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.

  • 2
    $\begingroup$ 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

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.

| cite | improve this answer | |
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.