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Featured I Gradient vector without a metric

  1. Oct 22, 2018 #1
    Is it possible to introduce the concept of a gradient vector on a manifold without a metric?
     
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  3. Oct 22, 2018 #2

    Orodruin

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    If by "gradient vector" you mean a tangent vector ##\nabla f## such that ##df/ds = g(\nabla f,V)##, where ##V## is the tangent to the curve you are following, then no.

    However, sometimes you will find the differential ##df## (which is a one-form) also being referred to as "the gradient".
     
  4. Nov 11, 2018 #3

    stevendaryl

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    To expand on what Orodruin just said:

    Think about the way that people use gradients. The usual purpose of considering ##\nabla f## is that it tells you how ##f## varies as you move around in your space. If you have a test particle that is moving at velocity ##\overrightarrow{V}##, then you can compute the rate of change of ##f## experienced by that particle via:

    ##\frac{df}{dt} = \overrightarrow{\nabla f} \cdot \overrightarrow{V}##

    But now, let's see how the metric tensor comes into play in this expression. First,

    ##(\nabla f)^j = \sum_i g^{ij} \frac{\partial f}{\partial x^i}##

    where ##g^{ij}## is the inverse of the metric, defined by: ##\sum_i g^{ij} g_{ik} = \delta^j_k## (##\delta^j_k## is 1 when ##j=k## and 0 otherwise).

    Then ##\overrightarrow{\nabla f} \cdot \overrightarrow{V} = \sum_{jk} g_{jk} (\nabla f)^j V^k##

    Putting these things together, you get:
    ##\overrightarrow{\nabla f} \cdot \overrightarrow{V} = \sum_{jk} g_{jk} \sum_i g^{ij} \frac{\partial f}{\partial x^i} V^k##
    ## = \sum_{ik} (\sum_j g_{jk} g^{ij}) \frac{\partial f}{\partial x^i} V^k##
    ## = \sum_{ik}\delta^i_k \frac{\partial f}{\partial x^i} V^k##
    ## = \sum_{i} \frac{\partial f}{\partial x^i} V^i##

    So the metric drops out. That implies that the whole thing could have been done without a metric at all. Just define a covector/one-form ##df## via:

    ##(df)_i = \frac{\partial f}{\partial x^i}##

    And then the directional derivative for velocity ##V## is given by:

    ##\sum_i (df)_i V^i##

    No metric involved at all.

    As far as I know, there is no good reason to use the vector ##\nabla f## that couldn't just as well be served using the covector ##df##. My preference is to use ##\nabla f## to mean the covector/one-form ##df##. That contradicts the usual meaning of ##\nabla f## as a vector, but since I never want to work with the vector, I hate to see a nice symbol like ##\nabla## go to waste.
     
  5. Nov 11, 2018 #4

    Orodruin

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    I can see that there is some point in defining the tangent vector ##\nabla f## using the inverse metric. It tells you the direction of maximal change per step length, with the step length defined by the line element. Of course, without a metric, "step length" lacks any meaning and a metric is therefore necessary for this purpose. It is also natural that this ##\nabla f## depends on the metric introduced since what direction it refers to is going to be dependent on how the step length is defined, and therefore the metric.
     
  6. Nov 11, 2018 #5

    stevendaryl

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    Something that clarified to me the need for a metric was considering an abstract manifold where the dimensions are not even the same units. For example, the state of a quantity of an ideal gas can be characterized as a point ##(P,T)## on a two-dimensional manifold, where ##P## is the pressure of the gas, and ##T## is the temperature. If the state is changing as a function of time, we can meaningfully talk about the "velocity" ##\overrightarrow{V}## of the state in this manifold:

    ##V^P = \frac{dP}{dt}##
    ##V^T = \frac{dT}{dt}##

    So that's a kind of vector, a tangent vector. However, lacking a metric, there is absolutely no sense to the notion of the "length" of ##\overrightarrow{V}##. Naively, you might think that ##|\overrightarrow{V}| = \sqrt{(V^P)^2 + (V^T)^2}##, but that makes no sense, because ##V^P## and ##V^T## aren't even in the same units.

    If you have two different tangent vectors: ##\overrightarrow{V}, \overrightarrow{U}##, it doesn't make any sense to try to take the dot-product of the two, for the same reason: the units don't work out.

    One thing that does make sense for tangent vectors is to say that one tangent vector is a linear multiple of another: If there is some constant ##\lambda## such that ##V^P = \lambda U^P## and ##V^T = \lambda U^T##, then we can say that ##\overrightarrow{U}## and ##\overrightarrow{V}## are parallel, or linear multiples of each other.

    We can also come up with another notion of a vector for this manifold. If we have some function, say the volume ##\mathcal{V}##, that depends on ##P## and ##T##, then we can express how ##\mathcal{V}## changes with state via a covector:

    ##(\nabla \mathcal{V})_P = \frac{\partial \mathcal{V}}{\partial P}##
    ##(\nabla \mathcal{V})_T = \frac{\partial \mathcal{V}}{\partial T}##

    Once again, because of the lack of metric, we can't make any sense of the length of ##\nabla \mathcal{V}##, nor can we make sense of the product of two such covectors. Also, there is no way to make sense of saying that a covector is parallel to a vector, because there cannot be a constant ##\lambda## relating the components, for dimensional reasons.

