"Homogeneity" of the Hopf fibration $S^7\to S^{15}\to S^8$

2012-11-28T23:20:02Z

My question has to do with an apparent contradiction I get regarding the Hopf fibration $S^7\to S^{15}\to S^8$. Namely, the two following statements cannot be true at the same time (but I do not see any problem with any of them):

The Hopf fibration $S^7\to S^{15}\to S^8$ is not homogeneous, i.e., there is no isometric group action on the round sphere $S^{15}$ whose orbits are the Hopf fibers. This is claimed by Guijarro-Walschap, Corollary 3.2; and was also previously observed, e.g., by Gromoll-Grove.
The Hopf fibration $S^7\to S^{15}\to S^8$ is of the form $K/H \to G/H \stackrel{\pi}{\to} G/K$, where $\pi(gH)=gK$ and $H< K < G$ are the groups $Spin(7)$, $Spin(8)$ and $Spin(9)$, respectively. The inclusion of $K$ in $G$ is the usual one; however, the inclusion of $H$ in $K$ is the usual one followed by a nontrivial triality automorphism of $Spin(8)$, see, e.g., Section 4 of this paper or Besse's book "Einstein manifolds", p. 258, 9.84 Example 4. Such an automorphism is outer, and is not the restriction of any automorphism of $Spin(9)$. Now, consider the $K$-action on $G/H$ given by $k\cdot gH:=gk^{-1} H$. Its orbits are clearly of the form $gKH\subset G/H$, and I claim these are exactly the fibers of $G/H\to G/K$, i.e., the Hopf fibers. Indeed, if both $aH$ and $bH$ are mapped to $gK$ under the projection $G/H\to G/K$, then $aK=bK$, i.e., $b^{-1}a\in K$, which means $a\in bK$; so the subset of $G/H$ that get mapped to $gK$ is exactly $(gK)H$; and these were the $K$-orbits. The above construction (for abstract Lie groups $H, K, G$; not specifically for the Hopf fibration as above) shows up in many places, e.g., in Ziller's survey, p. 16.

Edit: Shortly after posting this question, I received an email from Ziller in which he answers the question. Statement (1) is correct, and the problem is in statement (2), as suspected. In fact, the claim that the $K$-orbits are always fibers of $G/H\to G/K$ is false in general, unless $H$ is a normal subgroup of $K$. This is because the action by multiplication (by the inverse) on the right on cosets defined above is only well-defined (i.e., independent of choice of coset representative) if $H$ is normal in $K$, as was also pointed out in Emerton's comment. The rest of the claims in (2) are correct. As a side note, for the other Hopf fibrations $S^1\to S^{2n+1}\to \mathbb CP^n$ and $S^3\to S^{4n+3}\to \mathbb H P^n$ the corresponding subgroup $H$ is normal in $K$ -- after all the fiber $K/H$ is a group -- and the $K$-action is hence well-defined, so statement (2) holds in full for these cases. In general, however, the $K$-orbits are not fibers of $G/H\to G/K$, as the $S^7\to S^{15}\to S^8$ example illustrates.

I am not sure how useful these comments are, but here is some more information about a $Spin(8)$ action on $S^{15}$. The representation $\rho_8\oplus\Delta^\pm_8$ of $Spin(8)$ in $R^{16}$ gives a cohomogeneity one action on the unit sphere $S^{15}$, and the orbit space is the interval $[0,\pi/2]$. This action has two singular orbits, whose isotropy is $Spin(7)$, and the principal isotropy is $G_2$. The inclusion of the singular isotropies is the usual one followed by a nontrivial triality automorphism that is $\pm$, according to the choice $\Delta^\pm_8$ of spinorial representation. The principal orbits of this $Spin(8)$ action have dimension $14$, and are of course not Hopf fibers. The singular orbits, however, give a pair of antipodal $S^7$'s inside $S^{15}$ and these are Hopf fibers. [Note this action is different from the action $k\cdot gH=gk^{-1}H$ described in (2) above.] From what I have heard, the only subactions of the transitive $Spin(9)$ action on $S^{15}$ that preserve a fixed Hopf fiber (and hence its antipodal fiber as well) are actions by some $Spin(8)\subset Spin(9)$ conjugate to the one I have just described. The $Spin(9)$ action on $S^{15}$ is $g_1\cdot g_2H=g_1g_2H$, so the action described in (2) is not a restriction of it; however the statements in (1) seem to not specify any particular action on $S^{15}$, i.e., to my understanding, they show no group can act isometrically and have orbits that are precisely the Hopf fibers.

"Homogeneity" of the Hopf fibration $S^7\to S^{15}\to S^8$ - MathOverflow [closed]

"Homogeneity" of the Hopf fibration $S^7\to S^{15}\to S^8$