Post Closed as "no longer relevant" by Ryan Budney, Allen Knutson, Renato G Bettiol, Deane Yang, Lee Mosher
2 Added reason why apparent contradiction is false; will now close the question
• 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; as well as papers of Schwachhofer-Tapp, Kerr-Kollross and others.

• Although

Edit: Shortly after posting this question, I am almost sure that received an email from Ziller in which he answers the statement question. Statement (1) is correct, and that the problem with is in statement (2) is 2), as suspected. In fact, the claim that somehow this abstract construction with nested Lie groups the $H\subset K\subset G$ does not apply to K$-orbits are always fibers of$Spin(7)\subset Spin(8)\subset Spin(9)$G/H\to G/K$ is false in general, I still see no reason why this unless $H$ is a normal subgroup of $K$. This is because the case. The suspicion lies on action by multiplication (by the triality automorphism that follows inverse) on the usual inclusion to give right on cosets defined above is only well-defined (i.e., independent of choice of coset representative) if $Spin(7)\subset Spin(8)$, but as long as this H$is normal in$K$, as was also pointed out in Emerton's comment. The rest of the claims in (2) are correct. As a fixed automorphism side note, for the other Hopf fibrations$\phi$, one could pick S^1\to S^{2n+1}\to \mathbb CP^n$ and $S^3\to S^{4n+3}\to \mathbb H P^n$ the corresponding subgroup $H$ as is normal in $K$ -- after all the image under fiber $\phi$ of K/H$is a group -- and the usual$Spin(7)$K$-action is hence well-defined, so statement (2) holds in full for these cases. In general, however, the $Spin(8)$, and this seems a perfectly valid subgroup K$-orbits are not fibers of$K$to proceed with G/H\to G/K$, as the construction$S^7\to S^{15}\to S^8$ example illustrates.What am I missing here?

1

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

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):

1. 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.

2. 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; as well as papers of Schwachhofer-Tapp, Kerr-Kollross and others.

Although I am almost sure that the statement (1) is correct, and that the problem with statement (2) is that somehow this abstract construction with nested Lie groups $H\subset K\subset G$ does not apply to $Spin(7)\subset Spin(8)\subset Spin(9)$, I still see no reason why this is the case. The suspicion lies on the triality automorphism that follows the usual inclusion to give $Spin(7)\subset Spin(8)$, but as long as this is a fixed automorphism $\phi$, one could pick $H$ as the image under $\phi$ of the usual $Spin(7)$ in $Spin(8)$, and this seems a perfectly valid subgroup of $K$ to proceed with the construction. What am I missing here?

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.