What is the group of transformations $\subset SO(4)$ that sends every fibre circle to another fibre circle?

I think the Lie algebra might be generated by the bivectors


or, in terms of matrices,


but this is just a guess. And I don't know much about Lie algebras, or what the corresponding Lie group would be.

...Is it the intersection of $SO(4)$ with the symplectic group $Sp(4,\mathbb R)$ ? Does that have a name?


This was solved by Gluck, Warner, and Ziller in

Gluck, Herman & Warner, Frank & Ziller, Wolfgang. (1986). The geometry of the Hopf fibrations. L’Enseignement Mathématique. IIe Série. 32. 10.5169/seals-55085.


Here's a summary:

The symmetry group of the Hopf fibration $S^{2n+1}\rightarrow \mathbb{C}P^n$ is $U(n+1) \cup c U(n+1)$ where $c:\mathbb{C}^{n+1}\rightarrow \mathbb{C}^{n+1}$ is conjugation: $c(z_1,.., z_{n+1}) = (\overline{z}_1,..., \overline{z}_{n+1})$. For "the" Hopf fibration $S^3\rightarrow S^2$, the symmetry group is $U(2)\cup cU(2)\subseteq SO(4)$.

The symmetry group of the Hopf fibration $S^{4n + 3}\rightarrow \mathbb{H}P^n$ is $Sp(n+1)\times S^3/\langle (-I, -1)\rangle$.

The symmetry group of the Hopf fibration $S^{15}\rightarrow S^8$ is $Spin(9)$ acting by its spin representation.

  • $\begingroup$ In terms of real matrices, $U(2)=O(4)\cap Sp(4,\mathbb R)$. (This is exactly what I guessed, except I ignored orientation-reversal.) What is $cU(2)$ ? $\endgroup$ – mr_e_man Nov 7 '18 at 19:59
  • $\begingroup$ The symplectic group has determinant $1$, so this intersection is the same as $U(2)=SO(4)\cap Sp(4,\mathbb R)$. The conjugate $c$ with $n+1=2$ rotates 180 degrees, reversing the direction of the circles, or negating the symplectic form. I was thinking of transformations that preserve the symplectic form. That explains my missing $cU(2)$. $\endgroup$ – mr_e_man Nov 7 '18 at 20:39
  • $\begingroup$ It seems like you figured it out, but $cU(n)$ means the composition of something in $U(n)$ with $c$. (In fact, you can also write this as $U(n)c$ since $c$ normalizes $U(n)$ in the sense that for any $A\in U(n)$, $c\circ A \circ c^{-1} \in U(n)$ $\endgroup$ – Jason DeVito Nov 7 '18 at 20:58
  • $\begingroup$ Yes, I knew that it meant composition. I was asking what it would look like with real instead of complex matrices. $\endgroup$ – mr_e_man Nov 7 '18 at 21:13

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