# “Homogeneity” of the Hopf fibration $S^7\to S^{15}\to S^8$ [closed]

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.

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.

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## closed as no longer relevant by Ryan Budney, Allen Knutson, Renato G Bettiol, Deane Yang, Lee MosherNov 29 '12 at 16:57

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Dear Renato, I am probably just confused, but how does $k \cdot g H := g k^{-1} H$ give a well-defined action? If we replace $g$ by $gh$ for some $h$ in $H$ (which doesn't change the coset), then you would need $g h k^{-1}$ and $g k^{-1}$ to represent the same $H$-coset, so you would need $k h k^{-1}$ to be an element of $h$. But surely $H$ is not normal in $K$. Regards, –  Emerton Nov 28 '12 at 23:36
@Emerton: You are right, that is precisely the problem! A few minutes after posting the question on MO I got an email from Wolfgang Ziller (to whom I asked the question earlier) and he pointed out that the statement in his survey is actually imprecise, in that this action is well-defined only if $K$ is in the normalizer of $H$. That is not the case for the Spin groups, solving the mystery. Sorry for the mess -- I suppose the question can be closed then, I am not sure if I should delete it or if it should be kept for future reference... If any moderator sees this, please let me know what to do. –  Renato G Bettiol Nov 28 '12 at 23:43
There's no harm keeping this question around as simply closed, if you're okay with that. The question contains interesting mathematics so some people might eventually find it useful. If you prefer it to be deleted, we can do that. It's up to you. –  Ryan Budney Nov 28 '12 at 23:48
@Ryan: Thank you! I will vote to close the question then, and, if possible, leave it here for future reference. This issue confused me for a couple days, and I suppose it might be helpful for someone else... –  Renato G Bettiol Nov 29 '12 at 2:20
Instead of closing, Renato might answer it himself, or ask Ziller to do it. –  Mariano Suárez-Alvarez Nov 29 '12 at 3:13