# Deformations of the 4-sphere with 8-dimensional isometry groups

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

• 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? Mar 19 at 8:41
• Yes, exactly. Thanks Mar 19 at 8:44
• $SO(5)$, 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 $7$. Mar 21 at 3:07
• @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 $\mathrm{SO}(5)$. 2 days ago
• 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? 2 days ago

There cannot be an 8-dimensional 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 10-dimensional Lie group. [If you don't like relying on Borel's classification, here is a 'low-tech' 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 $$4$$.