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Duality/Equivalence Between V.Fields and Forms (Sorry for Previous)

  1. Dec 12, 2011 #1

    WWGD

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    Hello, Everyone:

    My apologies for not including a descriptive title; I was just very distracted:

    In the page:

    http://en.wikipedia.org/wiki/Closed_...erential_forms [Broken]

    there is a reference to the form dw= (xdx/(x^2+y^2) -ydy/(x^2+y^2) ) , next to which
    there is the graph of " the vector field associated with dw" , which I think is just the vector
    field V(x,y)=(x/(x^2+y^2),-y/(dx^2+y^2) )., so that it just seems that the assignment is:
    V(x,y)=(f(x,y),g(x,y))->fdx+gdy is the assignment.

    Now, I know forms are dual to vector fields, and, both being finite-dimensional spaces, the vector space of fields ( in a tangent space) is isomorphic to the space of forms (cotangent bundle), but I am not aware of any special correspondence between the two.

    Any Ideas? What Would Gauss Do (WWGD)?

    Thanks.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 12, 2011 #2
    Using the Euclidean metric in this case (or any metric in general), one maps a form a to the unique vector field X satisfying a(v)=<X,v> for all tangent vectors v.
     
  4. Dec 12, 2011 #3

    quasar987

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    Let's discuss this on the linear level first since this is where all the action happens.

    On a vector space V, there is no canonical isomorphism with the dual V*. But as soon as you are given a nondegenerate bilinear form B on V (ex: an inner product or a symplectic bilinear form), then you get an isomorphism in the person of V-->V*:v-->B(v, ). Indeed, the nondegeneracy condition is precisely saying that the ker of this thing is 0.

    On the manifold level, this implies for instance that given a smooth field of bilinear forms (ex: a riemannian metric or a symplectic form), there is a vector bundle isomorphism TM-->T*M (a fiber-preserving diffeomorphism which is linear on the fibers), and this thing, in turn, yields in the obvious way a vector space isomorphism between the vector fields on M (sections of TM) and the 1-forms on M (sections of T*M).

    As an exercice, check that the standard riemannian metric on R² yields the correspondance btw fields and forms that you wrote.
     
  5. Dec 12, 2011 #4

    quasar987

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    On R³, it is also possible to find a correspondance between...

    0) 0-forms and functions (the identity!)
    1) 1-forms and vector fields (the obvious one)
    2) 2-forms and vector fields (this is the less obvious one)
    3) 3-forms and functions (f dx dy dz -->f)

    such that the exterior derivative corresponds respectively to grad, curl and div. In the words of Bott and Tu, this partly illustrates that the exterior derivative is "the ultimate abstract extension" of the well-known differential operators of calculus. :)

    Final remark (promise!): in the case of a riemannian metric, the isomorphism TM-->T*M is called a musical isomorphism and is denoted by the symbol "flat" used in music sheets. The inverse isomorphism T*M-->TM is denoted by the symbol "sharp" (#) also used in music sheets. In the physics literature, TM-->T*M is simply called "lowering the index" since you go from Xii to Xidxi (where Xi=gijXj) while T*M-->TM is called "raising the index" for the same reason. And the fun thing is that these 2 sets if name are related because in music, flat means "lower in pitch" while sharp means to "higher in pitch"!
     
    Last edited: Dec 12, 2011
  6. Dec 17, 2011 #5

    Bacle2

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    Look up Hodge duality; I think Quasar was using it for n=3.
     
  7. Dec 19, 2011 #6

    lavinia

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    A finite dimensional vector space and its dual have the same dimension and so are isomorphic. However there is no canonical isomorphism between them. This means that one must choose an isomorphism. There is none just lying around waiting to be used.

    When the vector space has a positive definite inner product, then as ZHentl wrote, the inner product determines an isomorphism. The dual of a vector is the linear map, <v,>. Different metrics determine different duals, i.e. different isomorphisms.

    In your example, the inner product is the standard inner product on Euclidean space

    <(a,b,),(c,d)> = ac + bd


    Using this inner product, what is the dual of V(x,y)=(x/(x^2+y^2),-y/(dx^2+y^2) )?

    What about if you use the inner product, <(a,b,),(c,d)> = 5(ac + bd) ?

    For fields and forms one has a dual isomorphism at every point of space. At each point an inner product determines an isomorphism between vectors and dual vectors at that point. These fir together across the space to give a duality between forms and fields. This duality usually only is useful when the inner product is smooth so that the duals of smooth fields are smooth forms.
     
