Question: Consider the product Riemannian manifold $(M,g)=(S^n\times H^k,g_{round}\oplus g_{hyp})$ of a round sphere $(S^n,g_{round})$ of curvature $1$ and hyperbolic space $(H^k,g_{hyp})$ of curvature $-1$. Is it true that the only conformal transformations of $(M,g)$ are isometries? In other words, is $\mathrm{Conf}(M,g)=\mathrm{Iso}(M,g)$?

Although I am mainly interested in the case $n>k>1$, it would also be nice to know the answer for all $n,k\geq2$. Note that by a result of Ferrand (see Schoen), the action of $\mathrm{Conf}(M,g)$ is proper, hence there exists a metric $g_*$ conformal to $g$, such that $\mathrm{Conf}(M,g)=\mathrm{Iso}(M,g_*)$.

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    $\begingroup$ They are all isometric, you can see this by observing that the product metric is conformally flat with conformal structure realized as the complement of a round k-1 sphere in the round k+n sphere. $\endgroup$
    – Misha
    Jan 6, 2016 at 0:03
  • $\begingroup$ @Misha: I understand that $(M,g)$ is conformally flat and also conformally equivalent to the open manifold $S^{k+n}\setminus S^{k-1}$ with the round (incomplete) metric. But how does this imply that all of its conformal transformations are isometric? Can you please elaborate? $\endgroup$ Jan 6, 2016 at 1:12
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    $\begingroup$ My reading of what Misha said is that $S^{k-1}$ is removable, i.e. the conformal maps extend to conformal automorphisms of $S^{k+n}$ that stabilize $S^{k-1}$, and aren' those exactlly the isometries of $S^n\times H^k$? $\endgroup$ Jan 6, 2016 at 4:24
  • $\begingroup$ I was also thinking the same, and counting dimensions everything seems to match. Namely, the dimension of the group of conformal transformations of $S^{n+k}$ that leaves $S^{k-1}$ invariant is $k(k-1)/2+n(n+1)/2+k$, where the last $k$ come from dilations. At the same time, the dimension of the isometry group of $S^n\times H^k$ is $n(n+1)/2+k(k+1)/2$, so these numbers agree. But there are still some gaps in my understanding of the situation: (1) why do these conformal transformations extend across $S^{k-1}$? (2) Why aren't there other connected components? $\endgroup$ Jan 6, 2016 at 5:21
  • $\begingroup$ @RenatoG.Bettiol: This is just Liouville's theorem (conformal = restriction of Moebius), since your dimension $k+n$ is at least 3. $\endgroup$
    – Misha
    Jan 6, 2016 at 9:55

1 Answer 1


Corollary 1 here https://projecteuclid.org/download/pdf_1/euclid.jmsj/1261148578 answer positively your question.


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