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I define a subset $M$ of $\mathbb R^n$ to be a "homogeneous Euclidean manifold" if:

  • it is a closed connected smooth submanifold of $\mathbb R^n$,
  • for every $p, q$ in $M$, there is a Euclidean isometry $f$ of $\mathbb R^n$ sending $p$ to $q$ fixing $M$ (i.e., $f(M)=M$ and $f(p)=q$).

The problem is to classify the "homogeneous Euclidean manifolds".

I know there are several definitions of symmetric and homogeneous and that these spaces have been more or less classified so it may well be that my question is just a particular case.

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    $\begingroup$ These are called homogeneous spaces. See Tom Dieck's book on "Representations of compact Lie groups". His manifolds are abstract, i.e. not embedded in Euclidean space. But in the book he shows homogeneous spaces admit equivariant embeddings into Euclidean space, so they're the same class of objects. $\endgroup$
    – Ryan Budney
    Commented Feb 18, 2022 at 19:21
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    $\begingroup$ @RyanBudney: Just one small clarification request: in the OP's post, $f$ is an ambient isometry, that does not necessarily restrict to an isometry of $M$. Doesn't this muddle your argument? For instance, it seems impossible to conceive the OP's definition intrinsically, without an ambient space. Furthermore, a given manifold might be a "symmetric submanifold" with respect to some embedding, and not be so with respect to another one. It seems that the OP's property depends heavily on the ambient space. $\endgroup$
    – Alex M.
    Commented Feb 18, 2022 at 19:57
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    $\begingroup$ @RyanBudney Two minor cavils: 1. Not all homogeneous spaces are symmetric. For example, all the left-invariant metrics on $\mathrm{SU}(2)$ are homogeneous, but only the ones with constant sectional curvature are symmetric. 2. That embedding result requires compactness. For example, the Poincaré disk (which is abstractly symmetric) cannot be embedded equivariantly with respect to its isometry group into Euclidean space of any (finite) dimension. $\endgroup$
    – Robert Bryant
    Commented Feb 18, 2022 at 21:55
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    $\begingroup$ @AlexM.: I suppose my comment should be seen as more of a coarse equivalence of these two notions. As Robert Bryant points out, the argument in Dieck's book is for homogenous spaces, so we are using the induced metric from them being a homogenous space. And the proof there requires compactness. Andrea Aveni: do let us know if the content of Tom Dieck's book is insufficient for your purposes. It would seem to get at the heart of your question. But perhaps I have misunderstood your intent. $\endgroup$
    – Ryan Budney
    Commented Feb 18, 2022 at 23:47
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    $\begingroup$ The symmetric space $\mathbb{CP}^2 = \mathrm{SU}(3)/\mathrm{U}(2)$ is not a product of spheres, and it is isometrically embedded in ${\frak su}(3)\simeq\mathbb{R}^8$ as a symmetric Euclidean submanifold. $\endgroup$
    – Robert Bryant
    Commented Feb 19, 2022 at 10:56

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Such $M$ are called extrinsically homogeneous submanifolds of euclidean space.

Reference: for example the book by Berndt-Console-Olmos, Submanifolds and holonomy.

The set of all euclidean isometries that fix $M$ is a closed subgroup $G$ of the group of euclidean motions $E(n)$, and $M$ is an orbit of $G$. So the question is to classify all (compact connected) orbits of all closed subgroups of $E(n)$.

Since $M$ is assumed to be bounded, it has a euclidean center of mass, which one can assume to be the origin. That center of mass is a common fixed point of $G$. So $G$ is a closed subgroup of the orthogonal group, and the orbit $M$ is contained in a sphere around the origin. These are examples of isoparametric submanifolds of the sphere (Edit: This is not correct, see the comment by Claudio Gorodsky).

Classification is probably asking too much. For example, all compact homogeneous spaces occur as such orbits in the sphere (Mostow).

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  • $\begingroup$ Thank you a lot! $\endgroup$
    – Andrea Aveni
    Commented Feb 27, 2022 at 5:21
  • $\begingroup$ Not every orbit contained in a sphere is an isoparametric submanifold. All principal orbits of a compact connected $G$ will be isoparametric in the sphere if and only if these orbits are orbits of an isotropy representation of a symmetric space, see the book cited in the answer. $\endgroup$
    – Claudio Gorodski
    Commented Mar 13, 2022 at 0:34

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