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One form <-> vector field

  1. Jun 18, 2010 #1
    One form <-----> vector field

    How exactly does having a one form yield a vector field in a smooth way? I understand it's a duality relationship, but can anyone give me some more insight into this?
     
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  3. Jun 18, 2010 #2

    Hurkyl

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    Re: One form <-----> vector field

    There isn't any intrinsic way; you need an additional gadget to do the transpose.

    The most common way is to select an inner product (but any nondegenerate bilinear form will do), and the transpose is partial evaluation -- [itex]v^T = \langle v, \_ \rangle[/itex]. i.e. [itex]v^T(w) = \langle v, w \rangle[/itex].

    Note that if you did have the transpose, reversing the above defines a bilinear form: [itex]B(v, w) := v^T(w)[/itex]
     
  4. Jun 20, 2010 #3
    Re: One form <-----> vector field

    But, Hurkyl, I would believe that if one can define forms w_i on a space, one has chosen
    a basis {w1,..,wn} for the space of forms. Once the basis of forms has been chosen, the dual basis is determined uniquely as the collection of linear maps L_i* with
    L*_i(wj) =Del_i^j , where Del_i^j=1 if i=j, and it is zero otherwise. Then the
    basis represetation of w in terms of the w_n's would uniquely determine w*. Or
    did I miss or misunderstood ( or misunderestimate : ) ) something.?

    Maybe the choice of inner product underlies the definition of L* .
     
  5. Jun 20, 2010 #4
    Re: One form <-----> vector field

    Bacle, it's true that if you have a basis for the vector space then the usual dual basis is easily defined (as you mentioned using that delta), and so for any one-form you could potentially write out a vector that corresponds to it by having the same components (in terms of the basis and dual basis).

    But if you then choose a new basis, and the corresponding new dual basis, then generally (I think except for bases related by orthogonal transforms) each such choice would relate a different vector to your same original one-form. We could say your procedure is not coordinate-invariant, and more specifically, it is equivalent to always assuming the canonical inner product two-form (or presuming every basis to be constructed of perpendicular normalised vectors). You may need to think beyond Cartesian space to see the issue with this.
     
  6. Jun 21, 2010 #5
    Re: One form <-----> vector field

    I got a question about L*_i.

    L*_i is a function that takes a form wj and maps it to a number. L*_i is also a vector. Would it be true that L*_i(wj) =wj(L*_i)?

    I'm used to forms being linear functions of vectors. I'm not used to vectors being linear functions of forms.

    Sometimes I see a vector V acting on a 0-form f as:

    [tex]V[f]=V^i\frac{\partial f }{\partial x^i} [/tex]

    So is it safe to say that if f is a function of the coordinates, then V[f] is as above. If f is a linear function of vectors, then

    [tex]V[f]=V^i\frac{\partial f }{\partial x^i}=f(V) [/tex] ?
     
  7. Jun 21, 2010 #6
    Re: One form <-----> vector field

    Maybe it's just terminology?
    [tex]df[V] = \frac{\partial f }{\partial x^i} dx^i \left[ V^j \frac{\partial }{\partial x^j} \right]
    = \frac{\partial f }{\partial x^i} V^j \delta^i_j = V^i\frac{\partial f }{\partial x^i} = V[f] [/tex]
     
  8. Jun 21, 2010 #7
    Re: One form <-----> vector field

    That makes sense, but doesn't that only work for a perfect differential df, whose coefficients are [tex] \frac{\partial f }{\partial x^i}[/tex]? If df is not the differential of some function f, but instead a general 1-form, would you still write

    [tex]df[V]= V[f][/tex]

    only this time V[f] would not make sense as f is not defined, so write it as V[df], which is just the same as df[V]?
     
  9. Jun 23, 2010 #8
    Re: One form <-----> vector field

    If a one form is not the differential of a function, you don't write it as [tex]df[/tex]. You use a single symbol such as [tex]\omega[/tex]. Note that the product [tex] \langle V,w \rangle = \langle w,V \rangle [/tex] is just an algebraic multiplication, and that by convention we say that [tex] \langle \partial_x, dx \rangle = \langle dx, \partial_x \rangle = 1 [/tex] (i.e. the order does not matter).

