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Everywhere I'm looking a lot of authors claim (without any proof, but I'm sure they're right) that the group of isometries of $\mathbb{S}^2 \times \mathbb{R}$ and $\mathbb{H}^2 \times \mathbb{R}$ are $4$ dimensional and equal to $ \text{Isom}(\mathbb{S}^2) \times \text{Isom}(\mathbb{R})$ and $ \text{Isom}(\mathbb{H}^2) \times \text{Isom}(\mathbb{R})$, respectively. How can I prove those claims and what's the geometric intuition behind it?

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    $\begingroup$ $Isom(X \times Y)\cong Isom(X)\times Isom(Y)$ is not true in general, see here. $\endgroup$ Mar 11, 2019 at 19:52
  • $\begingroup$ @DietrichBurde I'm aware of that, but I know that in this case it's true. $\endgroup$ Mar 11, 2019 at 19:54
  • $\begingroup$ So you have a link? Then you could add it. $\endgroup$ Mar 11, 2019 at 19:58
  • $\begingroup$ @DietrichBurde books.google.com.br/… $\endgroup$ Mar 11, 2019 at 20:12

2 Answers 2

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Let $f$ be an isometry of $\Bbb S^2\times \Bbb R$.

Wlog $f(N,0)=(N,0)$. The only point $(z,t)$ such that there exist two shortest geodesics of length $\pi$ from $(N,0)$ to $(z,t)$, is $(S,0)$. Hence $f(S,0)=(S,0)$. Now $\Bbb S^2\times \{0\}$ is characterized as the set of points such that the sum of distances to $(N,0)$ and $(S,0)$ equals $\pi$. We conclude that $f$ maps $\Bbb S^2\times\{0\}$ to itself.

The rest is then easy.

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  • $\begingroup$ Thanks, but could you elaborate some more on the rest? It's not that easy yet to me. Also, could one make a similar argument to yours for the $\mathbb{H}^2 \times \mathbb{R}$ case? $\endgroup$ Mar 11, 2019 at 21:31
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I am totally late to the party, but just to have it explained clearly on the Internet, here is a proof of a more general statement that covers your question for both $\mathbb{S}^2 \times \mathbb{R}$ and $\mathbb{R}H^2 \times \mathbb{R}$. Observe that whenever you have Riemannian manifolds $M_1, \ldots, M_k$ you have a natural embedding of groups $I(M_1) \times \ldots \times I(M_k) \hookrightarrow I(M_1 \times \ldots \times M_k)$. If all $M_i$'s have finitely many connected components, these are Lie groups and the embedding is smooth. In general, it may fail to be surjective (in the worst case, the dimension of the RHS may be greater than that of the LHS). However, we have

Proposition. Let $M$ be a connected Riemannian manifold with nonzero constant sectional curvature $R$ (for example, a round sphere, or a real hyperbolic/projective space, or a hyperbolic manifold; no assumption on the sign of $R$ here) and $N$ a connected flat manifold (e.g., a flat torus, or a Euclidean space, or a product of those). Then the natural embedding $I(M) \times I(N) \hookrightarrow I(M \times N)$ is an isomorphism.

Proof. Take an arbitrary isometry $\varphi$ of $M \times N$. Let's take a point $(m,n) \in M \times N$ and look at the linear isometry $d\varphi_{(m,n)} \colon T_{(m,n)}(M \times N) \cong T_mM \oplus T_nN \to T_{(m',n')}(M \times N) = T_{m'}M \oplus T_{n'}N$, where $(m',n') = \varphi(m,n)$. I claim that if we prove that it sends $T_mM$ onto $T_{m'}M$, then we're done. Indeed, if it does, then it sends $(T_mM)^\perp = T_nN$ onto $(T_{m'}M)^\perp = T_{n'}N$. But since an isometry maps geodesics to geodesics, $\varphi$ sends the totally geodesic submanifold $M \times \{n\}$ onto $M \times \{n'\}$, and similarly $\{m\} \times N$ onto $\{m'\} \times N$. (Completeness is not needed here, but it is essential that these slices are properly embedded submanifolds, so a geodesic in $M \times N$ that tangentially touches, say, $M \times \{n\}$ even once, will stay in it throughout the whole interval where it is defined.) Denote:

\begin{align} \varphi_1 \colon M \to M \times \{n\} \xrightarrow{\varphi} M \times \{n'\} \to M, \\ \varphi_2 \colon N \to \{m\} \times N \xrightarrow{\varphi} \{m'\} \times N \to N. \end{align}

