# wedge product example

When bivectors are defined by $\beta^i\wedge\beta^j:=\beta^i\otimes\beta^j-\beta^j\otimes\beta^i$,

so, for two generic covectors $\theta=a\beta^1+b\beta^2+c\beta^3$ and $\phi=d\beta^1+e\beta^2+f\beta^3$,

we have the bivector $\theta\wedge\phi=(bf-ce)\beta^2\wedge\beta^3+(cd-af)\beta^3\wedge\beta^1+(ad-be)\beta^1\wedge\beta^2$.

Otherwise,

Cf. this with the data $\left(\begin{array}{c}a\\b\\c\end{array}\right)$ and $\left(\begin{array}{c}d\\e\\f\end{array}\right)$ to construct the famous $\left(\begin{array}{c}a\\b\\c\end{array}\right)\times\left(\begin{array}{c}d\\e\\f\end{array}\right)=\left(\begin{array}{c}bf-ce\\cd-af\\ad-be\end{array}\right)$

So, nobody should be confused about the uses of the symbol $\wedge$ dans le calcul vectoriel XD

### 2 responses to “wedge product example”

1. ianmarqz
2. Ray

I would like to mention that using the wedge product, with understanding, allows one to determine dependency. For instance moving b^2 /\ b^3 to the right is only valid if
bf-ce is not zero. This of course applies in the linear type case where the b,f,c,e are not functions of b2,b3.
This is important when b1,b2,b3 are defined implicitly; say
F(b1,b2,b3)=G(b1,b2,b3) =0 then dF/\dG can be expressed as the wedge products of the differentials dF/db1 db^1 ….dG/db3 db^3 and determine which variable are dependent/independent.
By realizing that the wedge products talk about something like area (independence) one then gets a truly geometric viewpoint on things that looked like mysterious matrix manipulations.