# The isometry group of a product of two Riemannian manifolds

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.

• 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. – Wlod AA Jan 31 at 20:41

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.
• 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. – user44191 Feb 1 at 3:52