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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, 2020 at 20:41

3 Answers 3

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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, 2020 at 3:52
  • $\begingroup$ I think, I don't understand the argument. Where does it use that $A$ and $B$ are compact? $\endgroup$
    –  V. Rogov
    Jul 12, 2021 at 13:09
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I was trying to understand Igor Rivin's answer and finally wrote a post on Thuses (a very nice young platform for math blogging, which I enthusiastically recommend). Actually, perhaps this is more or less the same proof, but with less things along the lines, at least, one can see explicitly, where irreducibility and compactness are used.

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  • $\begingroup$ I tried commenting on Thuses, but it won't let me register. You define a map $\phi_b : A \to A$ by $\phi_b(a) = \pi_A(A'_b)$ which does not seem to be an element of $A$. Did you mean $\phi_b(a) = \pi_A(f(a,b))$? $\endgroup$ Jul 17, 2021 at 19:38
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You may find this answer of mine of some interest. There, I explain a special case when one can prove that $I(M \times N) \cong I(M) \times I(N)$ with simple differential geometric arguments involving sectional curvatures. And then (in the edit) I sketch a proof of a much more general result using some holonomy techniques taken from Kobayashi & Nomizu. In a nutshell, this result says that if you have a product of some irreducible Riemannian manifolds and at most one flat Riemannian manifold, then the isometry group of the product is the product of the isometry groups of the factors plus permutations of isometric factors (no compactness or even completeness assumption required). One can also try to drop the irreducibility assumption, but then to say something interesting about the isometry group and how it relates to those of the factors, one needs to add the assumptions of completeness and simply connectedness (at least I don't see how to proceed otherwise). Here is the resulting statement:

Proposition. Let $M = M_1 \times \ldots \times M_k$ be a Riemannian product of complete simply connected Riemannian manifolds. We have an obvious embedding of Lie groups $i \colon I(M_1) \times \ldots \times I(M_k) \hookrightarrow I(M)$. The following are equivalent:

  1. $i$ is a local isomorphism, i. e. it gives an isomorphism on the identity components of the above groups: $I^0(M_1) \times \ldots \times I^0(M_k) \simeq I^0(M)$;
  2. $\dim I(M) = \dim I(M_1) + \cdots + \dim I(M_k)$;
  3. At most one of $M_i$'s has a nontrivial Euclidean de Rham factor.

Moreover, if these conditions are satisfied, then $i$ is an isomorphism if and only if there does not exist $i \ne j$ such that $M_i$ and $M_j$ share isometric de Rham factors. (If all $M_i$'s are irreducible, it means that no two of them are isometric.)

In order to prove this, just de Rham decompose each factor, stack all Euclidean subfactors together, and apply the second proposition from my answer linked above.

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  • $\begingroup$ There is also the following perspective: assume that $M$ and $N$ are homogenous and denote by $H_M \trianglelefteq I(M)$ (resp. $H_N \trianglelefteq I(N)$) the isotropy group (a stabiliser of a point). Then $I(M) \times I(N)$ acts on $M\times N$ transitively and thus $I(M \times N) / I(M) \times I(N)$ maps to $H_{M\times N}$. In fact, it further projects to $H_{M \times N} / H_M \times H_N$, and I think this is an isomorphism. What you are actually proving is that if $M$ and $N$ have different curvature, $H_{M \times N} =H_N \times H_M$, right? $\endgroup$
    –  V. Rogov
    Jul 15, 2021 at 14:39
  • $\begingroup$ (Is it important that $M$ is of non-negative curvature and $N$ is flat? I'd suspect, the same must hold whenever $M$ and $N$ have curvature of different signs) $\endgroup$
    –  V. Rogov
    Jul 15, 2021 at 14:40
  • $\begingroup$ (upd: of course $H_M \subset I(M)$ and $H_N \subset I(N)$ are not normal, unless $M$ itself is a Lie group $\endgroup$
    –  V. Rogov
    Jul 16, 2021 at 11:00

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