Let $f(x,y,z) := (xy, xz, yz)$. Compute the pullback $f^*\omega$, where $\omega$ runs through all wedge products of basis forms $dx$, $dy$, $dz$.
The definition of pullback that I know is:
The definition for the pullback of a $\mathcal{C}^{r+1}$ mapping $g: U \rightarrow V$ and a $\mathcal{C}^r$-smooth differential $p$-form $\omega$ in $V$ is
$$ (g^*\omega)_x(v_1,\ldots,v_p) = \omega_{g(x)}(Dg_x(v_1),\ldots,Dg_x(v_p))$$
First of all, I am not quite sure if I understand the statement of the problem correctly as the part
"where $\omega$ runs through all wedge products"
is a bit ambiguous to me. Does this mean that $\omega = dx \wedge dy \wedge dz$? If yes, then I think since $\omega$ is a $3$-form and the Jacobi-matrix is given by
$$(Df)_{(x,y,z)} =\begin{bmatrix}y &x &0\\ z &0 &x\\ 0 &z &y \end{bmatrix}$$
we have for $v_i = (x_i,y_i,z_i)$ and
$$a_i := D_f(v_i) =\begin{bmatrix} x_i+y_i \\x_i+z_i \\ y_i+z_i \end{bmatrix}$$
by definition of the pullback that
$$(f^*\omega)(v_1,v_2,v_3) = ((d_x \wedge d_y \wedge d_z))(v_1,v_2,v_3) = \det(D_f(v_1),D_f(v_2),D_f(v_3)). $$
, where the last equation follows from the fact that the wedge product of the canonical basis forms is the determinant.
Could you please tell me if this is correct so far?
EDIT: Using Masacroso's and peek-a-boo's suggestions I came up with the following:
$$f^*(dx) = d(f^*x) = d(x \circ f) = d(xy) = ydx + xdy$$
$$f^*(dy) = d(f^*y) = d(y \circ f) = d(xz) = zdx + xdz$$
$$f^*(dz) = d(f^*z) = d(z \circ f) = d(yz) = zdy + ydz$$
\begin{align}
f^*(dx \wedge dy) &= f^*(dx) \wedge f^*(dy) = (ydx + xdy) \wedge (zdx + xdz) \\
&= ydx \wedge zdx + ydx \wedge xdz + xdy \wedge zdx + xdy \wedge xdz
\end{align}
\begin{align}
f^*(dx \wedge dz) &= f^*(dx) \wedge f^*(dz) = (ydx + xdy) \wedge (zdy + ydz) \\
&= ydx \wedge zdy + ydx \wedge ydz + xdy \wedge zdy + xdy \wedge ydz
\end{align}
\begin{align}
f^*(dy \wedge dz) &= f^*(dy) \wedge f^*(dz) = (zdx + xdz) \wedge (zdy + ydz) \\
&= zdx \wedge zdy + zdx \wedge ydz + xdz \wedge zdy + xdz \wedge ydz
\end{align}
$$f^*(dx \wedge dy \wedge dz) = f^*(dz) \wedge f^*(dy) \wedge f^*(dz)$$
Is this correct or can we do more here?
Best Answer
HINT: use the fact that $$ \begin{align*} f^*\left(\bigwedge_{k=1}^n dx^k\right)&=\bigwedge_{k=1}^n f^*dx^k\\&=\bigwedge_{k=1}^n d(f^*x^k)\\&=\bigwedge_{k=1}^n d(x^k \circ f)\\&=\bigwedge_{k=1}^ndf^k\\ &=\bigwedge_{k=1}^n\left(\sum_{j=1}^m \partial_j f^k dx^j\right) \end{align*} $$ for $f:=(f^1,\ldots ,f^n)$. Example of solution for one case: using the above we have that $$ \begin{align*}f^*(dx\wedge dy)&=(d(xy))\wedge d(xz))=(xdy+ydx)\wedge (xdz+zdx)\\ &=-xzdx\wedge dy+x^2dy\wedge dz-xydz\wedge dx \end{align*} $$