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In the book "Discrete groups, expanding graphs and invariant measures" by Alexander Lubotzky, page 37, the author says that all finitely generated discrete Kazhdan subgroups of $SO(3)$ are finite, and the same follows for $SO(4)$, since $SO(4)$ is locally isomorphic to $SO(3) \times SO(3)$.

I know that $SO(4)$ is locally isomorphic to $SO(3) \times SO(3)$ but I really can't see why this would imply that a finitely generated discrete Kazhdan subgroup of $SO(4)$ must be finite. Indeed, let $U \subset SO(4)$ be the domain of the local isomorphism, and let $\Gamma \leq SO(4)$ be a finitely generated discrete Kazhdan subgroup. First of, I don't see how the local isomorphism can relate $\Gamma$ to a Kazhdan subgroup of $SO(3) \times SO(3)$, and even with that, I don't see why the fact that $\Gamma \cap U$ is finite should imply that $\Gamma$ is finite. I mean, $\Gamma$ is discrete, so if we take $U$ small enough $\Gamma \cap U$ is a singleton, which gives no new information...

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I assume that you take for granted that no infinite subgroup of $\mathrm{SO}(3)$ has Property T when endowed with the discrete topology.

Suppose by contradiction that $\mathrm{SO}(4)$ has an infinite subgroup $\Gamma$ having Property T (when endowed with the discrete topology). Then the image $\bar{\Gamma}$ of $\Gamma$ in $\mathrm{PSO}(4)=\mathrm{SO}(4)/\{\pm I_4\}$ is still infinite, and still has Property T. Then $\mathrm{PSO}(4)$ is isomorphic to $\mathrm{SO}(3)^2$, and it follows that the image of $\bar{\Gamma}$ in at least one of the two $\mathrm{SO}(3)$ factors is infinite and this is a contradiction.

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  • $\begingroup$ Ok so indeed it does not follow from the local isomorphism of $SO(4)$ with $SO(3)^2$, but from the stronger fact that $PSO(4) \cong SO(3)^2$. How does one prove that fact? I deduced the local isomorphism from the isomorphism of Lie algebras, I think one needs something more explicit here $\endgroup$ – frafour Nov 29 '18 at 22:53
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    $\begingroup$ They are isomorphic because they have the same Lie algebra and have trivial center. (An explicit argument is also possible.) $\endgroup$ – YCor Nov 29 '18 at 22:56

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