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Using Forms to Define Orientation of Curves. 
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#1
Nov1312, 11:18 PM

P: 531

Hi, All:
There is a standard method to construct a nowherezero form to show embedded (in R^n ) manifolds are orientable ( well, actually, we know they're orientable and we then construct the form). Say M is embedded in R^n, with codimension 1. Then we can construct a nowhere zero top form by selecting the vector N(x) normal to the manifold ( say, using the Riemann metric inherited from R^n), and choosing an orthonormal frame {v1,v2,..,v(n1)} for M . Then the form: w:=Det  N(x) v1 v2....vn1  , where we write the vectors as columns, is nowherezero, since any two vectors are perpendicular, and so the collection is linearlyindependent. **Now** how do we construct a form when: i) M is embedded in R^n, and the codimension is larger than 1. ii) For a curve, say a smooth curve. iii) Can this be done/does it make sense when M is not embedded? Thanks. 


#2
Nov1412, 07:43 AM

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Well, for any immersed or embedded curve c(t), just take the vector field c'(t) (immersed precisely means c'(t) never vanishes). Then take it's dual (co)vector field and this is your nowhere vanishing globally defined top form.



#3
Nov1412, 08:41 AM

P: 350

I am confused about your construction for codimension 1. What about a Mobius strip? You are assuming orientability, and you want to show that implies the existence of a volume form? Also your construction with the determinant doesn't quite make sense to me, though I think I get what you mean. But actually I am still confused because if you assume that the manifold is orientable, then there is no need to use an embedding into Rn, a volume form exists without that.
Nevertheless, for higher codimension you could generalize your same idea: At each point p of the manifold pick an orthonormal basis N1, ..., Nd for the orthogonal complement in such a way that if v1.....vm is a basis for TpM with positive orientation, then v1 ..... vm, N1, ...., Nd has positive orientation in Rn. Then for any m vectors w1,...., wm in TpM, define Omega(w1,....,wm) = det(w1, ..., wm, N1, ...., Nd) You then have to prove smoothness. Fill in the details for that part and check whether you had to use the embeddedness. 


#4
Nov1412, 09:46 AM

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Using Forms to Define Orientation of Curves.
What you need is to start with a globally defined frame for the normal bundle. I.e. you need the normal bundle to be trivial for the construction to work. In codimension 1, normality of the trivial bundle is equivalent to the submanifold being orientable, so that's why it always work. In higher codimension, the construction works if the normal bundle is trivial. Orientability does not imply triviality of the normal bundle however. 


#5
Nov1412, 01:07 PM

P: 350

I'm not sure I follow you. I think you misunderstood my language. Let N_pM denote the orthogonal complement of T_pM. I did not say to choose an arbitrary orthonormal basis for N_pM. Rather I said choose an orthonormal basis N1,...., Nd according to the restriction that
v1, ..., vm, N1, ...., Nd is oriented (as a basis for R^n) for any oriented basis v1,....,vm of TpM. For any fixed point p, that is a well defined construction when you are a priori given orientations on R^n and T_pM (which is true here). In the special case that d=1, there are two possible orthonormal bases for N_pM . But only one of them satisfies the restriction that I posed. The construction of Omega is done pointwise and is entirely independent of the choice of the vectors N1,...,Nd (as long as they are orthonormal and oriented correctly). If you locally construct smooth vector fields N_1(p),....,N_d(p) in such a way that they are orthonormal basis for N_pM at each point p in some neighborhood, you conclude that Omega is smooth. Whether the vector fields N_i(p) extend globally is irrelevant because their only role is to establish local smoothness of Omega. You could argue that their wedge product extends globally, and that is essentially the crux of the issue because it is really the wedge product of those vectors which is being substituted inside the determinant. 


#6
Nov1412, 02:49 PM

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Oh, yes sorry, I got confused with the situation in which we are trying to prove orientability of the submanifold by constructing a volume form for it from a volume form on the ambiant manifold and global sections of the normal bundle.
But here the submanifold is known to be orientable, so of course the procedure you describe works. Again, I apologize for wasting your time. 


#7
Nov1412, 03:55 PM

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P: 1,716

For compact manifolds without boundary  this excludes the Mobius strip  a codimension 1 manifold  i.e. a hypersurface  of Euclidean space is orientable although I am not sure how to prove that. The volume form is easily constructed by contracting the volume form on Euclidean space with the unit normal to the hypersurface. The two possible orientations are obtained by contracting in adjacent slots.
A codimenion 1 submanifold of another manifold may not be orientable even if the ambient manifold is orientable. For example the real projective plane inside projective 3 space. An orientable codimension 1 submanifold may not have a well defined unit normal if the ambient manifold is not orientable. For instance take the equatorial circle in the Klein bottle. It is oriented by a parameterization but is does not have a unit normal field. For higher codimensions in Euclidean space, submanifolds may not be orientable and so will not have volume forms. In fact any nonorientable manifold can be embedded in some Euclidean space. If the manifold and the ambient manifold are both orientable then one can always find a differential form that is Poincare dual to the fundamental class of the manifold. One should be able to find the volume form from this dual form  not sure how off the top of my head. 


#8
Nov1412, 04:09 PM

P: 350

Oh I see what you mean. No need to apologize, we just interpreted the original post in two different ways :) Actually, your interpretation would clear up all the points I was confused by in my original answer.



#9
Nov1512, 03:53 PM

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So here is a proof of orientability for hypersurfaces, one that I am not happy with but it works. See Milnor's Characteristic Classes for details.
The proof shows that the normal bundle is trivial by showing that its Whitney class is zero  it only has one Whitney class, the first, since it is a line bundle. Over Z2 the normal bundle is orientable and the maps H[itex]^{1}[/itex](R[itex]^{n+1}[/itex],R[itex]^{n+1}[/itex]S) > H[itex]^{1}[/itex](R[itex]^{n+1}[/itex]) >H[itex]^{1}[/itex](S) maps the orientation class into the first StiefelWhitney class of the normal bundle. This fact is a special case of a theorem that applies to a normal bundle of any dimension. The two maps are the restriction in cohomology and the pull back of the inclusion of the surface as the zero section of the normal bundle. The image of the orientation class is the top StiefelWhitney class of the normal bundle. Since the cohomology of Euclidean space is zero in all positive dimensions, the image of the orientation class is zero so the normal bundle of a hypersurface has zero first StiefelWhitney class and is therefore orientable. I would love to see a more constructive proof. 


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