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Riemannin generalization of the Taylor expansion

  1. Aug 30, 2015 #1
    I thought about the Taylor expansion on a Riemannian manifold and guess the Taylor expansion of ##f## around point ##x=x_0## on the Riemannian manifold ##(M,g)## should be something similar to:
    [itex]f(x) = f(x_0) +(x^\mu - x_0^\mu) \partial_\mu f(x)|_{x=x_0} + \frac{1}{2} (x^\mu - x_0^\mu) (x^\nu - x_0^\nu) H_{\mu\nu}(f)|_{x=x_0}+ \dots[/itex]
    here H is Hessian
    [itex]H_{\mu\nu} (f)=\nabla_\mu\partial_\nu f=\partial_\mu\partial_\nu f-\Gamma_{\mu\nu}^\rho\partial_\rho f[/itex]
    Is this generalization true? Unfortunately, I couldn't find any proper references.
  2. jcsd
  3. Sep 1, 2015 #2
    Hey Shooride,

    First of all this seems more like an adaption than a generalization since the Taylor expansion already makes perfect sense on a Riemannian manifold. Indeed it is a local approximation and locally the smooth structure on the manifold is simply that of the real Euclidean space. So note that regardless which Riemannian structure I pick on the manifold the Taylor expansion is the same, it only depends on the smooth structure. So simply pick some local coordinates and Taylor expand as usual to get the Taylor expansion in those coordinates.

    Of course one could think of an adaption that takes into account the Riemannian structure of the manifold. This seems to be what you are aiming for here. However in order to ask whether such a generalization is "true" it is necessary to indicate what you want this expansion to satisfy.

    It also wouldn't hurt to give an indication of the rest of the series actually.
  4. Sep 1, 2015 #3

    Thanks for reply. I am not familiar with structure and in particular the smooth structure, but I assumed ##f## is a continues function. You mean I can use Riemann normal coordinates around point ##x_0##?! Since ##x_0## is near to ##x##?! Suppose someone want to figure out the Taylor expansion of a simple function ##x^2##, how could he write it? I was wondering if you could explain it more..
  5. Sep 1, 2015 #4

    Excuse me for carelessly throwing weird terms around. Since you were considering Riemannian manifolds I figured you would be familiar with "smooth structure".

    Note that in the definition of Taylor expansion on Euclidean space you use that the function is differentiable (as often as you need) at the point where you take the Taylor expansion. So simply asking the function to be continuous will in general not be enough. For instance the derivative of the function x is ill-defined at x=0, although this function is definitely continuous at x=0.

    So in general when we consider Taylor expansions we consider these infinitely differentiable functions, instead of "infinitely differentiable" people often say "smooth". There are notions around for less differentiable functions, but they are not widely used and definitely not standard (by which I mean I don't know enough of them to tell you anything coherent).

    In calculus (as developed by Newton or Leibniz or whatever other smart fellow) we learn how to differentiate functions f:ℝn → ℝ. A Riemannian manifold is first of all a "smooth" manifold. This means that we can find coordinates around every point such that the coordinate transformations on overlaps of these coordinates are smooth, i.e. infinitely differentiable. This allows us to define differentiation, and thus smoothness, of functions on the manifold. We simply do it around each point as we do it in calculus.

    There are many excellent books on this topic. I used the book introduction to manifold by Tu.

    The Riemann normal coordinates are a special kind of local coordinates that respect certain properties of the Riemannian metric on the Riemannian manifold. So you could indeed use these. My point above was however that the Taylor expansion of a function (which should then be smooth) will not depend on the Riemannian metric. You could calculate it in any local coordinate system around the point.

    Now about the simple function x2. Let's say we have a smooth function on our manifold and around the point x0 we can "center" a coordinate system such that this function looks like x2. By center I mean we have a coordinate system such that x0=0 in those coordinates. The the Taylor expansion is exactly x2. both the function and it's first derivative vanish at x0, while ½ times the second derivative is 1 at x0 and the higher derivatives also vanish. If we pick some coordinate system around x0 such that the function looks like x2 (but not a centered one) then we get the Taylor expansion [itex]x_0^2+ 2x_0(x-x_0) +(x-x_0)^2 [/itex]. Note that we could then have picked the coordinate system [itex]y=x-x_0[/itex] instead which is centered around [itex]x_0[/itex] only then the function looks like [itex](y+x_0)^2[/itex]. We could then Taylor expand with respect to [itex]y[/itex] instead.

    I hope this helps a bit, in the last paragraph I was a bit unclear since I used [itex]x_0[/itex] both to denote the point and the value of the point in the coordinate system x. If anything remains unclear or you would like some references to good sources on this topic let me know!
    (sorry for the bad typesetting I only figured out how to tex on this forum again after a while).
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