independent normal vector fields.
In Euclidean space you contract the standard volume element. If the normal bundle is not trivial i am not sure off of the top of my head but let's see if we can figure it out. It shouldn't be hard. The normal bundle of a hypersurface (closed without boundary, codimension 1) of Euclidean space is always trivial but I do not know a direct proof and again would like to work on it with you. I do know a nasty indirect proof if you would like to see it. 



#3
Nov3012, 05:10 PM

P: 391





#4
Dec112, 08:53 AM

Sci Advisor
P: 1,716

Orientation Forms in Different Codimension. 



#5
Dec212, 09:05 AM

Sci Advisor
P: 1,716

I think you can Just contract the volume element of the ambient manifold by a local orthonormal basis for the normal bundle over each coordinate chart. Over each chart you get a local volume form for the submanifold and since the coordinate transformations are in SO(n) they should piece together to give you a global orientation for on the submanifold. Is this right?




#6
Dec212, 12:38 PM

Sci Advisor
P: 1,716

BTW: it follows that if the normal bundle is orientable then the submanifold must be orientable as well.




#7
Dec212, 01:09 PM

Sci Advisor
HW Helper
PF Gold
P: 4,768

I think this is correct. Well done lavinia!



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