## Isometries of a product of Riemannian manifolds

Theorem. Let and be two compact Riemannian manifolds with irreducible holonomy groups. Let . Then

This result seems to be a folklore, probably well known to the specialists, although it is hard to find it in the literature. The only discussion which I managed to find on Mathoverflow contains an answer by Igor Rivin (due to Mahan Mj) which seems to me incomplete (in fact, I just didn’t manage to understand it). Before we turn to the proof, let me note that both irreducibility and compactness are essential. The first can be seen by the example of , which isometry group contains the symmetric group , acting by permutations of factors. The non-compact counter-example is given by .

Proof of the theorem: Let be an isometry. Denote by (resp. the projection of on factors. Assume that . We will also denote and (and similarly and ).

We have a family of smooth maps , parametrised by , defined as

Let . We have three possibilities:

1. ;

2. ;

3. .

In the first case the isometry sends the foliation to the orthogonal foliation . Since , this is only possible if . Therefore the leaves of the first foliation (which are ) are isometric to the leaves of the second, and . Twisting by an involution, which interchanges and , we reduce ourselves to the case 3.

Assume the second case.  We will show that this case is not possible, unless has reducible holonomy. More precisely, we will see that the families   and give local splitting of into product of two orthogonal foliations, which contradicts the irreducibility assumption.

Take , such that and let . If , the  tangent space decomposes as

and intersects by a subspace of dimension . By a standard linear algebra argument,

In other words, if we take a copy of   passing through ( in our notation), we will see that locally cuts out on it a submanifold of dimension , which is precisely and it is clearly totally geodesic, since it is an intersection of totally geodesic manifolds.

Similarly, one can take a submanifold passing through . Its tangent space is the orthogonal compliment to and it locally cuts out a totally geodesic submanifold of dimension on , which identifies with . Varying and (since we are working locally, we can assume that is also maximal), we get a local splitting of into a product of totally geodesic manifolds and , therefore it’s holonomy is reducible. A contradiction.

It is left to deal with the case 3.  In this case is surjective for at least one . Notice that for every vector tangent to we have

Assume that in at least one point the inequality is strict. Thus, is (non-strictly) decreasing distance. Let be the Riemannian volume form on A. In one hand

and in fact the inequality is strict, providing for at least one vector tangent to . In the other hand,

since is a non-zero integer. Of course, here we are sufficiently using the compactness of .

But this means that is a point for every and for a suitable choice of . In other words, is a lifting of some isometry .

Applying the very same arguments to , we get that decomposes as product of isometries , Q.E.D>

Recall, that the de Rham theorem says that each compact Riemannian manifold splits as a product of manifolds with irreducible holonomies. By induction, we get a simple description of the group of isometries of arbitrary compact Riemannian manifold in terms of its irreducible factors, namely:

Let be a compact Riemannian manifold. Let be it’s splitting into factors with irreducible holonomy. Consider the equivalence relation on the set of indices , for which . Let be the splitting into equivalence classes and put . Then

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1 year ago

Notation question: should “Let r=max_b \phi_b” instead define r as some maximal rank?
Silly nitpick: do we need the factors to be connected? If so, which step of the proof uses this hypothesis?
Elementary question: do I understand correctly that in the theorem we can relax “compact” to “finite volume”?

Last edited 1 year ago by moomitten
1 year ago