wedge product/cross product

Question

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?

You're having trouble seeing a connection, but this person cannot tell the difference. Maybe that post can give you some inspiration :) — rschwieb, Apr 16 '18 at 16:47
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 :) — BAYMAX, Apr 16 '18 at 16:49
observe that $\sigma^2\wedge\sigma^3(v,w)=v_2w_3-v_3w_2$ which is the first component of the cross $v\times w$ — janmarqz, Apr 16 '18 at 17:25
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 $\omega\wedge\eta$ — janmarqz, Apr 16 '18 at 17:29
Wedge product is not related to cross product unless you have an inner product and Hodge star on $V$. — edm, Apr 16 '18 at 17:58

score 3 · Accepted Answer · answered Apr 16 '18 at 20:32

3

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.

answered Apr 16 '18 at 20:32

janmarqz

10,538

1

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 – janmarqz Apr 17 '18 at 18:19
Is there any reasoning why the resulting vector of the cross product starts with the coefficients of $\sigma^2 \wedge \sigma^3$? – user57 Dec 01 '22 at 18:35
1

For the cross product always is as this. With 1 and 2 forms, there in vector calculus, the operation $d(Adx+Bdy+Cdz)$, the exterior derivative of the 1-form $Adx+Bdy+Cdz$ is $$d(Adx+Bdy+Cdz)=(C_y-B_z)dy\wedge dz+(A_z-C_x)dz\wedge dx+(B_x-A_x)dx\wedge dy$$ is present in the corresponding Stokes' Theorem. – janmarqz Dec 01 '22 at 22:49
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@mnp here $A_x=\dfrac{\partial A}{\partial x}$, etc. – janmarqz Dec 01 '22 at 22:51

wedge product/cross product

1 Answers1