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P-form associated with vectors

  1. Sep 22, 2014 #1
    if i have a vector field [itex]\vec{F}=(f_1,f_2,f_3)[/itex],i know which for obtain 1-form associated with it i do [itex]\vec{F}^{\flat}=f_1dx^1+f_2dx^2+f_2dx^3[/itex], but how can i get the 2-form and p-form associated with that vector field?

    And one more thing, the musician isomorphisms which i used is only valid in cartesian coordinates? in general i have to use in some way the metric to obtain the p-form associated?
  2. jcsd
  3. Sep 22, 2014 #2
    There are various ways to move between the various spaces of p-forms (exterior derivatives, interior products, wedge products etc.), however these are almost never isomorphisms so you can't identify a vector field with a p-form without losing information along the way.

    The musical isomorphisms, which are indeed valid in any coordinate system in a Riemannian manifold work because the metric gives an isomorphism of the tangent space with the cotangent space. In coordinate free form, this is simply given by [itex] v^\flat(w)=g(v,w) [/itex] for two vector fields [itex]v,w [/itex]. Since every manifold (at least finite dimensional) can be given a Riemannian metric using a partition of unity, it follows that the isomorphism is true for any manifold. However, a different choice of metric will give a different isomorphism so you need to specify the metric explicitly to say what the isomorphism actually is. By using the universal property of tensor products, this can of course be upgraded to an isomorphism
    [tex] (TM)^{\otimes n}\otimes (T^*M)^{\otimes m} \cong (TM)^{\otimes n-i}\otimes (T^*M)^{\otimes m+i}[/tex]
    whenever the indices make sense simply by using the musical isomorphisms in each factor. Be careful to note that the right factors here are tensor product of cotangent spaces, not the alternating tensor product so these are not p-forms (think of the difference between a metric and a symplectic form.)

    The way to get a 2-form associated with a 1-form in 3-dimensions is to use the Hodge star operator, which again depends on the existence of a metric. It can be defined as taking the 1-form [itex] \mu[/itex] to the unique 2-form [itex] *\mu[/itex] such that the equality [itex] \omega\wedge *\mu=g(\omega,\mu) \mathrm{vol}_n [/itex] holds. If you write this down explicitly in the case of [itex] \mathbb{R}^3 [/itex] and the standard Euclidean metric, this will give you the usual 2-form associated to a vector field that allows you to think of cross products in terms of the exterior derivative.

    For a general manifold, note that if [itex] \omega\in \Omega^n(M^d) [/itex], then [itex] *\omega\in \Omega^{d-n}(M^d) [/itex]. Hence, you can start with a vector field, convert it to a 1-form using the metric however if you now use the hodge star, you get a [itex] d-1 [/itex]-form not a 2-form. So, in this case you can no longer associate vector fields with 2-forms.

    In general there is no way to associate a p-form with a vector field, at least none that doesn't lose information along the way since these spaces are not isomorphic. Of course you can define a many maps from vector fields to p-forms using different combinations of the standard operations I mentioned above however there is no reason for the operator you create to be interesting.
  4. Sep 22, 2014 #3


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    Well, the inverse map of the musical isomorphism makes use of the matrix ## g^{ji} ## , which is the inverse of the metric ## g_{ij} ## , so in this sense the inverse map depends on your original choice of metric.
  5. Sep 23, 2014 #4
    The two form in cartesian coordinates is just [itex]\beta_{\vec{F}}= f_1 dx^{1} \wedge dx^{2}+ f_2 dx^{1}\wedge dx{2}+ f_3 dx^{2} \wedge dx^{3}[/itex]. The general formula for a vector [itex]v[/itex] associated to the n-1 form is given by [itex]\gamma^{n-1}:= \iota_{v}vol^{n}[/itex] for example [itex]\gamma^{2}=\iota_{v}vol^{3}[/itex] where [itex]\iota_{v}vol^{3}= \sqrt{g}(v_1 dx^{1} \wedge dx^{2}+ v_2 dx^{1}\wedge dx{2}+ v_3 dx^{2} \wedge dx^{3})[/itex]. And [itex]\iota_{v}[/itex] is the interior product. Hope that helps. If you want to convert it back, you just have to divide by [itex]\sqrt{g}[/itex].
    Last edited: Sep 23, 2014
  6. Sep 23, 2014 #5
    Actually, this is in general only true for pseudoforms. So to every vector there is an associated pseudoform in general ! Sry
  7. Oct 28, 2014 #6
    Hi JonnyMaddox, do you know some book which talks about pseudoforms?
  8. Nov 3, 2014 #7
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