# Symmetry group of Hopf fibration

https://en.wikipedia.org/wiki/Hopf_fibration

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

$$(e_1e_2),(e_3e_4),(e_1e_3+e_2e_4),(e_1e_4+e_3e_2)$$

or, in terms of matrices,

$$\begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix},\begin{bmatrix}0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&1&0\end{bmatrix},\begin{bmatrix}0&0&-1&0\\0&0&0&-1\\1&0&0&0\\0&1&0&0\end{bmatrix},\begin{bmatrix}0&0&0&-1\\0&0&1&0\\0&-1&0&0\\1&0&0&0\end{bmatrix}$$

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.

• 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 $cU(2)$ ? – mr_e_man Nov 7 '18 at 19:59
• 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 $cU(2)$. – mr_e_man Nov 7 '18 at 20:39
• 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 $A\in U(n)$, $c\circ A \circ c^{-1} \in U(n)$ – Jason DeVito Nov 7 '18 at 20:58
• Yes, I knew that it meant composition. I was asking what it would look like with real instead of complex matrices. – mr_e_man Nov 7 '18 at 21:13