Suggestion for mistake on computation of a pull-back $2$-form on the sphere

Question

Let $f : S^3 \to S^2$ be the function given by $$ f(x , y , z , t) = (p = (2 (x z + y t) , q = 2 (- x t + y z) , r = - x^2 - y^2 + z^2 + t^2). $$ Let $\omega = (p \, dq \wedge dr + q \, dr \wedge dp + r \, dp \wedge dq)$ be a $2$-form in $S^2$. How can I show that $f^*(\omega) = 4(dx \wedge dy + dz \wedge dt) \in {\Omega}^2(S^3)$? My attempt is the next: $$ f^*(\omega) = p f^*(dq) \wedge f^*(dr) + q f^*(dr) \wedge f^*(dp) + r f^*(dp) \wedge f^*(dq). $$ For instance, $$ f^*(dq) = \frac{\partial q}{\partial x} \, dx + \frac{\partial q}{\partial y} \, dy + \frac{\partial q}{\partial z} \, dz + \frac{\partial q}{\partial t} \, dt = 2 (- t \, dx + z \, dy + x \, dz + y \, dt). $$ If we continue with this process, do we reach that $f^*(\omega) = 4(dx \wedge dy + dz \wedge dt)$? I guess no, because $$ p f^*(dq) \wedge f^*(dr) = 2 (x z + y t) 2 (- t \, dx + z \, dy + x \, dz + y \, dt) \wedge 2 (- x \, dx - y \, dy + z \, dz + t \, dt) = $$ $$ = \ldots = 8 (x z + y t) (t y + x z) \, dx \wedge dy + \ldots $$ (the coefficient of $dx \wedge dy$ in the expression of $p f^*(dq) \wedge f^*(dr)$ is $8 (x z + y t) (t y + x z)$. If we take the coefficient of $dx \wedge dy$ from $q f^*(dr) \wedge f^*(dp)$ and $r f^*(dp) \wedge f^*(dq)$, we obtain $$ 8 (- x t + y z) (- x t + y z) \quad \text{and} \quad 4 (- x^2 - y^2 + z^2 + t^2) (z^2 + t^2). $$ Then using that $x^2 + y^2 + z^2 + t^2 = 1$ in $S^3$, we have that $$ f^*(\omega) = (8 (x z + y t) (t y + x z) + 8 (- x t + y z) (- x t + y z) + 4 (- x^2 - y^2 + z^2 + t^2) (z^2 + t^2)) \, dx \wedge dy + \ldots = \ldots = 4 (z^2 + t^2) \, dx \wedge dy + \ldots $$ if I am not wrong, and it is not compatible with $f^*(\omega) = 4(dx \wedge dy + dz \wedge dt)$. Where am I getting mistake? Also I have to keep in mind that $x \, dx + y \, dy + z \, dz + t \, dt = 0$ but it is because $x^2 + y^2 + z^2 + t^2 = 1$.

Answer 1

The result $f^*(\omega) = 4(dx\wedge dy + dz\wedge dt)$ is correct. This is actually a rather well-known fact, since your $f$ is just the Hopf fibration.

The next step in the calculation is to use the relation $x\, dx + y\, dy + z\, dz + t \, dt = 0$, as you've noted. Specifically, with the $dx\wedge dy$ term, you can convert the coefficient $z^2+t^2$ to $1-x^2-y^2$. The $- x^2$ term can be rewritten using this relation as $$ - x (x\, dx\wedge dy) = x (z\, dz\wedge dy + t\, dt\wedge dy) = - xz\, dy\wedge dz - xt\, dy \wedge dt\ . $$ Similarly the $-y^2$ term equals $yz \, dx\wedge dz + yt\, dx\wedge dt$. You can also work out that the coefficient of the $dz\wedge dt$ term is $4(x^2+y^2)$, to which we can apply the same procedure and generate a few more terms. If you finish computing the coefficients of the other terms (through blood sweat and tears, or alternatively, with Mathematica) in $f^*(\omega)$, you will find that they actually all get cancelled by these extra terms we generated, and what's left will be the desired result.

Answer 2

Thanks to Eliot Yu, I could build the answer for the question. The whole expression for $p f^*(dq) \wedge f^*(dr)$ is $$ p f^*(dq) \wedge f^*(dr) = 2 (x z + y t) 2 (- t \, dx + z \, dy + x \, dz + y \, dt) \wedge 2 (- x \, dx - y \, dy + z \, dz + t \, dt) = $$ $$ = 8 (x z + y t) ((t y + z x) \, dx \wedge dy + (- t z + x y) \, dx \wedge dz + (- t^2 - x^2) \, dx \wedge dt + $$ $$ + (z^2 + y^2) \, dy \wedge dz + (z t - x y) \, dy \wedge dt + (y t + x z) \, dz \wedge dt). $$ We display $(- t z + x y) \, dx \wedge dz$ using $x \, dx + y \, dy + z \, dz + t \, dt = 0$ and the multilineality from $\wedge$ as he said to get: $$ (- t z + x y) \, dx \wedge dz = - t \, dx \wedge (z dz) + y \, (x \, dx) \wedge dz = $$ $$ = - t \, dx \wedge (- x \, dx - y \, dy - t \, dt) + y (- y \, dy - z \, dz - t \, dt) \wedge dz = $$ $$ = y t \, dx \wedge dy + t^2 \, dx \wedge dy - y^2 \, dy \wedge dz + y t \, dz \wedge dy. $$ Similarly, $$ (z t - x y) \, dy \wedge dt = x z \, dx \wedge dy - z^2 \, dy \wedge dz + x^2 \, dx \wedge dt + x z \, dz \wedge dt. $$ If we replace these two values in the expression obtained for $p f^*(dq) \wedge f^*(dr)$ and computing we get \begin{equation}\label{1}\tag{1} p f^*(dq) \wedge f^*(dr) = 16 {(x z + y t)}^2 \, (dx \wedge dy + dz \wedge dz). \end{equation} Similarly we can obtain easier expressions for $q f^*(dr) \wedge f^*(dp)$, \begin{equation}\label{2}\tag{2} q f^*(dr) \wedge f^*(dp) = 16 {(- x t + y z)}^2 \, (dx \wedge dy + dz \wedge dz) \end{equation} and for $r f^*(dp) \wedge f^*(dq)$, \begin{equation}\label{3}\tag{3} r f^*(dp) \wedge f^*(dq) = 4 {(- x^2 - y^2 + z^2 + t^2)}^2 \, (dx \wedge dy + dz \wedge dz). \end{equation} Then using \eqref{1}, \eqref{2} and \eqref{3}, $$ f^*(\omega) = 4 \left({(x z + y t)}^2 + {(- x t + y z)}^2 + {(- x^2 - y^2 + z^2 + t^2)}^2\right) \, (dx \wedge dy + dz \wedge dz) = $$ $$ = 4 (p^2 + q^2 + r^2) \, (dx \wedge dy + dz \wedge dz) = 4 \, (dx \wedge dy + dz \wedge dz) $$ because $p^2 + q^2 + r^2 = 1$, as $f(S^3) \subset S^2$ (is well defined).