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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.

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

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  • $\begingroup$ 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 '15 at 10:32
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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).

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