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Let $(M,g)$ be a Riemannian manifold and let $I = Iso(M)$ be the group of isometries of $M$. Suppose that $I$ has no subgroups. What does this tell us about $M$?

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    $\begingroup$ $I$ always has the subgroups $I$ and $\{\operatorname{id}\}$. These may not be distinct. Do you mean no other subgroups, and the two trivially existing subgroups may be different, or $I = \{\operatorname{id}\}$? $\endgroup$ Sep 13, 2013 at 19:57
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    $\begingroup$ The isometry group is a Lie group. A Lie group which does not have non-trivial proper subgroups must be discrete, and then it must be of prime order or trivial. It is very easy to construct manifolds with such isometry groups, and I doubt you can get much information about the topology of M from that. $\endgroup$ Sep 13, 2013 at 19:58
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    $\begingroup$ Every nonempty manifold except for the circle, admits a Riemannian metric with trivial group of symmetries. Thus, the only conclusion you can make is that your manifold is not a topological circle or line. $\endgroup$ Sep 14, 2013 at 4:47
  • $\begingroup$ @studiosus Interesting fact. Do you have any reference for this statement? $\endgroup$
    – gofvonx
    Sep 14, 2013 at 10:10

1 Answer 1

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Here are some details for my comments.

Claim. Suppose that $M$ is a smooth manifold of dimension $\ge 2$. Then it admits a Riemannian metric with no symmetries.

Suppose that $M$ has dimension $\ge 3$. Then it is proven in

J. Kazdan, F. Warner, "Scalar curvature and conformal deformation of Riemannian structure," J. Differential Geometry 10 (1975), 113–134

that every $C^\infty$-function $K: M\to {\mathbb R}$ which is negative at some point, equals the scalar curvature of a $C^\infty$-Riemannian metric on $M$. Since isometries of a manifold have to preserve scalar curvature, we can just take a "random" smooth function $K$ on $M$: It will have no symmetries and, hence, the corresponding Riemannian metric will have no symmetries either.

For 2-dimensional manifolds there are more restrictions on scalar/Gaussian curvature. However, one can use the results of

J. Kazdan, F. Warner, "Curvature functions for open 2-manifolds", Ann. of Math. (2) 99 (1974), 203–219

and

J. Kazdan, F. Warner, "Curvature functions for compact 2-manifolds," Ann. of Math. (2) 99 (1974), 14–47.

to get the same conclusion.

In the 1-dimensional case (I am assuming the manifold is connected), if the Riemannian manifold is compact, then it is isometric to a circle of certain radius and, hence, has 1-dimensional group of symmetries. For noncompact manifolds, one can take the metric isometric to the half-line; such metric has no symmetries.

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