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A Differential forms and vector calculus

  1. Oct 29, 2016 #1
    Let

    ##0##-form ##f =## function ##f##

    ##1##-form ##\alpha^{1} =## covariant expression for a vector ##\bf{A}##

    Then consider the following dictionary of symbolic identifications of expressions expressed in the language of differential forms on a manifold and expressions expressed in the language of vector calculus in ##\mathbb{R}^{n}##.

    ##i_{\bf{v}}(\alpha^{1}) \iff \bf{v}\cdot{\bf{A}}##

    ##df \iff \text{grad}f##



    Now, with ##\alpha^{1} = a_{i}dx^{i}##,

    why is ##\textbf{d}i_{\bf{v}}(a_{i}dx^{i})\iff\text{grad}({\bf{v\cdot{A}}})\cdot{d\bf{x}}##

    and not ##\textbf{d}i_{\bf{v}}(a_{i}dx^{i})\iff\text{grad}({\bf{v\cdot{A}}})##?
     
    Last edited: Oct 29, 2016
  2. jcsd
  3. Nov 3, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Nov 14, 2016 #3

    lavinia

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    An exterior derivative is always a differential form not a vector.
     
  5. Nov 15, 2016 #4

    WWGD

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    Doesn't the vector space duality give you equality (up to non-natural isomorphism)?
     
  6. Nov 15, 2016 #5

    fresh_42

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    The exterior derivation is an operator that shifts k-forms to (k+1)-forms. But this non-natural between 1-forms and vectors drives my nuts. To be or not to be?
     
  7. Nov 15, 2016 #6

    lavinia

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    Yes but that is not the point.

    If the isomorphism were natural then it might be reasonable not to distinguish them. The different isomorphisms define all of the possible metrics - I think. One could also say that a vector is the same as the operator that takes inner products with that vector. But again this operator is not a vector. It is a dual vector.
     
  8. Nov 15, 2016 #7

    fresh_42

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    I don't think that this is a legal implication. At a short glimpse into my homology book, duality seems to be a natural functor.
     
  9. Nov 15, 2016 #8

    lavinia

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    Can you explain this further? It seems that if there is no natural isomorphism then identifying the spaces is not natural. The isomorphism between a vector space and its dual seems to require a metric. But the double does not.

    BTW: What is the title of your homology book?
     
    Last edited: Nov 15, 2016
  10. Nov 15, 2016 #9

    fresh_42

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    I can, but it's in the wrong language and from the 70's (G. Brunner: Homological Algebra; even amazon apparently doesn't have it). But as I said, I didn't really searched for an exact definition of "natural" in a sense as "universal" is defined. I simply stumbled upon the definition (and a commutative diagram) of a functorial morphism, which the author said is also called morphism of functors or natural transformation, resp. natural equivalence in case of isomorphisms. To me it made sense, since duality transforms objects and morphisms between two categories in a - which I find - natural way and I couldn't see, why this diagram shouldn't commute in the case of the ##*##-functor.
     
  11. Nov 15, 2016 #10
    "Doesn't the vector space duality give you equality (up to non-natural isomorphism)?"

    This amounts to nothing more than saying that the two real vector spaces have the same dimension.

    "... duality seems to be a natural functor."

    There is such a thing as a "functor", and there is such a thing as a "natural transformation". But I do not know the definition of a "natural functor".

    For why an isomorphism between a vector space and its (algebraic) dual is not natural, see Wikipedia: https://en.wikipedia.org/wiki/Natur...le:_dual_of_a_finite-dimensional_vector_space.
     
  12. Nov 15, 2016 #11

    fresh_42

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    Yes, this is nonsense, see the correction in #9 to "natural transformation". But, as I said, I took only a glimpse on it, without further examination. Therefore thank you for clarifying this. I wouldn't have expected it to be a counterexample. Something learnt. (Does this count for tuesday or wednesday? :smile:)
     
  13. Nov 15, 2016 #12

    lavinia

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    If there were a canonical isomorphism from the tangent bundle of a smooth manifold to the bundle of differential 1-forms then every manifold would have a canonical metric.
     
    Last edited: Nov 15, 2016
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