Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Extrinsic and Intrinsic Curvature

  1. Mar 17, 2005 #1

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Those of you who know me know that my formal education is in physics, not mathematics. So hopefully you'll excuse the dumb question, but in what course would one learn about extrinsic and intrinsic curvature? I have books on tensors, differential forms, topology, analysis, and advanced calculus. And yet the only 2 books I own that cover this material are 2 books on general relativity.

    What math books am I missing? :confused:
  2. jcsd
  3. Mar 17, 2005 #2
    I would call them Extrinsic and Intrinsic methods, because they have more to do with the point of view then the geometry itself.

    Extrinsic geometry is the description of curves, survaces, and generalizations viewed as embedded in a space of higher dimension then themselves.

    Intrinsic geometry is a way of describing curvature without appeal to higher dimensions, hence it is appropriate for GTR.

    These are covered in many texts on advanced calculus (Courant or Spivak, I believe).
  4. Mar 18, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    they sound like concepts from differential geometry to me tom. spivak's big set of volumes on diff geom, particularly volume 2, would do it. look for "curvature".

    extrinsic curvature i would guess would be curvature defiend in terms of an embedding. for instance tha gauss map of a surface in R^3 sending a pointp to the unit notmal vector to the surface at p, translated to the origin. That defines a natural map from your surface to the surface of a standard surface of constant curvature, name;ly the unit sphere. then the way that map contortsa area locally, i.e. the jacobian determinant of the totalt differential at p, measires the curvature of the surface at p, extrinsically.

    to measure curvature intrinsically is harder. one defiens a scalar product on the tanegnt spaces to the surface, either using an embedding to restrict the notion of scalar productf rom the ambient space, or simply imposes one locally via a coordinate system and then glues these together smoothly using a "partition of unity". Once you havea way to lengths of tangent vectors, you can define curvature "intrinsically" i.e. depending only on the notion of length of tangent vectors you have assumed given.

    this fact is gauss' theorema egregium. there are elermentary notes available for free from the following website, by an expert on the topic: Theodore Shifrin, student of Chern.


    now i myself am not an expert on this topic. my ignorant opinion would be however that there is only one notion of curvature, but there are extrinsic and intrinsic ways of defining it. gauss theorem would be phrased as saying that extrinsically defined curvature is in fact an intrinsic invariant, dependent only on the riemannian metric. i think i am in agreement here with crosson's first sentence.
    Last edited: Mar 18, 2005
  5. Mar 18, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    a few more remarks, gleaned from perusing works of experts:

    a "connection" is a method of differentiating vector fields on manifolds in the direction of tangent vectors. this differentiation is called "covariant" differentiation, [and is denoted by some authors, such as Nomizu, with a del symbol]. The fundamental lemma of riemannian geometry says there is a unique "symmetric" connection which is compatible with a given metric.

    Then the curvature can be defined as the extent to which the SECOND covariant derivative fails to be symmetric.

    Thus there is a method if defining curvature entirely from the datum of a riemannian metric, i.e. dot product on each tangent space.

    In shifrins notes above, the covariant derivative of a surface in three space is given a very simple definition, as ordinary differentiation in R^3, followed by orthogonal projection of the derivative onto the tangent space of the surface, using of course the induced metric from the embedding.

    the curvature, unlike the connection it seems, behaves like a "tensor". Moreover curvature can be viewed locally as matrix of 2 forms, just as the connection can be viewed locally as a matrix of one forms.

    Anyway, since matrices have various invariants, trace, determinant, etc.... so also one can apply these invariant polynomials to the curvature matrices, obtaining globally meaningful objects called "characteristic classes". they are named after famous people like whitney, chern, pontrjagin, euler etc...

    [example: at a given point of the surface, there are two mutually orthogonal length minimizing curves, called geodesics, with respectively largest and smallest curvature, i.e. approximable by circles with smallest and largest radii. the product of these two curvatures is called the gauss curvature at the point. locally the curvature form of an embedded surface in R^3, is the product of the gaussian curvature, with the standard oriented area form. the integral of the gaussian curvature over the manifold is the euler characteristic. (times 2pi or something).]

    recall the euler class is the primordial characteristic class. the general fact, called the gauss bonnet chern theorem is perhaps that the top chern (characteristic) class equals the euler class.

    References: milnor: morse theory pages 43-54, milnor - stasheff: chracteristic classes, appendix C.
  6. Mar 18, 2005 #5
    Just to clear this one up for the OP, Gaussian curvature and Riemannian curvature are not the same concept, as is emphasized further here:
    Note that the one of the principle extrinsic curvatures at any point on the two-dimensional infinite standard cylinder of some radius r would be by definition 1/r, while the intrinsic curvature would be zero (There is no distortion in the parallel transport of a vector on a closed curve embedded in the cylinder). The intrinsic curvature is non-zero on a sphere, where parallel-transport of a vector on a closed curve does result in a disagreement. The motivation is that one does not need to reference an external space or embedding in order to measure the intrinsic curvature of a manifold (ie., measuring the 4-dimensional curvature of spacetime).
    I'm probably less knowledgeable about the subject than mathwonk, but I always tend to bring up the cylinder and the sphere when talking about intrinsic vs. extrinsic curvature, so there it is. :biggrin:
  7. Mar 25, 2005 #6


    User Avatar

  8. Mar 25, 2005 #7


    User Avatar
    Science Advisor
    Homework Helper

    hypermorphism, you have me a little puzzled. i thought that riemannian curvature was the generalization of gaussian curvature to higher dimensions. in particualr that they agree in dimension 3. am i confused??

    in particular the curvature of a cylinder is zero in anybody's definition, since one of the principal curvatures is zero.
  9. Mar 26, 2005 #8
    So I think you need a books on Riemannian , spinorial geometry
  10. Mar 27, 2005 #9
    Good question. I'd say that since this concept is a mathematical concept then it would/should be found in math texts. However all books are limited in scope and it seems that this topic is so rare that its not included in most, if not all, math texts.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Extrinsic Intrinsic Curvature Date
Manifolds: extrinsic and intrinsic Sep 16, 2015
Conversions between intrinsic/extrinsic coordinates Feb 26, 2014
Extrinsic intrinsic Feb 8, 2010
Intrinsic extrinsic Apr 7, 2008
Extrinsic/Intrinsic curvature Jun 9, 2004