Observe that $\varphi_1 \in I(M)$ and $\varphi_2 \in I(N)$, so we have a product isometry $(\varphi_1, \varphi_2)$ of $M \times N$. The isometries $\varphi$ and $(\varphi_1, \varphi_2)$ coincide on the "cross" $(M \times \{n\}) \cup (\{m\} \times N)$, which surely implies that they coincide at $(m,n)$ but also that their differentials at this point coincide as well:

\begin{multline} d\varphi_{(m,n)} = d(\varphi_1)_m + d(\varphi_2)_n = d(\varphi_1, \varphi_2)_{(m,n)} \colon \\ T_{(m,n)}(M \times N) \cong T_mM \oplus T_nN \to T_{m'}M \oplus T_{n'}N \cong T_{(m',n')}(M \times N). \end{multline}

But now it is a well-known fact that if two isometries of a connected Riemannian manifold coincide at some point together with their differentials, then they coincide globally (just apply these two isometries to geodesics emanating from that point and use the theorem on uniqueness of geodesics). So we are left to show that $d\varphi_{(m,n)}$ sends $T_mM$ onto $T_{m'}M$. Since isometries preserve sectional curvatures, it will immediately follow from the following claim:

Claim. If $L$ is a $2$-plane in $T_{(m,n)}(M \times N)$, its sectional curvature lies between $0$ and $R$ and only reaches $R$ if $L \subseteq T_mM$.

Proof of the claim. Clearly, if $L \subseteq T_mM$, then its sectional curvature is just $R$. Assume $L \nsubseteq T_mM$. Then there are two possibilities:

Case 1: $L$ intesects $T_nN$ nontrivially. Then take a basis for $L$ of the form $X = A, Y = W + B$, where $A, B \in T_nN$ and $W \in T_mM$. Then

$$ K(L) = \frac{Rm(X,Y,Y,X)}{\|X\|^2 \|Y\|^2 - \langle X \,|\, Y\rangle^2}. $$

Since the product is Riemannian and $N$ is flat, we have

$$ Rm(X,Y,Y,X) = Rm(A,W+B,W+B,A) = Rm(A,B,B,A) + Rm(0,W,W,0) = 0, $$

so $K(L) = 0$.

Case 2: $L$ intersects $T_nN$ trivially. Then we can take a basis $X = V+A, Y = W+B$ for $L$ such that $V,W$ is an orthonormal basis of the $2$-plane $\mathrm{pr}_{T_mM}(L) \subseteq T_mM$ and $A,B \in T_nN$ with at least one of them nonzero (since we assume $L \nsubseteq T_mM$). We compute:

$$ K(L) = \frac{Rm(X,Y,Y,X)}{\|X\|^2 \|Y\|^2 - \langle X \,|\, Y\rangle^2} = \frac{Rm(V,W,W,V) + Rm(A,B,B,A)}{(1 + \|A\|^2)(1 + \|B\|^2) - \langle A \,|\, B\rangle^2} = \frac{K(\mathrm{pr}_{T_mM}(L)) + 0}{1 + \|A\|^2 + \|B\|^2 + (\|A\|^2 \|B\|^2 - \langle A \,|\, B\rangle^2)} = \frac{R}{1 + \|A\|^2 + \|B\|^2 + (\|A\|^2 \|B\|^2 - \langle A \,|\, B\rangle^2)}, $$

which lies strictly between $0$ and $R$. $\square$ $\square$

Remark. There is also a nice fact directly pertinent to this discussion that if $M = M_0 \times M_1 \times \ldots \times M_k$ is the de Rham decomposition of a simply connected complete Riemannian manifold, then the natural morphism on the isometry groups above restricts to an isomorphism on the connected components of the idenitities: $I^0(M_0) \times I^0(M_1) \times \ldots \times I^0(M_k) \to I^0(M)$. Here $I(M)$ and each $I(M_i)$ are Lie groups in the compact-open topology, so we can take the connected component of the identity in each of them. For a proof, see Chapter VI, theorem 3.5 of

Kobayashi, Sh.; Nomizu, K., Foundations of differential geometry. I, New York-London: Interscience Publishers, a division of John Wiley & Sons. XI, 329 p. (1963). ZBL0119.37502.


EDIT. After giving it a thought, I managed to generalize the above proposition and remark by employing some ideas about holonomy from Kobayashi and Nomizu. In a nutshell, the following result says that if you have the product of a bunch of irreducible Riemannian manifolds and at most one flat manifold, then the full isometry group of the product is the product of the full isometry groups of the factors plus permutations of isometric factors. I am leaving the above proposition here because it already answers the MO's question, and its proof is rather elementary and doesn't require any sophisticated arguments involving holonomy.

