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So this is my question :

Let $M$ be a smooth manifold. With any riemaniann metric $g$ on $M$ comes an isometry group $I(g)$. Intuition (well at least mine, which may be flawed...) suggests that there should be a metric $g$, not necessarily unique, that gives maximal symmetry to $M$ in the following sense :

if $g'$ is any riemannian metric on $M$ then there is an injective morphism from $I(g')$ to $I(g)$.

The example I have in mind is the sphere $S^2$ : it seems clear that the usual metric gives maximal symmetry and that any other metric would either give the same group of symmetry or one strictly smaller.

Does anyone know if this is true or not ? If so can you give me the argument or a reference ?

Thanks !

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up vote 3 down vote accepted

I realize that this is an old question, but here is a counter example. Consider metrics on two-dimensional tori. Namely, if you take the metric which is the product of two circles of equal length then the group of symmetries is the semidirect product of the abelian group, product of $T^2$ and the dihedral group of order 8. However, you can also take the flat torus whose isometry group is the semidirect product of $T^2$ and the dihedral group of order 12. (This metric corresponds to the tiling of the plane by equilateral triangles.) It is not hard to see, just from the classification of Euclidean crystallographic groups, that the torus does not admit a metric whose isometry group would contain both groups described above (the metric would have to be flat, which is then easy to rule out by the classification theorem).

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thanks a lot for digging through to this old question ! – Glougloubarbaki Sep 1 '14 at 11:41

This is not quite an answer to your question as you posed it, but there does exist a nice generalization of the standard metric on the 2-sphere: homogeneous spaces. Let's interpret "maximal symmetry" to mean the following: given any two points $x$ and $y$, there exists an isometry $\phi$ such that $\phi(x) = y$ (i.e. the group of isometries acts transitively). This is maximal symmetry in the sense that all the points look the same. Then it immediately follows that $M = G/H$, where $G$ is the group of isometries and $H$ is some subgroup of $G$ (to realize this, just pick $x \in M$ and let $H$ be the stabilizer of $x$). In the case of $S^2$, it is the homogeneous space $\mathrm{SO}(3)/\mathrm{SO}(2)$. However, it is not the case that every manifold is diffeomorphic (or even homeomorphic) to a homogeneous space (of finite-dimensional Lie groups), so this does not cover everything.

You might consider looking into some geometric flows (Ricci, Yamabe, etc.). Roughly, these flows try to produce metrics that are canonical in some sense, and this seems related to your question.

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As others have pointed out, the answer is definitely no. Here's a short sketch of another proof: For any dimension $n$, the metrics (on connected manifolds) with the isometry groups of the largest dimension are precisely the space forms, that is, the metrics of constant sectional curvature. Up to rescaling and passing to the universal cover (neither of which can increase the dimension of the isometry groups), these spaces are precisely

  • the sphere $\mathbb{S}^n$ with the round metric
  • Euclidean space $\mathbb{R}^n$ with the flat metric
  • hyperbolic space $\mathbb{H}^n$ with the hyperbolic metric.

The respective isometry groups are $SO(n + 1)$, $SO(n) \rtimes \mathbb{R}^n$, and $SO_0(n, 1)$, each of which has dimension $\frac{1}{2} n (n + 1)$.

Now, $\mathbb{R}^n$ and $\mathbb{H}^n$ are diffeomorphic to one another but there is no group of dimension $\frac{1}{2} n (n + 1)$ into which their respective isometry groups both inject.

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I'm not sure you're using the same definition of maximality as the OP. Note that the underlying manifold is fixed. – Qiaochu Yuan Dec 4 '14 at 8:10
Qiaochu: Thank you, I misread the OP's question. OP's actual question is actually easier to deal with, and I've adjusted my answer accordingly. – Travis Dec 4 '14 at 13:41

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