# Example of a symmetric manifold without rotation symmetry?

Consider a flat torus, in 2D. It has translation symmetry but not rotational symmetry.

Are there any manifolds that are not simply products like $$S_1 \times S_1$$ that do not have rotational symmetry but have either translational or a combination of a translation plus a rotation in their symmetry?

Edit: In other words a symmetric manifold with no rigid symmetries in which one point is fixed. (Would they all be flat?)

For example what about the projective plane or, homology sphere? What symmetries do those have?

• Can you be more precise? I guess by manifold you mean Riemannian manifold? Then you can define a rotation as an isometry that fixes a point and is orientation-preserving at that point, but what do you mean by translation? – Eric Wofsey May 26 at 20:37
• Basically what I mean is, a manifold in which there is no symmetry where you fix a point and do a rotation, but does have symmetry where you transport the point to another point. e.g. a sphere has rotational symmetry, but a flat torus doesn't. Maybe think of it as tranporting a little vector. Rotate a vector on a sphere and it's still the same. – zooby May 26 at 21:02
• What about a Klein bottle? – Jason DeVito May 27 at 2:07
• Even with your comment, you notion of a translation symmetry is undefined. "transporting a little vector" is not a mathematical concept. Each rotation of the sphere also "transports one point to another". Maybe what you mean is that a translation is an isometry that has no fixed points? Then, do you regard the map $v\mapsto -v$ of the sphere to itself as a translation? It has no fixed points on the sphere. – Moishe Kohan May 27 at 2:55
• Exactly no fixed points. – zooby May 27 at 14:07

Let me try to interpret your question.

It appears that for a Riemannian manifold $$M$$ a "translation" of $$M$$ is an isometry $$\phi: M\to M$$ which is either the identity or has no fixed points in $$M$$. Then to say that the isometry group of $$M$$ consists only of "translations" amounts to saying that the isometry group $$Isom(M)$$ of $$M$$ acts freely. (The latter is the standard terminology.)

Boring examples where $$Isom(M)$$ acts freely are when $$Isom(M)$$ is trivial, i.e. consists only of the identity element. One can prove that a "generic" Riemannian metric on a given manifold (of dimension $$\ge 2$$) has trivial group of isometries. For instance, for (connected) surfaces $$M$$ this comes from the fact one can prescribe the Gaussian curvature of the Riemannian metric $$M$$ (subject to some sign conditions). See this wikipdeai article for details and references.

A generic choice of such a function allows no (non-identity) diffeomorphisms $$\phi: M\to M$$ which preserve the function $$K$$ (i.e. $$K\circ \phi=\phi$$).

One can promote this to more interesting examples where $$Isom(M)$$ is nontrivial but acts freely. For instance, start with the projective plane $$P=RP^2$$, choose a generic curvature function $$K(x)>0$$ on $$P$$ and take the corresponding Riemannian metric $$g$$. Now, lift $$g$$ to a metric $$\tilde{g}$$ on the universal covering of $$P$$, which is $$M\sim S^2$$. Then $$(M, \tilde{g})$$ has a nontrivial isometry (order two involution $$\tau$$ generating the deck-transformation group of the covering $$M\to P$$). With a bit more work one verifies that $$Isom(M,\tilde{g})=\langle\tau\rangle= \{id, \tau\}$$. Thus, $$Isom(M,\tilde{g})$$ acts freely on $$M$$.

One can, however, ask for more, namely, that $$M$$ is homogeneous, i.e. the isometry group of $$M$$ acts freely and transitively (i.e. simply transitively). Such examples exist in dimension 3, they are called Berger spheres. Berger spheres are homogenous Riemannian manifolds diffeomorphic to $$S^3$$ (the 3-dimensional sphere). These manifolds come with some parameters.

In Theorem 4.1 in

P. Gadea, J. Oubiña, Homogeneous Riemannian Structures on Berger 3-Spheres, Proceedings of the Edinburgh Mathematical Society, 48 (2005), no. 2, 375–387.

the authors prove that for certain values of parameters of the Berger sphere $$M$$, its isometry group acts simply transitively on $$M$$. (The group is isomorphic to $$SU(2)$$.)

The manifolds in these examples are definitely not products and the metric is not flat. (Actually, a sphere of dimension $$\ge 2$$ admits no flat metric.)

Edit. One more thing: In geometry a "symmetric space" (or a manifold) usually means something different from the header of your post, see here. To make it more confusing, in topology, a "symmetric product" means yet something else.