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Connection needed to define derivative of a vector field?

  1. Mar 1, 2015 #1

    stevendaryl

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    Some context for my question: If you have a smooth manifold [itex]\mathcal{M}[/itex] you can define tangent vectors to parametrized paths in the following way: If [itex]\mathcal{P}(s)[/itex] is a parametrized path, then

    [itex]\frac{d}{ds} \mathcal{P}(s) = V[/itex]

    where [itex]V[/itex] is the differential operator that acts on scalar fields [itex]\psi(\mathcal{P})[/itex] to give

    [itex]V(\psi) = \frac{d}{ds} \psi(\mathcal{P}(s))[/itex]

    This actually defines a vector [itex]V(s)[/itex] at each point along the path [itex]\mathcal{P}(s)[/itex].

    The way that I know of defining a derivative of [itex]V(s)[/itex] involves a connection, or parallel transport:

    [itex]\frac{d}{ds} V|_{s_0} = lim_{\delta s \rightarrow 0} \frac{V'(s_0 + \delta s) - V(s_0)}{\delta s}[/itex]

    where [itex]V'(s_0 + \delta s)[/itex] is the result of parallel-transporting [itex]V(s_0 + \delta s)[/itex] along the path [itex]\mathcal{P}(s)[/itex] from [itex]s=s_0 + \delta s[/itex] back to [itex]s_0[/itex].

    My question is: can you define the derivative of a vector (or vector field) without using parallel transport?
     
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  3. Mar 1, 2015 #2

    Ben Niehoff

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    Yes and no.

    A connection, generally speaking, is just an object that tells you how neighboring tangent spaces are related to each other. Without this information, you cannot define a "derivative". In fact, that is essentially what a derivative is. So really, you will always have a connection defined, at least implicitly.

    However, there is a special kind of derivative which doesn't require any extra structure on your manifold: the Lie derivative. There is a catch, though: the Lie derivative is not a directional derivative; it is a derivative along a vector field X. This vector field X must be defined on a neighborhood containing ##\mathcal{P}##; not just at ##\mathcal{P}## itself. It is X that tells you how to relate neighboring tangent spaces, effectively telling you that "little arrows" emerging from ##\mathcal{P}## should flow along with X.
     
  4. Mar 1, 2015 #3

    stevendaryl

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    You can define the derivative of a scalar field along a parametrized path without the need of a connection, though.

    Actually, it's in the context of Lie derivatives that the question arose. The Wikipedia article on Lie derivative associates a vector field [itex]X(\mathcal{P})[/itex] with a "flow", [itex]\phi_t^X[/itex], which basically associates every point on the manifold with that parametrized path [itex]\mathcal{P}(t)[/itex] such that [itex] \frac{d}{dt} \mathcal{P}(t) = X(\mathcal{P}(t))[/itex]. But then the article goes on to talk about [itex]\frac{d}{dt} \phi^X_t[/itex]. I don't see how to define the derivative of [itex]\phi^X_t[/itex] without a connection.
     
  5. Mar 1, 2015 #4

    lavinia

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    The Wikipedia article doesn't do much explaining but what they mean is that a vector field may be thought of as determining a paramterized collection of diffemorphisms of the manifold with time parameter,t. (While this sounds fancy, all it really is doing is restating the fundamental theorem for solutions of ordinary differential equations. )

    So at each time,t, there is a diffeomorphism, [itex]\phi^X_t[/itex], of the manifold.

    The differential of this diffeomorphism maps the vector field Y at some unique point on the manifold to the original point,x, where the Lie derivative is being calculated.

    The difference between the image of Y under the differential and the value of Y at this point,x, divided by the time gives you a Newton quotient at the point x. This Newton quotient is comparing vectors at the same point. So you do not need a notion of parallel translation .
     