    However, one thing you can do that makes perfect sense: Take the product of a tangent vector and a covector:

    ##(\nabla \mathcal{V}) \cdot \overrightarrow{V} = V^P (\nabla \mathcal{V})_P + V^T (\nabla \mathcal{V})_T = \frac{dP}{dt} \frac{\partial \mathcal{V}}{\partial P} + \frac{dT}{dt} \frac{\partial \mathcal{V}}{\partial T} = \frac{d \mathcal{V}}{dt}##

    This is a meaningful quantity, and all the units work out. It is the rate of change of ##\mathcal{V}## as the system state changes with a rate given by the velocity vector ##\overrightarrow{V}##.
     
  7. Nov 13, 2018 #6

    lavinia

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    Here is an example of the use of gradients in mathematical theory.

    Gradients viewed as flows are dynamical systems. The gradient flow of a Morse function - a smooth function with only non-degenerate critical points - on a compact manifold reveals the manifold's topological structure, in particular its decomposition as a CW complex. The metric is not needed for length and angle measurements per se but rather to define the gradient flow. The geometry of the gradient flow is not what is important. Instead it is its topological properties as a dynamical system.

    Different metrics determine different flows for the same Morse function. Among them are Morse-Smale gradient flows. These flows allow one to define a chain complex which recovers the singular homology of the manifold. Again it is not the metric relations that matter so much as the the structure of the flow. Not all metrics determine a Morse-Smale flow.


    Classic references are Minor's Morse Theory and Milnor's Lectures on the H-cobordism Theorem. There are also the papers of Smale for instance the 1961 paper On Gradient Dynamical Systems which appeared in the Annals of Mathematics.
     
    Last edited: Nov 15, 2018
  8. Nov 15, 2018 #7
    A differential form, you can define on a continuous manifold.

    A gradient is more complicated. You have to be clear about what you call a gradient, and typically a few properties are like this: It has to become the ordinary derivative for scalar fields (i.e. just the form), has to follow the Leibniz rule, etc... Problem is on a general manifold the form $$\partial_i$$ doesn't transform like a tensor, i.e. it's not independent of the coordinates you used to define it.

    You can get around this difficulty by introducing a Christoffel symbol $$\Gamma$$ - the metric connection, which produces a differential tensor, commonly referred to as a gradient. If you don't set any additional constraints e.g. that the gradient tensor has to commute with the metric, you could define a gradient on a manifold without a metric.

    A big problem is that when you do have a metric, the definition of the connection can be unique, so you also have a unique definition of the gradient. However none of those definitions can be extended to the case where the connection is not metric compatible, or torsional, or the metric doesn't exist altogether.

    So in short you can define gradient like structures on manifolds without metric, but they're not the same as $$\nabla^i$$ or $$\nabla_i$$
     
    Last edited: Nov 15, 2018
  9. Nov 15, 2018 #8

    Orodruin

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    This is correct only if you assume that the connection is metric compatible and torsion free. Having a metric compatible connection is not sufficient to uniquely identify it as the Levi-Civita connection.
     
  10. Nov 15, 2018 #9
    There fixed it.
     
  11. Nov 16, 2018 #10

    lavinia

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    What you are calling a gradient is usually called a covariant derivative or with smoothness requirements it is called an affine connection. The gradient in this thread is meant to be the dual of the differential ##df## of a function with respect to a metric. It is a vector field with the property that ##df(X)= <∇f,X>##. ##∇f## is uniquely determined by the metric. Here ##<,>## denotes the metric.

    It has been also pointed out in this thread that the term gradient sometimes refers to ##df## rather than to ##∇f##. With this definition one does not need a metric.

    BTW: A continuous manifold may not admit an idea of differentiation. By continuous is meant that the transition functions are continuous but not necessarily differentiable. There are continuous manifolds that do not admit an idea of differentiation.
     
  12. Nov 16, 2018 #11

    stevendaryl

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    Is there a simple example of a manifold that is not differentiable?
     
  13. Nov 16, 2018 #12

    lavinia

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  14. Nov 16, 2018 #13

    stevendaryl

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    That would be interesting, but I was only asking if there was some simple example that could give an intuitive reason that some manifolds can't have a differentiable structure. It sounds like the example is not simple.
     
  15. Nov 16, 2018 #14

    lavinia

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    I don't know of one.
     
  16. Nov 16, 2018 #15

    stevendaryl

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    John Baez gave a tutorial showing how the 7-sphere (I think that was it) allowed multiple notions of "differentiable". I sort of understood it at the time.
     
  17. Nov 16, 2018 #16

    lavinia

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    This is different. The 7 sphere as a topological manifold admits an idea of differentiation. It is just that globally across the entire sphere two differentiable structures may not be diffeomorphic.

    Every smooth manifold can be smoothly triangulated. This means that the simplices can be extended to smooth maps from an open neighborhood in Euclidean space into the manifold. Kervaire's paper shows a triangulated manifold that can not be given a smooth structure.

    There are also manifolds that can not even be triangulated.
     
    Last edited: Nov 16, 2018
  18. Nov 16, 2018 #17

    martinbn

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    I personally think that the notation ##\nabla f## for the gradiant of ##f## is not a good one. A better one is ##\text{grad} f##. The first one is the same as the notation for the covariant differential of ##f##, which is a one form, not a vector.
     
  19. Nov 16, 2018 #18

    fresh_42

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    Well, it's not unusual to switch between a vector and its dual. And in the end, a one form is a vector, too! Notations are always a compromise, e.g. what is ##D_pf(v)\,##? The answer depends on what is considered to be the variable here, and notation does not say this. In this sense, the error is already to write a function as its image: ##f(x)## instead of ##x \mapsto f(x)\,.## And I do not see a fundamental difference between ##\nabla## and ##\operatorname{grad}## - these are only symbols and meaning has still to be attached. That one of them is a vector and the other a covector, is already your individual choice. It isn't part of the syntax.
     
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