Proposition. Let $M = M_0 \times M_1^{l_1} \times \ldots \times M_k^{l_k}$ be a Riemannian product, where all $M_i$'s are connected Riemannian manifolds, $M_0$ is flat, each $M_i, 1 \leqslant i \leqslant k,$ is irreducible, and $M_i^{l_i}$ simply means $\underbrace{M_i \times \ldots \times M_i}_{l_i \, \text{times}}$. (Here by irreducibility of $M_i$ we mean that it is not flat and its restricted holonomy representation is irreducible; if $M_i$ is complete, this is equivalent to asking that its universal Riemannian covering cannot be decomposed as a nontrivial Riemannian product.) We also assume that $M_i$ is not isometric to $M_j$ for $i \ne j$. Let $l = \sum_{i=1}^k l_i$, write $S_l$ for the symmetric group on $l$ elements, and let $S_l^{l_1, \ldots, l_k} \cong S_{l_1} \times \ldots \times S_{l_k}$ stand for the subgroup of permutations that permute the first $l_1$ elements with each other, the next $l_2$ elements with each other, and so on. Then the obvious map $$ I(M_0) \times I(M_1)^{l_1} \times \ldots \times I(M_k)^{l_k} \rtimes S_l^{l_1, \ldots, l_k} \to I(M) $$ is an isomorphism of Lie groups.

One obvious example of such decomposition is the de Rham decomposition of a complete simply connected Riemannian manifold. Note that we don't require the factors to be complete in the proposition.

Sketch of the proof. Let $p = (p_0, (p^1_i)_{i=1}^{l_1}, \ldots, (p^k_i)_{i=1}^{l_k}) \in M$, and consider the decomposition

$$ T_pM = T_{p_0}M_0 \oplus \bigoplus_{i=1}^{l_1} T_{p^1_i} M_1 \oplus \ldots \oplus \bigoplus_{i=1}^{l_k} T_{p^k_i} M_k. $$

Let $\mathrm{Hol}^0(M,p)$ stand for the restricted holonomy group of $M$ at $p$, which is defined using parallel transport along contractible loops based at $p$ and is the identity component of the full holonomy group $\mathrm{Hol}(M,p)$. The decomposition of $T_pM$ above is obviously orthogonal, the first summand $T_{p_0}M_0$ is the subspace of invariants of $\mathrm{Hol}^0(M,p)$ in $T_pM$, and all the other summands are irreducible $\mathrm{Hol}^0(M,p)$-subrepresentations in $T_pM$ (see Chapter IV, Section 5 in Kobayashi&Nomizu for details on that). A decomposition of $T_pM$ satisfying these three properties is called a canonical decomposition in K&N. Now, part (4) of theorem 5.4 in Chapter IV, Section 5 of K&N says that a canonical decomposition of $T_pM$ is unique up to reordering of its factors. Caveat: K&N actually talk about decompositions with respect to the full holonomy group, and theorem 5.4(4) requires $M$ to be simply connected. But note that the restricted holonomy representation of $M$ is isomorphic to the full holonomy representation of its universal Riemannian covering $\widetilde{M}$ (just by lifting contractible loops). So the uniqueness of a canonical $\mathrm{Hol}(\widetilde{M}, \widetilde{p})$-decomposition for $\widetilde{M}$ ($\widetilde{p} \in \widetilde{M}$ any point over $p$) translates into the uniqueness of a canonical $\mathrm{Hol}^0(M, p)$-decomposition for $M$. It is also not hard to show that if $\varphi$ is an isometry of $M$, then $d\varphi$ sends a canonical decomposition of $T_pM$ to a canonical decomposition of $T_{\varphi(p)}M$ (basically because 'isometries commute with parallel transport'). Now, given such $\varphi$, a small adaptation of the first part of the proof of the proposition above (the one with geodesics) shows that $\varphi$ 'permutes the factors of the de Rham decomposition of $M$', or, formally, lies in the image of the morphism of isomerty groups above. $\square$

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    $\begingroup$ I see- thanks for clarifying! Is this the only care where the isometry group of the product has larger dimension than the sum of dimensions of the factors? $\endgroup$ Feb 16, 2021 at 17:51
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    $\begingroup$ @C.F.G. I didn't read in super detail, but the argument seems fine, and I upvoted. It's probably even more general. Riemannian products have lots of zero curvature planes, so to have hope of getting a larger isometry group on the priduct (in terms of dimension), some factors must have a lot of zero curvature planes. $\endgroup$ Feb 16, 2021 at 17:57
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    $\begingroup$ I agree the simply connected case is clear. For non-simply connected, it's much less clear. And one cannot just pull things back to the universal cover since this can have an isometry group of strictly larger dimension (e.g., $T^2$ vs $\mathbb{R}^2$.) $\endgroup$ Feb 16, 2021 at 22:04
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    $\begingroup$ Thank you so much! I will definitely have to take a more serious look at this when I have the time. By the way, you're not late to the party at all, you're more than welcome to it! Help is always appreciated. Thank you for taking the time to write such a detailed answer. $\endgroup$ Feb 17, 2021 at 2:45
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    $\begingroup$ @MatheusAndrade, no probem at all, this issue with isometry groups has been in my to-do list for a while, so writing an anwer to your question was a great way to clear a few things up! $\endgroup$ Feb 17, 2021 at 10:06

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