    Last edited: Mar 2, 2015
  6. Mar 2, 2015 #5
    Maybe look here? (I'm well out of my league here) :)
    Covariant derivative with respect to a tangent vector page:
    Lie derivative is an intrinsic notion depending only on the differentiable structure, a connection is an additional piece of geometric structure.
    Acceleration page:
    Whereas the notion of tangent vector, C', (i.e. the \velocity" of the curve) makes sense on any manifold, the acceleration ( C'' ) only makes sense when we are given a connection. So to formulate some variant of Newton's equation F = ma on a manifold we need the concept of a connection.
     
  7. Mar 3, 2015 #6

    Matterwave

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    A scalar field does not require any definition of a tangent space to be defined. It exists on the manifold itself and the relationship between its values between two different points is independent of any affine structure, and there's naturally no reason to require a connection to define its derivative. I certainly don't need to understand anything about affine connections in order to know that 5-3=2.

    Lavina answered this question perfectly. I do think that the notion Wikipedia uses is somewhat confusing though. We can define the Lie derivative very simply by: $$\left.\mathcal{L}_X \xi\right|_t = \lim_{h\rightarrow 0} \frac{(\phi^X_{t+h})^{-1} \xi(t+h)-\xi(t)}{h}$$

    And I think this is very clear. You pull back (infinitesimally) the vector field from the point at parameter t+h to the original point at parameter t and look at the difference between the vector (in the vector field) that was already at the point at parameter t and the vector you pulled back to parameter t to find the derivative.
     
  8. Mar 3, 2015 #7

    stevendaryl

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    In terms of your notation, I guess I'm asking what is the definition of "pulling back" the vector field. [itex]\phi^X_t[/itex] is a function that applies to points on the manifold. How is that extended to vector fields?
     
  9. Mar 3, 2015 #8

    martinbn

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    The pullback is the usual pullback of thangent vectors.
     
  10. Mar 3, 2015 #9

    stevendaryl

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    Yes, that's what I thought. But it seems that one of the definitions of "Lie derivative" given in Wikipedia http://en.wikipedia.org/wiki/Lie_derivative#The_Lie_derivative_of_a_vector_field is in terms of the second derivative of a flow (which is the same thing as an acceleration).

    Specifically, they have the following definition:

    [itex]\mathcal{L}_X(Y) = \frac{1}{2} \frac{d^2}{dt^2}|_{t=0}\ \phi^Y_{-t} \circ \phi^X_{-t} \circ \phi^Y_t \circ \phi^X_t[/itex]

    The first derivative produces a vector field (or "velocity" field). The second derivative would seem to require a connection.

    On the other hand, for the specific case of defining the Lie derivative, it could be the case that the result is independent of WHICH connection you use, because the connection terms all cancel.
     
  11. Mar 3, 2015 #10

    stevendaryl

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    Okay, that's what I was fishing around for---I didn't realize there was a "usual notion". How do you relate vector fields at different points on a manifold without using parallel transport?

    But I see that there is a standard definition in Wikipedia http://en.wikipedia.org/wiki/Pullback_(differential_geometry)#Pullback_of_bundles_and_sections:

    If [itex]E[/itex] is a vector field, and [itex]\phi[/itex] is a function of type [itex]\mathcal{M} \rightarrow \mathcal{M}[/itex] then

    [itex](\phi * E)_x = E_{\phi(x)}[/itex]

    But I don't understand what equality means here, without parallel transport. [itex]E_{\phi(x)}[/itex] is a vector at the point [itex]\phi(x)[/itex], while [itex](\phi * E)_x[/itex] is a vector at point [itex]x[/itex]. So how are they related? How are they equal?
     
  12. Mar 3, 2015 #11

    lavinia

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    It applies to vector fields via its differential.
    The differential of a smooth map determines a map on tangent spaces.
     
  13. Mar 3, 2015 #12

    martinbn

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    No, that is the definition of the whole fibre for the pullback of the bundle, not just a section.
     
  14. Mar 3, 2015 #13

    Ben Niehoff

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    Remember that ##t## is just a parameter here, not a coordinate. You can imagine the flow ##\phi^X_t## just like a fluid flow. The vector field X is its velocity field; the streamlines of the fluid are the integral curves of X.

    The flow ##\phi^X_t## carries along tangent vectors in a natural way. You can literally imagine little arrows; the base of the arrow follows one streamline, and the head of the arrow follows another.