    Remember though that the differential of the inner product [tex] d \langle V, \omega \rangle [/tex] and Lie derivative [tex]L_V \omega = d \langle V, \omega \rangle + \langle V,d \omega \rangle [/tex] are not the same for general one forms, whereas, the result is compatible if the one form is closed or exact (i.e. can be expressed as [tex]df[/tex]), since the second term in the Lie derivative becomes zero ([tex] d df = 0[/tex] - so [tex]\langle V, d df \rangle = 0[/tex]).
     
    Last edited: Jun 23, 2010
  10. Jun 23, 2010 #9

    lavinia

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    Re: One form <-----> vector field

    A couple of points.

    First there may not be a global basis for the 1 forms on a manifold.

    If there is, the choice of basis is equivalent to the choice of a metric.

    In general one passes - as Hurky says - from the 1 forms to vector fields through a metric.
     
  11. Jul 20, 2010 #10
    Re: One form <-----> vector field

    Apparently there is a technique to get the equivalence without using the Riemannian metric, but just using a Partition of Unity. Having said that, I know that having a Partition of Unity implies the existence of a Riemannian metric, but the book I am using, a Riemannian metric is not even mentioned, it's a pure topological proof.

    Any insight into this?
     
  12. Jul 20, 2010 #11
    Re: One form <-----> vector field

    I think you only need paracompactness --which you get for free on a manifold,

    by 2nd-countability and Local Compactness, I think-- to be able to define partitions of

    unity. If your manifold is C<sup>oo</sup> , then it does admit a Riemannian

    metric, tho I think this condition is too strong, i.e., it is sufficient but not necessary.

    The proof I know for the existence of a Riemannian metric makes use of the PO unity:

    you pullback the inner-product patch-wise and put it together globally using the POUnity.
     
  13. Jul 21, 2010 #12
    Re: One form <-----> vector field

    I have a question relating to this discussion that I've been wondering about, and it pertains just to R3. Maybe this restriction is why I'm having problems with some subtle points.

    If I define a general non-orthogonal and/or curvilinear coordinate system in R3 (i.e. a collection of basis vectors at every point in R3), there is a reciprocal (dual) set of basis vectors defined by the relation with the kronecker delta. In R3 at least, it is easily defined by an expression that looks like a cross product over a triple product for each of the dual e1, e2, e3. One could then use the Euclidian dot product to extract a covariant or contravariant component of each of these vectors.

    My question is, how do we get the idea of duality between a vector field <-----> one form, from this description of using some frame field and its reciprocal to extract covariant and contravariant components of a vector in R3? Is the duality between a basis set in R3 and its reciprocal basis the same concept of duality discussed between a vector field and one form?
     
  14. Jul 26, 2010 #13
    Re: One form <-----> vector field

    I replied to one of your earlier posts, and some of the concepts I covered may be useful for your question above, although I don't address the question completely. Note that a covector basis is essentially the same thing as a one form basis, and that often we define a set of coordinate functions, take the differential of the coordinate functions to get a one form basis, and use the kronecker delta relation you refer to to calculate the associated coordinate line basis. The duality relationship is more natural once you see how it is connected to the duality between coordinate functions and coordinate lines. The definition of the coordinate line and a discussion of the relation to coordinate functions is given the the post below:

    Look https://www.physicsforums.com/showpost.php?p=2815042&postcount=6".

    Hope this is useful because it really helps to be able to understand the relationship between the coordinate chart that is being used and the differential bases that are derived from the coordinate charts. It also helps to see how the chain rule plays a part in defining vectors and covectors and in the the transformations between coordinate charts.
     
    Last edited by a moderator: Apr 25, 2017
  15. Jul 27, 2010 #14
    Re: One form <-----> vector field

    please tell me what is meaning of dimension i knew only three dimension what about four ,five.............. ialso want to know that any point in the three dimension is a vector than why we writing vector as a special case suppose u(2,3,5) is any point in the three dimension than why we writing it as a u=2i+3j+5K please reply me via email indetai
     
  16. Aug 13, 2010 #15
    Re: One form <-----> vector field

    Alright everyone, I think I mostly understand my question above. Can anyone explain to me, however, the origin of the reason that the bilinear form from the matrix equation J(transpose)J gives you the Euclidian dot product of two vectors under a change of coordinates?
     