    Last edited by a moderator: May 5, 2017
  8. Dec 19, 2011 #7

    Bacle2

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    If your basis {ei} is orthogonal, then the dual of a vector v is just the coefficient of w the expansion v= Ʃaiei. If the basis is symplectic, then you just pick up the coefficient of the "symplectic dual" , i.e., if w(xi,yj)=1 , will be a similar issue, just by linearity.
     
  9. Dec 19, 2011 #8

    Bacle2

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    Right, I am also seeing now how we can use it to relate it to the existence of symplectic forms: if there are no nonwhere-zero vector fileds, I imagine it means we cannot have non-degenerate (symplectic) forms, in relation to my question below.
     
  10. Dec 19, 2011 #9

    quasar987

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    I have no idea what you're saying Bacle :) Do you have a question? Or are you just making statements?
     
  11. Dec 19, 2011 #10

    Bacle2

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    Quasar: Sorry myself for kind of mumbling in writing. My idea was that , if the vector field has zeros, trhen I think its dual (under the isomorphism V to V*) must be degenerate, but I never bothered to check this claim. I am kind of running, but I will look into it later. Then, relating this to the post on symplectic manifolds, this would mean that it may be necessary to be able to define a nowhere-zero field in M, in order for M to be symplectic. But I haven't checked this carefully, and I am kind of running now; will look into it tomorrow.
     
  12. Dec 19, 2011 #11

    Bacle2

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    I was trying to connect/relate this thread with the one on symplectic manifolds ( symplectic but not complex). How can we (if at all) relate this duality result
    , re properties of forms, to the definability of a global non-degenerate form?

    I meant, given that the duality between vector fields and forms given a bilinear form B is given by:

    X -->B(X,. )

    It would seem that if there are no nowhere-zero V.Fields, there would be no candidate for B that would be non-degenerate, so that a symplectic manifold would be one where we can define nowhere-zero V.Fields. But I have not had enough time to think it thru carefully-enough; will look into it later.

    Nice posts in this thread, BTW, Quasar.

    So this one is itself a question.
     
    Last edited: Dec 19, 2011
  13. Dec 19, 2011 #12

    Bacle2

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    And here I was referring to the fact that a basis for the space of forms is given by the projections dxi , which project into the ith coordinate, so that if we have a vector field
    X=Ʃcnen, then:

    dxi(X)=dxi(Ʃ.cnei)= dxi(ci)ei=ci , so that we recover the coefficient of the vector field with respect to the original basis. And something similar happens with symplectic bases.

    Just a comment.
     
  14. Dec 20, 2011 #13

    quasar987

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    Here is a disproof of the bolded statement: S². (S² is an orientable surface, hence any area form is a symplectic form for it, and S² has Euler caracteristic 2, so by Poincaré-Hopf, any vector field on it has at least 1 zero)

    Now, I think if you rethink carefully about your argument, you will see that the existence of a somewhere vanishing vector field, or the nonexistence of a nowehere vanishing vector field does in no way contradict the nondegeneracy condition on a symplectic form (if X vanishes at p, then we have B((X,.)=0 at p... but X=0 at p, so it's all good!).

    Maybe your confusion stems from thinking that the nondegeneracy condition is related somehow to vector fields, but no: to say a 2-form B is nondegenerate is merely to say that the bilinear B_p form at each tangent space is nondegenerate. So its a purely pointwise condition.
     
  15. Dec 20, 2011 #14

    Bacle2

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    Right, Quasar, I was just careless and just threw it out there without thinking it through carefully. I understand the issue is a pointwise issue with forms, but I
    was using (incorrectly, it turns out) the isomorphism between the two. I need
    to make up for the years fighting with my landlord, when I was taking geometry
    and topology; I am learning this now on my own. Thanks.
     
    Last edited: Dec 20, 2011
  16. Dec 22, 2011 #15

    Bacle2

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    Excellent answer again, Quasar. I'm looking up the Da Silva book. Take it easy.
     
  17. Dec 22, 2011 #16

    quasar987

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    I love the Silva book! You can learn the basics of a lot of symplectic-related subjects really fast and in more than a superficial depth with this book. It's unbelievably well-written imho.
     
  18. Dec 22, 2011 #17

    Bacle2

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    Just curious, Quasar, weren't you using Hodge duality in (a last one; don't mean to beat this one to death):

    ?


    BTW, Merry Christma-Hannu-Kwanza-Dan, or Happy Festivus to all (sorry if I left someone out).

    I myself am waiting for a Festivus miracle to have my computer fixed.
     
  19. Dec 22, 2011 #18

    quasar987

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    Let us pray before the holy pole of Festivus for a miracle!

    (If I was using Hodge-duality, it was unbeknownst to me.)

    And a merry christmas to you as well dear Bacle. :)
     
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