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Differential forms and differential operators

  1. Apr 1, 2014 #1
    After read this stretch https://en.wikipedia.org/wiki/Closed_and_exact_forms#Vector_field_analogies, my doubts increased exponentially...
    1. A scalar field correspond always to a 0-form?
    1.1. The laplacian of 0-form is a 2-form?
    1.2. But the laplacian of sclar field is another scalar field.
    1.3. If yes, we have a scalar field 2-form, thus scalar field can be k-form.
    2. Exist (-1)-form?
    3. Exist 4-form?
    4. How know if a vector field (and a scalar field too) is a 0-form, 1-form, 2-form, 3-form or k-form?
    5. If the gradient of a 0-form is zero, so, this 0-form is closed?
    5.1. Which the name given for a scalar field that have gradient zero?
    5.3. Which are your the properties?
    6. A 3-form can be exact?
    7. A 3-form can be closed?
    8. Why the laplacian wasn't mentioned in the stretch? The laplacian can be used in the theory of closed and exact forms?
    8.1. If the laplacian of a vector field is zero, the vector field is called harmonic; if the laplacian of a scalar field is zero, so the scalar field is called harmonic, correct?
    9. d²=0 but ∇²≠0
    10. If a vector field F can be expressed as F=∇²G, so, what is G?
    11. If a scalar field f can be expressed as f=∇²g, so, what is g?
  2. jcsd
  3. Apr 1, 2014 #2


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    Yes, this is true.

    No, this is not true.

    Yes, it is the trace of a 2-form, not a 2-form.

    Some texts will consider the trivial vector space that contains only the 0 vector to be a "-1 form", but it has no useful application.

    If you are working only with R3, then no, but yes, you can have any integer "n-forms", if the underlying space is of high enough dimension.

    4. How know if a vector field (and a scalar field too) is a 0-form, 1-form, 2-form, 3-form or k-form?[/quote]


    A "closed" field.

    Well, I own one property in Alabama and another in .... wait, what?

    Yes, if there is a 2 form of which it is the gradient.

    Yes, if its gradient is 0.

    The Laplacian is, as above, the trace of a second derivative operator. It's not going to be directly connected with first derivative operators.


    ??? What?

    If you mean [itex]\nabla^2[/itex] to indicate the Laplacian, a vector field cannot be expressed in such a way.

    Again, if you mean [itex]\nabla^2[/itex] to indicate the Laplacian, g is some smooth scalar field.
    g is some scalar field.
  4. Apr 1, 2014 #3


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    Halls, in physics, electrodynamics to be precise, we have laplacians of vector fields (like the electric field intensity E), not necessarily of scalar fields.

    11. g = (∇2)-1 f
    Last edited: Apr 1, 2014
  5. Apr 1, 2014 #4
    So no make sense says that a 0-form is exact?

    I said in the sense of mathematical property, for example, if the curl of a vector field is zero, so this vector fields is irrotational, if the divergent is zero, so is incompressible. In the same way, if the gradient of a scalar field is zero, this scalar field have some special property?

    But, like you said, scalar field is always a 0-form, and you just compute gradient of scalar field, and the gradient of 0-form is always an 1-form, therefore, how can a 3-form be the result of a gradient of a 0-form?

    I think that a 3-form is exact if exist a 2-form such that the divergent of this 2-form results the 3-form...

    But the gradient is a operator for differentiate scalar fields, so, if is possible to apply the gradient in a 3-form, thus 3-form is a scalar field, but scalar field can be only 0-form. This is a contradiction!

    And the Hessian, is connected with the theory of differential form?

    Why not? The laplacian can be applied in a scalar and in a vector too.

    EDIT: for this question, consider F and G too as vectors!

    For, 10 and 11, I thought this way: given this equation ##\vec{f}=\vec{\nabla}f##, ##\vec{f}## is called conservative and ##f## is called scalar potential; for this equation ##\vec{F} =\vec{\nabla}\times \vec{f}##, ##\vec{F}## is called solenoidal, and ##\vec{f}## is called vector potential.

    So, analogously, given this equation ##f=\nabla^2 g##, ##f## f is called harmonic and ##g## I don't know; now for this equation: ##\vec{f}=\nabla^2\vec{g}##, ##\vec{f}## is called harmonic and ##\vec{g}## I don't know. But I'd like to know the name those guys for that I can search about.
  6. Apr 10, 2014 #5
  7. Apr 10, 2014 #6


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    Right, it doesn't make sense.

    A scalar field who's gradient is 0 is a constant field.

    A 3 form is exact if it is the exterior derivative of a 2-form. Gradient might be a little bit confusing terminology wise.

    Replace "gradient" with "exterior derivative". "Gradient" is, in my opinion, confusing terminology to use here.

    I've never seen the Hessian generalized through the use of differential forms. (Doesn't mean such a generalization doesn't exist, I just have never seen it).

    I'll let Halls answer this one, as I don't even know where you're trying to go with your original question #10.

    The names you give them "vector potential" and "scalar potential" are exclusive for electrodynamics (and sometimes used in other fields where an analogy to electrodynamics is drawn). They are not general names given to these vectors and scalars. In general, these things don't really have designated names other than "scalar field" and "vector field".
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