Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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. –  Eric O. Korman Jan 30 '14 at 15:16
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 $y=|z_1|^2-|z_2|^2, w=2z_1\bar z_2.$ –  Gil Bor Jan 31 '14 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.

share|improve this answer
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. –  ScroogeMcDuck Jan 31 '14 at 0:46
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. –  user72694 Jan 31 '14 at 8:39
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... –  ScroogeMcDuck Jan 31 '14 at 10:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.