This question was asked in my assignment on Linear Algebra and I am struck on it because I am not very good in Wedge products .
Question: Suppose the standard co-ordinate on $\mathbb{R}^3$ are x,y,z. Now using spherical co-ordinates write $x=\rho cos(\theta) cos(\phi)$, $y =cos(\theta) sin(\phi)$ and $z= sin (\theta)$. Calculate dx, dy, dz and $dx\wedge dy \wedge dz$ in terms of $d \rho $ , $d\theta$ , $d \phi$.
I have calculated the dx , dy and dz but I am not sure how should I proceed to find the wedge product. In my notes it is given that if $w \in A^k (v)$ , $n \in A^r(v) $ then $w\wedge n = Alt(w \otimes n) \times \frac{(k+r)!} { k! r!} $ and related results but I am having a hard time implementing it.
Can you please help me with calculating wedge product?
Given $\displaystyle\;\left\{\begin{align} x &= \rho\cos\theta\cos\phi\\ y &= \cos\theta\sin\phi\\ z &= \sin\theta \end{align}\right.$, we have
$$\left\{\begin{array}{rlll} dx &= \cos\theta\cos\phi\; d\rho &- \rho\sin\theta \cos\phi\; d\theta &- \rho\cos\theta\sin\phi\; d\phi\\ dy &= &-\sin\theta\sin\phi\;d\theta &+ \cos\theta\cos\phi\;d\phi\\ dz &= &+\cos\theta\;d\theta \end{array}\right.$$ So $$\require{cancel} \begin{align} dy \wedge dz &= \left( -\sin\theta\sin\phi\; d\theta + \cos\theta\cos\phi\; d\phi \right) \wedge ( \cos\theta\; d\theta )\\ &=-\color{red}{\cancelto{0}{\color{gray}{\sin\theta\sin\phi\cos\theta\; d\theta\wedge d\theta}}} + \cos^2\theta\cos\phi\; d\phi \wedge d\theta\\ &= - \cos^2\theta \cos\phi\; d\theta\wedge d\phi \end{align} $$ and hence $$\begin{align} dx \wedge dy \wedge dz &= (\cos\theta\cos\phi\; d\rho - \rho\sin\theta \cos\phi\; d\theta - \rho\cos\theta\sin\phi\; d\phi )\wedge (-\cos^2\theta\cos\phi\; d\theta \wedge d\phi)\\ &= -\cos^2\theta\cos\phi \left( \begin{array}{ll} + \cos\theta\cos\phi\; d\rho \wedge d\theta \wedge d\phi \\ - \color{red}{\cancelto{0}{\color{gray}{\rho\sin\theta \cos\phi\; d\theta \wedge d\theta \wedge d\phi}}}\\ - \color{red}{\cancelto{0}{\color{gray}{\rho\cos\theta\sin\phi\; d\phi \wedge d\theta \wedge d\phi}}} \end{array} \right)\\ &= -\cos^3\theta\cos^2\phi\;d\rho \wedge d\theta \wedge d\phi \end{align} $$
In actual computation, one typically won't show all those terms which vanish because it contains some $d(\cdot)$ twice.
For an example, since $dz$ only contains $d\theta$, in computing $dy \wedge dz$, we will ignore the $d\theta$ component of $dy$ and directly writes down
$$dy \wedge dz = \cos^2\theta\cos\phi d\phi \wedge d\theta = \cdots$$
Similarly, since $dy \wedge dz$ only contains $d\theta \wedge d\phi$, in computation of $dx \wedge dy \wedge dz$, we only need to consider the $d\rho$ component in $dx$ and directly write down $$dx \wedge dy \wedge dz = (\cos\theta \cos\phi d\rho) \wedge (-\cos^2\theta\cos\phi \;d\theta \wedge d\phi) = \cdots$$