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

1. Nov 26, 2011

### Rasalhague

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 $(\varepsilon^i)$.

2. Nov 26, 2011

### 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.

Last edited: Nov 26, 2011