    So, to take the Lie derivative along X of a vector field Y, you do the following:

    1. Look at the vector ##Y_{\mathcal{P} + X \, \delta t}## that already sits some infinitesimal distance away from ##\mathcal{P}## along the flow of X.

    2. Run the flow backwards by an amount ##\delta t## in order to bring this vector back to point ##\mathcal{P}##.

    3. Compare this to the vector ##Y_{\mathcal{P}}## that already sits at ##\mathcal{P}##.
     
  15. Mar 3, 2015 #14

    stevendaryl

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    Then what is the appropriate definition?
     
  16. Mar 3, 2015 #15

    stevendaryl

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    It's step #2 that I don't understand without the use of parallel transport. Naively, it sounds like it involves parallel transport, to bring the vector [itex]Y_{\mathcal{P} + X \, \delta t}[/itex] back to [itex]\mathcal{P}[/itex]
     
  17. Mar 3, 2015 #16

    stevendaryl

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    But what is the definition of that map?
     
  18. Mar 3, 2015 #17

    Ben Niehoff

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    One doesn't need connections in the above expression, because ##t## is a parameter and ##\phi^Y_t, \; \phi^X_t## are flows along vector fields. The tangent spaces as you move along the flow are related by the pushforward (or "differential") ##d\phi^X_t##.

    Remember that Lie derivatives are not directional derivatives. ##\mathcal{L}_X Y## at ##\mathcal{P}## is not a derivative of ##Y_{\mathcal{P}}## in the direction of ##X_{\mathcal{P}}## (that notion would certainly require connections!). Rather, it is the derivative of ##Y## along the flow generated by ##X##. You must know ##X## in an open neighborhood of ##\mathcal{P}## for it to be meaningful.
     
  19. Mar 3, 2015 #18

    Ben Niehoff

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    Any diffeomorphism ##\varphi : M \to N## induces a differential (or pushforward) map ##d\varphi : TM \to TN##. In coordinates, ##\varphi## is effectively a map from ##\mathbb{R}^n \to \mathbb{R}^n##. The differential map ##d \varphi## is then the Jacobian matrix of partial derivatives ##\partial \varphi^a / \partial x^b##. Read this for more detail:

    http://en.wikipedia.org/wiki/Pushforward_(differential)#The_differential_of_a_smooth_map
     
  20. Mar 3, 2015 #19

    Matterwave

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    Ah, so here is your confusion, I see. I admit the notation is sometimes overloaded as ##\phi^X_t## is sometimes used to mean the flow itself in some contexts and the push forward in others. The ##\phi^X_t## you see in my post is the pushforward so that the inverse map is the pull back.

    Anyways, a vector field defines a family of diffeomorphisms of points on the manifold to other points on the manifold, this is the flow. A point P will be mapped to a point Q a parameter t distance forward along the integral curves of the vector field X. This map, being a diffeomorphism (and not just an isomorphism) of the manifold, will map curves into other curves (e.g. you can't map a curve into disjoint points). Now, you seem to be dealing with vectors using the definition using the 1-1 isomorphism between vectors and directional differential operators on scalar functions at the some point P, but it's much more easily to visualize vectors in this case as tangent vectors to curves. A parametrized curve in the manifold passing through the point P will define a vector (but this mapping is many to one since different curves can have the same tangent, so we have to define vectors based on an equivalence classes of curves, but that's a complication that doesn't give us any insight anyways here). Now recall that the family of diffeomorphisms of the manifold naturally induced by a vector field maps curves into other curves, therefore, a vector field ##X## will induce a family of maps which maps the integral curves of a vector field ##\xi## into different sets of integral curves. Imagine one integral curve of ##\xi##. What ##\phi^X_t \xi## is is the vector field one obtains from the integral curves of ##\xi## being pushed a parameter ##t## forward along the integral curves of ##X##. The pull back is just going the other direction.

    There is absolutely no need to define a connection for us to do these push forwards and pull backs.

    Is this clear?
     
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