I am currently practising the wedge product, but I don't quite understand the structer overall. There is a task in my textbook marked "easy". Could anyone help me with this? I think an example would help me a lot.
Let $V$ be a real vector space, $\dim V=3, \ \ \sigma_1,\sigma_2,\sigma_3$ a basis of $V^*$, $\omega=\sum a_i \sigma_i, \ \ \eta=\sum b_i \sigma_i$ two random elements of $V^*$.
Calculate $\omega \wedge \eta$ and give reasons why the wedge product is a generalization of the cross product.
Let $v,w \in V$. I know that then: $$(\omega \wedge \eta)(v,w)=\omega(v)\eta(w)-\omega(w)\eta(v)$$ But where is the connection to the cross product?
If $$\omega=a_1\sigma^1+a_2\sigma^2+a_3\sigma^3$$ and $$\eta=b_1\sigma^1+b_2\sigma^2+b_3\sigma^3$$ then $$\omega\wedge\eta= (a_2b_3-a_3b_2)\sigma^2\wedge\sigma^3 +(a_3b_1-a_1b_3)\sigma^3\wedge\sigma^1 +(a_1b_2-a_2b_1)\sigma^1\wedge\sigma^2 ,$$ which has another meaning of $$ \left(\begin{array}{c}a_1\\a_2\\a_3\end{array}\right)\times\left(\begin{array}{c}b_1\\b_2\\b_3\end{array}\right)= \left(\begin{array}{c}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{array}\right),$$ however both have the same components.