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Basic diff. geometry question - Gradient of F

  1. Feb 10, 2012 #1
    Hi folks,

    I have a basic question I would like to ask.
    I ll start from the Euclidean analogue to try to explain what I want.

    Suppose we have a smooth function (real valued scalar field)

    [itex]F(x,y)=x^2+y^2[/itex], with [itex]x,y \in ℝ[/itex].

    We also have the gradient [itex] \nabla F=\left( \frac{\partial F}{\partial x},\frac{\partial F}{\partial y} \right)=\left( 2x,2y \right)[/itex]

    Now lets imagine that F is defined on a Riemannian manifold [itex]M[/itex] with a metric [itex]g_{ij}[/itex].

    I would like to calculate the gradient of [itex]F[/itex] for a point [itex] x,y \in M[/itex].

    I read that in this case, that the local form of the gradient is

    [itex]\nabla F = g^{ij}\frac{∂F}{∂x^k}\frac{∂}{∂x^i}[/itex]

    But I do not understand what exactly this formulation means. I have an anyltic expresion for [itex]F, g^{i,j} [/itex] but I am not sure how to calculate the gradient in this case. Can someone perhaps explain the above expresion in layman's terms to me? (I do understand Einstein notation btw).

    Also, does the gradient vector live in the tangent space of the point at [itex] x,y[/itex]? because somewhere I read about [itex]\hat{F}=F \circ R [/itex] as the pull back of [itex]F[/itex] through the retraction function [itex]R[/itex] onto the tangent space, and I was confused which "version" lie on the tangent space exactly.

    Many thanks
    Last edited: Feb 10, 2012
  2. jcsd
  3. Feb 10, 2012 #2


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    On a manifold a smooth function has a differential which is defined by the formula df(X) = X.f

    where X is a tangent vector at some point and X.f is the directional derivative of f with respect to X. This definition does not require a Riemannian metric. df is a 1 form, That is df(aX + bY) = adf(X) + bdf(Y) for any linear combination of the tangent vectors X and Y.

    With a Riemannian metric there is a tangent vector at each point which is the dual to df with respect to the metric. Different metrics give different gradients.

    the defining formula is df(X) = <gradf,X>
  4. Feb 10, 2012 #3
    Hi, ok so practically so I can understand this.
    What is [itex]grad(F)[/itex] for this specific case? Is it as in the Euclidean case [itex](2x,2y)[/itex]? or it is dependent on some arbitrary tangent vector and the Riemannian metric? I didnt understand your answer completely.

    Lets try a practical simple example.
    Say I have [itex]g_{ij}=[ax, b; b, cy][/itex] and a tangent vector [itex]t_{P}=[2,0][/itex] at some arbitrary point [itex]P=(x_0,y_0)[/itex]. The scalar function [itex] F [/itex] is as before.

    What is [itex]gradF[/itex] for this example?

    And is [itex] gradF \in T_PM [/itex] ?
  5. Feb 10, 2012 #4


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    The gradient is naturally a one form (the one that lavinia defined). To turn it into a vector, you should use the metric.

    This is why, for example, you have those funky r and rsin(theta) terms in the formula for a gradient in spherical polar coordinates.

    For example, the gradient (one form) in spherical coordinates is still simply:

    [itex]df=\frac{\partial f}{\partial r}dr+\frac{\partial f}{\partial \theta}d\theta+\frac{\partial f}{\partial \phi}d\phi[/itex]

    Now, to convert this into a vector gradient, we then have very simply:

    [tex]\nabla f=g^{ij}(df)_j \frac{\partial}{\partial x^i}=\frac{\partial f}{\partial r} \frac{\partial}{\partial r}+\frac{1}{r^2} \frac{\partial f}{\partial \theta} \frac{\partial}{\partial \theta}+\frac{1}{r^2\sin\theta^2} \frac{\partial f}{\partial \phi} \frac{\partial}{\partial \phi}[/tex]

    Like we usually expect in vector calculus.

    EDIT: I think the expression I have is off by our usual definition in the coefficients. This is because the basis vectors I used are not ortho-normal bases but coordinate bases. If I switch to an orthonormal set of bases with the definitions: [tex]\frac{\partial}{\partial r'}=\frac{\partial}{\partial r}[/tex] [tex]\frac{\partial}{\partial \theta'}=\frac{1}{r}\frac{\partial}{\partial \theta}[/tex] [tex]\frac{\partial}{\partial \phi'}=\frac{1}{r\sin\theta} \frac{\partial}{\partial \phi}[/tex], then I will get the usual gradient.
    Last edited: Feb 10, 2012
  6. Feb 10, 2012 #5


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    BTW:Your metric only works for x and y positive

    The differential of F is 2xdx + 2ydy
    Its value on a vector (a,b) is 2xa + 2yb

    The gradient satisfies the equations

    <gradF,(1,0)> = 2x <gradF,(0,1)>= 2y

    If gradF = (s,t) then

    2x = axs + bt 2y = bs + cyt

    The gradient is a tangent vector by definition. It is defined to be that tangent vector that is dual to the differential with respect to the metric.
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