Differential form computation

Question

I was reading a text on differential geometry and I noticed this: "Given $\omega=xdy\wedge dz+ydz\wedge dx+zdx\wedge dy\in\Omega^2(S^2)$ with $S^2=\{p\in\mathbb{R}^3:\lvert p\rvert=1\}$, show that $A^{*}\omega=(\det A)\omega$ for $A\in O(3)$".

I know some basic computation rules of differential forms but this really stacks me. Any help?

Answer 1

The function $A:S^2\to S^2$ is a rotation or reflection. In any case, $A$ is given by multiplication by some orthogonal matrix

$$ M = \begin{bmatrix} m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ m_{31} & m_{32} & m_{33} \end{bmatrix} $$ That is, $$A(a,b,c)=(m_{11}a+m_{12}b+m_{13}c,m_{21}a+m_{22}b+m_{23}c,m_{31}a+m_{32}b+m_{33}c)$$

Now the funcion $x:S^2\to\mathbb R$ is the "extract the first coordinate" function, i.e. $x(a,b,c)=a$. Hence $(x\circ A)(a,b,c)=x(A(a,b,c))$ is the first coordinate of $A(a,b,c)$: $$(x\circ A)(a,b,c)=m_{11}a+m_{12}b+m_{13}c$$.

Does that help?

Answer 2

You have abused notation since really $\beta = x \,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy \in \Omega^{2}(\mathbb{R}^3)$. What is probably meant is $\omega = i_{S^2}^*\beta$. Anyway, it is convenient to compute $A^*\beta$ first. The formula for the interior product $⌋$ yields that $$\beta = (dx \wedge dy \wedge dz) ⌋ R,$$ where $R : \mathbb{R}^3 \to \mathbb{R}^3$ is the radial vector field $R(x, y, z) = (x, y, z)$. Using a simple identity for the pullback of an interior product, $$A^*\beta = A^*(dx \wedge dy \wedge dz) ⌋ A_{\#}R,$$ where $A_{\#}R$ is the "pullback" of the vector field $R$ by $A$. We have $A_{\#}R(v) = A^{-1}R(Av) = A^{-1}Av = v$. Also $A^*(dx \wedge dy \wedge dz) = \det(A)\,dx \wedge dy \wedge dz$. Hence $$A^*\beta = \det(A) \beta.$$ Now pull back both sides by $i_{S^2}$, and after some manipulation of the abusive notation you get the result $$A^*\omega = \det(A)\omega.$$ Here $A$ is being regarded as a map $A : S^2 \to S^2$, whereas in the previous equation $A$ was being regarded as a map $A : \mathbb{R}^3 \to \mathbb{R}^3$. Really the $A$ in the above equation is $A \circ i_{S^2} : S^2 \to S^2$.