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I feel like the following statement is well-known, but I have been unable to find a reference for it:

Let $(M_{i},g_{i})$ $1\leq i \leq n$ be isometric Riemannian manifolds and let $G$ denote the isometry group of one, and hence any, of the $(M_{i},g_{i})$'s. Then the isometry group of $(M_{1}\times\ldots M_{n}, g_{1}\oplus\ldots \oplus g_{n})$ is $\prod_{i=1}^{n}G\rtimes S_{n}$ where $S_{n}$ is the symmetric group on $n$ letters and it acts on the product by permuting its factors.

My questions are:

Is this true?

If true, could someone provide a reference?

Thanks!

Daniel Mckenzie
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1 Answers1

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This is not true. For example $Iso(\mathbb{R}^2)=O(2)\times \mathbb{R}^2\neq \mathbb{R}^2\times \mathbb{Z}/(2)^2=Iso(\mathbb{R})^2$. For a compact example, take $Iso(S^1\times S^1)=D_8\rtimes S^1\times S^1\neq (S^1\times \mathbb{Z}/(2))^2=Iso(S^1)^2$. Even $Iso(\mathbb{Z}/(2))^2=\mathbb{Z}/(2)^2\neq D_4=Iso(\mathbb{Z}/(2)^2)$ provides examples with discrete isometry group.

Pax
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    Hmmm. You may be right, but it seems that the metric you are considering on $\mathbb{R}^{2}$ is not the usual Euclidean metric, and not the direct sum of the metrics on the two copies of $\mathbb{R}$, which would be the taxi cab metric? – Daniel Mckenzie Oct 28 '15 at 18:19
  • @DanielMckenzie On the contrary, the euclidean metric on $\mathbb{R}^2$ has $O(2)\rtimes \mathbb{R}^2$ as isometry group (orthogonal transformations $\rtimes$ translations). – Llohann May 27 '16 at 19:40
  • @Kamina Is it always the case that the isometry group of a product of manifolds coincides with the product of isometry groups up to a discrete factor, if there is no euclidean factor on the product? – Llohann May 27 '16 at 19:44
  • The motivation for my question, although it may be poorly phrased, is the uniqueness part of the De Rham decomposition theorem. Llohann, I think you're right that the euclidean factors case is different, although I would point out that the euclidean metric on $\mathbb{R}^{2}$ is not the product metric you get from considering $\mathbb{R}^{2} = \mathbb{R}^{1}\times \mathbb{R}^{1}$. Similarly, Kamina's example of $S^{1}\times S^{1}$ is using the flat metric, which is not (I think?) the product of the metrics on the two copies of $S^{1}$ – Daniel Mckenzie Jun 22 '16 at 15:42