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?

  • $\begingroup$ You're having trouble seeing a connection, but this person cannot tell the difference. Maybe that post can give you some inspiration :) $\endgroup$ – rschwieb Apr 16 '18 at 16:47
  • $\begingroup$ Yes I also found a related post as posted by rschwieb, but @newbie can you state the book you are following, just out of curiosity and resource collector :) $\endgroup$ – BAYMAX Apr 16 '18 at 16:49
  • $\begingroup$ observe that $\sigma^2\wedge\sigma^3(v,w)=v_2w_3-v_3w_2$ which is the first component of the cross $v\times w$ $\endgroup$ – janmarqz Apr 16 '18 at 17:25
  • 1
    $\begingroup$ and ordering $$\sigma^2\wedge\sigma^3$$ $$\sigma^3\wedge\sigma^1$$ $$\sigma^1\wedge\sigma^2$$ for basics bivector you would have $$a_2b_3-a_3b_2$$ as a first component of $\omega\wedge\eta$ $\endgroup$ – janmarqz Apr 16 '18 at 17:29
  • $\begingroup$ Wedge product is not related to cross product unless you have an inner product and Hodge star on $V$. $\endgroup$ – edm Apr 16 '18 at 17:58

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.

  • 1
    $\begingroup$ also it could be thought as the known use $$(a_1i+a_2j+a_3k)\times(b_1i+b_2j+b_3k)=c_1i+c_2j+c_3k,$$ where $$c_1=a_2b_3-a_3b_2,$$ $$c_2=a_3b_1-a_1b_3,$$ $$c_3=a_1b_2-a_2b_1,$$ as commonly used in the "vector-calculus" symbology $\endgroup$ – janmarqz Apr 17 '18 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.