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Under what conditions is the isometry group of a product of two Riemannian manifolds the product of the isometry groups of each one of the components?

One counterexample is a product of two isometric manifolds, since the isometry contains the involution which cannot be a product. I was wondering if there are some general criterion.

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    $\begingroup$ Of course, $\ M^a\times M^b\ $ will have a bunch of additional isometries (where $M^n := M\times\ldots\times M$; -- $n$-fold product). Possibly, there are no other non-trivial examples. $\endgroup$ – Wlod AA Jan 31 at 20:41
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The argument here is due to Mahan Mj:

If $A, B$ are compact and not isometric, then the isometry group is the product.

Here is the point: Let $X=A \times B$ be the product manifold with the product metric. Let $f: X \to X$ be an isometry. Then each $f(A \times \{b\})$ is totally geodesic for every b. so that $X$ fibers over $B$ with fibers $f(A).$ Further there is a section of the bundle map through every point. These sections are isometric to $B.$ So unless the factors are interchanged (when $A$ and $B$ are isometric), the isometry must be a bundle map. covering some isometry of the base (you can prove this from the existence of sections). It follows that if $A, B$ are not isometric, the isometry group is just the product. else there is a quotient map of $Isom(X)$ onto $\mathbb{Z}/2\mathbb{Z}$ with kernel $Isom(A) \times Isom (A)$ (when $A, B$ are isometric).

The argument fails when $A, B$ are not compact. Notice that the isometry group of the plane is far larger than the isometry group of the line squared.

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    $\begingroup$ I think you might be assuming that $A$ and $B$ are indecomposable when saying "unless the factors are interchanged (when $A$ and $B$ are isometric)", as pointed out by Wlod AA's comment above. $\endgroup$ – user44191 Feb 1 at 3:52

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