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Checking my understanding (re. wedge product) of a passage in Bishop & Goldberg

  1. Nov 26, 2011 #1
    Given that Bishop and Goldberg's definition of the exterior product is

    [tex]\alpha \wedge \beta := (\alpha \otimes \beta)_a[/tex]



    giving, in this case,


    am I right in thinking that the coefficients [itex]b_{ij}[/itex] in the quote above are not [itex]b(e_i,e_j)[/itex], but rather [itex]b_{ij} = 2b(e_i,e_j)[/itex]?

    By [itex](e_i)[/itex], I mean the basis for [itex]V[/itex] with dual basis [itex](\varepsilon^i)[/itex].
  2. jcsd
  3. Nov 26, 2011 #2
    Ah, no, looking again at this, I think they must mean the usual coefficient [itex]b_{ij}=b(e_i,e_j)[/itex] after all.

    A factor of 2 would be needed to convert components with respect to a tensor product basis [itex](\varepsilon^i\otimes\varepsilon^j)[/itex] for [itex]T^0_2[/itex] (all permutations of the indices) to the corresponding wedge product basis [itex](\varepsilon^i\wedge\varepsilon^j)[/itex] (only increasing permutations) for [itex]\wedge^2 V^*[/itex]. But in this case, I think, they're not restricting the indices to only increasing permutations.

    So, for example, if [itex]\text{dim} V = 3[/itex] and [itex]b=b_{12} \varepsilon^1\otimes\varepsilon^2 + b_{21} \varepsilon^2\otimes\varepsilon^1[/itex], where [itex]b_{12}=-b_{21}[/itex], we have

    [tex]b=b_{12} \varepsilon^1\otimes\varepsilon^2 + b_{21} \varepsilon^2\otimes\varepsilon^1[/tex]

    [tex]=b_{12} \varepsilon^1\otimes\varepsilon^2 - b_{12} \varepsilon^2\otimes\varepsilon^1[/tex]

    [tex]=b_{12} (\varepsilon^1\otimes\varepsilon^2 - \varepsilon^2\otimes\varepsilon^1)[/tex]

    [tex]=\frac{1}{2}b_{12} (2\varepsilon^1\otimes\varepsilon^2 - 2\varepsilon^2\otimes\varepsilon^1)[/tex]

    [tex]=\frac{1}{2}b_{12} (\varepsilon^1\otimes\varepsilon^2 - \varepsilon^2\otimes\varepsilon^1 - \varepsilon^2\otimes\varepsilon^1 + \varepsilon^1\otimes\varepsilon^2)[/tex]

    [tex]=\frac{1}{2}b_{12} (\varepsilon^1\otimes\varepsilon^2 - \varepsilon^2\otimes\varepsilon^1) - \frac{1}{2}b_{12}(\varepsilon^2\otimes\varepsilon^1-\varepsilon^1\otimes\varepsilon^2)[/tex]

    [tex]=b_{12} \, \varepsilon^1\wedge\varepsilon^2 - b_{12} \, \varepsilon^2\wedge\varepsilon^1[/tex]

    [tex]=b_{12} \, \varepsilon^1\wedge\varepsilon^2 + b_{21} \, \varepsilon^2\wedge\varepsilon^1,[/tex]

    just as they claim.
    Last edited: Nov 26, 2011
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