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How are 1-forms related to infinitesimal displacements?

  1. Mar 17, 2012 #1
    Consider a point p in a manifold with coordinates [tex]x^\alpha[/tex] and another point nearby with coordinates [tex]x^\alpha + dx^\alpha[/tex] where dx are infinitesimal or arbitrarily small. Suppose we have a function f on this manifold. Then we can write

    [tex]df=f(x^\alpha + dx^\alpha)-f(x^\alpha)=\partial_\mu f dx^\mu[/tex]

    Notationally this is identical to the expression for the d operator acting on the scalar function f to give the 1-form

    [tex]df=\partial_\mu f dx^\mu[/tex]

    where the dx^mu here are the coordinate basis of the 1-forms at p. Surely the similarity in notation is meaningful in some way. But how?

    Now I can easily repeat back to you the definition of a tangent vector as a map from functions on the manifold to real numbers, and the definition of 1-forms as the dual space to the tangent space. But I can't see the conceptual relationship to small displacements.

    Furthermore, since tangent vectors at a point p can also be thought of as directional derivatives to curves passing through p, visually it is tangent vectors which seem to me more like small displacements, rather than their dual the 1-forms.
    Last edited: Mar 17, 2012
  2. jcsd
  3. Mar 23, 2012 #2
    Well, since 400 people looked at this thread, there must be some mildly curious, so I will share what I have so far.

    There is a relationship between the two expressions above. Let the product of a 1-form omega and a vector X be written

    [tex]<\omega,X>= \omega_\mu X^\mu [/tex]

    Suppose we have a parametrized curve whose tangent vector at t is X. If the coordinates of the curve at t are x^mu, then the components of X at t are dx^mu/dt. Given a function f at x^mu, we have

    [tex]<\text{d}f,X>=\partial_\mu f X^\mu = \frac{\partial f}{\partial x^\mu}\frac{dx^\mu}{dt} =\frac{df}{dt}[/tex]

    We get even closer if X, instead of having components dx^mu/dt, simply has components dx^mu:

    [tex]<\text{d}f,X>=<\partial_\nu f \text{d}x^\nu,dx^\mu\partial_\mu>=\partial_\mu f dx^\mu = \frac{\partial f}{\partial x^\mu}dx^\mu=df[/tex]

    where df on the RHS is the small change in f described in the OP. Readers are invited to expound on the conceptual significance. I don't get it yet myself.
  4. Mar 23, 2012 #3
    Some one forms ("exact" 1-forms) are essentially gradients of functions. So they just describe the linearization of a function at a point. Just as in ordinary 1-variable calculus, if you have take a derivative, you could think of it as making an infinitesimal displacement dx, and in response df, which is an infinitesimal displacement. But if you want to think of the df as a 1-form, the df is now a linear functional on the tangent space, so it's a bit different. Now, to get the infinitesimal displacement that was df in ordinary calculus, you have to evaluate the 1-form df on the tangent vector that you are displacing by.

    But not all 1-forms are the exterior derivative of some function. Thus, you can't really think of them as always having to do with infinitesimal displacements, as I discussed above.

    However, if you think of tangent vectors as infinitesimal displacements, then one forms are just linear functions whose domain is the vector space of infinitesimal displacements.

    One definition of a tangent vector, the one you mentioned, is all the "directional derivatives" at a point. But there are other equivalent definitions, some of which more closely resemble the idea of infinitesimal displacements. I have to go, so I can't get into it right now.
  5. Mar 23, 2012 #4


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    This is exactly right. Even more telling is how they interact with functions: if f is a differentiable function M --> N, and v is a tangent vector on M, then [itex]f_*(v)[/itex] is a tangent vector on N. However, 1-forms work the other way: if [itex]\omega[/itex] is a 1-form on N, then [itex]f^*(\omega)[/itex] is a 1-form on M.

    I like the very suggestive notation "P+v" for the tangent vector v at the point P. The biggest reason is because you have this formula:
    [tex]f(P + v) = f(P) + f'(P) v[/tex]​
    (well, technically, I should have written [itex]f_*(P+v)[/itex] for the left hand side. But if I'm using this notation, I prefer just using f)

    If the x's are scalar fields denoting coordinate functions on your manifold, then the dx's can be thought of coordinate functions on the tangent spaces.

    Recall, even in the general vector space setting, that the function that returns the n-th coordinate of a vector (relative to a chosen basis) is a covector. (and conversely, for any non-zero covector, you can find a basis that it is a coordinate function for)
  6. Mar 26, 2012 #5
    Thanks for the info, guys.
  7. Sep 12, 2012 #6
    I never liked this interpretetion and I think it is a little bit abusive. Tangent vectors are geometric objects and shouldn't be identified with operators.
  8. Sep 12, 2012 #7
    But it's easy to differentiate in the direction of a tangent vector, so I don't see what the problem is. I don't like how in Do Carmo's Riemannian Geometry, it is the ONLY definition of a tangent vector. That's abusive. But as long as the other definitions are also given, I don't have a problem with identifying them with operators because they are closely associated to the geometric objects.
  9. Sep 13, 2012 #8
    The transition from looking like a linear map to actually being one, is important to avoid misunderstandings. Even the wiki article doesn't make it clear, that we are dealing with another definition and not with a different point of view of the same entity.
    Anyway, my objection is that people/textbooks (especially non-mathematicians) don't bother to explain the difference.
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