I embed the 2-sphere and the 3-sphere in $\mathbb{R}^3$ and $\mathbb{R}^4$ respectively. Then denote by $\{x_1,x_2,x_3,x_4\}$ the coordinates on the 3-sphere and $\{y_1,y_2,y_3\}$ on the 2-sphere. So $\sum_{i=1}^4 x_i^2 = 1 = \sum_{i=1}^3 y_i^2$.

Now consider the 2-forms $\omega=dx_1\wedge dx_2+dx_3\wedge dx_4$ on $S^3$ and $\alpha=y_1 dy_2\wedge dy_3+y_2 dy_3\wedge dy_1 + y_3 dy_1\wedge dy_2$ on $S^2$.

I wish to show that when pulling back $\alpha$ to $S^3$ using the Hopf map $H:S^3\rightarrow S^2$ given by

$(x_1,x_2,x_3,x_4)\mapsto \left(x_1^2+x_2^2-x_3^2-x_4^2, 2(x_2x_3 - x_1 x_4), 2(x_2x_4+x_1x_3)\right)$,

the result is proportional to $\omega$, so:

$H^*\alpha=c\omega$ for some $c\in\mathbb{R}$.

I tried starting from the left hand side, so by definition of the pullback I just wrote down the form $\alpha$ with the $y_i$ replaced by the $i^{th}$ component of $H(x_1,x_2,x_3,x_4)$, and then I used $\sum_{i=1}^4 x_i^2 = 1$ and $\sum_{i=1}^4 x_idx_i = 0$. This however became a huge mess and nearly impossible to do by hand...

Is there an easier way? Is there a way to take $H$ to the right hand side, using some properties of the pull back? Then I could start from the right hand side, which might be easier.

  • 2
    $\begingroup$ Since $S^3 \to S^2$ is a principal $S^1$ bundle, the map $H^* : \Omega^\bullet(S^2)\to \Omega^\bullet(S^3)$ gives an iso between $\Omega^\bullet(S^2)$ and the basic forms on $S^3$ (those forms $\mu$ which satisfy $i_v \mu = 0$ for all $v \in \ker H_*$ and $R_z^* \mu = \mu$ for all $z \in S^1$). $S^2$ is oriented and two-dimensional, so there is a unique basic 2-form on $S^3$ up to multiplication by functions pulled back from a function on $S^2$. So if you can show that $\omega$ is basic then you get $H^* (f \alpha) = \omega$ for some $f \in C^\infty(S^2)$. Not sure how to show f is const. $\endgroup$ Commented Jan 30, 2014 at 15:16
  • $\begingroup$ Try using complex coordinates: $z_1=x_1+ix_2, z_2=x_3+ix_4$, $w=y_3+iy_2$, and say $y=y_1$. Then $y=|z_1|^2-|z_2|^2, w=2z_1\bar z_2.$ $\endgroup$
    – Gil Bor
    Commented Jan 31, 2014 at 4:00

1 Answer 1


Your 2-form on the 2-sphere is the area form, while the 2-form on the 3-sphere is the contact form. The point is that both are invariant under a suitable group of transformations. In the case of the 2-sphere it is SO(3) (at least), whereas in the case of the 3-sphere it is the restriction of SU(2) (at least). Both of these act transitively. Use this to reduce the general calculation to the calculation at a single point, say $(0,0,1)$ of the 2-sphere.

  • $\begingroup$ I get the idea, I think... Though I'm not familiar with contact forms and I only have an $SO(2)$ action given on $S^3$, under which $\omega$ is indeed invariant. I think I can use this to make one coordinate on $S^3$ 0, or maybe even 2, at any rate it will simplify things... Thanks. $\endgroup$ Commented Jan 31, 2014 at 0:46
  • $\begingroup$ The 2-form on $S^3$ is the restriction of the standard symplectic 2-form on $\mathbb{C}^2$ defined by the complex structure, so it is invariant under the unitary transformations. $\endgroup$ Commented Jan 31, 2014 at 8:39
  • $\begingroup$ I'm not sure if I understand. How exactly is $\alpha$ invariant under the transitive $SO(3)$ action? Clearly for $(0,0,1)$ and $(1,0,0)$ $\alpha$ looks different, right? I'm missing something, I only know what it means for a form in $\Omega(M,\mathfrak{g})$ to be $G$-invariant, but $\alpha$ is not an $\mathfrak{SO(3)}$ valued form I think... $\endgroup$ Commented Jan 31, 2014 at 10:13

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