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I Do derivative operators act on the manifold or in R^n?

  1. Jul 26, 2016 #1
    I am really struggling with one concept in my study of differential geometry where there seems to be a conflict among different textbooks. To set up the question, let M be a manifold and let (U, φ) be a chart. Now suppose we have a curve γ:(-ε,+ε) → M such that γ(t)=0 at a ∈ M. Suppose further that we have a function ƒ defined on M.

    Now my question: Consider the directional derivative Dv. Some authors (e.g. J.M. Lee) imply that Dva acts at point a∈M. Other authors say that the derivative makes no sense on a manifold and that Dvφ(a) acts at φ(a)∈ℝn.

    Which is the correct understanding?

    Thanks in advance.
     
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  3. Jul 26, 2016 #2

    stevendaryl

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    Well, the definition of derivative used in multivariable calculus doesn't directly make sense in a manifold if you are taking the derivative of a vector field (or tensor field). That definition would say something like:

    [itex]D_v^a \vec{A} = lim_{\lambda \rightarrow 0} \dfrac{\vec{A}(a + \lambda v) - \vec{A}(a)}{\lambda}[/itex]

    There are two reasons that this definition fails for a curved manifold:
    1. For a curved manifold, you can't add a vector [itex]v[/itex] to a location [itex]a[/itex].
    2. For a curved manifold, you can't subtract vectors at different locations.
    That might be what was meant. You can define a directional derivative on a curved manifold, but it requires a slightly different definition of derivative than in [itex]R^n[/itex]
     
  4. Jul 26, 2016 #3

    andrewkirk

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    The directional derivative of a vector field on a manifold at a point P is a function from the tangent space at P to itself. The 'input' to the function is a vector that specifies the direction in which the directional derivative of the field is to be calculated, and the 'output' of the function is another vector. I would say that the directional derivative 'acts' neither on the manifold nor on ##\mathbb R^n## but on the tangent space.

    But note that 'acts on' and 'acts at' are verbal flourishes used to help understanding, and have no strict technical meaning. So there is no strictly correct answer and individual mathematicians are free to adopt whatever interpretation they find helpful.
     
  5. Jul 26, 2016 #4
    That's kind of what I'm getting at. Without modifying our typical understanding of derivatives, then I gather from what you have written that we do the calculus in ℝn. But I have found something that goes even further and says the ordinary derivative of an ordinary (non-vector) function.
    Here is a direct quote for some notes I found on the internet:
    "Unfortunately if M is not embedded in any ℝN the derivative γ'(0) does not make sense." From here.
    γ is a curve on M. And I have seen a similar statement elsewhere. I believe it has something to do with subtracting points does not make sense on a manifold not embedded in ℝn.

    To carry this a bit further to clear up my confusion, why do we call the tangent space, TaM and not Tφ(a)M?
     
  6. Jul 26, 2016 #5
    We don't even need to consider the directional derivative. It may be I chose a bad example. Just consider the simple derivative as in my last post.
     
  7. Jul 26, 2016 #6

    andrewkirk

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    Because the chart is not part of the definition of the definition of a tangent space, it is just one way of picturing it. There are a number of formal ways of defining the tangent space, none of which use charts. The one I like is that ##T_pM## it is the set of equivalence classes of curves in ##M## through ##p## under the equivalence relation that two curves ##\mu,\gamma## through ##p## are equivalent if ##\frac d{dt} \phi(\gamma(t))|_{\gamma^{-1}(p)}=\frac d{dt} \phi(\mu(t))|_{\mu^{-1}(p)}## for every chart ##(U,\phi)## such that ##p\in U##.
     
  8. Jul 26, 2016 #7
    Ok, but your derivatives are taken in ##\mathbb{R}^n## because ##\phi(\gamma(t)) \subset \mathbb{R}^n## which is exactly what I am trying to get at.

    Can you ever say, ##\gamma'=\frac{d\gamma}{dt}## without using the chart to bring it into ##R^n##?
     
  9. Jul 26, 2016 #8
    This is something that I like a lot and it does include the charts and the transition functions between charts in the definition of the tangent space.
     
  10. Jul 26, 2016 #9

    andrewkirk

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    It depends on what you mean by 'bring it into ##R^n##'. As noted above, this is imprecise wording that is used to help intuition. There is no right or wrong answer for it.

    If 'bring it into ##R^n##' means 'use a concept that somewhere along the line refers to ##R^n##' then the answer is no, because manifolds are defined as topological spaces that are locally homeomorphic to ##R^n##.

    There are two other definitions of tangent spaces that you could look into, if you feel uncomfortable with the one involving equivalence classes. They are as 'derivations' or as an 'algebra of germs'. I have never looked into these. As far as I know they could even be the same thing. But they will inevitably have some eventual connection with ##R^n## because of the point in the previous para.

    I'm glad you like the ANU presentation you linked. ANU had a very good maths program when I studied there in the early eighties, and I have very fond memories of it. However I doubt the person who taught me calculus on manifolds back then is still lecturing, or wrote that.
     
  11. Jul 26, 2016 #10
    What I mean by bring it into ##\mathbb{R}^n## is this: Suppose we have a curve ##\gamma:I \to M## with ##I## being an interval ##(-\varepsilon, +\varepsilon)## and that ##\gamma(0)=p##. Then is it true that the correct way of taking the derivative of the curve ##\gamma## at ##p## is not ##\frac{d\gamma}{dt}\big\rvert_p## but rather ##\frac{d\phi(\gamma)}{dt} \big\rvert##.
     
