I am looking for deformations of the 4sphere with 8dimensional isometry group, like a 4dimensional Berger sphere.

1$\begingroup$ Are you looking for a continuous family of Riemann metrics near the round metric on $S^4$ such that the isometry group is $8$ dimensional? $\endgroup$– Ryan BudneyMar 19 at 8:41

$\begingroup$ Yes, exactly. Thanks $\endgroup$– Thomas SchuckerMar 19 at 8:44

$\begingroup$ $SO(5)$, which is the (identity component of) the isometry group of $S^4$, doesn't have any closed subgroups of dimension $8$: any such subgroup $K$ would give rise to an effective presentation ($SO(5)$ is simple) of a compact homogeneous space $M^2 = SO(5)/K$. The same argument works for $7$. $\endgroup$– Ramiro LafuenteMar 21 at 3:07

$\begingroup$ @RamiroLafuente: It's not clear that there isn't some Riemannian metric on $S^4$ whose isometry group has dimension $8$ but is not isomorphic to a subgroup of $\mathrm{SO}(5)$. $\endgroup$– Robert Bryant2 days ago

$\begingroup$ I agree, your answer is much more general, that’s why I only include this as a comment. But for deformations of the round metric the inclusion in $SO(5)$ should hold, right? $\endgroup$– Ramiro Lafuente2 days ago
1 Answer
There cannot be an 8dimensional group $G$ acting effectively on $S^4$ by Riemannian isometries. The following argument may not be the best, but it explains why this is true. (I will assume that $G$ acts effectively, since, otherwise, we can quotient by the closed subgroup $K\subset G$ that acts trivially, and work with $G/K$, which does act effectively, instead.)
First, such a $G$ cannot act transitively on $S^4$, since the effective transitive Lie group actions on spheres were classified by Borel, and the only effective transitive compact connected group action on $S^4$ is the standard one with $G = \mathrm{SO}(5)$, a 10dimensional Lie group. [If you don't like relying on Borel's classification, here is a 'lowtech' argument to rule out the transitive case: If a connected compact $d$dimensional group $G$ acts transitively and effectively on $S^4$, then we can write $S^4 = G/K$, where $K$ is the $G$stabilizer of a point. This $K$ must be a compact connected subgroup of $\mathrm{SO}(4)$ of dimension $d{}4$. All the proper connected subgroups of $\mathrm{SO}(4)$ are conjugate to a subgroup of $\mathrm{U}(2)$. So, if $K$ were of dimension less than $6$, then $S^4$ would have a $G$invariant, almost complex structure. However, $S^4$ does not support an almost complex structure.]
Now look at a generic $G$orbit $\Sigma$, which will be a smooth submanifold. If it is codimension $1$, then $G$ acts by isometries on $\Sigma$ and the normal subgroup $H\subset G$ that fixes $\Sigma$ pointwise is discrete. However, the dimension of $G/H$ can be at most $6$ (since it acts effectively on a compact Riemannian $3$manifold), so $G$ has dimension at most $6$. Similarly, if $G$ fixes a point of $S^4$, then $G$ has dimension at most $6$.
If all of the $G$orbits were of dimension $1$ or of dimension $2$ then this would define a nontrivial splitting of the tangent bundle of $S^4$, which does not exist.
Thus, the $G$orbits of an $8$dimensional Lie group acting by isometries on $S^4$ endowed with some Riemannian metric would have dimension either $1$ or $2$ and there would have to be some of each dimension. A 'generic' $G$orbit $\Sigma$ would be of dimension $2$. If $H\subset G$ is the normal subgroup that fixes $\Sigma$ pointwise, then $H$ has dimension at most $1$ and $G/H$ has dimension at most $3$, so $G$ has dimension at most $4$.