3
$\begingroup$

Good day.

Let (M,g) be an n-dimensional Riemannian manifold (complete, if you wish), and suppose that there exists an n-1 dimensional Abelian group acting by isometries on M. Or locally, near a point p in M, let A be an Abelian Lie algebra of Killing vector fields of M. Local coordinates can be chosen so that A is generated by d/dx_1,..., d/dx_{n-1}, and g=g(x_n).

In many cases it turns out that A is a codimension one Lie subalgebra of the full isometry algebra, which is bigger. Can anyone provide a counterexample, please? That is, a Riemannian n-manifold of which the full isometry group is Abelian and n-1 dimensional. I did not manage to investigate this thoroughly, but intuitively it seems that (though g(x_n) is arbitrary) you can find one more Killing vector which involves d/dx_n as well ( similar to reparameterization of a one dimensional Riemannian manifold to make the metric constant). If this or the opposite is obvious for experts or can be found readily somewhere in the literature, I would not like to spend hours thinking on it. Otherwise, I will delve into it.

Thank you in advance.

$\endgroup$
2
  • 2
    $\begingroup$ A surface of revolution of non constant curvature, isn'it ? $\endgroup$
    – Holonomia
    Dec 15, 2015 at 10:04
  • $\begingroup$ Good remark. Isometries preserve the curvature. That's probably the answer. $\endgroup$
    – Bedovlat
    Dec 15, 2015 at 10:27

2 Answers 2

3
$\begingroup$

Consider $M=T^{n-1}\times(a,b)$, then $TM$ comes with an obvious trivialisation. Let $x_n\in(a,b)$ be the last coordinate. If you simply define a metric $g$ that depends only on $x_n$ in this trivialisation, then generically, the isometry group will be $T^{n-1}$. You can produce compact examples by letting one eigenvalue of $g$ go to $0$ as $x_n\to a$ or $x_n\to b$ in a controlled way. You can also get open examples by taking $(a,b)=\mathbb R$ and $g$ bounded.

$\endgroup$
1
  • $\begingroup$ Thanks for the answer. Of course, but the question is how to quickly prove that the isometry group is exactly T^n-1 for a generic g. The mere fact that g is generic does not do. As I mentioned, on R, a generic metric has local isometries. Holonomia's remark is probably the quickest test, I think. $\endgroup$
    – Bedovlat
    Dec 15, 2015 at 10:32
2
$\begingroup$

This question is answered in much greater generality in

Isometry groups of proper metric spaces
Author: Piotr Niemiec 
Journal: Trans. Amer. Math. Soc. 366 (2014), 2597-2623 
MSC (2010): Primary 37B05, 54H15; Secondary 54D45 
Published electronically: October 28, 2013 
MathSciNet review: 3165648

See corollary 1.2 (the examples are closely related to Sebastian's answer).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.