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
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
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
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”?
>Notation question
Yes, thank you, that’s a typo;
>Silly nitpick
I think, we don’t need the connectedness.
>Elementary question
Hmm… this is not so elementary, because I am implicitly using the existence of the fundamental class in the top homology, saying that the volume is multiplied by degree. The question is if any non-compact finite volume manifold admits a shrinking diffeomorphism. I don’t know, but intuitively there must be counterexamples (one can look for hyperbolic ones or something like solids of revolution of graphs of rapidly decreasing functions…). At least, it is not clear to me how to generalise this last step to manifolds with finite volume.
Thanks!
Re: Connectedness — what if A has n connected components and B has m connected components, and the m×n resulting components are isometric? For simplicity, let’s work with 0 dimensional manifolds. Then Isom(A×B) has (m×n)! elements while Isom(A)×Isom(B) has only m!×n! components — am I missing something?
Ah, you are right, this happens because
might be not isometric to
, but isometric to it’s connected component (or a collection of those). Let’s assume connectedness indeed, otherwise we’ll need just some additional boring analysis of cases
Very cool! I am still trying to digest the proof. The use of compactness seems (to my naive mind) a pleasant surprise! Absent the compactness hypothesis, can we say that
for
some subgroup of
? Which subgroups are so realized?
Actually, here is an answer by Ivan Solonenko on a similar (but more particular) question. In fact he proves that if
is a complete Riemanian manifold with de Rham decomposition
, then
where
is a subgroup of the symmetric group on
indices, which preserves the subsets
,
…
BUT! one has to be careful with the terminology: here
is the maximal flat factor of
. That is, if
, one decomposes
not like
, but rather
.
A few remarks:
You define the map
by the formula
which is (1) independent of
, (2) valued in subsets of
rather than valued in
itself. Did you mean
instead?