Isometries of a product of Riemannian manifolds

Theorem. Let $(A, g_A)$ and $(B, g_B)$ be two compact Riemannian manifolds with irreducible holonomy groups. Let $(X, g):= (A \times B, g_A \oplus g_B)$ . Then

$\operatorname{Isom}(X) = \begin{cases} \operatorname{Isom}(A) \times \operatorname{Isom}(B), \ \text{ if } A \text{ is not isometric to } B \\ \left ( \operatorname{Isom}(A) \times \operatorname{Isom}(B) \right ) \rtimes \mathbb{Z}/2\mathbb{Z}, \ \text{ otherwise } \end{cases}$

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 $T^3 = S^1 \times S^1 \times S^1$ , which isometry group contains the symmetric group $\mathfrak{S}_3$ , acting by permutations of factors. The non-compact counter-example is given by $\mathbb{R}^2$ .

Proof of the theorem: Let $f \colon X \to X$ be an isometry. Denote by $\pi_A$ (resp. $\pi_B)$ the projection of $X$ on factors. Assume that $n =\dim A \ge \dim B$ . We will also denote $A_b:= A \times \{b\}$ and $A'_b := f(A \times \{b\}$ (and similarly $B_a$ and $B'_a$ ).

We have a family of smooth maps $\phi_b \colon A \to A$ , parametrised by $b \in B$ , defined as

$\phi_b(a) = \pi_A(A'_b).$

Let $r = \max_b \operatorname{rk} \phi_b$ . We have three possibilities:

1. $r = 0$ ;

2. $0 < r < n=\dim A$ ;

3. $r = n$ .

In the first case the isometry $f$ sends the foliation $\pi_A^* TA$ to the orthogonal foliation $\pi_B^* TB$ . Since $\dim A \ge \dim B$ , this is only possible if $f(\pi_A^*TA) = \pi_B^*TB$ . Therefore the leaves of the first foliation (which are $A_b$ ) are isometric to the leaves of the second, and $(A, g_A) \simeq (B, g_B)$ . Twisting $f$ by an involution, which interchanges $A$ and $B$ , we reduce ourselves to the case 3.

Assume the second case. We will show that this case is not possible, unless $A$ has reducible holonomy. More precisely, we will see that the families $\pi_A(A'_b)$ and $\pi_A(B'_a)$ give local splitting of $A$ into product of two orthogonal foliations, which contradicts the irreducibility assumption.

Take $b$ , such that $\operatorname{rk} \phi_{b} = r$ and let $x \in A'_b$ . If $x =(a',b') \in A \times B$ , the tangent space $T_x X$ decomposes as

$T_x X = T_{a'}A \oplus T_{b'} B,$

and $T_x A'_b \subset T_x X$ intersects $T_b' B =\ker D\pi_A$ by a subspace of dimension $n-r$ . By a standard linear algebra argument,

$\dim \left (T_x f(A \times \{b\}) \cap T_{\pi_A(x)}A \right ) = r.$

In other words, if we take a copy of $A$ passing through $x$ ( $A_{b'}$ in our notation), we will see that $A'_b$ locally cuts out on it a submanifold of dimension $r$ , which is precisely $\phi_{b}(A)$ and it is clearly totally geodesic, since it is an intersection of totally geodesic manifolds.

Similarly, one can take a submanifold $B'_a$ passing through $X$ . Its tangent space $T_xB'_a$ is the orthogonal compliment to $T_x A'_b$ and it locally cuts out a totally geodesic submanifold of dimension $n-r$ on $A_{b'}$ , which identifies with $\pi_A(B'_a)$ . Varying $a$ and $b$ (since we are working locally, we can assume that $\operatorname{rk }\pi_A(B'_a)$ is also maximal), we get a local splitting of $A$ into a product of totally geodesic manifolds $\pi_A(A'_b)$ and $\pi_A(B'_a)$ , therefore it’s holonomy is reducible. A contradiction.

It is left to deal with the case 3. In this case $\phi_b \colon A \to A$ is surjective for at least one $b$ . Notice that for every vector $v$ tangent to $A'_b$ we have

$g(v,v) = \pi_A^*g_A(v,v) + \pi_B^*g_B(v,v) \ge g_A(D\pi_A(v), D\pi_A(v)).$

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

$\int_A \phi_b^* \omega_A \le \int_A \omega_A = vol(A),$

and in fact the inequality is strict, providing $D\pi_B(v) \neq 0$ for at least one vector $v$ tangent to $A'_b$ . In the other hand,

$\int_A \phi_b^*\omega_A = |\deg (\phi_b) |\cdot vol(A) \ge vol(A),$

since $\deg(\phi_b)$ is a non-zero integer. Of course, here we are sufficiently using the compactness of $A$ .

