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.