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D/dt g(t) and d/dt D_X Y

  1. Apr 27, 2013 #1
    Hi!

    I'm reading this script* and I fail to understand a rather simple calculation. I assume the problem lies in me not understanding the notation that is used, and I was unable to figure it out or find it in literature.

    We have a smooth family of metrics [itex]g = g_t[/itex] on a Riemannian manifold, and we set [itex]h := \frac{\partial}{\partial t}g_t[/itex].

    First question:

    [itex]\frac{\partial}{\partial t} \nabla_X Y[/itex]: Does this mean [itex]\frac{\partial}{\partial t} \nabla_X^t Y[/itex], where [itex]\nabla^t[/itex] is the Levi-Civita connection w.r.t [itex]g_t[/itex]?

    Second question:

    The script says:

    [itex]\langle \frac{\partial}{\partial t} \nabla_X Y, Z\rangle = \frac{\partial}{\partial t}g(\nabla_X Y, Z\rangle - h(\nabla_X Y, Z)[/itex]

    I don't understand this step. Also I don't see the difference between the two terms

    [itex]\frac{\partial}{\partial t}g(\nabla_X Y, Z\rangle[/itex] and

    [itex]h(\nabla_X Y, Z)[/itex]

    In class we defined

    [itex]\frac{\partial g}{\partial t}(X, Y) := \frac{\partial}{\partial t}(g_t(X, Y))[/itex].

    Therefore those two terms seem the same to me.

    I would appreciate any help :)

    * http://homepages.warwick.ac.uk/~maseq/RFnotes.html , p. 32.
     
  2. jcsd
  3. Apr 27, 2013 #2

    Fredrik

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    That's what I would guess too. (I don't see what else it could mean).

    I haven't done this sort of thing in a while, but I think that the first term on the right is equal to a sum of three terms (as if we're taking the derivative of a product of three functions), with the d/dt acting on a different "factor" in each term. The term with d/dt acting on the Z is zero.

    I interpret the first one as $$\frac{\partial}{\partial t}\left(g(\nabla_X Y, Z)\right)$$ and the second one as
    $$\left(\frac{\partial}{\partial t}g\right)(\nabla_X Y, Z).$$
     
  4. Apr 27, 2013 #3
    https://www.youtube.com/watch?v= Thank you!

    So what exactly is the difference between them?

    I would interpret:

    $$\frac{\partial}{\partial t}\left(g(\nabla_X Y, Z)\right)$$

    as $$\frac{\partial}{\partial t} (t \mapsto \left(g_t(\nabla_X Y, Z)\right))$$

    But then, how is the other one different? As I said, we defined:

    [itex]\frac{\partial g}{\partial t}(X, Y) := \frac{\partial}{\partial t}(g_t(X, Y))[/itex]

    Which would lead me to the same equation as the first one?
     
  5. Apr 27, 2013 #4

    Fredrik

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    ##\left(\frac{d}{d t}g\right)(\nabla_X Y, Z)## is just one of the three terms you get when you compute ##\frac{d}{d t}\left(g(\nabla_X Y, Z)\right)##.

    I don't understand your definition. Are you sure the left-hand side isn't supposed to be ##\frac{d}{dt}g(X,Y)##? That would make sense if it's not really a definition, and your teacher was just trying to explain that the notation ##\frac{d}{dt}g(X,Y)## is to be interpreted as ##\frac{d}{dt}\left(g(X,Y)\right)## and not as ##\left(\frac{d}{dt}g\right)(X,Y)##.
     
  6. Apr 27, 2013 #5
    Mhh, the definition is definitely the same in my notes. It was supposed to define what exactly is meant when writing ##\frac{\partial g}{\partial t}## in the Ricci Flow-equation.

    Would you mind telling me what those three terms are when I compute ##\frac{d}{d t}\left(g(\nabla_X Y, Z)\right)##? Maybe my problem is that I don't understand how this derivative is calculated.
     
  7. Apr 27, 2013 #6

    Fredrik

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    For all vector fields X,Y, we have
    $$g(X,Y)=g(X,Y)=g(X^\mu\partial_\mu,Y^\nu\partial_\nu) =g_{\mu\nu}X^\mu Y^\nu,$$ where the ##\partial_\mu## are the tangent vector fields associated with an arbitrary coordinate system. The right-hand side is just a product of three functions, so if we want to compute a derivative of the left-hand side, we can use the product rule on the right-hand side.
    $$\frac{d}{dt}g(X,Y) =\left(\frac{d}{dt}g_{\mu\nu}\right) X^\mu Y^\nu + g_{\mu\nu}\left(\frac{d}{dt}X^\mu\right) Y^\nu + g_{\mu\nu}X^\mu \left(\frac{d}{dt}Y^\nu\right).$$
     
  8. Apr 27, 2013 #7
    Ah, I didn't think of that. Thanks for all your help!
     
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