P-form associated with vectors

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In summary: The two form in cartesian coordinates is just \beta_{\vec{F}}= f_1 dx^{1} \wedge dx^{2}+ f_2 dx^{1}\wedge dx{2}+ f_3 dx^{2} \wedge dx^{3}. The general formula for a vector v associated to the n-1 form is given by \gamma^{n-1}:= \iota_{v}vol^{n} for example \gamma^{2}=\iota_{v}vol^{3} where \iota_{v}vol^{3}= \sqrt{g}(v_1 dx^{1} \wedge dx^{2
  • #1
davi2686
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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?
 
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  • #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.
 
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  • #3
davi2686 said:
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?

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.
 
  • #4
davi2686 said:
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?

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].
 
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  • #5
JonnyMaddox said:
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].

Actually, this is in general only true for pseudoforms. So to every vector there is an associated pseudoform in general ! Sry
 
  • #6
JonnyMaddox said:
Actually, this is in general only true for pseudoforms. So to every vector there is an associated pseudoform in general ! Sry

Hi JonnyMaddox, do you know some book which talks about pseudoforms?
 

Related to P-form associated with vectors

1. What is "P-form associated with vectors"?

The "P-form associated with vectors" refers to a mathematical concept in which a vector space is associated with a P-form, which is a multilinear map that takes in P vectors and outputs a scalar value.

2. What is the significance of P-form associated with vectors in science?

P-form associated with vectors is significant in science as it allows for the manipulation and analysis of vector quantities in a more general and abstract manner. This is useful in various fields such as physics, engineering, and computer science.

3. How is P-form associated with vectors different from traditional vector operations?

P-form associated with vectors differs from traditional vector operations in that it allows for the manipulation of multiple vectors at once, rather than just two or three as in traditional vector operations. It also considers the direction and magnitude of the vectors, rather than just their components.

4. Can P-form associated with vectors be applied to real-world problems?

Yes, P-form associated with vectors can be applied to real-world problems in fields such as fluid dynamics, electromagnetism, and quantum mechanics. It provides a powerful tool for analyzing and solving complex problems involving multiple vector quantities.

5. Are there any limitations to using P-form associated with vectors in scientific research?

While P-form associated with vectors is a useful and versatile tool, it does have some limitations. It may not be suitable for all types of problems and may require advanced mathematical knowledge to fully utilize its capabilities. Additionally, it may not always provide intuitive or easily interpretable solutions.

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