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Manifolds: extrinsic and intrinsic

  1. Sep 16, 2015 #1
    Dear all
    We all agree that a manifold is globally non euclidean but locally it is. So we can find near each point a hemeomorphic to an open set of euclidean space of the same dimension as the manifold. This is a general definition for all manifold to follow. Then what is the difference between extrinsic way of constructing a manifold and intrinsic one. I know that the first is related to embedded surface but still i didnt understand. Thank you
     
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  3. Sep 16, 2015 #2

    andrewkirk

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    'Homeomorphic', not 'hemeomorphic', just in case that wasn't a typo.

    The extrinsic view of a manifold is a subspace of a larger space - usually a space with more dimensions. In that case the manifold can be defined by an equation that tells us what points in that larger space it occupies. For instance the unit sphere, viewed extrinsically as a subset of Euclidean 3-space ('embedded' in that space), is the set of points ##(x,y,z)## that satisfies the equation ##x^2+y^2+z^2=1##. The intrinsic properties of the manifold - the curvature - can be derived from the extrinsic definition.

    The intrinsic view does not define the manifold as embedded in another space. Rather, it defines it via the metric or related information that tells us about curvature. The metric for the above unit sphere is

    $$
    \left( \begin{array}{cc}
    1 & 0 \\
    0 & sin^2\theta \end{array} \right)
    $$

    where the coordinates are ##\theta## and ##phi## with ranges ##[0,\pi)## and ##[0,2\pi)## respectively and correspond to the spherical coordinate angles in the extrinsic formulation.

    For any extrinsic definition there is only one possible matching intrinsic definition. But for any intrinsic definition there will usually be many different possible compatible extrinsic definitions, which are different ways of embedding the manifold in different parts of different host spaces.
     
  4. Sep 16, 2015 #3
    Than
    Thank you for this reply yes for sure it is a typing error :).
    I just need to check something. If the dimension of the manifold is n then we say that the neighborhood of each point in the manifold is Homeomorphic to an open subset of euclidean space of dimension n ( same dimension). My question is suppose i am looking to the manifold from an extrinsic point view lets say a 2- sphere embbeded in a 3 dimension euclidean space. Then do i say that the neighborhood of each point in the manifold( dimension 2 here) is Homeomorphic to an open subset of euclidean space of dimension 3 or 2??
    Thank you for your reply
     
  5. Sep 16, 2015 #4

    andrewkirk

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    It's 2. The dimension of the space in which the manifold is embedded is not relevant.
     
  6. Sep 16, 2015 #5
    you need to look at the 2-sphere in the subspace topology of the euclidean 3-space. you can remove the north pole from the sphere and lay it flat through projection because it is locally homeomorphic to flat 2-space.
     
  7. Sep 16, 2015 #6

    WWGD

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    It would be 2, because the sphere is a 2-manifold. Notice a patch of the sphere "looks more" like a copy of ##\mathbb R^2 ## than of ## \mathbb R^3 ## -- you can flatten it to look like a square. And notice that every point has a neighborhood homeomorphic _to an open ball_ which is itself homeomorphic to ## \mathbb R^n ##, stating only an open set allows possibilities like an open annulus, etc.
     
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