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Tangent vectors in affine spaces

  1. Mar 4, 2008 #1

    mma

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    Affine spaces can be regarded as smooth manifolds if we take the natural topology and affine coordinate charts as atlas. So, if M is an n-dimensional affine space, then the tangent vector of a curve [tex] C: [0,1] \rightarrow M [/tex]in a point [tex]p = C(t_0)[/tex] can be defined as a derivation (as in any smooth manifold):
    [tex]\dot C(t_0): C^\infty(M) \rightarrow \mathbb{R}, f \mapsto \frac{d f(C(t))}{d t}\bigg|_{t=t_0} [/tex]

    On the other hand, for M is an affine space, the tangent vector of the curve C in the point [tex]p = C(t_0)[/tex] can be defined as an element of the underlying vectorspace V:
    [tex]C'(t_0) = \lim_{t \to t_0} \frac {C(t) - C(t_0)}{t - t_0}[/tex]

    [tex]C'(t_0)[/tex] and [tex]\dot C(t_0)[/tex] is related simply as
    [tex]\dot C(t_0)(f) = \frac{d f(C(t))}{d t}\bigg|_{t=t_0} =f'(C(t_0))C'(t_0)[/tex]
    where [tex]f'(C(t_0))[/tex] is the derivative of f at [tex]C(t_0)[/tex], that is a linear functional on V which satisfy:

    [tex]f(C(t))-f(C(t_0))= f'(C(t_0)) (C(t) - C(t_0)) + \mathcal{O}(\|{C(t) - C(t_0)\|^2})[/tex]

    But this works only if a norm is also defined on V. Evidently, the relation between [tex]C'(t_0)[/tex] and [tex]\dot C(t_0)[/tex] can also described using coordinates and then showing that the relation is independent from the coordinates chosen.

    My question is: How can be related [tex]C'(t_0)[/tex] to[tex]\dot C(t_0)[/tex] without using norm or coordinates?
     
    Last edited: Mar 4, 2008
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  3. Mar 9, 2008 #2

    mma

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    Perhaps [tex]f'(a)[/tex] can be defined alternatively on affine spaces as follows.

    Let [tex]\Phi: M \rightarrow\mathbb{R}^n [/tex] be an affine chart on [tex]M[/tex] with origin [tex]a \in M [/tex], i.e.,
    [tex] p - a = \sum_{i=1}^{n} \mathrm{proj}_i(\Phi(p))e_i [/tex]
    [tex](p \in M)[/tex]
    with some basis [tex]\{e_i\} [/tex] of [tex] V [/tex].
    Clearly, this is equivalent with a linear map [tex]\tilde \Phi [/tex] between [tex]V [/tex] and [tex]\mathbb{R}^n[/tex]: [tex]\Phi(p) = \tilde{\Phi}(p-a)[/tex].
    Let [tex]g := f \circ \Phi^{-1} : \mathbb{R}^n \rightarrow \mathbb{R}[/tex],
    and let [tex]g'(0) = A [/tex], the derivative of [tex]g[/tex] at [tex]\Phi^{-1}(a) = 0 [/tex] in the usual sense.

    We define now [tex]f'(a):= A \circ \tilde \Phi [/tex]. This definition doesn't require a norm on [tex]V[/tex] and can easily shown that the definition is unambiguous, i.e. independent of the choice of the [tex]\Phi[/tex] chart.
     
    Last edited: Mar 10, 2008
  4. Mar 14, 2008 #3

    mma

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    Of course the original problem persists here. We have chosen a coordinate chart for defining the derivative and then see that the definition is independent of our choice.

    Of course we could have used the original definition of the derivative using an arbitrarily chosen norm on V and then we could have shown that the any norm yields the same result.

    But why do we use something in a definition what is really irrelevant? Couldn't we avoid such objects? What would Mr. Occam say about this?
     
  5. Mar 14, 2008 #4

    Hurkyl

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    Tangent vectors define directional derivatives. Directional derivatives are derivations.
     
    Last edited: Mar 14, 2008
  6. Mar 14, 2008 #5

    mma

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    Exactly. My problem about it is the following. The definition of the directional derivative on affine spaces needs a metric. But it doesn't matter what metric we use. The metric can be arbitrary: any choice of it leads to the same directional derivative. So the definition of the directional derivative really doesn't need a metric. Still we use this unnecessary metric in the definition. But if it is unnecessary, why do we use it?

    How can we define directional derivatives without any reference of any metric?
     
  7. Mar 14, 2008 #6

    Hurkyl

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    It doesn't need a metric; it only needs a topology.
     
    Last edited: Mar 14, 2008
  8. Mar 14, 2008 #7

    mma

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    OK, C' needs only topology. And directional derivative too.
    But how do you define the total derivative of an [tex]M \rightarrow \mathbb{R} [/tex] function without metric (norm)? That is, the linear functional what maps the directional derivative of on [tex]M \rightarrow \mathbb{R} [/tex] function to the tangent vectors.
     
  9. Mar 14, 2008 #8

    Hurkyl

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    I just realized the directional derivative doesn't even use the topology on M! (Because it's defined as the derivative of a particular real-valued function of the reals) But that's getting into irrelevant detail.

    Defining [itex]d_p f[/itex] is easy; a functional is just a fancy sort of function, and you can define it 'pointwise'. e.g.

    [tex](d_p f)(v) := (\nabla_v f)(p)[/tex]

    Which is exactly what the English definition says: it's the functional that maps tangent vectors to the directional derivative of f at p in that direction.

    (which I think is what you were trying to say here:)
     
  10. Mar 15, 2008 #9

    mma

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    Really. Only C' requires the topology.

    Really. We have to take only

    [tex](\nabla_v f)(p) : = \frac{df(p+tv)}{dt}\bigg|_{t=0} [/tex]

    Thank you, Hurkyl!
     
  11. Apr 9, 2008 #10
    The differential of a function requires no norm. Its value on a tangent vector is the derivative of the function composed with any curve whose velocitiy equals that vector.

    It is important to realize that metrics are not part of the definition of calculus. What they allow you to do is to interpret differentials (1 forms) as vectors (gradients)
     
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