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Vectors and associated 2-forms

  1. Aug 19, 2014 #1
    In Frankel's book he writes that in [itex]R^{3}[/itex] with cartesian coordinates, you can always associate to a vector [itex]\vec{v}[/itex] a 1-form [itex]v^{1}dx^{1}+v^{2}dx^{2}+v^{3}dx^{3}[/itex] and a two form [itex]v^{1}dx^{2}\wedge dx^{3}+v^{2}dx^{3}\wedge dx^{1} +v^{3}dx^{1} \wedge dx^{2}[/itex], now in general this is not possible and you have to convert the components of a vector with a metric to get covariant components, for example like this [itex]v_{i}=g_{ij}v^{j}[/itex]

    Then he asks what is in general the associated 2-form for [itex]\vec{v}[/itex] ? He then proofs that to a vector [itex]\vec{v}[/itex] one does associate a pseudo-2-form [itex]\beta^{2}:= \iota_{\vec{v}}vol^{3}[/itex] Later when he discusses the cross product he writes that one would like to say that [itex]v^{1} \wedge \omega^{1}[/itex] is the 2-form associated to the vector [itex]\vec{v} \times \vec{w}[/itex], but we only have a pseudo-2-form associated to a vector thus the pseudovector [itex]\vec{v} \times \vec{w}[/itex] is associated to the 2-form [itex]v^{1} \wedge \omega^{1}[/itex] (which is just a flip flop of words I think).

    Now is it true that if we have other coordinates than cartesian, one can only associate pseudo-forms to a vector? Because in the text he calls the forms in cartesian coordiantes just forms, but in general he says pseudo-forms.

    For example (everything in cartesian coord.) when I have a simple vector field [itex]v=3x \partial_{x}+4x \partial_{y}[/itex] then according to the formula [itex]\beta^{1}= \iota_{v}vol^{2}= 3xdy-4ydx[/itex] is the associated pseudo-1-form. Now what about the form [itex]\gamma^{1}=3x dx + 4y dy[/itex] isn't that the 1-form associated to the vector field? And [itex]\gamma^{2}= 3x dy \wedge dz+4y dz \wedge dx[/itex] should be the associated 2-form, but at the same time there should also be an associated pseudo-2-form [itex]3x dx\wedge dy -4y dx \wedge dz[/itex] according to the formular and if I calculated right. Now does this in general mean (none-cart.) that I have only pseudo-forms associated to vectorfields ??
     
    Last edited: Aug 19, 2014
  2. jcsd
  3. Aug 27, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
     
  4. Sep 3, 2014 #3
    Possibly my answer will be too abstract, but you do not need Euclidean structure in three dimensions to have a correspondence between vectors and (pseudo)-2-forms. You need a weaker thing. Depending on how it deals with orientation, it may be called either volume form (that identifies vectors with 2-forms) or density (that identifies vectors with pseudo-2-forms). Euclidean/Riemannian metric induces a density, but one can’t recover the metric from a density only.

    In other words, if in certain coordinates you see a familiar correspondence between vectors and pseudo-2-forms, you can’t be sure that these coordinates are not skew.
     
  5. Sep 3, 2014 #4

    WWGD

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    AFAIK, a pseudometric is enough to define a natural isomorphism between a space--say vectors here--and their duals, being the forms.
     
  6. Sep 3, 2014 #5
    WWGD: you completely miss the point. You think about lowering an index: ωk = gkℓX. It makes an 1-form from a vector, and ιX ω = 1. It corresponds to scalar product (dot product in Euclidean space, or whatever, not necessarily positive-definite).

    Original poster asks about ωjk = εjkℓX, a different thing making a 2-form from a vector. It corresponds to vector product in 3 dimensions and gives ιX ω = 0.
     
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