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

[Rasalhague]

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 b on V as being in \wedge^2V^*. If b can be written in terms of \varepsilon^1...\varepsilon^r, then we may discard any independent \varepsilon^i's and extend to a basis, getting

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

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

so it does not matter, in the definition of rank, whether the mode of expressing b 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

\alpha \wedge \beta := (\alpha \otimes \beta)_a

where

\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}),

giving, in this case,

\alpha\wedge\beta=\frac{1}{2}(\alpha\otimes\beta-\beta\otimes\alpha),

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

By (e_i), I mean the basis for V with dual basis (\varepsilon^i).

[Rasalhague]

Ah, no, looking again at this, I think they must mean the usual coefficient b_{ij}=b(e_i,e_j) after all.

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

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

b=b_{12} \varepsilon^1\otimes\varepsilon^2 + b_{21} \varepsilon^2\otimes\varepsilon^1

=b_{12} \varepsilon^1\otimes\varepsilon^2 - b_{12} \varepsilon^2\otimes\varepsilon^1

=b_{12} (\varepsilon^1\otimes\varepsilon^2 - \varepsilon^2\otimes\varepsilon^1)

=\frac{1}{2}b_{12} (2\varepsilon^1\otimes\varepsilon^2 - 2\varepsilon^2\otimes\varepsilon^1)

=\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)

=\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)

=b_{12} \, \varepsilon^1\wedge\varepsilon^2 - b_{12} \, \varepsilon^2\wedge\varepsilon^1

=b_{12} \, \varepsilon^1\wedge\varepsilon^2 + b_{21} \, \varepsilon^2\wedge\varepsilon^1,

just as they claim.