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Let $X$ be a complete Riemannian space. Let us denote by $Iso(X)$ the group of isometries of $X$. It is a well-known fact that the group $Iso(X)$, when endowed with the compact-open topology, is a Lie group. Let $X\times X$ be the cartesian prodcut of $X$ with itself endowed with the product metric. Let $G:=(Iso(X)\times Iso(X))\rtimes S_2$, where $S_2$ is the symmetric group of order $2$ which permutes the two coordinates. We have a natural inclusion of $G\leq Iso(X\times X)$.

Q: Is there a general criterion on $X$ which allows us to determine when is $G=Iso(X\times X)$ ?

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  • $\begingroup$ In what examples is this true? Note that this formula does not hold if $X=T^n:=\mathbb{R}^n/\mathbb{Z}^n$, at least for $n\ge 2$. $\endgroup$
    – Mikhail Borovoi
    Commented Feb 7, 2017 at 18:16
  • $\begingroup$ It is true for example if $X$ is the Poincare half-plane with the Poincare metric. $\endgroup$
    – Hugo Chapdelaine
    Commented Feb 7, 2017 at 21:15

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There is a simple answer in the case that $X$ is simply connected. In this case, your claim is true if and only if $X$ is not flat and does not split as a Riemannian product.

"$\Longleftarrow$". If $X$ is as above, each isometry of $X\times X$ has to permute the de Rham factors. There are only two, so they are swapped or they aren't.

"$\Longrightarrow$". If there is a flat de Rham factor $\mathbb R^k$, then the isometry group of $\mathbb R^{2k}$ is larger than the group in the question. If there is more than one de Rham factor, an isometry can choose to swap the corresponding de Rham factors in the two copies of $X$ for each de Rham factor separately, so again the group is larger.

EDIT: In the nonsimply connected case, a similar statement holds: the isometry group of $X\times X$ is exactly $\mathrm{Iso}(X)^2\rtimes(\mathbb Z/2)$ if and only if $X$ is indecomposable in the sense of Eschenburg and Heintze and not isometric to flat $\mathbb R^k$ for $k\ge 1$. The argument is as above, using the main theorem of op. cit. (thanks to Holonomia for the hint).

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  • $\begingroup$ Thanks Sebastian for bringing to my attention the so-called "de Rham decomposition" for a simply connected and complete Riemannian manifold. The key result which you are using may be found in the statement of Theorem III of de Rham's article entitled "Sur la reductibilite d'un espace de Riemann". Since I had in mind simply connected manifolds, this solves my question! $\endgroup$
    – Hugo Chapdelaine
    Commented Feb 7, 2017 at 22:24
  • $\begingroup$ @Sebastian Goette: Do you think the claim is true if $X$ is irreducible (as Riemannian manifold) and $ dim(X) > 1 $ ? If not do you have an example? I know "things are more complicated" but I wonder if you have something to say in the difficult case. I was thinking the case when $X$ is an irreducible flat 2-dimensional torus and it seems to me that the claim is true (but maybe I made mistakes). $\endgroup$
    – Holonomia
    Commented Feb 8, 2017 at 11:11
  • $\begingroup$ @Holonomia. I think the claim is true if $X$ is irreducible and dim $X>1$. Everything hinges on the question which Riemannian manifold have more than one decomposition as a product of irreducible manifolds. It seems to me that a product of two non-rectangular 2-tori splits in exactly one way. $\endgroup$
    – Sebastian Goette
    Commented Feb 8, 2017 at 12:07
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    $\begingroup$ @Sebastian Goette: After a memory effort I remember a paper and find it in Internet: ams.org/journals/proc/1998-126-10/S0002-9939-98-04630-9/… $\endgroup$
    – Holonomia
    Commented Feb 8, 2017 at 12:36
  • $\begingroup$ @Holonomia. Thanks, I've added the link to the answer above. $\endgroup$
    – Sebastian Goette
    Commented Feb 8, 2017 at 19:14

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