But this means that $\pi_B(A'_b)$ is a point for every $b$ and $A'_b = A_{b'}$ for a suitable choice of $b' \in B$ . In other words, $f \colon A \times B \to A \times B$ is a lifting of some isometry $f_B \colon B \to B$ .

Applying the very same arguments to $\phi_a \colon B \to B_a \to B'_a \to B$ , we get that $f$ decomposes as product of isometries $f_A \times f_B$ , 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 $(X,g)$ be a compact Riemannian manifold. Let $(X,g) = (X_1, g_1) \times \ldots \times (X_n, g_n)$ be it’s splitting into factors with irreducible holonomy. Consider the equivalence relation on the set of indices $\{1, \ldots, n\}$ , for which $i \sim j \Leftrightarrow (X_i, g_i) \simeq (X_j, g_j)$ . Let $\{1, \ldots, n\} = \bigsqcup C_k$ be the splitting into equivalence classes and put $\lambda_k:= |C_k|$ . Then

$\operatorname{Isom}(X) = \prod_{k} \left ( \prod_{i \in C_k} \operatorname{Isom}(X_i, g_i) \right) \rtimes \mathfrak{S}_{\lambda_k}.$

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moomitten

3 years 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 3 years ago by moomitten

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Vasily Rogov

Reply to moomitten

>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.

Reply to Vasily Rogov

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 $A$ might be not isometric to $B$ , 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 $\text{Isom}(A\times B) = (\text{Isom}(A)\times \text{Isom}(B)) \rtimes G$ for $G$ some subgroup of $O(\text{dim}(A)+\text{dim}(B))$ ? 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 $M$ is a complete Riemanian manifold with de Rham decomposition $M = M_0 \times M_1^{l_1} \times \ldots \times M_n^{l_n}$ , then

$\operatorname{Isom}(M) = \left ( \prod_{i=0}^n \operatorname{Isom}(M_i)^{l_i} \right ) \rtimes \mathfrak{S}_{l_1, \ldots, l_n},$

where $\mathfrak{S}_{l_1, \ldots, l_n}$ is a subgroup of the symmetric group on $l = l_1+\ldots +l_n$ indices, which preserves the subsets $\{1, \ldots, l_1\}, \ \{l_1+1, \ldots, l_1+l_2\}, \ \ldots, \{l-l_n+1, \ldots, l\}$ ,

…

BUT! one has to be careful with the terminology: here $M_0$ is the maximal flat factor of $M$ . That is, if $M=\mathbb{R}^n$ , one decomposes $M$ not like $\mathbb{R} \times \mathbb{R} \times \ldots \times \mathbb{R}$ , but rather $M=M_0 = \mathbb{R}^n$ .

A few remarks:

Vanya’s proof is a bit sketchy, but before it he presents a very detailed proof for a particular case (of a product of two homogenous manifolds, where one is flat and the second is of nonzero curvature), and one could easily adopt it to the required generality;
He also assumes $\pi_1(M) = \{1\}$ , because he is referring to a statement from Kobayashi-Nomizu textbook, which is formulated for simply connected manifolds. But it seems to me, this is not needed indeed.
Philosophically, the ‘non-compact counter-example’ $\mathbb{R}^n$ is the only possible example, where after taking product you get much bigger isometry group then the product of isometry groups of the factors. The reason for that is that for a model space of constant curvature $\mathbb{X}$ (which ate in some sense ”the most symmetric” manifolds) one has $\mathbb{X}^n \times \mathbb{X}^m \neq \mathbb{X}^{n+m}$ , unless it is not $\mathbb{R}^n$ . Namely, $\mathbb{S}^n \times \mathbb{S}^m \neq \mathbb{S}^{n+m}$ and $\mathbb{H}^n \times \mathbb{H}^m \neq \mathbb{H}^{n+m}$ .
It is a complete mystery for me why is it so hard to find such a basic and powerful fact (an immediate consequence of de Rham decomposition theorem) in the literature. I guess, the reason is that the world of basic textbooks in differential geometry is an obscure, dark antique labyrinth, which is almost impossible to go through (or may be I was too unlucky not to find a good book).

Last edited 3 years ago by Vasily Rogov

Omar Antolín Camarena

You define the map $\phi_b :A \to A$ by the formula $\phi_b(a) = \pi_A(A'_b)$ which is (1) independent of $a$ , (2) valued in subsets of $A$ rather than valued in $A$ itself. Did you mean $\phi_b(a) = \pi_A(f(a,b))$ instead?