  17. Aug 13, 2010 #16
    Re: One form <-----> vector field

    Okay, a few things here. I'm going to presume that you're at least a little bit familiar with what a vector space, dual space, differentiable manifold, tangent space, and cotagent space are (If not, Christopher Isham's book on differential geometry for physicists covers the topics rather concisely).


    To review (I'll get to your question in a moment):

    1.) A tangent vector is simply an intrinsic direction, which a smooth curve can "point to" in at a given point in the manifold.
    2.) The set of tangent vectors (all different ways to "point"), at a point, forms a vector space.
    3.) You can, at every point, create another vector space which is dual to the tangent space at each point. These are called the cotangent spaces, and elements of cotangent spaces are called "one-forms."



    Now, as it has been pointed out, you can construct a dual space from any given vector space by taking the set of linear transformations satisfying certain properties (this has been done in greater detail in Hurkyl's post), which is the usual way of doing it in linear algebra. However, as in linear algebra, there's a much more natural way of relating (via a natural linear isomorphism) using a non-degenerate bilinear form --the most common being inner products and symplectic forms.

    Something analoguous is true for the the collections of tangent spaces defined on points on a manifold, only because now manifold deals with collections of vector spaces and dual spaces (tangent spaces and cotagent spaces) that are parameterized via points, we need to have them vary smoothly from point to point (to respect the differentiable structure). This is done using (psuedo-)Riemannian metrics and also symplectic forms.


    1.) However, we're not interested in individual tangent vectors or one-forms; we're interested in new objects called vector fields and differential forms. Vector fields are objects that take in points on a manifold and return vectors on the corresponding tangent spaces of those points, which vary smoothly from point to point. Now because of the rules of linear algebra and the natural isomorphism between vector spaces and their dual spaces gives a non-degenerate bilinear form (and a Riemannian metric, at each point, gives such a creature for each vector space and dual space at each point), it follows then that there is a natural relationship between each differential form and each vector field (given by the linear isomorphism, or more explicitly stated in RedX's post).

    There's not really a "further insight" into this, other than to point out the difference in the transformation properties of one-forms and tangent vectors are all related via the geometry of the manifold (given by the Riemannian metric). They have other properties that are different but in certain sense, 'dual' (differential forms are necessary for integrating on manifolds and vector fields are necessary for taking derivatives of smooth functions on manifolds), but I'm afraid this sort of thing just is the way that it is. It's a matter of logic and cleverly chosen definitions.
     
    Last edited: Aug 13, 2010
  18. Aug 16, 2010 #17
    Re: One form <-----> vector field

    I think the formula [tex] J^T J[/tex] is a little misleading. I will write this out using index notation since matrix format doesn't clarify the difference between upper and lower indices (vectors and covectors).

    A metric can be represented as a symmetric matrix (or tensor) [tex]g_{j k} = g_{k j}[/tex] (i.e. switching the indexes is the equivalent of a transpose in matrix notation). If we want to convert this to a new coordinate system, we need to convert each index. The Jacobian is the equivalent of the chain rule transformation, and we need a Jacobian [tex] J_{i}^{j}[/tex] to convert the first index, and another Jacobian [tex] J_{m}^{k}[/tex] to convert the second index. The formula to transform the metric is [tex]J_{i}^{j} g_{jk} J_{m}^{k} = \hat{g}_{i m}[/tex] where [tex]\hat{g}_{i m}[/tex] is the metric on the new space.

    For euclidean space, the metric is simply the Identity matrix [tex]\delta_{jk}[/tex], so we have [tex]J_{i}^{j} \delta_{jk} J_{m}^{k} = \hat{g}_{i m}[/tex]. For euclidean metric, the identity matrix can be removed leading to [tex]J_{i}^{j} J_{m}^{j} = \hat{g}_{i m}[/tex]. This is the [tex]J^T J[/tex] formula you identified. The index formula is not really correct because two upper indices can't be multiplied directly. The [tex]\delta_{j k}[/tex] is really needed, but this is dropped in the matrix formula you used. Anyway, it is really just a transformation of the euclidean metric (the identity matrix) to a new set of coordinates. Dropping the actual metric (the identity matrix) leads to some confusion about what you are doing.
     
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