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Tensor components of a Hodge dual

  1. Oct 21, 2008 #1
    If A is a p-vector and B is a (n-p)-vector, then the hodge dual, *A, is defined by:
    [TEX] A\ \wedge\ B = (*A,B)E \ \ \forall B\in \Lambda ^{(n-p)} [/TEX], where E=[TEX]e_1 \wedge\ ... \ \wedge e_n [/TEX]

    I am having trouble in deriving the tensor components of the dual (n-p)-vector - *A.
    Specifically, I am getting stuck when I write down the components of the (n,0)-tensor on both sides and then comparing the coefficients - because, the LHS involves the antisymmetrization, [TEX] A^{[i_1 ... i_p}B^{j_1 ... j_{n-p} ] } [/TEX].
    I got stuck with the LHS even when I took B to be just a simple (n-p)-vector of basis vectors. Because, when I do that, I get the following (n,0)-tensor on LHS...
    [TEX]\frac{1}{n!} \sum_\sigma\ (-1)^\sigma\ A^{i_{\sigma (1)} ... i_{\sigma (p)}}\ \epsilon ^ {j_{\sigma (1)} ... j_{\sigma (n-p)}}\ \ e_{i_1}\otimes ... \otimes e_{i_p}\otimes e_{j_1} \otimes ... e_{j_{n-p}} [/TEX]
     
  2. jcsd
  3. Dec 2, 2008 #2
    Just a repost of the above message with clearer typesetting (and some more):

    If A is a p-vector, then the hodge dual, [tex]*A[/tex] is a (n-p)-vector and is defined by:
    [tex] A\ \wedge\ B = (*A,B)E \ \ ,\ \forall B\in \Lambda ^{(n-p)} [/tex]

    [tex]\ where,\ \ E = e_1 \wedge\ ... \ \wedge e_n [/tex]

    I am having trouble in deriving the tensor components of the dual (n-p)-vector - [tex]*A[/tex] in an orthonormal basis.
    Specifically, I am getting stuck when I write down the components of the (n,0)-tensor on both sides and then comparing the coefficients - because, the LHS involves the antisymmetrization, [tex] A^{[i_1 ... i_p}B^{j_1 ... j_{n-p} ] }\ .[/tex]

    I proceed as follows, taking B to be a simple (n-p)-vector of orthonormal basis vectors...
    [tex] i.e.\ \ Let\ B\ =\ e_{i_{p+1}}\ \wedge\ e_{i_{p+2}}\ \wedge\ ... \wedge\ e_{i_n}
    \ =\ \frac{1}{n!} \epsilon^{i_{p+1}, ... ,i_n}\ e_{i_{p+1}} \otimes ... \otimes e_{i_n} [/tex]
    where [tex] \epsilon [/tex] (epsilon) is the Levi-Civita symbol and [tex] e_{i_x} [/tex] (subscripted e) are the o.n. basis vectors.

    [tex] LHS [/tex]
    [tex]=\ A\ \wedge\ B\ [/tex]
    [tex]=\ A^{[i_1 ... i_p}B^{j_1 ... j_{n-p} ] }\ e_{i_1}\otimes ... \otimes e_{i_p}\otimes e_{i_{p+1}} \otimes ... \otimes e_{i_n} [/tex]
    [tex] =\ \left(\ \frac{1}{n!} \sum_\sigma\ (-1)^\sigma\ A^{i_{\sigma (1)} ... i_{\sigma (p)}}\ \epsilon ^ {i_{\sigma (p+1)} ... i_{\sigma (n)}}\ \right)\ \ e_{i_1}\otimes ... \otimes e_{i_p}\otimes e_{i_{p+1}} \otimes ... \otimes e_{i_n} [/tex]

    [tex] RHS [/tex]
    [tex] =\ (*A,B)\ E [/tex]
    [tex] =\ (*A,B)\ e_1 \wedge\ ... \ \wedge e_n [/tex]
    [tex] =\ \left(\ (*A,B)\ \frac{1}{n!}\ \epsilon^{i_1, ... ,i_p,i_{p+1}, ... ,i_n}\ \right) e_{i_1}\otimes ... \otimes e_{i_p}\otimes e_{i_{p+1}} \otimes ... \otimes e_{i_n} [/tex]
    [tex] =\ \left(\ \frac{1}{n!}\ \left(\ (n-p)!\ *A^{j_1,...,j_{n-p}} \epsilon_{j_1, ... ,j_{n-p}}\ \right)\ \epsilon^{i_1, ... ,i_p,i_{p+1}, ... ,i_n}\ \right) e_{i_1}\otimes ... \otimes e_{i_p}\otimes e_{i_{p+1}} \otimes ... \otimes e_{i_n} [/tex]

    Any pointers on how to proceed further?
     
  4. Dec 14, 2008 #3
    It might be easier just to try to figure out how a basis of k-forms transforms. I.e. what is the dual of e_1 ^ e_2 ^ ... ^ e_k? Then you can show that that the Hodge dual is linear and write your form A in terms of the basis.
     
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