  12. Jul 27, 2016 #11

    Orodruin

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    If this were true it would be sufficient to know the vector field at P. In order to compute the derivative you need to know the vector field in some neighbourhood of P. There are several vector fields that may have the same value at P but different derivatives. The directional derivative maps a vector field in some neighbourhood to another vector field in the same region.
    No. Depending on your choice of definition of a tangent vector, we either define the tangent vector ##\dot\gamma## as a particular equivalence class or as the tangent vector ##X## that satisfies ##X(f) = df(\gamma(t))/dt##, where ##f## is a function on the manifold. There is no need to bring coordinate charts into the game unless you actually need them to compute something.
     
  13. Jul 27, 2016 #12

    andrewkirk

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    In the first sentence of the post I deliberately omitted the words 'in direction v', thereby leaving that as the one thing that needs to be specified - so that we have a unary function. Based on that set-up, the vector field in the neighbourhood is part of the identification of the function rather than an input to it (ie there is a different such function for every vector field around P), and the only input the function needs is a direction - a single vector at P.

    If one wishes, one can alternatively define 'The directional derivative' at p (without reference to a vector field). That will be a binary function, taking as inputs (1) a vector field and (2) a directional vector.
     
  14. Jul 27, 2016 #13


    So what do I make of authors who say that the derivative doesn't make sense on the manifold? And other authors who bring charts into the definition? I have even read at least one author who has said that "we don't know how to take derivatives on manifolds, but we do know how to do calculus in ##\mathbb{R}^n##." Or is this just one of those things that depends on the author's philosophical point of view that has no bearing on the mathematics?
     
  15. Jul 27, 2016 #14

    Orodruin

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    There is no derivative operation on the manifold involved in this definition of the tangent vector. Only how the vector acts on a function in terms of how the curve equivalence class is defined (##f(\gamma(t))## is a function from an interval to the real line). You can then verify that tangent vectors have all the required properties for actually being a differential operator at the point they are defined, that they have the expected coordinate expressions, and work as expected when the manifold actually is R^n.
     
  16. Jul 27, 2016 #15

    stevendaryl

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    Indirectly, charts are involved in determining which functions are "smooth". But given the notion of a smooth function as a primitive, we can define:
    • A (real) scalar field [itex]f[/itex] is a smooth function from [itex]M[/itex] into [itex]R[/itex].
    • A parametrized path [itex]\gamma[/itex] is a smooth function from [itex]R[/itex] (or an interval of [itex]R[/itex]) into [itex]M[/itex].
    • A tangent vector [itex]v = \frac{d \gamma}{dt}_{t=0}[/itex] to path [itex]\gamma[/itex] at the point [itex]t=0[/itex] is an operator which acts on any scalar field [itex]f[/itex] and returns [itex]v(f) = \frac{d f(\gamma(t))}{dt}_{t=0}[/itex]
    With this way of defining "tangent vector", the directional derivative as applied to a scalar field is completely trivial:

    [itex]D^{\gamma(0}_v f = v(f)[/itex]

    To extend the directional derivative to apply to vector fields and tensor fields requires a notion of parallel transport (Or you can do it in the other way around---just let the directional derivative be an operator that obeys the Liebniz rules for derivatives, and for scalar fields acts as above. Then you can define parallel transport in terms of the directional derivative.)
     
  17. Jul 27, 2016 #16
    This is very helpful. Thank you. It clears a lot up.

    I'd still like to ask why some make the absolute statement that even the ordinary derivative on a manifold doesn't make sense unless the manifold is a submanifold of ##\mathbb{R}^n##. Is it because a manifold may not have a metric? By "ordinary derivative" I mean,
    $$ f'(0)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \bigg\rvert_{x=0} $$
     
  18. Jul 27, 2016 #17

    andrewkirk

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    It is because that formula does not mean anything, because if ##x## is a point on the manifold then ##x+h## does not mean anything unless an operation of either addition or affine translation is defined for points on the manifold. In general, such operations are not defined on manifolds. But they do have a meaning in ##R^n## so if the manifold is embedded in ##R^n## we can use that meaning.

    Alternatively, if it is not ##x## but ##f(x)## that is a point on the manifold (eg if ##f## is a curve) then the ##f(x+h)-f(x)## has no meaning, for the same reason given in the previous paragraph.

    The simple guiding principle is that any arithmetic operations can only be performed on objects for which such operations are defined.
     
  19. Jul 27, 2016 #18
    Thank you. That clears up some of my confusion. Now I think I'm getting somewhere, but maybe not. Maybe I'm backing up.

    So this leads me to again think that the proper way to define the tangent space is by using the homeomorphism and doing the derivatives in ##\mathbb{R}^n## since derivatives only make sense in ##\mathbb{R}^n##, i..e. ##\phi(\gamma(t))' = d\phi(\gamma)/dt \rvert_{t=0}## makes sense but that ##\gamma'=d\gamma/dt \rvert_{t=0}## is essentially meaningless.

    Thank you for all your help and patience in cutting through my confusion.
     
  20. Jul 27, 2016 #19

    Orodruin

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    What do you mean by "proper"? There are several equivalent definitions of the tangent space and you do not need to refer to a coordinate chart to define it.
     
  21. Jul 28, 2016 #20
    By proper I mean that given the fact that derivatives on a general manifold that is not a submanifold of ##\mathbb{R^n}## do not make sense then you should be using the chart to define things like velocity of a curve, etc. Some authors do point that out as in the references I posted, for example, here.

    As to definitions of the tangent space, I personally prefer the definition using derivations, but even if you use the equivalence class of tangent curves definition, you still end up taking derivatives because you define the equivalence relation between two curves through:
    ##\gamma_1'(0) = \gamma_2'(0) \implies \gamma_1 \sim \gamma_2##.

    I appreciate your help. I'm not trying to argue just to argue, I'm just trying to resolve what I perceive to be a contradiction that is really confusing me.
     
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