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

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The rank of a skew-symmetric bilinear form is the minimum number of vectors in terms of which it can be expressed. We may think of a skew-symmetric bilinear form [itex]b[/itex] on [itex]V[/itex] as being in [itex]\wedge^2V^*[/itex]. If [itex]b[/itex] can be written in terms of [itex]\varepsilon^1...\varepsilon^r[/itex], then we may discard any independent [itex]\varepsilon^i[/itex]'s and extend to a basis, getting

[tex]b = b_{ij}\varepsilon^i \otimes \varepsilon^j, \enspace\enspace \text{where } b_{ij} = 0 \text{ unless } i,j \leq r,[/tex]

[tex]=b_{ij}\varepsilon^i \wedge \varepsilon^j, \enspace\enspace \text{since } b_{ij} = -b_{ji},[/tex]

so it does not matter, in the definition of rank, whether the mode of expressing [itex]b[/itex] is in terms of tensor products or exterior products.

- Bishop & Goldberg: Tensor Analysis on Manifolds, Dover 1980, §2.23, pp. 111-112
Given that Bishop and Goldberg's definition of the exterior product is

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

where

[tex]\omega_a(v_1,...,v_s):=\frac{1}{s!}\sum_{(i_1,...,i_s)}\text{sgn}(i_1,...,i_s)A(v_{i_1},...v_{i_s}),[/tex]

giving, in this case,

[tex]\alpha\wedge\beta=\frac{1}{2}(\alpha\otimes\beta-\beta\otimes\alpha),[/tex]

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].
 

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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.
 
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