This is a question about cohomogeneity one actions of a compact Lie group $G$ on a Riemannian manifold $M$, such that $G$ acts via isometries. Many articles have been published about classifications of such actions. Different articles classify actions up to different notions of equivalence. Here are two examples:

  1. Classification up to $G$-diffeomorphism (i.e. there exists a diffeomorphism $M \rightarrow M$ that is $G$ equivariant, i.e. $f(g \cdot_1 x)=g \cdot_2 f(x)$, where $\cdot_1$ and $\cdot_2$ denote two different actions of $G$ on $M$): Karsten Grove, Burkhard Wilking & Wolfgang Ziller: POSITIVELY CURVED COHOMOGENEITY ONE MANIFOLDS AND 3-SASAKIAN GEOMETRY

  2. Classification up to isometric orbit equivalence (i.e. there exists an isometry $f:M \rightarrow M$ such that $G \cdot_1 f(x)=f(G \cdot_2 x)$): J. Berndt M. Domínguez-Vázquez: COHOMOGENEITY ONE ACTIONS ON SOME NONCOMPACT SYMMETRIC SPACES OF RANK TWO

  3. The notion of "$G$-isometry" (replace the diffeomorphism in point 1 by an isometry) would be the most natural to me, but I haven't found it in the literature.

Question 1: Are these notions identical?

Question 2: One application of a classification of cohomogeneity one actions was to find an exotic nearly Kähler structure on $S^6$ (Lorenzo Foscolo, Mark Haskins: New $G_2$-holonomy cones and exotic nearly Kähler structures on $S^6$). They used a classification of cohomogeneity one actions on $S^6$ up to $G$-diffeomorphism. Could it happen that there is another, hidden exotic nearly Kähler structure on $S^6$ which arises from a $G$-diffeomorphic cohomogeneity one action on $S^6$ that is not $G$-isometric? (That would lead to two nearly Kähler structures that are not isometric)


First, for the isometry vs diffeomorphism issue, diffeomorphism is more natural, in some sense. For example, if one considers the standard $S^1$ action on $S^2$ by rotations, there are many metrics on $S^2$ for which this action is isometric - if one deforms $S^2$ into a football shape, the $S^1$ action is isometric.

More generally, any way of getting something diffeomorphic to $S^2$ as a surface of revolution admits an isometric cohomogeneity one action by $S^1$.

On the other hand, all these seemingly different pictures are equivariantly diffeomorphic.

This same idea works on any cohom 1 manifold $M$ which is compact. Specifically, the quotient space $M/G$ is either $S^1$ or $[0,1]$. In the first case, $M$ looks like a $G/H$ bundle over $S^1$ for some homogeneous space $G/H$. One can scale the metric on the $G/H$ fibers by different amounts as one traverses the circle, so there are always many $G$-invariant metrics. In the second case, $M$ is a union of two disc bundles with boundary a homogeoneous space $G/H$. That is, $M = G/H\times [0,1]$ with some quotienting happening at both end points. Then one must be careful with the metric near the boundary (to keep everything smooth), but can freely scale the metric on $G/H$.

(In addition, often $G/H$ has many families of deformations of metrics with are all $G$-invariant. This would have to be dealt with in any classification up to $G$-isometry).

Now, for question 1, no, these are not the same. For a really dumb example, suppose $G$ acts on $M$ via a cohomogeneity 1 action: $g\ast m = gm$. Consider the two different $G\times G$ actions on $M$: $(g_1,g_2)\ast_1 m = g_1 m$ and $(g_1,g_2)\ast_2 m = g_2 m$. Trivially, these actions are orbit equivalent, but they are not $G\times G$-equivariant.

To see that they are not $G$-equivariant, let $f$ be a $G$-equivariant diffeomorphism. Choose $m\in M$ for which $f(m)$ is not fixed by $G$, and let $e\in G$ denote the identity. Then for any $g_2\in G$, $f(m) = f(m) = f((e,g_2) \ast_1 m) = (e,g_2)\ast_2 f(m)$, so $f(m)$ is fixed by $G$.

For a slightly less dumb idea, consider the $S^1$ action on $S^2$ via rotations compared to the $S^1$ action on $S^2$ via rotations going twice as fast. Then these are clearly orbit equivalent (orbits are lines of latitude) but they are not $G$-equivariant since, for example, the kernels of the actions are different.

(I do not know of a non-"dumb" idea in the cohom 1 case which distinguishes the notions of $G$-equivariant from orbit-equivalent.)

For question 2, I don't know.


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