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Understanding Orientated Manifolds

  1. Aug 18, 2011 #1
    I am having some trouble understanding the notion of an orientated manifold. But first let me get some preliminary definitions out of the way:

    A diffeomorphism is said to be orientation-preserving if the determinant of its Jacobian is positive. A k-manifold M in [itex] \mathbb{R}^n [/itex] is said to be orientable if there is an atlas of coordinate patches, [itex] \vartheta = \{\alpha_i : U_i \rightarrow V_i \} [/itex] covering M such that the transition functions are orientation-preserving. This atlas is said to be an orientation on M.

    I am trying to geometrically visualize this. My book gives some help in the case where k = 1, k = n, or k = n - 1.

    Let us suppose for the moment that k = n - 1, because that is what I am going to do most of my visualizing in. And let us use in our example the 2-sphere, [itex] S^2 [/itex], which is a 2-manifold in [itex] \mathbb{R}^3 [/itex].

    Let's picture the unit normal field to [itex] S^2 [/itex], corresponding to some orientation that we give [itex] S^2 [/itex]. Now, because [itex] S^2 [/itex] is orientable, we can picture the unit normal field corresponding to the given orientation, as say, all of the normal vectors to the sphere that are pointing outwards. The only other possible unit normal field to [itex] S^2 [/itex] is the one where all of the vectors which are normal to the sphere point inward, toward the origin.

    Now, tell me if my reasoning is correct here. The ONLY reason that we can find a unit normal field to [itex] S^2 [/itex] such that all of the normal vectors point in the same direction (i.e. outward in this example) is because [itex] S^2 [/itex] is orientable. If we had another manifold, say the Mobius Strip, then we cannot find a unit normal field such that each normal vector is pointing outwards, because as we travel around the strip, there will be normal vectors starting to point in the opposite direction. So because of this, the Mobius strip is not orientable.

    Is my reasoning correct here? And also, why do we care if a manifold is orientable or not? What purpose does it serve? And suppose that we do indeed have an orientable manifold, M, with two possible choices of orientation. How does the choice of orientation affect the structure of the manifold?
     
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  3. Aug 18, 2011 #2

    hunt_mat

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    You can find a normal for non-orientable manifolds too (I am thinking here of a Klein bottle but the difference is that for an orientable manifold the normal is unique whereas the Klein bottle isn't.
     
  4. Aug 18, 2011 #3
    Okay, let me see if I get what you mean. My textbook touched on some of this...

    Given an (n-1)-manifold M in [itex] \mathbb{R}^n [/itex], with an orientation [itex] \vartheta = \{ \alpha_i \} [/itex], take any [itex] p \in M [/itex]. Let (p;n) be a unit vector in [itex] T_p(\mathbb{R}^n) that is orthogonal to [itex] T_p(M) [/itex]. Then n is uniquely determined, but only up to sign.

    So take a coordinate patch in the atlas of M, [itex] \alpha: U \rightarrow V [/itex] and let [itex] \alpha(x) = p [/itex]. Then the colum vectors [itex] \frac{\partial \alpha}{\partial x_i} [/itex] of the matrix [itex] D\alpha(x) [/itex] give a basis [itex] (p; \frac{\partial \alpha}{\partial x_1}), (p; \frac{\partial \alpha}{\partial x_2}), ... , \frac{\partial \alpha}{\partial x_{n-1}}) [/itex] for [itex] T_p(M) [/itex]. Then we specify that the determinant of the matrix [itex] \begin{pmatrix} n & \frac{\partial \alpha}{\partial x_1} & \cdots & \frac{\partial \alpha}{\partial x_{n-1}} \end{pmatrix}[/itex] be positive, and this is how we determine the sign of n.

    So what you are saying is that given an orientation of M, there is one and only one unit normal vector (p;n) such that [itex] det \begin{pmatrix} n & \frac{\partial \alpha}{\partial x_1} & \cdots & \frac{\partial \alpha}{\partial x_{n-1}} \end{pmatrix} > 0 [/itex].

    BUT, if M were non-orientable, then given [itex] p \in M [/itex] and a coordinate patch [itex] \alpha [/itex] about p, it is impossible to choose a unit normal vector n at p such that [itex] det \begin{pmatrix} n & \frac{\partial \alpha}{\partial x_1} & \cdots & \frac{\partial \alpha}{\partial x_{n-1}} \end{pmatrix} > 0 [/itex].

    So the point of an orientation is to see if we can get a consistent notion of right-handedness on the manifold.

    Is all of my reasoning correct here? Especially my reasoning concerning non-orientable manifolds?
     
  5. Aug 19, 2011 #4

    hunt_mat

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    I think that you're beginning to get the idea of orientable manifolds. I think for a non-orientable manifold there are normal vectors where the determinant can be either sign.
     
  6. Aug 19, 2011 #5
    This cannot be. If M is non-orientable, and my reasoning in my previous post is correct, then for every normal vector n at p, of unit length, we must have [itex] det \begin{pmatrix} n & \frac{\partial \alpha}{\partial x_1} & \cdots & \frac{\partial \alpha}{\partial x_{n-1}} \end{pmatrix} = 0 [/itex].

    For if [itex] det \begin{pmatrix} n & \frac{\partial \alpha}{\partial x_1} & \cdots & \frac{\partial \alpha}{\partial x_{n-1}} \end{pmatrix} < 0 [/itex] then [itex] det \begin{pmatrix} -n & \frac{\partial \alpha}{\partial x_1} & \cdots & \frac{\partial \alpha}{\partial x_{n-1}} \end{pmatrix} > 0 [/itex] and vice versa. That is, if the vector n causes the determinant of our matrix to be negative, then we can just take -n to make the determinant positive, which is all we need. So we MUST have that the determinant is 0. This is assuming that what I said in my second post is correct though. But something just feels off from this...
     
  7. Aug 19, 2011 #6
    Ok I think I have finally got the idea of an orientated manifold. Now, what applications do orientated manifolds have? What useful properties does orientability give us?
     
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